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2023-10-05T12:34:16Z

Rosso: /* Tutorials */

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

* Jacopo De Nardis [https://www.jacopodenardis.com/]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

T-I-1

2023-09-27T13:18:18Z

Rosso: /* Harmonic functions and hydrodynamics in the plane */

__FORCETOC__

=Analytical functions: conformal map and applications to hydrodynamics=

This homework deals with the application of conformal maps to the study of two-dimensional hydrodynamics. A conformal map
is a geometrical transformation which preserves all (oriented) crossing angles between lines. In dimension <math>d \geq 3</math> a conformal map is necessarily composed
from the following limited number of transformations: translations, rotations, homothetic transformation and special conformal transformation
(which is the composition of a reflection and an inversion in a sphere). However in two dimensions, <math> d=2 </math>, the space of conformal mappings is
much larger and one can show that, given an open set <math> \Omega \in {\mathbb{C}}</math>, any holomorphic function <math> f : \Omega \rightarrow {\mathbb{C}} </math> such that
<math> f'(z) \neq 0 </math>, <math>\forall z \in \Omega </math> defines a conformal map from <math>\Omega</math> to <math>f(\Omega)</math>. The aim of this homework is to exploit this property to study some hydrodynamic flows in two spatial dimensions.

= Joukovski's transformation =

The Joukovski's transformation is defined by the following application

<center><math>
J:
\begin{array}[t]{ccc}
\mathbb{C} \setminus \{ 0 \} &\to &\mathbb{C} \\
z &\mapsto & z + \displaystyle \frac{1}{z}
\end{array}
</math></center>

* Compute <math>J'(z)</math> and deduce from it the maximal ensemble on which <math>J</math> is a conformal map. Show that <math> J </math> is always surjective. Under which condition on the set <math>\Omega</math> the application <math>J</math> on <math>\Omega</math> is injective ? Give some examples of such (maximal) set <math>\Omega</math>.

* Give the image by <math>J</math> of the following sub-sets: (a) the half-line passing through the origin <math>O</math> and making an angle <math>\alpha </math> with the <math>x</math>-axis, (b) the circle centered at the origin of radius <math>R</math> (analyze in particular the case <math>R=1</math>). What is the image, by <math>J</math>, of the outside of the unit circle <math>|z| > 1</math>.

Hint: you will find useful to write the Cartesian coodinates of <math>f(z)</math> in terms of the polar coordinates of <math>z</math> writing <math>z = r e^{i \theta}</math>.

Get a better idea of this Joukowski's transformation using the following code in Mathematica:

- for the half-line passing through the origin:
Jouk[z_] := z + 1/z
Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {u -> 0.5}, {R Cos[u], R Sin[u]} /. {u -> 0.5}}, {R, .01, 10}]

- for the circle centered at the origin of radius <math>R</math>:

Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 0.79}, {R Cos[u], R Sin[u]} /. {R -> 0.79}}, {u, 0, 2 \[Pi]},
PlotRange -> {{-3, 3}, {-1.5, 1.5}}]

* Study the conformal map <math>J</math> in the vicinity of <math> z = 1</math>: we consider a "smooth" curve <math>\gamma</math> passing through <math>z=1</math>, with a well defined tangent. Show that the image of <math>\gamma</math> exhibits a cusp in <math>J(1)</math>. In this purpose, we parametrize this curve <math>\gamma</math> by <math>z(t) </math> with <math>z(0)=1</math> and <math>z'(0) \neq 0</math>. Write then the Taylor expansion of <math>z</math> in <math>t=0</math> up to first order and the expansion of <math>J</math> close to <math>1</math> up to second order.

Joukowski showed that the image of a circle passing through <math>z=1</math> and containing the point <math>z=-1</math> is mapped onto a curve shaped like the cross section of an airplane wing. We call this curve the Joukowski airfoil.

*Convince yourself that the parametric curve
<center><math> 1 - R \left( \cos(u) + \sin(\alpha) \right) + i R \left( \cos(\alpha) + \sin(u) \right) \quad \quad \text{with} \quad 0<u<2 \pi </math></center>
identifies a circle of radius <math>R</math>, passing through <math>z=1</math>. <math>\alpha</math> being the angle between the real axis and the tangent at <math>z=1</math>. You can now visualize the Joukowski airfoil using the following code:
Jouk[z_] := z + 1/z
Jouk[1 - R Sin[\[Alpha]] + R Cos[u] + I (R Cos[\[Alpha]] + R Sin[u])];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 1.15, \[Alpha] -> 1.3}, {{1 - R Sin[\[Alpha]] + R Cos[u]}, {R Cos[\[Alpha]] + R Sin[u]}} /. {R -> 1.15, \[Alpha] -> 1.3}}, {u, 0, 2 \[Pi]}]

=Harmonic functions and hydrodynamics in the plane =

We recall that a function of differentiability class <math> C^2 </math>, <math> \varphi : \Omega \to \mathbb{R}</math> or <math>\mathbb{C}</math> (<math>\Omega</math> being an open set of <math>\mathbb{C} </math>) is
called a "harmonic function" if it satisfies the Laplace equation
<center>
<math>

\Delta \varphi = 0
\qquad \text{where} \qquad
\Delta \varphi \equiv \frac{\partial^2 \varphi}{\partial
x^2}+\frac{\partial^2 \varphi}{\partial y^2} \;,

</math>
</center>
in all point <math> z = x + i y \in \Omega</math>. Similarly to conformal maps, harmonic functions in two dimensions, are closely related to holomorphic functions.

* Let us consider <math>g: \Omega \to \mathbb{C} </math> a holomorphic function. Show that <math>g, \varphi = \mathrm{Re}\, g </math> and <math> \psi =
\mathrm{Im}\, g </math> are harmonic functions.

* Geometric interpration of <math>\varphi </math> and <math>\psi </math>: show that the streamlines of <math>\nabla \varphi </math> are the level curves of <math> \psi </math>.

* Show that, if <math> \varphi: \Omega \to \mathbb{R} </math> is a harmonic function and <math> f: \Omega' \to \Omega </math> a conformal map, then <math>\Phi = \varphi \circ f </math> is also a harmonic function.

We now consider the irrotational flow of a non-viscous and incompressible fluid in some region of the plane. We denote by <math>\vec{v}=(v_x,v_y) </math> its velocity field.
* Show that <math>\vec{v}=(v_x,v_y) </math> is the gradient of a scalar potential <math>\varphi(x,y)</math> which satisfies <math>\Delta \varphi=0</math> .
* Show that you can construct <math>\psi(x,y)</math> such that <math>g=\varphi +i \psi</math> is holomorphic and <math>v_x+i v_y= \overline{g'(z)}</math>. <math>g(z)</math> is the complex potential associated to the 2-dimensional fluid flow.

= Back to the Joukovski's transformation=

* Consider a constant and uniform flow, parallel to the real axis and with velocity <math>V_0</math>. Show that the complex potential writes <math>g_0(z)=V_0 z</math>.
* Consider a fluid in presence of an obstacle. The obstacle is a circle with <math>R=1</math>. Far from the circle the velocity is <math>V_0</math>. Use the Joukovski's transformation to show that the complex potential writes
<center><math>g(z)=V_0(z+\frac{1}{z}) \;.</math></center>
* Compute the velocity along the real and the imaginary axis. Draw the streamlines (<math>\psi(z)=\text{const.}</math>) of the flow.

ContourPlot[Im[(x + I y) + 1/(x + I y)], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> Red, RegionFunction -> Function[{x, y}, x^2 + y^2 > 1], Contours -> 100]

* Explain (without calculation) how you can use the Joukovski's transformation to study the flow if the circle is replaced by the airfoil.

T-I-1

2023-09-27T13:17:37Z

Rosso: /* Harmonic functions and hydrodynamics in the plane */

__FORCETOC__

=Analytical functions: conformal map and applications to hydrodynamics=

This homework deals with the application of conformal maps to the study of two-dimensional hydrodynamics. A conformal map
is a geometrical transformation which preserves all (oriented) crossing angles between lines. In dimension <math>d \geq 3</math> a conformal map is necessarily composed
from the following limited number of transformations: translations, rotations, homothetic transformation and special conformal transformation
(which is the composition of a reflection and an inversion in a sphere). However in two dimensions, <math> d=2 </math>, the space of conformal mappings is
much larger and one can show that, given an open set <math> \Omega \in {\mathbb{C}}</math>, any holomorphic function <math> f : \Omega \rightarrow {\mathbb{C}} </math> such that
<math> f'(z) \neq 0 </math>, <math>\forall z \in \Omega </math> defines a conformal map from <math>\Omega</math> to <math>f(\Omega)</math>. The aim of this homework is to exploit this property to study some hydrodynamic flows in two spatial dimensions.

= Joukovski's transformation =

The Joukovski's transformation is defined by the following application

<center><math>
J:
\begin{array}[t]{ccc}
\mathbb{C} \setminus \{ 0 \} &\to &\mathbb{C} \\
z &\mapsto & z + \displaystyle \frac{1}{z}
\end{array}
</math></center>

* Compute <math>J'(z)</math> and deduce from it the maximal ensemble on which <math>J</math> is a conformal map. Show that <math> J </math> is always surjective. Under which condition on the set <math>\Omega</math> the application <math>J</math> on <math>\Omega</math> is injective ? Give some examples of such (maximal) set <math>\Omega</math>.

* Give the image by <math>J</math> of the following sub-sets: (a) the half-line passing through the origin <math>O</math> and making an angle <math>\alpha </math> with the <math>x</math>-axis, (b) the circle centered at the origin of radius <math>R</math> (analyze in particular the case <math>R=1</math>). What is the image, by <math>J</math>, of the outside of the unit circle <math>|z| > 1</math>.

Hint: you will find useful to write the Cartesian coodinates of <math>f(z)</math> in terms of the polar coordinates of <math>z</math> writing <math>z = r e^{i \theta}</math>.

Get a better idea of this Joukowski's transformation using the following code in Mathematica:

- for the half-line passing through the origin:
Jouk[z_] := z + 1/z
Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {u -> 0.5}, {R Cos[u], R Sin[u]} /. {u -> 0.5}}, {R, .01, 10}]

- for the circle centered at the origin of radius <math>R</math>:

Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 0.79}, {R Cos[u], R Sin[u]} /. {R -> 0.79}}, {u, 0, 2 \[Pi]},
PlotRange -> {{-3, 3}, {-1.5, 1.5}}]

* Study the conformal map <math>J</math> in the vicinity of <math> z = 1</math>: we consider a "smooth" curve <math>\gamma</math> passing through <math>z=1</math>, with a well defined tangent. Show that the image of <math>\gamma</math> exhibits a cusp in <math>J(1)</math>. In this purpose, we parametrize this curve <math>\gamma</math> by <math>z(t) </math> with <math>z(0)=1</math> and <math>z'(0) \neq 0</math>. Write then the Taylor expansion of <math>z</math> in <math>t=0</math> up to first order and the expansion of <math>J</math> close to <math>1</math> up to second order.

Joukowski showed that the image of a circle passing through <math>z=1</math> and containing the point <math>z=-1</math> is mapped onto a curve shaped like the cross section of an airplane wing. We call this curve the Joukowski airfoil.

*Convince yourself that the parametric curve
<center><math> 1 - R \left( \cos(u) + \sin(\alpha) \right) + i R \left( \cos(\alpha) + \sin(u) \right) \quad \quad \text{with} \quad 0<u<2 \pi </math></center>
identifies a circle of radius <math>R</math>, passing through <math>z=1</math>. <math>\alpha</math> being the angle between the real axis and the tangent at <math>z=1</math>. You can now visualize the Joukowski airfoil using the following code:
Jouk[z_] := z + 1/z
Jouk[1 - R Sin[\[Alpha]] + R Cos[u] + I (R Cos[\[Alpha]] + R Sin[u])];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 1.15, \[Alpha] -> 1.3}, {{1 - R Sin[\[Alpha]] + R Cos[u]}, {R Cos[\[Alpha]] + R Sin[u]}} /. {R -> 1.15, \[Alpha] -> 1.3}}, {u, 0, 2 \[Pi]}]

=Harmonic functions and hydrodynamics in the plane =

We recall that a function of differentiability class <math> C^2 </math>, <math> \varphi </math> : <math> \Omega \to \mathbb{R}</math> or <math>\mathbb{C}</math> (<math>\Omega</math> being an open set of <math>\mathbb{C} </math>) is
called a "harmonic function" if it satisfies the Laplace equation
<center>
<math>

\Delta \varphi = 0
\qquad \text{where} \qquad
\Delta \varphi \equiv \frac{\partial^2 \varphi}{\partial
x^2}+\frac{\partial^2 \varphi}{\partial y^2} \;,

</math>
</center>
in all point <math> z = x + i y \in \Omega</math>. Similarly to conformal maps, harmonic functions in two dimensions, are closely related to holomorphic functions.

* Let us consider <math>g: \Omega \to \mathbb{C} </math> a holomorphic function. Show that <math>g, \varphi = \mathrm{Re}\, g </math> and <math> \psi =
\mathrm{Im}\, g </math> are harmonic functions.

* Geometric interpration of <math>\varphi </math> and <math>\psi </math>: show that the streamlines of <math>\nabla \varphi </math> are the level curves of <math> \psi </math>.

* Show that, if <math> \varphi: \Omega \to \mathbb{R} </math> is a harmonic function and <math> f: \Omega' \to \Omega </math> a conformal map, then <math>\Phi = \varphi \circ f </math> is also a harmonic function.

We now consider the irrotational flow of a non-viscous and incompressible fluid in some region of the plane. We denote by <math>\vec{v}=(v_x,v_y) </math> its velocity field.
* Show that <math>\vec{v}=(v_x,v_y) </math> is the gradient of a scalar potential <math>\varphi(x,y)</math> which satisfies <math>\Delta \varphi=0</math> .
* Show that you can construct <math>\psi(x,y)</math> such that <math>g=\varphi +i \psi</math> is holomorphic and <math>v_x+i v_y= \overline{g'(z)}</math>. <math>g(z)</math> is the complex potential associated to the 2-dimensional fluid flow.

= Back to the Joukovski's transformation=

* Consider a constant and uniform flow, parallel to the real axis and with velocity <math>V_0</math>. Show that the complex potential writes <math>g_0(z)=V_0 z</math>.
* Consider a fluid in presence of an obstacle. The obstacle is a circle with <math>R=1</math>. Far from the circle the velocity is <math>V_0</math>. Use the Joukovski's transformation to show that the complex potential writes
<center><math>g(z)=V_0(z+\frac{1}{z}) \;.</math></center>
* Compute the velocity along the real and the imaginary axis. Draw the streamlines (<math>\psi(z)=\text{const.}</math>) of the flow.

ContourPlot[Im[(x + I y) + 1/(x + I y)], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> Red, RegionFunction -> Function[{x, y}, x^2 + y^2 > 1], Contours -> 100]

* Explain (without calculation) how you can use the Joukovski's transformation to study the flow if the circle is replaced by the airfoil.

T-I-1

2023-09-27T13:16:47Z

Rosso: /* Harmonic functions and hydrodynamics in the plane */

__FORCETOC__

=Analytical functions: conformal map and applications to hydrodynamics=

This homework deals with the application of conformal maps to the study of two-dimensional hydrodynamics. A conformal map
is a geometrical transformation which preserves all (oriented) crossing angles between lines. In dimension <math>d \geq 3</math> a conformal map is necessarily composed
from the following limited number of transformations: translations, rotations, homothetic transformation and special conformal transformation
(which is the composition of a reflection and an inversion in a sphere). However in two dimensions, <math> d=2 </math>, the space of conformal mappings is
much larger and one can show that, given an open set <math> \Omega \in {\mathbb{C}}</math>, any holomorphic function <math> f : \Omega \rightarrow {\mathbb{C}} </math> such that
<math> f'(z) \neq 0 </math>, <math>\forall z \in \Omega </math> defines a conformal map from <math>\Omega</math> to <math>f(\Omega)</math>. The aim of this homework is to exploit this property to study some hydrodynamic flows in two spatial dimensions.

= Joukovski's transformation =

The Joukovski's transformation is defined by the following application

<center><math>
J:
\begin{array}[t]{ccc}
\mathbb{C} \setminus \{ 0 \} &\to &\mathbb{C} \\
z &\mapsto & z + \displaystyle \frac{1}{z}
\end{array}
</math></center>

* Compute <math>J'(z)</math> and deduce from it the maximal ensemble on which <math>J</math> is a conformal map. Show that <math> J </math> is always surjective. Under which condition on the set <math>\Omega</math> the application <math>J</math> on <math>\Omega</math> is injective ? Give some examples of such (maximal) set <math>\Omega</math>.

* Give the image by <math>J</math> of the following sub-sets: (a) the half-line passing through the origin <math>O</math> and making an angle <math>\alpha </math> with the <math>x</math>-axis, (b) the circle centered at the origin of radius <math>R</math> (analyze in particular the case <math>R=1</math>). What is the image, by <math>J</math>, of the outside of the unit circle <math>|z| > 1</math>.

Hint: you will find useful to write the Cartesian coodinates of <math>f(z)</math> in terms of the polar coordinates of <math>z</math> writing <math>z = r e^{i \theta}</math>.

Get a better idea of this Joukowski's transformation using the following code in Mathematica:

- for the half-line passing through the origin:
Jouk[z_] := z + 1/z
Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {u -> 0.5}, {R Cos[u], R Sin[u]} /. {u -> 0.5}}, {R, .01, 10}]

- for the circle centered at the origin of radius <math>R</math>:

Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 0.79}, {R Cos[u], R Sin[u]} /. {R -> 0.79}}, {u, 0, 2 \[Pi]},
PlotRange -> {{-3, 3}, {-1.5, 1.5}}]

* Study the conformal map <math>J</math> in the vicinity of <math> z = 1</math>: we consider a "smooth" curve <math>\gamma</math> passing through <math>z=1</math>, with a well defined tangent. Show that the image of <math>\gamma</math> exhibits a cusp in <math>J(1)</math>. In this purpose, we parametrize this curve <math>\gamma</math> by <math>z(t) </math> with <math>z(0)=1</math> and <math>z'(0) \neq 0</math>. Write then the Taylor expansion of <math>z</math> in <math>t=0</math> up to first order and the expansion of <math>J</math> close to <math>1</math> up to second order.

Joukowski showed that the image of a circle passing through <math>z=1</math> and containing the point <math>z=-1</math> is mapped onto a curve shaped like the cross section of an airplane wing. We call this curve the Joukowski airfoil.

*Convince yourself that the parametric curve
<center><math> 1 - R \left( \cos(u) + \sin(\alpha) \right) + i R \left( \cos(\alpha) + \sin(u) \right) \quad \quad \text{with} \quad 0<u<2 \pi </math></center>
identifies a circle of radius <math>R</math>, passing through <math>z=1</math>. <math>\alpha</math> being the angle between the real axis and the tangent at <math>z=1</math>. You can now visualize the Joukowski airfoil using the following code:
Jouk[z_] := z + 1/z
Jouk[1 - R Sin[\[Alpha]] + R Cos[u] + I (R Cos[\[Alpha]] + R Sin[u])];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 1.15, \[Alpha] -> 1.3}, {{1 - R Sin[\[Alpha]] + R Cos[u]}, {R Cos[\[Alpha]] + R Sin[u]}} /. {R -> 1.15, \[Alpha] -> 1.3}}, {u, 0, 2 \[Pi]}]

=Harmonic functions and hydrodynamics in the plane =

We recall that a function <math> \varphi </math> of differentiability class <math> C^2 </math> : <math> \Omega \to \mathbb{R}</math> or <math>\mathbb{C}</math> (<math>\Omega</math> being an open set of <math>\mathbb{C} </math>) is
called a "harmonic function" if it satisfies the Laplace equation
<center>
<math>

\Delta \varphi = 0
\qquad \text{where} \qquad
\Delta \varphi \equiv \frac{\partial^2 \varphi}{\partial
x^2}+\frac{\partial^2 \varphi}{\partial y^2} \;,

</math>
</center>
in all point <math> z = x + i y \in \Omega</math>. Similarly to conformal maps, harmonic functions in two dimensions, are closely related to holomorphic functions.

* Let us consider <math>g: \Omega \to \mathbb{C} </math> a holomorphic function. Show that <math>g, \varphi = \mathrm{Re}\, g </math> and <math> \psi =
\mathrm{Im}\, g </math> are harmonic functions.

* Geometric interpration of <math>\varphi </math> and <math>\psi </math>: show that the streamlines of <math>\nabla \varphi </math> are the level curves of <math> \psi </math>.

* Show that, if <math> \varphi: \Omega \to \mathbb{R} </math> is a harmonic function and <math> f: \Omega' \to \Omega </math> a conformal map, then <math>\Phi = \varphi \circ f </math> is also a harmonic function.

We now consider the irrotational flow of a non-viscous and incompressible fluid in some region of the plane. We denote by <math>\vec{v}=(v_x,v_y) </math> its velocity field.
* Show that <math>\vec{v}=(v_x,v_y) </math> is the gradient of a scalar potential <math>\varphi(x,y)</math> which satisfies <math>\Delta \varphi=0</math> .
* Show that you can construct <math>\psi(x,y)</math> such that <math>g=\varphi +i \psi</math> is holomorphic and <math>v_x+i v_y= \overline{g'(z)}</math>. <math>g(z)</math> is the complex potential associated to the 2-dimensional fluid flow.

= Back to the Joukovski's transformation=

* Consider a constant and uniform flow, parallel to the real axis and with velocity <math>V_0</math>. Show that the complex potential writes <math>g_0(z)=V_0 z</math>.
* Consider a fluid in presence of an obstacle. The obstacle is a circle with <math>R=1</math>. Far from the circle the velocity is <math>V_0</math>. Use the Joukovski's transformation to show that the complex potential writes
<center><math>g(z)=V_0(z+\frac{1}{z}) \;.</math></center>
* Compute the velocity along the real and the imaginary axis. Draw the streamlines (<math>\psi(z)=\text{const.}</math>) of the flow.

ContourPlot[Im[(x + I y) + 1/(x + I y)], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> Red, RegionFunction -> Function[{x, y}, x^2 + y^2 > 1], Contours -> 100]

* Explain (without calculation) how you can use the Joukovski's transformation to study the flow if the circle is replaced by the airfoil.

T-I-1

2023-09-27T13:14:49Z

Rosso: /* Harmonic functions and hydrodynamics in the plane */

__FORCETOC__

=Analytical functions: conformal map and applications to hydrodynamics=

This homework deals with the application of conformal maps to the study of two-dimensional hydrodynamics. A conformal map
is a geometrical transformation which preserves all (oriented) crossing angles between lines. In dimension <math>d \geq 3</math> a conformal map is necessarily composed
from the following limited number of transformations: translations, rotations, homothetic transformation and special conformal transformation
(which is the composition of a reflection and an inversion in a sphere). However in two dimensions, <math> d=2 </math>, the space of conformal mappings is
much larger and one can show that, given an open set <math> \Omega \in {\mathbb{C}}</math>, any holomorphic function <math> f : \Omega \rightarrow {\mathbb{C}} </math> such that
<math> f'(z) \neq 0 </math>, <math>\forall z \in \Omega </math> defines a conformal map from <math>\Omega</math> to <math>f(\Omega)</math>. The aim of this homework is to exploit this property to study some hydrodynamic flows in two spatial dimensions.

= Joukovski's transformation =

The Joukovski's transformation is defined by the following application

<center><math>
J:
\begin{array}[t]{ccc}
\mathbb{C} \setminus \{ 0 \} &\to &\mathbb{C} \\
z &\mapsto & z + \displaystyle \frac{1}{z}
\end{array}
</math></center>

* Compute <math>J'(z)</math> and deduce from it the maximal ensemble on which <math>J</math> is a conformal map. Show that <math> J </math> is always surjective. Under which condition on the set <math>\Omega</math> the application <math>J</math> on <math>\Omega</math> is injective ? Give some examples of such (maximal) set <math>\Omega</math>.

* Give the image by <math>J</math> of the following sub-sets: (a) the half-line passing through the origin <math>O</math> and making an angle <math>\alpha </math> with the <math>x</math>-axis, (b) the circle centered at the origin of radius <math>R</math> (analyze in particular the case <math>R=1</math>). What is the image, by <math>J</math>, of the outside of the unit circle <math>|z| > 1</math>.

Hint: you will find useful to write the Cartesian coodinates of <math>f(z)</math> in terms of the polar coordinates of <math>z</math> writing <math>z = r e^{i \theta}</math>.

Get a better idea of this Joukowski's transformation using the following code in Mathematica:

- for the half-line passing through the origin:
Jouk[z_] := z + 1/z
Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {u -> 0.5}, {R Cos[u], R Sin[u]} /. {u -> 0.5}}, {R, .01, 10}]

- for the circle centered at the origin of radius <math>R</math>:

Jouk[R Cos[u] + I R Sin[u]];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 0.79}, {R Cos[u], R Sin[u]} /. {R -> 0.79}}, {u, 0, 2 \[Pi]},
PlotRange -> {{-3, 3}, {-1.5, 1.5}}]

* Study the conformal map <math>J</math> in the vicinity of <math> z = 1</math>: we consider a "smooth" curve <math>\gamma</math> passing through <math>z=1</math>, with a well defined tangent. Show that the image of <math>\gamma</math> exhibits a cusp in <math>J(1)</math>. In this purpose, we parametrize this curve <math>\gamma</math> by <math>z(t) </math> with <math>z(0)=1</math> and <math>z'(0) \neq 0</math>. Write then the Taylor expansion of <math>z</math> in <math>t=0</math> up to first order and the expansion of <math>J</math> close to <math>1</math> up to second order.

Joukowski showed that the image of a circle passing through <math>z=1</math> and containing the point <math>z=-1</math> is mapped onto a curve shaped like the cross section of an airplane wing. We call this curve the Joukowski airfoil.

*Convince yourself that the parametric curve
<center><math> 1 - R \left( \cos(u) + \sin(\alpha) \right) + i R \left( \cos(\alpha) + \sin(u) \right) \quad \quad \text{with} \quad 0<u<2 \pi </math></center>
identifies a circle of radius <math>R</math>, passing through <math>z=1</math>. <math>\alpha</math> being the angle between the real axis and the tangent at <math>z=1</math>. You can now visualize the Joukowski airfoil using the following code:
Jouk[z_] := z + 1/z
Jouk[1 - R Sin[\[Alpha]] + R Cos[u] + I (R Cos[\[Alpha]] + R Sin[u])];
ParametricPlot[{{Re[%], Im[%]} /. {R -> 1.15, \[Alpha] -> 1.3}, {{1 - R Sin[\[Alpha]] + R Cos[u]}, {R Cos[\[Alpha]] + R Sin[u]}} /. {R -> 1.15, \[Alpha] -> 1.3}}, {u, 0, 2 \[Pi]}]

=Harmonic functions and hydrodynamics in the plane =

We recall that a function <math> \varphi </math> in <math> C^2 </math> : <math> \Omega \to \mathbb{R}</math> or <math>\mathbb{C}</math> (<math>\Omega</math> being an open set of <math>\mathbb{C} </math>) is
called a "harmonic function" if it satisfies the Laplace equation
<center>
<math>

\Delta \varphi = 0
\qquad \text{where} \qquad
\Delta \varphi \equiv \frac{\partial^2 \varphi}{\partial
x^2}+\frac{\partial^2 \varphi}{\partial y^2} \;,

</math>
</center>
in all point <math> z = x + i y \in \Omega</math>. Similarly to conformal maps, harmonic functions in two dimensions, are closely related to holomorphic functions.

* Let us consider <math>g: \Omega \to \mathbb{C} </math> a holomorphic function. Show that <math>g, \varphi = \mathrm{Re}\, g </math> and <math> \psi =
\mathrm{Im}\, g </math> are harmonic functions.

* Geometric interpration of <math>\varphi </math> and <math>\psi </math>: show that the streamlines of <math>\nabla \varphi </math> are the level curves of <math> \psi </math>.

* Show that, if <math> \varphi: \Omega \to \mathbb{R} </math> is a harmonic function and <math> f: \Omega' \to \Omega </math> a conformal map, then <math>\Phi = \varphi \circ f </math> is also a harmonic function.

We now consider the irrotational flow of a non-viscous and incompressible fluid in some region of the plane. We denote by <math>\vec{v}=(v_x,v_y) </math> its velocity field.
* Show that <math>\vec{v}=(v_x,v_y) </math> is the gradient of a scalar potential <math>\varphi(x,y)</math> which satisfies <math>\Delta \varphi=0</math> .
* Show that you can construct <math>\psi(x,y)</math> such that <math>g=\varphi +i \psi</math> is holomorphic and <math>v_x+i v_y= \overline{g'(z)}</math>. <math>g(z)</math> is the complex potential associated to the 2-dimensional fluid flow.

= Back to the Joukovski's transformation=

* Consider a constant and uniform flow, parallel to the real axis and with velocity <math>V_0</math>. Show that the complex potential writes <math>g_0(z)=V_0 z</math>.
* Consider a fluid in presence of an obstacle. The obstacle is a circle with <math>R=1</math>. Far from the circle the velocity is <math>V_0</math>. Use the Joukovski's transformation to show that the complex potential writes
<center><math>g(z)=V_0(z+\frac{1}{z}) \;.</math></center>
* Compute the velocity along the real and the imaginary axis. Draw the streamlines (<math>\psi(z)=\text{const.}</math>) of the flow.

ContourPlot[Im[(x + I y) + 1/(x + I y)], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> Red, RegionFunction -> Function[{x, y}, x^2 + y^2 > 1], Contours -> 100]

* Explain (without calculation) how you can use the Joukovski's transformation to study the flow if the circle is replaced by the airfoil.

Main Page

2023-09-25T10:13:36Z

Rosso:

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

* Jacopo De Nardis [https://www.jacopodenardis.com/]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

T-I-3

2022-11-24T12:46:55Z

Rosso: /* Definition of Radon transform */

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[File:Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:Binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:Radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

T-I-3

2021-12-01T11:13:36Z

Rosso: /* No calculation needed */

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[File:Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t'}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:Binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:Radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

File:Back projection.png

2021-12-01T11:11:39Z

Rosso:

File:Images radon.png

2021-12-01T11:10:17Z

Rosso:

File:Radon transform.png

2021-12-01T11:09:21Z

Rosso:

File:Binary images.png

2021-12-01T11:06:31Z

Rosso:

File:Line parametrisation.png

2021-12-01T10:57:22Z

Rosso:

Main Page

2021-11-30T22:46:13Z

Rosso: /* Tutorials */

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

Main Page

2021-11-30T22:45:03Z

Rosso: /* Tutorials */

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3draft|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

T-I-3

2021-11-30T22:42:34Z

Rosso: /* Preliminaries: parametrisation of a line in the plane */

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[File:Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t'}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

T-I-3

2021-11-30T22:42:07Z

Rosso: /* Preliminaries: parametrisation of a line in the plane */

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t'}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

T-I-3

2021-11-30T22:21:14Z

Rosso: /* Radon transform and X-ray tomography */

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[File:Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t'}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

T-I-2

2021-11-10T15:47:59Z

Rosso: /* Causality and Kramers-Kronig relations */

=Linear response theory and Kramers-Kronig relations=

The goal of this homework is to introduce the theory of linear response and, among the related properties, such as causality, the so-called Kramers-Kronig relations. We will consider the damped harmonic oscillator as a prototypical example to discuss this topic.

== Reminder on the Fourier transform ==

Throughout this assignment, we will use the symmetric convention of Fourier transform. The Fourier transform of a function <math>f(t)\in L^1(\mathbb{R})</math> is defined as:
<center><math>
\widehat f(\omega):=\frac{1}{\sqrt{2\pi}}\int d t \,f(t)\,e^{i\omega t}.
</math></center>
With this convention, the Fourier transform of a product results in a convolution, with the following pre-factor: <math>
\sqrt{2\pi}\,\widehat{(f\, g)}=\widehat f \ast \widehat g .</math>

The repeated application of the transform gives <math>
\widehat{\widehat f}(t)=f(-t).</math>

The conjugate of a Fourier transform is equal to the transform of the conjugate, with a change of sign in the argument: <math>
\overline{\widehat f}(\omega)=\widehat{\overline{f}}(-\omega).</math>

Furthermore, we can exchange the Fourier transform under the integral sign: <math>
\int _\mathbb{R}dt \, \hat f(t)\, g(t)=\int _\mathbb{R}dt \, f(t)\,\hat g(t) .
</math>

The last property implies the result <math>
\int _\mathbb{R}d\omega \, \overline{\hat f}(\omega)\,\hat g(\omega)=\int _\mathbb{R}dt \, \overline{f(t)}\,\hat g(t) ,
</math>
which gives directly the Plancherel theorem (see lecture notes, page 45).

== Example: damped harmonic oscillator in one dimension ==
We consider a damped harmonic oscillator <math>x(t)</math>, which is described by the following equation of motion:
<center><math>
\ddot x + 2 \gamma \,\dot x + \omega _0^2\,x=f(t).
</math></center>
First, we will compute its response to an harmonic perturbation, and then its response to a generic perturbation by introducing its Green function.

=== Response to an harmonic perturbation ===

We assume that the oscillator is excited by an harmonic perturbation of the form:
<math>
f(t)=F_\omega\,e^{-i\omega t}.
</math>
We look for a solution of the form:
<math>
x(t)=X_\omega\,e^{-i\omega t}.
</math>

'''Q1:''' Show that
<center><math>
X_\omega=\hat g(\omega)\, F_\omega,
</math></center>
where
<center><math>
\hat g(\omega)=-\frac{1}{(\omega-\omega_1)(\omega-\omega_2)}.
</math></center>

Write the explicit form of the poles that we have indicated as <math>\omega_1</math> and <math>\omega_2</math>.

=== Response to a generic perturbation ===

We consider a generic perturbation represented by <math>f(t)</math>, whose Fourier transform is given by <math>\hat f (\omega)</math>.

'''Q2:''' Write <math>x(t)</math> as a function of <math>\hat g(\omega)</math> and <math>\hat f(\omega)</math>.

'''Q3:''' Show that it follows that
<center><math>
x(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}d t' \, g(t-t')f(t'),
</math></center>
where <math>g(t)</math> is the function having <math>\hat g(\omega)</math> as Fourier transform. The above equation is nothing but the convolution of <math>f</math> with <math>g</math>: <math>\sqrt{2\pi}x(t)=(f\ast g)(t)</math>.

=== Holomorphism of the response and consequences ===
'''Q4:''' Show that <math>g(t)</math> is the response of the system to an impulsive perturbation <math>f(t)=\sqrt{2\pi}\delta(t)</math>.
What is the effect of a perturbation applied at time <math>s</math> on the solution at time <math>t<s</math>?

'''Q5:''' We have defined <math>\hat g (\omega)</math> for <math>\omega\in\mathbb{R}</math>, i.e. on the real axis. Show that its extension to the complex plane is holomorphic everywhere (except at the poles). Using the explicit expression for <math>\widehat g (\omega)</math> derived above, compute explicitly <math>g(t)</math>. You will have to treat separately the case of positive time and the case of negative time. Use the residue theorem.

'''Q6:''' Deduce the expression of the response:
<center><math>
x(t)=\int_{-\infty}^t dt' \, e^{-\gamma (t-t')}\,\frac{\sin\left[(\omega_0^2-\gamma^2)^{1/2} (t-t')\right]}{(\omega_0^2-\gamma^2)^{1/2}}f(t').
</math></center>
The above expression indicates the Green function of the system. Notice that the response of the system is causal.

It is worth highlighting the importance of the holomorphism in this proof. Indeed, we have seen that we had to apply Jordan's lemma. Therefore, we need a relatively fast decay of the response function at infinity, i.e. a system that does not diverge.

We also point out that, if <math>\gamma</math> was negative, we would have lost causality: we can guess a relation between causality and the dissipation of energy in a passive system.

== General properties of linear response ==
We look for a generalization of the properties that we have seen in the previous example. We discuss an equation of motion (or possibly another physical equation)
<center><math>
\mathcal{L}x(t)=f(t),
</math></center>
with a linear operator <math>\mathcal{L}</math> (of differentiation, multiplication, integration, etc.), and with an inhomogeneity <math>f(t)</math> (external stimulation, perturbation, etc.). The generic solution is unknown since the explicit form of the operator is not specified. However, we know that this solution must have some physical properties:
* Linearity: <math>x</math> is a function or a linear functional of <math>f</math>, which reads <math>
x(t)=\int _\mathbb{R}ds \, g(t-s)f(s). </math>
* Causality: in the sense that the effect cannot precede the cause: <math>
g(t)=0</math> for <math> t<0</math>.
* The total response to a finite perturbation must be finite.
* The response to a real perturbation, <math>f(t)\in \mathbb{R}</math>, must be real as well, <math>x(t)\in\mathbb{R}</math> (at least in the case we are not considering quantum mechanics effects).

=== Link between odd/even and real/imaginary ===
'''Q7:''' Show the following properties of the Fourier transform:
* if <math>g\in \mathbb{R} </math> is an odd function, then <math> \hat g\in i\mathbb{R} </math> is also an odd function;
* if <math> g\in i\mathbb{R}</math> is an odd function, then <math>\hat g\in \mathbb{R}</math> is also an odd function;
* if <math> g\in \mathbb{R} </math> is an even function, then <math> \hat g\in \mathbb{R} </math> is also an even function;
* if <math>g\in i\mathbb{R} </math> is an even function, then <math> \hat g\in i\mathbb{R} </math> is also an even function.

'''Q8:''' Conclude that, if <math>g</math> is real, its Fourier transform is written as <math>\hat g=\hat g ' +i\hat g ''</math>, with <math>\hat g '</math> even, and <math>\hat g '' </math> odd.

'''Q9:''' Derive the same properties for the case of the oscillator.

=== Preliminaries ===
We define the sign function as
<center><math>
S(t)=\begin{cases}
-1 & t<0\\
+1 &t\geq 0
\end{cases},
</math></center>
and we want to show that its Fourier transform is
<center><math>
\hat S(\omega)=\sqrt{\frac{2}{\pi}}\frac{i}{\omega}.
</math></center>
Since <math>S</math> is not summable, we have to use an auxiliary function <math>S_\epsilon (t):=e^{-\epsilon |t|}S(t)</math>.

'''Q10:''' Show that <math>\lim_{\varepsilon\rightarrow 0}\hat S_\epsilon (\omega)=\sqrt{\frac{2}{\pi}}\frac{i}{\omega}</math>.

'''Q11:''' Discuss the possibility of using <math>\hat S</math> instead of <math>\hat S_\epsilon</math> in a computation (e.g., in the example below).

=== Causality and Kramers-Kronig relations ===
'''Q12:''' Using the expression of the Fourier transform of a convolution and the sign function defined above, show the following property of Fourier transform:
<center> If <math>g(t)=0</math> <math>\forall t<0</math>, then <math>\quad\hat g(\omega)=\frac{1}{i\pi}\,{\rm PV} \int_\mathbb{R}d\xi \,\frac{ \hat g(\xi)}{\xi -\omega}</math>.
</center>
By separating the real and imaginary part of this equation, we obtain Kramers-Kronig relations:
<center><math>
\operatorname{Re}\hat g(\omega)=\frac{1}{\pi}\, {\rm PV}\int_\mathbb{R}d\xi \,\frac{\operatorname{Im}\hat g(\omega)}{\xi-\omega},
</math></center>
<center><math>
\operatorname{Im} \hat g(\omega)=-\frac{1}{\pi} \,{\rm PV}\int_\mathbb{R}d\xi \,\frac{\operatorname{Re}\hat g(\omega)}{\xi-\omega}.
</math></center>

'''Q13:''' Show that Kramers-Kronig relations express the fact that the extension <math>\hat g (\omega + i \nu)</math> to <math>\hat g (\omega)</math> on the complex upper-half plane <math>(\nu>0)</math> is holomorphic, without any poles, and that it decreases sufficiently fast as <math>|\omega +i\nu|\rightarrow \infty</math>.

We will start from a weaker result:

Show that the Kramers-Kronig relation is valid for all functions <math>\hat g (\omega +i\nu)</math> with the aforementioned properties. Apply the residue theorem integrating along the path <math>\gamma</math> displayed in the figure below, that can be reproduced via the following code:

<pre>
gamma = ParametricPlot[{{R Cos[t], R Sin[t]}, {-R*(1 - t/Pi) + (x - r)*t/Pi, 0}, {r Cos[t + Pi] + x, r Sin[t + Pi]}, {(x + r)*(1 - t/Pi) + R*t/Pi, 0}} /. {R -> 2, r -> 0.5, x -> 1}, {t, 0, Pi}, Ticks -> {{{-R, "-R"}, {x - r, "x-r"}, {x, "x"}, {x + r, "x+r"}, {R, "R"}} /. {R -> 2, r -> 0.5, x -> 1}, None}, TicksStyle -> Directive[Red, 20], AxesLabel -> {None}] /. Line[u_] :> Sequence[Arrowheads[Table[.05, {1}]], Arrow@Line[u]]
</pre>
[[File:Gamma_contour.png|400px|thumb|center|Sketch of the integration path <math>\gamma</math>.]]

Integrate the function <math>\hat g (z)/(z-\omega)</math> along this path. Compute the contribution of the small half-circle <math>C_r</math> in the limit <math>r\rightarrow 0</math> using its Taylor expansion in the vicinity of <math>\omega</math> and performing the change of variable <math>z'=\omega + r\,e^{i\theta}</math>.

The last result suggests the existence of a relation with the causality at <math>g(t)=0</math> <math>\forall t<0</math> and the fact that <math>g</math> is holomorphic without poles in the upper-half plane. This was detailed in the Titchmarsch theorem.

T-I-3

2021-11-10T14:52:14Z

Rosso: /* Principles of X-ray tomography */

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.
The second part of the homework consists of a documentation work to be conducted in pairs: each pair of students should prepare a blackboard presentation of approximately five minutes on this part.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[File:Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t'}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

T-I-3

2021-11-10T14:50:03Z

Rosso: Created page with "=Radon transform and X-ray tomography= The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is inverti..."

=Radon transform and X-ray tomography=
The goal of this homework is to introduce the Radon transform of a two-dimensional function. We will show that this transform is invertible and the inverse involves the Fourier transform in two dimensions. From a practical point of view, the Radon transform is the basis of X-ray tomography (as well as X-ray scanning), applied in the medical context in order to obtain cross-section images of different organs.
The second part of the homework consists of a documentation work to be conducted in pairs: each pair of students should prepare a blackboard presentation of approximately five minutes on this part.

== Radon transform ==
=== Preliminaries: parametrisation of a line in the plane ===
[[File:Line_parametrisation.png |thumb|right| '''Figure 1:''' Parametrisation <math>(t,\Phi)</math> of a line in the plane.]]
'''Q1:''' In a two-dimensional space, how many parameters are needed in order to define a line? Provide some examples of equations that define a unique line in the plane.

'''Q2:''' In the context of Radon transform, we choose to define a line via the parameters <math>t</math> and <math>\Phi</math>, where <math>(t,\Phi)\in \mathbb{R}\times [0;2\pi[</math>, as displayed in Figure 1. Each angle <math>\Phi</math> is associated to a unique unit vector <math>\mathbf{u_t}</math>:
<center><math>\mathbf{u_t} = \cos(\Phi) \mathbf{u_x}+ \sin(\Phi)\mathbf{u_y}.</math></center>
Show that for each given line there exist two possible pairs of values <math>(t,\Phi)</math>.

'''Q3:''' We choose to ''orient'' the line positively along the unit vector <math>\mathbf{u_{\Phi}}</math>, defined by:
<center><math> \mathbf{u_{\Phi}} = -\sin(\Phi)\mathbf{u_x} + \cos(\Phi)\mathbf{u_y}.</math></center>
Show that for each pair <math>(t,\Phi)</math> there exists one unique ''oriented'' line.

'''Q4:''' A natural pair of coordinates, associated to the family of lines obtained from a given <math>\Phi</math>, is the pair <math>(t,s)\in \mathbb{R}^2</math> of coordinates of a point in the basis <math>(\mathbf{u_t},\mathbf{u_\Phi})</math> related to the line that passes through that point. Provide the expression for <math>(x,y)</math> as a function of <math>(t,s)</math>, as well as the expression for <math>(t,s)</math> as a function of <math>(x,y)</math>. Deduce the relation between the surface elements <math>\textrm{d}x \,\textrm{d}y</math> and <math>\textrm{d}s \,\textrm{d}t</math>.

=== Definition of Radon transform ===
''' Definition: ''' The Radon transform of a function <math> f : \mathbb{R}^2\rightarrow \mathbb{R}</math> is the function <math>\hat f : \mathbb{R}^2\rightarrow \mathbb{R}</math> defined by the following expression

<center><math>
\hat f(t,\mathbf{u_t})=\int_{-\infty}^{+\infty}f(t\mathbf{u_t}+s\mathbf{u_\Phi})\,\mathrm{d}s.
</math></center>

''' Use of the <math>\delta-</math>distribution: ''' The definition of Radon transform can be elegantly written by means of the Dirac-delta distribution. This formulation provides the advantage of generalising the definition of Radon transform to arbitrary dimension.

'''Q5:''' Using the relation
<center><math>
f(t\mathbf{u_t}+s\mathbf{u_\Phi})=\int_{-\infty}^{+\infty}f(t' \mathbf{u_{t'}}+s\mathbf{u_\Phi})\delta(t'-t)\mathrm{d}t',
</math></center>
propose a definition of Radon transform in the form of a surface integral.

'''Q6:''' Propose a definition of the Radon transform of a function <math>f:\mathbb{R}^n\rightarrow \mathbb{R}</math> (<math>n\geq 2</math>). Give a geometrical interpretation of the Radon transform in dimension <math>n=3</math>.

From now on, we will always consider the Radon transform in the two-dimensional case <math>n=2</math>.

=== Some examples ===

==== No calculation needed ====
''' Q7: '''In Figure 2, match each image with the corresponding Radon transform. Images represent functions <math>f: \mathbb{R}^2\rightarrow \mathbb{R}^+</math>. Specify the meaning of the grey scales in the different figures. Draw the axes <math>t,\Phi</math> on the Radon transforms, knowing that in the images in the direct space (plane <math>x,y</math>), the origin of axes <math>x,y</math> is located at the center of the figure.
<center> Some binary images...</center>
[[File:binary_images.png | 800px|thumb|center]]
<center> ... and their Radon transform (in random order):</center>
[[File:radon_transform.png | 800px| thumb|center| ''' Figure 2:''' Qualitative illustration of the Radon transform.]]

'''Q8:''' Indicate (without doing any calculation) the profile of the Radon transform of a constant function on a quasi-point-like support. Similarly, indicate (without any calculation) the profile of the Radon transform of a constant function on a line, and on a line segment.

==== Hand calculations ====
'''Q9:''' Compute the Radon transform of a function that is constant on a disc of radius <math>R</math>, and null outside.

'''Q10:''' Compute the Radon transform of the function <math>f(x,y)=a\,\exp\left(-\frac{x^2+y^2}{\sigma^2}\right)</math>.

=== Expression for radial functions ===
'''Q11:''' Show that in the case of radial function <math>f(x,y)=F(r)=F(\sqrt{x^2+y^2})</math> the Radon transform can be written as a simple integral transform of the function <math>F(r)</math>.

== Principles of X-ray tomography==
This section is about a documentation project that should be done in pairs. Each pair of students will present their work at the blackboard in ~5 minutes.

What are the physical principles underlying X-ray tomography? In particular, you should specify which physical quantity is measured, and which quantity we try to reconstruct in form of an image. You should verify that we can express the measured quantities as a Radon transform of the parameter that we try to image. X-ray tomography is based on the inversion of the Radon transform, knowing how to deduce <math>f</math> from a measure of <math>\hat f</math>.

The following section addresses the principle at the basis of this inversion.

== Inversion of the Radon transform==
=== Back-projection formula===
'''Q12:''' Explain why the following function
<center><math>
g(\mathbf{r})=\frac{1}{2\pi}\int_0^{2\pi}\hat f (\mathbf{r}\cdot \mathbf{u_t},\mathbf{u_t})\,\mathrm{d}\Phi,
</math></center>
obtained via an angular mean of the Radon transform of a function <math>f</math>, is likely to resemble to the function <math>f</math>. This formula is called ''back-projection formula''.
<center>Some images...</center>
[[File:images_radon.png | 800px| thumb|center]]
<center>... and the back-projection of their Radon transform.</center>
[[File:back_projection.png | 800px| thumb|center| ''' Figure 3:''' Illustration of the back-projection formula.]]
Figure 3 displays the application of the back-projection formula to the examples that we have previously considered (binary images), as well as to an example from medical imaging context. Does the back-projection formula allow to inverse the Radon transform? Explain.

=== Projection-slice theorem ===
Show that, for a given <math>\Phi</math>, the one-dimensional Fourier transform of <math>\hat f (t,\mathbf{u_t})</math> over the variable <math>t</math> is equal to the two-dimensional transform of <math>f(\mathbf{r})</math>:
<center><math>
\int_{-\infty}^{+\infty}\hat f (t,\mathbf{u_t})\, e^{-ikt}\,\mathrm{d}t =\int \int f(\mathbf{r})\, e^{-i(k\mathbf{u_t})\cdot \mathbf{r}}\mathrm{d}\mathbf{r}
</math></center>
or, similarly,
<center><math>
\textrm{TF}_{1D(t)}\left(\hat f(t,\mathbf{u_t})\right)[k]=\textrm{TF}_{2D}\left( f(x,y)\right)[k\mathbf{u_t}].
</math></center>
In the context of medical imaging, this relation is called ''projection-slice theorem''. It relates the two-dimensional Fourier transform of a function to its Radon transform.

Recall the meaning of the Fourier transform of a function of one variable and of two variables. From the projection-slice theorem, deduce an interpretation of the two-dimensional Fourier transform in terms of the one-dimensional Fourier transform.

=== Inversion formula ===
For brevity, we choose to denote the one-dimensional Fourier transform (notation: <math>\tilde f</math>) of the Radon transform (notation: <math>\hat f</math>) over the variable <math>t</math> as follows:
<center><math>
\widetilde{(\hat f)}\,(k,\mathbf{u_t})=\mathrm{TF}_{1D(t)}\left(\hat f (t,\mathbf{u_t})\right)[k].
</math></center>
From the projection-slice theorem, show that the inversion formula of the Radon transform is:
<center><math>
f(\mathbf{r})=\frac{1}{(2\pi)^2}\int_0^\pi\int_{-\infty}^{+\infty}|k|\widetilde{(\hat f)}\,(k,\mathbf{u_t})\,e^{+i(\mathbf{r}\cdot\mathbf{u_t})k}\,\mathrm{d}k\mathrm{d}\Phi.
</math></center>

T-I-2

2021-11-09T14:01:54Z

Rosso:

=Linear response theory and Kramers-Kronig relations=

The goal of this homework is to introduce the theory of linear response and, among the related properties, such as causality, the so-called Kramers-Kronig relations. We will consider the damped harmonic oscillator as a prototypical example to discuss this topic.

== Reminder on the Fourier transform ==

Throughout this assignment, we will use the symmetric convention of Fourier transform. The Fourier transform of a function <math>f(t)\in L^1(\mathbb{R})</math> is defined as:
<center><math>
\widehat f(\omega):=\frac{1}{\sqrt{2\pi}}\int d t \,f(t)\,e^{i\omega t}.
</math></center>
With this convention, the Fourier transform of a product results in a convolution, with the following pre-factor: <math>
\sqrt{2\pi}\,\widehat{(f\, g)}=\widehat f \ast \widehat g .</math>

The repeated application of the transform gives <math>
\widehat{\widehat f}(t)=f(-t).</math>

The conjugate of a Fourier transform is equal to the transform of the conjugate, with a change of sign in the argument: <math>
\overline{\widehat f}(\omega)=\widehat{\overline{f}}(-\omega).</math>

Furthermore, we can exchange the Fourier transform under the integral sign: <math>
\int _\mathbb{R}dt \, \hat f(t)\, g(t)=\int _\mathbb{R}dt \, f(t)\,\hat g(t) .
</math>

The last property implies the result <math>
\int _\mathbb{R}d\omega \, \overline{\hat f}(\omega)\,\hat g(\omega)=\int _\mathbb{R}dt \, \overline{f(t)}\,\hat g(t) ,
</math>
which gives directly the Plancherel theorem (see lecture notes, page 45).

== Example: damped harmonic oscillator in one dimension ==
We consider a damped harmonic oscillator <math>x(t)</math>, which is described by the following equation of motion:
<center><math>
\ddot x + 2 \gamma \,\dot x + \omega _0^2\,x=f(t).
</math></center>
First, we will compute its response to an harmonic perturbation, and then its response to a generic perturbation by introducing its Green function.

=== Response to an harmonic perturbation ===

We assume that the oscillator is excited by an harmonic perturbation of the form:
<math>
f(t)=F_\omega\,e^{-i\omega t}.
</math>
We look for a solution of the form:
<math>
x(t)=X_\omega\,e^{-i\omega t}.
</math>

'''Q1:''' Show that
<center><math>
X_\omega=\hat g(\omega)\, F_\omega,
</math></center>
where
<center><math>
\hat g(\omega)=-\frac{1}{(\omega-\omega_1)(\omega-\omega_2)}.
</math></center>

Write the explicit form of the poles that we have indicated as <math>\omega_1</math> and <math>\omega_2</math>.

=== Response to a generic perturbation ===

We consider a generic perturbation represented by <math>f(t)</math>, whose Fourier transform is given by <math>\hat f (\omega)</math>.

'''Q2:''' Write <math>x(t)</math> as a function of <math>\hat g(\omega)</math> and <math>\hat f(\omega)</math>.

'''Q3:''' Show that it follows that
<center><math>
x(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}d t' \, g(t-t')f(t'),
</math></center>
where <math>g(t)</math> is the function having <math>\hat g(\omega)</math> as Fourier transform. The above equation is nothing but the convolution of <math>f</math> with <math>g</math>: <math>\sqrt{2\pi}x(t)=(f\ast g)(t)</math>.

=== Holomorphism of the response and consequences ===
'''Q4:''' Show that <math>g(t)</math> is the response of the system to an impulsive perturbation <math>f(t)=\sqrt{2\pi}\delta(t)</math>.
What is the effect of a perturbation applied at time <math>s</math> on the solution at time <math>t<s</math>?

'''Q5:''' We have defined <math>\hat g (\omega)</math> for <math>\omega\in\mathbb{R}</math>, i.e. on the real axis. Show that its extension to the complex plane is holomorphic everywhere (except at the poles). Using the explicit expression for <math>\widehat g (\omega)</math> derived above, compute explicitly <math>g(t)</math>. You will have to treat separately the case of positive time and the case of negative time. Use the residue theorem.

'''Q6:''' Deduce the expression of the response:
<center><math>
x(t)=\int_{-\infty}^t dt' \, e^{-\gamma (t-t')}\,\frac{\sin\left[(\omega_0^2-\gamma^2)^{1/2} (t-t')\right]}{(\omega_0^2-\gamma^2)^{1/2}}f(t').
</math></center>
The above expression indicates the Green function of the system. Notice that the response of the system is causal.

It is worth highlighting the importance of the holomorphism in this proof. Indeed, we have seen that we had to apply Jordan's lemma. Therefore, we need a relatively fast decay of the response function at infinity, i.e. a system that does not diverge.

We also point out that, if <math>\gamma</math> was negative, we would have lost causality: we can guess a relation between causality and the dissipation of energy in a passive system.

== General properties of linear response ==
We look for a generalization of the properties that we have seen in the previous example. We discuss an equation of motion (or possibly another physical equation)
<center><math>
\mathcal{L}x(t)=f(t),
</math></center>
with a linear operator <math>\mathcal{L}</math> (of differentiation, multiplication, integration, etc.), and with an inhomogeneity <math>f(t)</math> (external stimulation, perturbation, etc.). The generic solution is unknown since the explicit form of the operator is not specified. However, we know that this solution must have some physical properties:
* Linearity: <math>x</math> is a function or a linear functional of <math>f</math>, which reads <math>
x(t)=\int _\mathbb{R}ds \, g(t-s)f(s). </math>
* Causality: in the sense that the effect cannot precede the cause: <math>
g(t)=0</math> for <math> t<0</math>.
* The total response to a finite perturbation must be finite.
* The response to a real perturbation, <math>f(t)\in \mathbb{R}</math>, must be real as well, <math>x(t)\in\mathbb{R}</math> (at least in the case we are not considering quantum mechanics effects).

=== Link between odd/even and real/imaginary ===
'''Q7:''' Show the following properties of the Fourier transform:
* if <math>g\in \mathbb{R} </math> is an odd function, then <math> \hat g\in i\mathbb{R} </math> is also an odd function;
* if <math> g\in i\mathbb{R}</math> is an odd function, then <math>\hat g\in \mathbb{R}</math> is also an odd function;
* if <math> g\in \mathbb{R} </math> is an even function, then <math> \hat g\in \mathbb{R} </math> is also an even function;
* if <math>g\in i\mathbb{R} </math> is an even function, then <math> \hat g\in i\mathbb{R} </math> is also an even function.

'''Q8:''' Conclude that, if <math>g</math> is real, its Fourier transform is written as <math>\hat g=\hat g ' +i\hat g ''</math>, with <math>\hat g '</math> even, and <math>\hat g '' </math> odd.

'''Q9:''' Derive the same properties for the case of the oscillator.

=== Preliminaries ===
We define the sign function as
<center><math>
S(t)=\begin{cases}
-1 & t<0\\
+1 &t\geq 0
\end{cases},
</math></center>
and we want to show that its Fourier transform is
<center><math>
\hat S(\omega)=\sqrt{\frac{2}{\pi}}\frac{i}{\omega}.
</math></center>
Since <math>S</math> is not summable, we have to use an auxiliary function <math>S_\epsilon (t):=e^{-\epsilon |t|}S(t)</math>.

'''Q10:''' Show that <math>\lim_{\varepsilon\rightarrow 0}\hat S_\epsilon (\omega)=\sqrt{\frac{2}{\pi}}\frac{i}{\omega}</math>.

'''Q11:''' Discuss the possibility of using <math>\hat S</math> instead of <math>\hat S_\epsilon</math> in a computation (e.g., in the example below).

=== Causality and Kramers-Kronig relations ===
'''Q12:''' Using the expression of the Fourier transform of a convolution and the sign function defined above, show the following property of Fourier transform:
<center> If <math>g(t)=0</math> <math>\forall t<0</math>, then <math>\quad\hat g(\omega)=\frac{1}{i\pi}\,{\rm PV} \int_\mathbb{R}d\xi \,\frac{\operatorname{Im} \hat g(\xi)}{\xi -\omega}</math>.
</center>
By separating the real and imaginary part of this equation, we obtain Kramers-Kronig relations:
<center><math>
\operatorname{Re}\hat g(\omega)=\frac{1}{\pi}\, {\rm PV}\int_\mathbb{R}d\xi \,\frac{\operatorname{Im}\hat g(\omega)}{\xi-\omega},
</math></center>
<center><math>
\operatorname{Im} \hat g(\omega)=-\frac{1}{\pi} \,{\rm PV}\int_\mathbb{R}d\xi \,\frac{\operatorname{Re}\hat g(\omega)}{\xi-\omega}.
</math></center>

'''Q13:''' Show that Kramers-Kronig relations express the fact that the extension <math>\hat g (\omega + i \nu)</math> to <math>\hat g (\omega)</math> on the complex upper-half plane <math>(\nu>0)</math> is holomorphic, without any poles, and that it decreases sufficiently fast as <math>|\omega +i\nu|\rightarrow \infty</math>.

We will start from a weaker result:

Show that the Kramers-Kronig relation is valid for all functions <math>\hat g (\omega +i\nu)</math> with the aforementioned properties. Apply the residue theorem integrating along the path <math>\gamma</math> displayed in the figure below, that can be reproduced via the following code:

<pre>
gamma = ParametricPlot[{{R Cos[t], R Sin[t]}, {-R*(1 - t/Pi) + (x - r)*t/Pi, 0}, {r Cos[t + Pi] + x, r Sin[t + Pi]}, {(x + r)*(1 - t/Pi) + R*t/Pi, 0}} /. {R -> 2, r -> 0.5, x -> 1}, {t, 0, Pi}, Ticks -> {{{-R, "-R"}, {x - r, "x-r"}, {x, "x"}, {x + r, "x+r"}, {R, "R"}} /. {R -> 2, r -> 0.5, x -> 1}, None}, TicksStyle -> Directive[Red, 20], AxesLabel -> {None}] /. Line[u_] :> Sequence[Arrowheads[Table[.05, {1}]], Arrow@Line[u]]
</pre>
[[File:Gamma_contour.png|400px|thumb|center|Sketch of the integration path <math>\gamma</math>.]]

Integrate the function <math>\hat g (z)/(z-\omega)</math> along this path. Compute the contribution of the small half-circle <math>C_r</math> in the limit <math>r\rightarrow 0</math> using its Taylor expansion in the vicinity of <math>\omega</math> and performing the change of variable <math>z'=\omega + r\,e^{i\theta}</math>.

The last result suggests the existence of a relation with the causality at <math>g(t)=0</math> <math>\forall t<0</math> and the fact that <math>g</math> is holomorphic without poles in the upper-half plane. This was detailed in the Titchmarsch theorem.

Main Page

2021-11-09T13:52:39Z

Rosso: /* Tutorials */

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

Main Page

2021-11-09T13:51:34Z

Rosso:

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2draft| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

File:Gamma contour.png

2021-10-13T18:11:32Z

Rosso: Rosso uploaded a new version of File:Gamma contour.png

File:Gamma contour.png

2021-10-13T18:10:20Z

Rosso:

T-I-2

2021-10-13T17:33:31Z

Rosso:

=Linear response theory and Kramers-Kronig relations=

The goal of this homework is to introduce the theory of linear response and, among the related properties, such as causality, the so-called Kramers-Kronig relations. We will consider the damped harmonic oscillator as a prototypical example to discuss this topic.

== Reminder on the Fourier transform ==

Throughout this assignment, we will use the symmetric convention of Fourier transform. The Fourier transform of a function <math>f(t)\in L^1(\mathbb{R})</math> is defined as:
<center><math>
\widehat f(\omega):=\frac{1}{\sqrt{2\pi}}\int d t \,f(t)\,e^{i\omega t}.
</math></center>
With this convention, the Fourier transform of a product results in a convolution, with the following pre-factor: <math>
\sqrt{2\pi}\,\widehat{(f\, g)}=\widehat f \ast \widehat g .</math>

The repeated application of the transform gives <math>
\widehat{\widehat f}(t)=f(-t).</math>

The conjugate of a Fourier transform is equal to the transform of the conjugate, with a change of sign in the argument: <math>
\overline{\widehat f}(\omega)=\widehat{\overline{f}}(-\omega).</math>

Furthermore, we can exchange the Fourier transform under the integral sign: <math>
\int _\mathbb{R}dt \, \hat f(t)\, g(t)=\int _\mathbb{R}dt \, f(t)\,\hat g(t) .
</math>

The last property implies the result <math>
\int _\mathbb{R}d\omega \, \overline{\hat f}(\omega)\,\hat g(\omega)=\int _\mathbb{R}dt \, \overline{f(t)}\,\hat g(t) ,
</math>
which gives directly the Plancherel theorem (see lecture notes, page 45).

== Example: damped harmonic oscillator in one dimension ==
We consider a damped harmonic oscillator <math>x(t)</math>, which is described by the following equation of motion:
<center><math>
\ddot x + 2 \gamma \,\dot x + \omega _0^2\,x=f(t).
</math></center>
First, we will compute its response to an harmonic perturbation, and then its response to a generic perturbation by introducing its Green function.

=== Response to an harmonic perturbation ===

We assume that the oscillator is excited by an harmonic perturbation of the form:
<math>
f(t)=F_\omega\,e^{-i\omega t}.
</math>
We look for a solution of the form:
<math>
x(t)=X_\omega\,e^{-i\omega t}.
</math>

'''Q1:''' Show that
<center><math>
X_\omega=\hat g(\omega)\, F_\omega,
</math></center>
where
<center><math>
\hat g(\omega)=-\frac{1}{(\omega-\omega_1)(\omega-\omega_2)}.
</math></center>

Write the explicit form of the poles that we have indicated as <math>\omega_1</math> and <math>\omega_2</math>.

=== Response to a generic perturbation ===

We consider a generic perturbation represented by <math>f(t)</math>, whose Fourier transform is given by <math>\hat f (\omega)</math>.

'''Q2:''' Write <math>x(t)</math> as a function of <math>\hat g(\omega)</math> and <math>\hat f(\omega)</math>.

'''Q3:''' Show that it follows that
<center><math>
x(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}d t' \, g(t-t')f(t'),
</math></center>
where <math>g(t)</math> is the function having <math>\hat g(\omega)</math> as Fourier transform. The above equation is nothing but the convolution of <math>f</math> with <math>g</math>: <math>x(t)=(f\ast g)(t)</math>.

=== Holomorphism of the response and consequences ===
'''Q4:''' Show that <math>g(t)</math> is the response of the system to an impulsive perturbation <math>f(t)=\delta(t)</math>.
What is the effect of a perturbation applied at time <math>s</math> on the solution at time <math>t>s</math>?

'''Q5:''' We have defined <math>\hat g (\omega)</math> for <math>\omega\in\mathbb{R}</math>, i.e. on the real axis. Show that its extension to the complex plane is holomorphic everywhere (except at the poles). Using the explicit expression for <math>\widehat g (\omega)</math> derived above, compute explicitly <math>g(t)</math>. You will have to treat separately the case of positive time and the case of negative time. Use the residue theorem.

'''Q6:''' Deduce the expression of the response:
<center><math>
x(t)=\int_{-\infty}^t dt' \, e^{-\gamma (t-t')}\,\frac{\sin\left[(\omega_0^2-\gamma^2)^{1/2} (t-t')\right]}{(\omega_0^2-\gamma^2)^{1/2}}f(t').
</math></center>
The above expression indicates the Green function of the system. Notice that the response of the system is causal.

It is worth highlighting the importance of the holomorphism in this proof. Indeed, we have seen that we had to apply Jordan's lemma. Therefore, we need a relatively fast decay of the response function at infinity, i.e. a system that does not diverge.

We also point out that, if <math>\gamma</math> was negative, we would have lost causality: we can guess a relation between causality and the dissipation of energy in a passive system.

== General properties of linear response ==
We look for a generalization of the properties that we have seen in the previous example. We discuss an equation of motion (or possibly another physical equation)
<center><math>
\mathcal{L}x(t)=f(t),
</math></center>
with a linear operator <math>\mathcal{L}</math> (of differentiation, multiplication, integration, etc.), and with an inhomogeneity <math>f(t)</math> (external stimulation, perturbation, etc.). The generic solution is unknown since the explicit form of the operator is not specified. However, we know that this solution must have some physical properties:
* Linearity: <math>x</math> is a function or a linear functional of <math>f</math>, which reads <math>
x(t)=\int _\mathbb{R}ds \, g(t-s)f(s). </math>
* Causality: in the sense that the effect cannot precede the cause: <math>
g(t)=0</math> for <math> t<0</math>.
* The total response to a finite perturbation must be finite.
* The response to a real perturbation, <math>f(t)\in \mathbb{R}</math>, must be real as well, <math>x(t)\in\mathbb{R}</math> (at least in the case we are not considering quantum mechanics effects).

=== Link between odd/even and real/imaginary ===
'''Q7:''' Show the following properties of the Fourier transform:
* if <math>g\in \mathbb{R} </math> is an odd function, then <math> \hat g\in i\mathbb{R} </math> is also an odd function;
* if <math> g\in i\mathbb{R}</math> is an odd function, then <math>\hat g\in \mathbb{R}</math> is also an odd function;
* if <math> g\in \mathbb{R} </math> is an even function, then <math> \hat g\in i\mathbb{R} </math> is also an even function;
* if <math>g\in i\mathbb{R} </math> is an even function, then <math> \hat g\in \mathbb{R} </math> is also an even function.

'''Q8:''' Conclude that, if <math>g</math> is real, its Fourier transform is written as <math>\hat g=\hat g ' +i\hat g ''</math>, with <math>\hat g '</math> odd, and <math>\hat g '' </math> even.

'''Q9:''' Derive the same properties for the case of the oscillator.

=== Preliminaries ===
We define the sign function as
<center><math>
S(t)=\begin{cases}
-1 & t<0\\
+1 &t\geq 0
\end{cases},
</math></center>
and we want to show that its Fourier transform is
<center><math>
\hat S(\omega)=\sqrt{\frac{2}{\pi}}\frac{i}{\omega}.
</math></center>
Since <math>S</math> is not summable, we have to use an auxiliary function <math>S_\epsilon (t):=e^{-\epsilon |t|}S(t)</math>.

'''Q10:''' Show that <math>\lim_{\varepsilon\rightarrow 0}\hat S_\epsilon (\omega)=\sqrt{\frac{2}{\pi}}\frac{i}{\omega}</math>.

'''Q11:''' Discuss the possibility of using <math>\hat S</math> instead of <math>\hat S_\epsilon</math> in a computation (e.g., in the example below).

=== Causality and Kramers-Kronig relations ===
'''Q12:''' Using the expression of the Fourier transform of a convolution and the sign function defined above, show the following property of Fourier transform:
<center> If <math>g(t)=0</math> <math>\forall t<0</math>, then <math>\quad\hat g(\omega)=\frac{1}{i\pi}\,{\rm PV} \int_\mathbb{R}d\xi \,\frac{\operatorname{Im} \hat g(\xi)}{\xi -\omega}</math>.
</center>
By separating the real and imaginary part of this equation, we obtain Kramers-Kronig relations:
<center><math>
\operatorname{Re}\hat g(\omega)=\frac{1}{\pi}\, {\rm PV}\int_\mathbb{R}d\xi \,\frac{\operatorname{Im}\hat g(\omega)}{\xi-\omega},
</math></center>
<center><math>
\operatorname{Im} \hat g(\omega)=-\frac{1}{\pi} \,{\rm PV}\int_\mathbb{R}d\xi \,\frac{\operatorname{Re}\hat g(\omega)}{\xi-\omega}.
</math></center>

'''Q13:''' Show that Kramers-Kronig relations express the fact that the extension <math>\hat g (\omega + i \nu)</math> to <math>\hat g (\omega)</math> on the complex upper-half plane <math>(\nu>0)</math> is holomorphic, without any poles, and that it decreases sufficiently fast as <math>|\omega +i\nu|\rightarrow \infty</math>.

We will start from a weaker result:

Show that the Kramers-Kronig relation is valid for all functions <math>\hat g (\omega +i\nu)</math> with the aforementioned properties. Apply the residue theorem integrating along the path <math>\gamma</math> displayed in the figure below, that can be reproduced via the following code:

<pre>
gamma = ParametricPlot[{{R Cos[t], R Sin[t]}, {-R*(1 - t/Pi) + (x - r)*t/Pi, 0}, {r Cos[t + Pi] + x, r Sin[t + Pi]}, {(x + r)*(1 - t/Pi) + R*t/Pi, 0}} /. {R -> 2, r -> 0.5, x -> 1}, {t, 0, Pi}, Ticks -> {{{-R, "-R"}, {x - r, "x-r"}, {x, "x"}, {x + r, "x+r"}, {R, "R"}} /. {R -> 2, r -> 0.5, x -> 1}, None}, TicksStyle -> Directive[Red, 20], AxesLabel -> {None}] /. Line[u_] :> Sequence[Arrowheads[Table[.05, {1}]], Arrow@Line[u]]
</pre>
[[File:Gamma_contour.png|400px|thumb|center|Sketch of the integration path <math>\gamma</math>.]]

Integrate the function <math>\hat g (z)/(z-\omega)</math> along this path. Compute the contribution of the small half-circle <math>C_r</math> in the limit <math>r\rightarrow 0</math> using its Taylor expansion in the vicinity of <math>\omega</math> and performing the change of variable <math>z'=\omega + r\,e^{i\theta}</math>.

The last result suggests the existence of a relation with the causality at <math>g(t)=0</math> <math>\forall t<0</math> and the fact that <math>g</math> is holomorphic without poles in the upper-half plane. This was detailed in the Titchmarsch theorem.

T-I-2

2021-10-13T08:11:15Z

Rosso: Created page with "http://lptms.u-psud.fr/wiki-cours/index.php/T-I-4"

http://lptms.u-psud.fr/wiki-cours/index.php/T-I-4

Main Page

2021-10-13T08:10:55Z

Rosso:

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco [https://sites.google.com/view/personal-webpage-of-francesca/home]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

Main Page

2021-10-13T07:56:12Z

Rosso:

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

* Francesca Mignacco

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

Main Page

2020-09-21T13:04:03Z

Rosso:

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the LPTMS's[http://lptms.u-psud.fr/] website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

Among the years some of our members have been involved in teaching at ESPCI:

* Alberto Rosso [http://lptms.u-psud.fr/alberto_rosso/]

* Raoul Santachiara [http://lptms.u-psud.fr/raoul_santachiara/]

* Gregory Schehr [http://lptms.u-psud.fr/gregory-schehr/]

* Veronique Terras [http://lptms.u-psud.fr/veronique-terras/]

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

Main Page

2020-09-21T12:58:26Z

Rosso:

Main Page

2020-09-21T12:55:29Z

Rosso: /* Tutorials */

[[Image:Gumbel.png|600px|right|thumb| Gumbel distribution. [[T-II-3| Extreme value Statistics]] ]]

[[Image:cylflow.jpg|300px|left|thumb|Streamlines (lignes de courant) around a disk in two dimensions. [[T-I-1| Conformal map ]]]]

This is the interactive website of the Tutorials for the ESPCI Mathematical methods [http://www.espci.fr/enseignement/maths/Enseignement/index.html].

These Tutorials are conducted by [[User:rosso| Alberto Rosso]] and [[User:Gregory| Gregory Schehr]].

=Tutorials=

{| class="wikitable" border="1"
|-
! width="100"|Date
! width="500"| First Year : 15h45-17h45
! width="500"| Second Year : 13h15-15h15

|-valign="top"

| First Tutorial
|
* [[T-I-1| Analytic Functions: Conformal map]]
|
* [[T-II-1| Partial Differential Equations: Wave breaking, Burgers and Shocks]]
[[Complements | Complements]]
|-valign="top"
| Second Tutorial
|
* [[T-I-2| Residue Theorem: Green Functions]]

|
* [[T-II-2new| Variational Methods: Sound waves in solids and fluids]]
|-valign="top"
| Third Tutorial
|
* [[T-I-3|Integral Transform: Radon Transform]]
|
* [[T-II-3| Probability: Extreme value Statistics]]
|}

T-II-2new

2019-11-16T11:29:00Z

Rosso: /* Displacement fields and strain tensor */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Here we review the basic objects of elastic theory that you have done last year (e.g. see pages 62-66 of '' Mécanique du solide et des matérieaux, partie 1, Pascal Kurowski'').

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at a distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order in the deformation it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>.

Tip: it is convenient to write the element volume in the principal coordinate around the point <math>\vec r</math>. There the symmetric deformation tensor is diagonal.

Conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-16T11:28:37Z

Rosso: /* Displacement fields and strain tensor */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Here we review the basic objects of elastic theory that you have done last year (e.g. see pages 62-66 of '' Mécanique du solide et des matérieaux, partie 1, Pascal Kurowski'').
Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at a distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order in the deformation it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>.

Tip: it is convenient to write the element volume in the principal coordinate around the point <math>\vec r</math>. There the symmetric deformation tensor is diagonal.

Conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-16T11:27:32Z

Rosso: /* Displacement fields and strain tensor */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Here we review the basic objects of elastic theory that you have done last year (e.g. see pages 62-66 of '' Mécanique du solide et des matérieaux, partie 1, Pascal Kurowski''}
Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at a distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order in the deformation it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>.

Tip: it is convenient to write the element volume in the principal coordinate around the point <math>\vec r</math>. There the symmetric deformation tensor is diagonal.

Conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-16T11:26:45Z

Rosso: /* Compression: the (relative) density field */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Here we review the basic objects of elastic theory that you have done last year (e.g. see pages 62-66 of '' M\'ecanique du solide et des mat\'erieaux, partie 1, Pascal Kurowski''}
Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at a distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order in the deformation it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>.

Tip: it is convenient to write the element volume in the principal coordinate around the point <math>\vec r</math>. There the symmetric deformation tensor is diagonal.

Conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-16T11:26:00Z

Rosso: /* Displacement fields and strain tensor */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Here we review the basic objects of elastic theory that you have done last year (e.g. see pages 62-66 of '' M\'ecanique du solide et des mat\'erieaux, partie 1, Pascal Kurowski''}
Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at a distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order in the deformation it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>.

Tip: it is convenient to write the element volume in the principal coordinate around the point $\vec r$. There the symmetric deformation tensor is diagonal.

Conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-13T09:19:14Z

Rosso:

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-11T14:53:56Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = - \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at T=273 K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-11T14:53:36Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = - \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (331 m /s at $T=273$ K).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> (at T=273 K)) compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-11T14:45:50Z

Rosso: /* Equation of motion with the variational principle */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = - \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-11T14:23:08Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sound speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-11T14:12:40Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_l =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-11T12:21:07Z

Rosso: /* Kinetic energy */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \frac{\rho_0}{2} \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_t =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-10T17:11:24Z

Rosso: /* Equation of motion with the variational principle */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \rho_0 \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_k} \frac{d}{d (\partial_k u_i)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_t =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-11-10T15:40:58Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \rho_0 \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_i} \frac{d}{d (\partial_i u_k)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_t =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d V </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} (V-V_0) </math></center>

'''Q11:''' Show that
<center><math> K= - V_0 \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-10-13T20:37:05Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \rho_0 \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_i} \frac{d}{d (\partial_i u_k)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_t =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d \vec{r} </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} \delta V </math></center>

'''Q11:''' Show that
<center><math> K= - \frac{1}{V_0} \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the bulk modulus and the sound speed predicted by Laplace.

T-II-2new

2019-10-13T20:31:07Z

Rosso: /* Sound speed in perfect gas */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \rho_0 \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_i} \frac{d}{d (\partial_i u_k)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

 
 

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_t =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d \vec{r} </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} \delta V </math></center>

'''Q11:''' Show that
<center><math> K= - \frac{1}{V_0} \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

 
 

Newton idea was that the the compressed gas remain at the same temperature <math>T</math>, so that
<center><math>P V= </math> constant</center>

'''Q12:''' Show that in this case
<center><math> K= P_0 </math> </center>
and using that at 1 atmosphere (<math>101325</math> Pa) the density of air is <math>1.293</math> kg/m<math>^3</math> compute the sound speed predicted by Newton.

 
 

However, Laplace undestood that a sound wave gas oscillates sufficiently rapidly that there is no time to thermalize. The heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

'''Q13:''' Compute the sound speed predicted by Laplace.

T-II-2new

2019-10-13T20:20:56Z

Rosso: /* Elasticity in fluids */

=Variational methods: Sound waves in solid and fluid materials=

 
 
 

Consider a solid body or a fluid at rest. Under the action of an external force it will start to move or flow and the work done the force is transformed in kinetic energy. This is the case of a rigid body or an incompressible fluids, but in real materials part of the work can be used to induce a deformation.
A compression for example has an ergetic cost both for fluids and solids, while a change in shape (shear) has a cost only for solids.

If the deformation is not too large the work is stored in terms of elastic energy and it is reversibly released when the force is removed. As a consequence, after a sudden pertubation, elastic waves propagates in the material.
Similarly to light, which are waves associated to the electromagnetic field, the sound that we can ear are waves associated to the elastic field. But which field?

At variance with their electromagnetic cousins, elastic waves can propagate only inside materials and display a quite different nature in solids with respect to fluids. Here we derive their equations of motion in both kind of material and compute their speed.

==Part I: Elasticity in solids==
 
 
 

The first succesful attempt to describe deformations in solids is provided by the ''theory of elasticity''. The theory is based on the assumptions that (1) the solid body can modeled as a continuum medium and (2) the energetic cost of the deformation is kept at the lowest order in perturbation. The fundamental equations of elasticity have been established by Cauchy and Poisson in the twenties of the 19th century, well before the discovery of the atomic structure of matter.

 
 
 

==Displacement fields and strain tensor==

 
 
 

Consider a perfectly isotropic continuum medium with density mass <math>\rho_0</math> at equilibrium:. Assume that at time <math> t=0 </math> a body is at rest and no external forces are applied. Each point of the body can be identified with its postion in the space:

<center><math> \vec r = (x_1=x, x_2 =y, x_3 =z) </math></center>

Then under the action of external forces the body starts to be deformed: the point located in <math> \vec r </math> moves to a new location <math> \vec r_{\text{new}} </math> at time <math> t>0 </math> . The displacement, <math> \vec u= \vec r_{\text{new}}-\vec r </math>, differs from point to point of the body and is then a ''field'', namely
a function of the original position <math> \vec r </math>:

<center><math> \vec u( \vec r,t) = \left(u_1( \vec r,t), u_2( \vec r,t), u_3( \vec r,t)\right) </math> </center>

the ''displacement field'' encodes then all information about the deformation.

However large displacements does not always correspond to large deformation as the body can move rigidly without deformation. To quantify how much a body is deformed it is useful to introduce the symmetric ''strain tensor'' <math> \epsilon_{ik}</math> :
<center><math> \epsilon_{ik} = \frac12 \left( \partial_k u_i + \partial_i u_k \right)\, , \quad \text{with}\; \partial_k =\frac{\partial}{\partial x_k} \quad(1) </math></center>

'''Q1:''' Take two points at an infinitesimal distance <math> d \ell </math> and compute the distance <math> d \ell_{\text{new}} </math> after a small deformation. Show that at the lowest order it can be written as
<center><math> d \ell_{\text{new}}^2 \approx d \ell^2 + 2 \sum_{i,k} \epsilon_{ik} d x_i dx_k </math></center>

=== Compression: the (relative) density field===

 
 
 

Consider now a local compression at the point <math>\vec r</math>. The volume element around the point, <math> d V= dx_1 \, dx_2 \, dx_3 </math>, is squeezed to a smaller volume <math>d V_{\text{new}}</math> and the density mass will locally increases from the uniform value <math>\rho_0</math>:
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0 \left( 1+\sigma(\vec r,t)\right)</math> </center>
The ''scalar field'' <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

<center><math>\sigma(\vec r,t) = \frac{\rho(\vec r,t)-\rho_0}{\rho_0} =- \frac{d V_{\text{new}}-d V}{ d V} </math> </center>

'''Q2:''' Show that at the first order in perturbation one has

<center><math> d V_{\text{new}}\approx d V \left(1+ \text{Tr}(\epsilon)\right) \quad </math></center>

and conclude that
<center><math>\sigma(\vec r,t) =-\text{div}(\vec u) </math> </center>

 
 
 

In general all deformation can be represented as the sum of a uniform compression (change in volume without change of shape) and a shear (change of shape without change of volume):

<center><math> \epsilon_{ik}= \left(\epsilon_{ik} -\frac{1}{3} \delta_{ik}\text{Tr}(\epsilon)\right) + \frac{1}{3} \delta_{ik} \text{Tr}(\epsilon) \quad \text{with} \quad \delta_{ik}= \begin{cases}
0 \; \text{if} \; i\ne k\\
1 \; \text{if} \; i=k
\end{cases} \quad
(2) </math></center>
The first term is a shear as its trace is zero. The second term is a compression/rarefaction.

== Energy cost of the deformation==

 
 

A body at rest, in abscence of external force and at thermal equilibrium does not display any deformation, this means that when the strain tensor is zero for all points, the potential energy displays a minimum. As a consequence for small deformations the first terms of the expansion of the potential energy must be quadratic in the strain tensor (elastic approximation).
The more general quadratic form can be written as <math> \int d \vec r \sum_{ijkl} \epsilon_{ij} \, C_{ijkl} \, \epsilon_{kl} </math>, where the tensor <math>C_{ijkl}</math> contains all the elastic moduli. For an isotropic material the energy cost should be frame independent and we can write a simpler quadratic form that we will discuss in class:

<center><math>E_{\text{el}} =\int d \vec r \left[ \frac{K}{2} \text{Tr}(\epsilon)^2 + \mu \sum_{ik} \left( \epsilon_{ik}-\frac{1}{3}\delta_{ik} \text{Tr}(\epsilon) \right)^2\right] </math></center>
The first term accounts for the cost of the change in volume, the second for the change of shape. The positive constants <math> K , \mu </math> are respectively the bulk and the shear modulus and have dimensions of a pressure. These two constants are material dependent and fully describe its elastic properties.

 
 

==Kinetic energy==

 
 

The kinetic energy can then be written as
<center>
<math> T = \rho_0 \int d \vec r \left( \dot u_1^2 ( \vec r,t) +\dot u_2^2 ( \vec r,t)+\dot u_3^2 ( \vec r,t) \right) </math></center>
where <math> \rho_0</math> is the mass density of the material at rest. Here used the fact that the mass contained in an infinitesimal volume around the point located in <math> \vec r </math> is <math> \rho_0 \, d V= \rho_0 \, d \vec r </math>.

 
 

== Equation of motion with the variational principle==

 
 

'''Q3:''' Show that the potential elastic energy can be re-written in the following form:

<center><math> E_{\text{el}} = \int d \vec r \left[\frac{1}{2}\left( K-\frac{2}{3} \mu\right) \left( \sum_l \partial_l u_l \right)^2 +\frac{\mu}{4} \sum_{l,m} (\partial_l u_m +\partial_m u_l)^2 \right] </math> </center>

 

The associated action is given by
<center><math> S= \int d t \int d \vec{r} \, \mathcal{L} = \int d t \int d \vec{r} \, \left( \mathcal{T}_{\text{kin}} - \mathcal{E}_{\text{el}} \right) </math></center>
where <math> {\mathcal L} </math> is the Lagrangian density, <math>\mathcal{T} </math> is the kinetic energy density, and <math> \mathcal{E}_{\text{el}} </math> is the potential energy density

 

'''Q4:''' Write explicitly <math> \mathcal{T}, \mathcal{E}_{\text{el}} </math>.

'''Q5:''' Using the variational methods show that the equation of motion for a given component <math> u_i(\vec{r},t)</math> of the displacement writes
<center><math> \rho_0 \ddot u_i = \sum_k
\frac{d}{d x_i} \frac{d}{d (\partial_i u_k)} \mathcal{E}_{\text{el}} </math></center>
'''Q6:''' Show that the 3 equations of motion take the form

<center><math> \rho_0 \ddot u_i = \mu \sum_k
\frac{\partial^2 u_i}{\partial x_k^2} + ( K+\frac{\mu}{3}) \sum_k \frac{\partial^2 u_k}{\partial x_i \partial x_k} </math></center>
and organize them in the vectorial form:

<center><math> \ddot {\vec u} = c_t^2 \nabla^2 {\vec u} + (c_l^2-c_t^2) \vec{\text{grad}} \;\text{div}\,(\vec u)</math></center>
'''Q7:''' Provide the expression of the two constants <math> c_t, c_l</math>
To make progress we will consider plane waves, namely deformations that are fonction of <math>(x, t)</math> only.
<center><math> {\vec u}({\vec r}, t)={\vec u}(x,t)</math></center>
This means that all <math>y, z</math> derivatives
are zero.

'''Q8:''' Write the explicit form of the three wave equations. They are D'Alembert equations, explain why <math> c_t, c_l</math> are called transverse mode velocity and longitudinal mode velocity.

 

It is possible to show that the result obtained for plain waves is very general: in solids elastic waves has two transverse components propagating at velocity <math> c_t</math> and one longitudinal component propagating at velocity <math> c_l > c_t </math>.

Conclude that the elastic field in fluids is the displacement vectorial field.

<gallery widths=500px heights=300px mode="packed">
Image:earthquake.png|Seismic waves recorded by a seismograph after an earthquake. Longitudinal waves are the quickest and for this reason called primary (P-waves), then arrive the transverse waves, for this reason called secondary (S-waves). Finally it's the turn of Rayleigh waves which are also solution of the D'Alembert equation but propagate only on the Earth surface. They are slower and less damped then bulk waves.
Image:earthquake2.png| The time delay between the P and S waves grows with seismograph-epincenter distance. The measure of three independent seismographs allows thus to identify the location of the epicenter.
</gallery>

==Elasticity in fluids==

 
 

A fluid does not oppose a change in its shape with any internal resistance, but displays a finite energy cost to compression. Using the same reasoning used before we can write the Lagrangian density:

<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 - \frac{K}{2} (\text{div}(\vec{u}))^2 </math></center>
'''Q9''' Show that the associated equation of motion
<center><math> \rho_0\, \ddot \vec{u} = K \vec{\text{grad}}\left(\text{div}(\vec{u})\right) </math></center>

'''Q10''' Show that the previous equation of motion corresponds to the D'Alembert Equation for the scalar field <math> \sigma(\vec r,t) </math>.
Conclude that the elastic field in fluids is the scalar relative density field.

<gallery widths=500px heights=400px mode="packed">
Image:Voyage.png|Men landed on the Moon 50 years ago, but a Journey to the Center of the Earth remains science fiction. The deepest artificial holes go only 12 km into the Earth's crust.
Image:Earthsturcture.png|The chemical-physical structure of the 6000 km of our planet are known by the ''trip reports'' of seismic waves. For example we know the core is fluid as secondary waves cannot travel deeper than 3000 Km.
</gallery>

==Sound speed in perfect gas==

Generally speaking the sond speed (the transverse one for solids) takes the form

<center><math>c_t =\sqrt{\frac{K + (4 \mu)/3}{\rho_0}}</math> </center>

the constant K is much bigger in solid then liquids and much bigger in liquids then gases. For this reason sound propagates much faster in solids (6 km /s in steal) then in water (1 km /s) or, even worst, in air (343 m /s).

For perfect gases it is possible to evaluate the bulk modulus and then the speed sound velocity in gases. Newton did the first attempt, but he made a mistake corrected later by Laplace. Consider the gas at equilibrium at temperature <math>T</math>, pressure <math>P_0</math> and density <math>\rho_0</math>. We consider a sudden expansion/compression from a volume <math>V_0</math> around <math>\vec r</math> to <math>V_0+\delta V</math> and evaluate again the stored elastic energy. It can be written as the work associated to the expansion/compression:
<center><math> E_{\text{el}} = \frac{K}{2}\int d \vec r \text{Tr}(\epsilon)^2 = \frac{K}{2} \frac{(\delta V)^2}{V_0} =- \int_{V_0}^{V_0+\delta V} \, P(V) \, d \vec{r} </math> </center>
the pressure can be expanded around the equlibrium
<center> <math> P(V) \approx P_0 + \left(\frac{\partial P }{\partial V}\right)_{V=V_0} \delta V </math></center>

'''Q11:''' Show that
<center><math> K= \frac{1}{V_0} \left(\frac{\partial P }{\partial V}\right)_{V=V_0} </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.

, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.