2
$\begingroup$

I am working on a problem where I have been given the following Lagrangian density for describing sound vibrations in a gas:

$\mathcal{L}=\frac{1}{2}[\rho_0\dot{\eta}^2+2P_0\nabla\cdot\eta-\gamma P_0(\nabla\cdot\eta)^2]$

where $\rho_0$ is the equilibrium density, $P_0$ is the equilibrium pressure, $\gamma$ is the gas constant, and $\eta=(\eta_1,\eta_2,\eta_3)$ is the displacement vector at a point in space.

So far, I have calculated the conjugate momentum density vector (hopefully correctly), as:

$\pi=(\pi_1,\pi_2,\pi_3)$ where $\pi_j=\frac{\partial\mathcal{L}}{\partial\dot{\eta}_j}=\rho_0\dot{\eta}_j$, and then used this to write the Hamiltonian density as:

$\mathcal{H}=\pi^{\rho}\eta_{\rho,0}-\mathcal{L}=\frac{1}{2}\rho_0\dot{\eta}^2-P_0\nabla\cdot\eta+\frac{1}{2}\gamma P_0(\nabla\cdot\eta)^2$

Now I would like to express the Hamiltonian $H=\int d^3 x\mathcal{H}$ in the momentum representation, where $\eta_j=\frac{1}{\sqrt{V}}\sum\limits_k q_{j,k}(t)e^{i\vec{k}\cdot\vec{r}}$ and $\pi_j=\frac{1}{\sqrt{V}}\sum\limits_k p_{j,k}(t)e^{-i\vec{k}\cdot\vec{r}}$. I am pretty lost as to how to do this...I've tried doing the integration, but am confused, since $\mathcal{H}$ is written in terms of $\eta$, which seems as if it deals with position. Should I make a change of coordinates to express $\eta$ in terms of $\pi$? Possibly as a Fourier transform? Any suggestions would be appreciated.

$\endgroup$
1
+50
$\begingroup$

I am not too sure of the homework policy on this forum so I won't answer your question directly, but I hope this helps you :)

Starting from the Lagrange density $\mathcal L$; \begin{equation} \mathcal L=\frac 12 \rho _0\dot \eta^2 +P_0\vec \nabla \cdot \eta -\frac 12 \gamma P_0(\vec\nabla \cdot \eta )^2 \end{equation} The equation of motion for the $\eta$ field is given by:The equation of motion for the $\eta$ field is given by: \begin{equation} \rho_0\ddot \eta -\gamma P_0 \vec \nabla (\vec \nabla \cdot \eta )=0 \end{equation} The Fourier conjugate to the $\eta (\boldsymbol x,t)$ field is the momentum space $\tilde \eta(\boldsymbol p,t)$; \begin{equation} \eta(\boldsymbol x,t)=\int \frac{d^3p}{(2\pi)^3}e^{ip\cdot x}\tilde \eta (\boldsymbol p,t) \end{equation}

\begin{equation} \rho _0\frac{d^2}{dt^2}\int \frac{d^3p}{(2\pi)^3}e^{ipx}\tilde \eta -\gamma P_0\frac{d}{dx}\bigg[\frac{d}{dx}\int \frac{d^3p}{(2\pi)^3}e^{ipx}\tilde \eta\bigg]=0 \end{equation} Then we multiply by $\exp (-ip')$ and integrating over spatial dimensions $\int d^3x$, \begin{equation} \rho_0\frac{d^2}{dt^2}\int d^3x\frac{d^3p}{(2\pi)^3}e^{i(p-p')x}\tilde \eta -\gamma P_0\frac{d}{dx}\bigg[\frac{d}{dx}\int d^3 x\frac{d^3p}{(2\pi )^3}e^{i(p-p')x}\tilde \eta \bigg]=0 \end{equation} Using $\int d^3x \exp\{i(p-p')x \}=\delta (p-p')$, \begin{equation} \rho_0\frac{d^2}{dt^2}\int \frac{d^3p}{(2\pi )^3}\delta(p-p')\tilde \eta-\gamma P_0\frac{d}{dx}\bigg[\frac{d}{dx}\int \frac{d^3p}{(2\pi )^3}\delta (p-p')\tilde \eta \bigg]=0 \end{equation} The equation of motion for the Fourier conjugate field is then given by, \begin{equation} \rho _0\frac{d^2}{dt^2}\tilde \eta -\gamma P_0\vec \nabla (\vec \nabla \cdot \tilde \eta )=0 \end{equation} The momentum conjugate to the $\eta$ field is given by: \begin{equation} \pi_\mu =\frac{\partial \mathcal L}{\partial (\partial _\mu \eta)} \end{equation} We can use this to compute two momenta $\pi _t$ and $\pi _x$ as follows, \begin{equation} \pi _t=\frac{\partial \mathcal L}{\partial \dot \eta}=\rho _0\dot \eta \end{equation} \begin{equation} \pi _x=\frac{\partial \mathcal L}{\partial (\partial _x\eta)}=P_0-\gamma P_0(\vec \nabla \eta ) \end{equation} The Hamiltonian density $\mathcal H$ is then given by: \begin{equation} \mathcal H=\pi _\mu \partial ^\mu \eta-\mathcal L=\pi _t\dot \eta +\pi _x\vec\nabla \eta-\mathcal L \end{equation} From here a repeat of the above calculation should give you the answer! Hope that helps :)

$\endgroup$
  • $\begingroup$ Thank you for your help! I was going off of my notes and from the Goldstein "Classical Mechanics" text for my definition of $\pi^j$, but possibly I misinterpreted things. Also, quick question: why would integrating the equation of motion give the Hamiltonian? Or are you suggesting that I use a similar trick when integrating the Hamiltonian density to find the Hamiltonian? $\endgroup$ – Ben Sheller Sep 14 '15 at 23:45
  • $\begingroup$ Do you happen to have a reference for where the conjugate momenta are defined in the way you used them? $\endgroup$ – Ben Sheller Sep 14 '15 at 23:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.