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.


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 :)

  • $\begingroup$ 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

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