I am trying to derive the out-of-plane phonon dispersion relation for a membrane. As far as I can tell, one of the simplest ways to do so is with a Lagrangian of the form:

$$L_{bending}=\frac{1}{2}\rho \dot{h}^2-\frac{1}{2}\kappa(\nabla^2h)^2$$

Where $h$ is the out-of-plane displacement, $\rho$ the mass density and $\kappa$ the bending rigidity.

I tried to proceed by using the Euler-Lagrange equations with the substitution:

$$h=h_o \exp\left[-i(\omega t + qa)\right]$$

as per standard-solid-state phonon treatments, but I don't know how to deal with the $(\nabla^2 h)$ term following the above substitution (perhaps that is the incorrect part).

I know the answer should turn out to be:

$$\omega^2=\frac{\kappa q^4}{\rho}$$

Note: Many of the papers I look at point to the following paper for the answer - but Google gives me nothing for it. But I feel the answer is simple and I am not doing the math correctly. There are various more complicated treatments in the literature, including QM formulations, none of which I can follow very well.

I. M. Lifshitz, Zh. Eksp. Teor. Fiz, vol. 22, p. 475, 1952.


Here we go :

Lets considere the lagrangian density of our free membrane. For convenience, we will use a cartesian set of coordinates, so that :

$$\mathcal{L}=\frac{\rho}{2}\dot{h}-\frac{\kappa}{2}\left[{\partial_x}^2 h+{\partial_y}^2 h\right]^2$$

Apparently, this lagrangian depends on a quite large number of variables :

$$\mathcal{L}(x,y,t,h,\dot{h},\partial^2_x h,\partial^2_y h)$$

But still nothing to be afraid of! :)

One can show (see this) that for this type of lagrangian density, Euler-Lagrange equation is :

$$\sum_{\alpha\beta}\frac{\partial}{\partial\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_\alpha h)}\right)-\frac{\partial^2}{\partial\alpha\partial\beta}\left(\frac{\partial\mathcal{L}}{\partial(\partial^2_{\alpha\beta} h)}\right)=\frac{\partial\mathcal{L}}{\partial h}$$

where $(\alpha,\beta)\in\{x,y,t\}$

Computing this equation for our membrane lagrangian, it follows the differential equation :

$$\rho\ddot{h}+\kappa\left(\partial^4_x h + \partial^4_y h\right)=0$$

Then, simply by taking the (space-time) Fourier transform $\tilde{h}$ of $h$ so that :

$$h(\vec{r},t)=\sum_\vec{q}\int\frac{d\omega}{2\pi} e^{i(\vec{q}\cdot\vec{r}-\omega t)}\tilde{h}(\vec{q},\omega)$$

where $\vec{r}=(x,y)$, and replacing it in the previous equation, you obtain the dispersion relation :

$$\omega^2=\frac{\kappa q^4}{\rho}$$

For a quantum treatment, it feels like you will have to compute the associated hamiltonian $\mathcal{H}$ of $\mathcal{L}$ and then quantized it, using the correspondence $\{\cdot\}\rightarrow\frac{1}{i\hbar}\left[\cdot\right]$ between Poisson brackets and commutators.

  • $\begingroup$ Im not sure I understand where $\tilde{h}$ comes into this. Whenever you have the differential equation can't you just make a substitution like I made above and get the answer? What is $\tilde{h}$ explicitly? $\endgroup$ – Fire Apr 14 '14 at 22:27
  • $\begingroup$ @Fire: It is a generalization of your solution, because any linear combination of such solutions (with different $q$ and $\omega^2=\kappa q^4/\rho$) is also a solution. Dolan's solution is the most general one (i.e. you can choose any function $\tilde h$ and it will give you a solution). Look at what a Fourier transform is (and how to use them to solve linear differential equations) if you don't understand this. $\endgroup$ – Adam Apr 14 '14 at 23:44

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