# Talk:Capillary waves

## Thermal capillary waves

Hello, now I'm writing the same article for Russian wikipedia. While I was deducing the expression for mean square amplitude I found my result to be two times less than common one (that is in Molecular Theory of Capillarity and refered to in many articles). Could you please tell me weather I am right or not.

I claim that the mean energy of each mode is $$k_B T$$ rather than $$\frac{1}{2} k_B T$$. That's because each mode has to degrees of freedom $$A_{mn}$$ and $$B_{mn}$$, since each wave is $$A_{mn} \cos(\frac{2\pi}{L}mx+\frac{2\pi}{L}ny) + B_{mn} \sin(\frac{2\pi}{L}mx+\frac{2\pi}{L}ny)$$, with the energy of each mode proportional to $$A_{mn}^2+B_{mn}^2$$. This obviously lead to the mean energy of each mode to be $$k_B T$$. That was the real notation and now lets turn to the complex notation.

Each mode with the fixed wave vector is presented as $$h_\mathbf{k} \exp(i \; \mathbf{k} \cdot \boldsymbol{\tau})$$, $$\mathbf{k}=(\frac{2 \pi}{L}m, \frac{2 \pi}{L}n), \; m,n \in \mathbb{Z}$$ — wave vector, $$\boldsymbol{\tau}$$ — $$(x,y)$$ vector. The energy is proportional to $$h_\mathbf{k}^*h_\mathbf{k}$$ (indeed it is $$E_\mathbf{k}=\frac{\sigma L^2}{2} \left( \frac{2}{a_c^2} + \mathbf{k}^2 \right) h^*_\mathbf{k} h_\mathbf{k}$$). According to equipartition: $\left \langle x_i \frac{\partial H}{\partial x_j} \right \rangle = \delta_{ij} k_B T,$ we obtain: $\left \langle h_\mathbf{k} \, \frac{\partial}{\partial h_\mathbf{k}} \left[ \frac{\sigma L^2}{2} \left( \frac{2}{a_c^2} + \mathbf{k}^2 \right) h^*_\mathbf{k} h_\mathbf{k} \right] \right \rangle = \left \langle h_\mathbf{k} \left[ \frac{\sigma L^2}{2} \left( \frac{2}{a_c^2} + \mathbf{k}^2 \right) h^*_\mathbf{k} \right] \right \rangle = \left \langle E_\mathbf{k} \right \rangle = k_B T.$ And again we get the same result. What do you think of it? Is there a mistake? Please help, I'm really stuck with it. Grigory Sarnitskiy. 91.76.179.101 19:45, 30 January 2009 (CET)

### A quick comment

I may be wrong, but looking at your derivation it seems the boundary conditions are not correctly described. If the system is fixed to some immobile frame, only $$\sin$$ terms should appear in the modes, not $$\cos$$. If, on the other hand, periodic boundary conditions are applied, the opposite applies: only $$\cos$$, not $$\sin$$. This may explain the factor of $$2$$ that's missing... but I still have to think more carefully about this. --Dduque 09:58, 3 February 2009 (CET)

Thank you. Yes, I didn't appreciate the importance of boundary conditions. I think it is quite reasonable to take something like $$\frac{\partial h(x,y)}{\partial \mathbf{n}} \Big|_{wall}=0$$, with $$\mathbf{n}$$ normal to the wall. I suppose it is valid at least for more or less long waves (several minimal wavelengths), which contribute most to $$\langle h \rangle$$. It is likely there is no need to study molecular interaction between liquid and solid surface in this case — boundary conditions will not be affected by the material of the walls at least for real-life systems. So we get $$\frac{1}{2} k_B T$$ for each mode. Still I'll think it over again and will wait for your reply. 91.76.179.246 12:53, 3 February 2009 (CET) Well now I'm not sure at all in the variant above. Have to dig the question. 91.76.179.246 15:29, 3 February 2009 (CET)