Talk:Capillary waves

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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. 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)