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 $h_{\mathbf{k}}\exp (i\mathbf{k}\cdot {\tau })$, $\mathbf{k}=(\frac{2\pi }{L}m,\frac{2\pi }{L}n),m,n\in \mathbb{Z}$ — wave vector, ${\tau }$ — $(x,y)$ vector. The energy is proportional to $h_{\mathbf{k}}^{*}{h}_{\mathbf{k}}$ (indeed it is $E_{\mathbf{k}}=\frac{\sigma L^2}{2}(\frac{2}{a_c^2}+{\mathbf{k}}^2)h_{\mathbf{k}}^{*}{h}_{\mathbf{k}}$). According to equipartition:

$⟨x_i\frac{\partial H}{\partial x_j}⟩={\delta }_{ij}{k}_{B}T,$

we obtain:

$⟨h_{\mathbf{k}}\frac{\partial }{\partial h_{\mathbf{k}}}[\frac{\sigma L^2}{2}(\frac{2}{a_c^2}+{\mathbf{k}}^2)h_{\mathbf{k}}^{*}{h}_{\mathbf{k}}]⟩=⟨h_{\mathbf{k}}[\frac{\sigma L^2}{2}(\frac{2}{a_c^2}+{\mathbf{k}}^2)h_{\mathbf{k}}^{*}]⟩=⟨E_{\mathbf{k}}⟩=k_{B}T.$

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)