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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \langle x_i \frac{\partial H}{\partial x_j} \right \rangle = \delta_{ij} k_B T, }

we obtain:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)