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)