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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h_{\mathbf {k} }\exp(i\;\mathbf {k} \cdot {\boldsymbol {\tau }})} , Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \mathbf {k} =({\frac {2\pi }{L}}m,{\frac {2\pi }{L}}n),\;m,n\in \mathbb {Z} } — wave vector, ${\boldsymbol {\tau }}$ — Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x,y)} vector. The energy is proportional to $h_{\mathbf {k} }^{*}h_{\mathbf {k} }$ (indeed it is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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)