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 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 k_B T} rather than 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 \frac{1}{2} k_B T} . That's because each mode has to degrees of freedom 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 A_{mn}} and 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 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 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 A_{mn}^2+B_{mn}^2} . This obviously lead to the mean energy of each mode to be 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 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 (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 h_\mathbf{k} \exp(i \; \mathbf{k} \cdot \boldsymbol{\tau})} , 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 \mathbf{k}=(\frac{2 \pi}{L}m, \frac{2 \pi}{L}n), \; m,n \in \mathbb{Z}} — wave vector, ${\boldsymbol {\tau }}$ — 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 (x,y)} vector. The energy is proportional to 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 h_\mathbf{k}^*h_\mathbf{k}} (indeed it is 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 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)