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
rather than
. That's because each mode has to degrees of freedom
and
, since each wave is
, with the energy of each mode proportional to
. This obviously lead to the mean energy of each mode to be
. That was the real notation and now lets turn to the complex notation.
Each mode with the fixed wave vector is presented as
,
— wave vector,
—
vector. The energy is proportional to
(indeed it is
). According to equipartition:

we obtain:
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca206b7e6243e9d369d4d3c3b74a90dba54822be)
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)