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I don't think it is correct to say that fluctuations are a source of dissipation. If you look at the fluctuation-dissipation theorem in the limit T → 0, you have no fluctuations yet you still have dissipation. I would say instead that both fluctuations and dissipation have a common source, which is the coupling of the system with the environment (a.k.a. thermal bath). --
Edgar.bonet (
talk)
13:50, 18 September 2010 (UTC)reply
Can you give me a specific example of a system with dissipation and no fluctuations? I don't think the fluctuation-dissipation theorem applies to such a case because quantum effects dominate near absolute zero. And in the quantum regime there are probably zero-point fluctuations. Perhaps you are right, though, that a more careful statement should be made about the relationship between fluctuations and dissipation.
RockMagnetist (
talk)
16:56, 18 September 2010 (UTC)reply
I was thinking about the (classical) magnetization of a Cobalt nanoparticle. The damping parameter of the
Landau–Lifshitz–Gilbert equation is roughly constant at low temperature. On the other hand, the power of the random field in the corresponding
Langevin equation scales roughly as T. In this case, both the fluctuations (Langevin field) and the dissipation (Gilbert’s damping) can be thought of as the consequence of the environment (phonons, magnons, electronic excitations...) applying a random torque on the magnetization. The ensemble distribution of this torque has some average (or expectation value) which is the Gilbert’s damping term, and some variance (statistical fluctuations) corresponding to the Langevin field. I would not say that “the variance is the source of the average”! Actually I think of the damping constant like a coupling constant between the magnetization and the environment. If the environment is very cold (say ~ 35 mK, in a dilution fridge) and the magnetization is in motion (say I just sent a few ns long microwave pulse to do some sort of pulsed FMR), then the magnetization will loose energy to the environment (damping) yet the environment will not give energy back because it’s just too cold to induce any significant fluctuations. --
Edgar.bonet (
talk)
18:36, 18 September 2010 (UTC)reply
Disagree. What does observation have to do with anything? Quarks can't be observed individually, but
quark still makes sense. The fact that you describe a "single fluctuation" seems to invalidate your own argument, about the necessity of using the plural.
70.250.177.191 (
talk)
04:34, 1 July 2012 (UTC)reply
Statistical vs. dynamic fluctuations
It's perhaps worth pointing out the distinction between statistical fluctuations that occur from system to system within an ensemble, versus dynamic fluctuations (
stationary processes) that occur during the evolution of a single system. For very long times and in ergodic systems the two correspond. However, for short times the dynamic fluctuations can be much smaller than the statistical fluctuations, and in non-ergodic or weakly ergodic systems the dynamic fluctuations may even never reach the size of the statistical fluctuations, or require the lifetime of the universe to do so. When one uses language like "the system fluctuates around X" or "thermal fluctuations provide the energy necessary for the atoms to occasionally hop from one site to a neighboring one", one is referring to dynamic fluctuations on (implicitly) a prompt timescale. On the other hand, formulae like give the statistical fluctuation.
Nanite (
talk)
14:18, 22 January 2014 (UTC)reply
Application of Central Limit Theorem
Right before the application of central limit theorem, in the article, it says the moments are finite and therefore, we can expand f(E) around <E>, and the result is to the lowest order Gaussian.
It was not immediately obvious to me why the expansion results in Gaussian. After thinking for a while, I came to the following conclusion, which I'm not sure if it is exactly what the author meant to say or not:
since E is the sum of order N contributions of approximately independent random variables (energies of individual degrees of freedom?), the central limit theorem applies, and therefore, f(E) is Gaussian?
In general statistical mechanics (which includes systems of finite degrees of freedom) the fluctuations are in general NOT gaussian. So indeed they would have to assume the
thermodynamic limit to prove they are gaussian, i.e., it is just as usual for the
central limit theorem. --
Nanite (
talk)
07:35, 11 April 2016 (UTC)reply
Error in section Multiple variables
The inverse quantity in the Gaussian exponent must be replaced by the inverse matrix. See Landau, v5.
Correct please who is better in English.
Luksaz (
talk) —Preceding
undated comment added
17:03, 6 June 2018 (UTC)reply
Max.kit and I disagree on whether angle brackets should be used around . The angle brackets refer to an average over the ensemble - something that should be clarified in the text - and is just another name for . Without the angle brackets, is just a variance for some unspecified state, not the energy dispersion for the system. RockMagnetist(
talk)18:45, 4 July 2020 (UTC)reply