I was thinking about fluctuations in energy for a system in thermal equilibrium. I think that the Boltzmann distribution itself has an standard deviation approximately equal to the mean, as it is approximately an exponential distribution.

From the microcanonical ensemble perspective, I think the reason the fluctuations in energy (the standard deviation) grow as $\sqrt{N}$ is because in statistics we know that for $N$ independent random variables, the variance of their sum scales with $N$.

However, if I consider the whole gas as the system (this is the macrocanonical ensmeble, isn't it?) for which I know the average energy (the same as before) and I can apply Boltzmann statistics directly, will I not get a different answer for the size of energy fluctuations? Will I not get an standard deviation on the same order as the mean energy, as expected for a plain Boltzmann distribution?

  • $\begingroup$ The Boltzmann distribution is not an exponential distribution: the Boltzmann weight has certainly an exponential form, but you're forgetting the entropy (that is, how many microstates of a given energy there are). The distribution of the energy density in the canonical ensemble concentrates sharply on its average (with fluctuations of the order of $1/\sqrt{N}$). This is why you get equivalent results for both microcanonical (fixed energy) and canonical (fluctuating energy) ensembles (well, at least in the absence pf phase transition). $\endgroup$ – Yvan Velenik May 25 '14 at 18:39
  • $\begingroup$ Yeah I was thinking it may have to do with that (number of microstates of a given energy), but I'm not sure how that'd solve it. Say there are M states of a give energy, wouldn't that appear both in the partition function and in the density of states when computing $\int E \frac{e^{-\beta E}}{MZ} M dE$ ? $\endgroup$ – guillefix May 25 '14 at 18:59
  • $\begingroup$ Consider a super simple example: you have $N$ objects each of which may be either in its "ground state" of energy $0$ or in an "excited state" of energy $\epsilon$. There is no interaction between these objects. The partition function is $\sum_{k=0}^N\binom{N}{k}e^{-\beta\epsilon k}=(1+e^{-\beta\epsilon})^N$, and thus the probability that the energy is equal to $E=M\epsilon$ is $\binom{N}{M}e^{-\beta\epsilon M}/(1+e^{-\beta\epsilon})^N$, which you can check develops a sharp peak (of width of order $\sqrt{N}$) around its expected value $N\epsilon/(1+e^{\beta\epsilon})$. $\endgroup$ – Yvan Velenik May 26 '14 at 6:51
  • $\begingroup$ Yeah I asked a professor today and I think I understand it now. The gas does have a Boltzmann distribution, but it does so over its microstates, not over its energy. To get the distribution over energy, we need to multiply by the density of (micro)states, and it is this density that develops a Dirac-delta kind of peak as $N$ grows! $\endgroup$ – guillefix May 26 '14 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.