Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking at a specific derivation on wikipedia relevant to statistical mechanics and I don't understand a step.

$$ Z = \sum_s{e^{-\beta E_s}} $$

$Z$ (the partition function) encodes information about a physical system. $E_s$ is the energy of a particular system state. Z is found by summing over all possible system states.

The expected value of E is found to be:

$$ < E > = -\frac{\partial ln Z}{\partial \beta} $$

Why is the variance of E simply defined as:

$$ <(E - <E>)^2> = \frac{\partial^2 ln Z}{\partial \beta^2} $$

just a partial derivative of the mean.

What about this problem links the variance and mean in this way?

share|cite|improve this question
What are the variables in the first equation. What is $E_s$ which is not used in the rest of the question. Unless you give more clarity in the question it would be difficult to answer. – Rajesh Dachiraju Mar 28 '11 at 6:30
what is $\beta$ ? – Rajesh Dachiraju Mar 28 '11 at 6:43
The expressions for mean and variance of some random variable in your complicated system turned out that way. There is no such concept. Also it is very wrong to interpret your result in this way for any general system because, where is $\beta$ defined for a random variable, I also think your question is not appropriate for this forum and I suggest you read fundamentals of Random Variables and mean and variance bfore even start thinking of statistical mechanics ! – Rajesh Dachiraju Mar 28 '11 at 6:52
@Rajesh: Dear Rajesh, I don't see why it's necessary to be so hostile; the OP seems to understand means and variances, and has a question on how to do a specific computation for a certain distribution. – Akhil Mathew Mar 28 '11 at 17:16
@Akhil : First of all I didn't intend to be hostile and I know it wouldn't be appropriate to get into any arguments on this. I felt that the OP didn't put effort into formulating the question in a readable form for a general audience. I think it would've been great if it was that way. – Rajesh Dachiraju Mar 29 '11 at 5:03
up vote 3 down vote accepted

The answer is valid for the partition sum $Z$ (which is closely related to the moment generating function). The reason is the special structure of the partition sum $$Z = \sum_s e^{-\beta E_s}.$$ The system is characterized with probability $$P_s=\frac{e^{-\beta E_s}}{Z}$$ that a state $s$ with energy $E_s$ is attained.

Given this definition it is easy to see that $$-\partial_\beta \ln Z = -\frac{\partial_\beta Z}{Z} = \sum_s E_s \frac{e^{-\beta E_s}}{Z}= \sum_s P_s E_s =\langle E \rangle .$$

Similarly, one can easily convince oneself that $$ \begin{align*} \partial_\beta^2 \ln Z &= -\partial_\beta \left[ \sum_s E_s \frac{e^{-\beta E_s}}{Z} \right] =\sum_s E_s^2 \frac{e^{-\beta E_s}}{Z} - \left[ \sum_s E_s \frac{e^{-\beta E_s}}{Z}\right] \left[\sum_{s'} E_{s'} \frac{e^{-\beta E_{s'}}}{Z}\right]\\ &= \langle E^2\rangle -\langle E\rangle^2 = \langle (E- \langle E\rangle)^2\rangle, \end{align*}$$ i.e., the variance is given by the second derivative of $\ln Z$.

share|cite|improve this answer

Have you seen the link of the definition of variance or expected value in wiki, here?

share|cite|improve this answer

The variance of a random variable $X$ is always defined as $<(X - <X>)^2>$; this is the expected square of the difference between the expected and actual values.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.