1
$\begingroup$

Let's say I have a canonical partition function for the canonical assemble related to the Helmholtz free energy $A$, given by $$A=-kT\ln Z$$

Now, I want to derive thermodynamical quantities, like the internal energy $E$, pressure $p$ and whichever thermodynamic quantity I want.

How do I go about this?

I know any thermodynamic quantity $X$ can be obtained by $$\langle X \rangle = \sum_{v} P_v X_v$$ where $v$ is an index of a permissible microstate.

For example, how would I get average energy $E$ or average pressure $p$ from such an equation?

So I know, from the above equation, I know $$ Z = \sum_{i} \exp (-\beta E_i - \beta p_i V) \implies P_i \propto \exp (-\beta E_i - \beta p_i V)$$

So, $$\langle E \rangle = \sum _i P_i E_i = \frac{-\frac{dZ}{d\beta}}{Z}$$

I can do the same for pressure, but the differentiation can be done by $\beta V$. How would I find say entropy $S$ for example?

$\endgroup$

1 Answer 1

2
$\begingroup$

The Helmholtz energy is related to entropy as $A = E - TS$ and hence, $dA = -SdT + \dots$, where I have ignored the other work terms. Therefore, to find the entropy just calculate

$S = -\frac{\partial A}{\partial T} = \frac{\partial (KT \,\mbox{ln} Z)}{\partial T} \,,$

where you take parameters other than temperature to be constant. This will give you the entropy.

$\endgroup$
3
  • $\begingroup$ Thanks! I have another side question, what if the canonical partition is some elementary function $f(N,V,T)$? Can I still use the methods above? $\endgroup$
    – megamence
    Sep 13, 2020 at 13:49
  • $\begingroup$ I did not get the question exactly. This is a general procedure. What do you mean by elementary function? $\endgroup$ Sep 13, 2020 at 14:11
  • $\begingroup$ Sorry, I understand now. Thank you again! $\endgroup$
    – megamence
    Sep 13, 2020 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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