1
$\begingroup$

In statistical mechanics, after finding the cannoncial partition function,

$$ Z = \frac{1}{N!h^{3N}}\int\mathrm{d}p \int\mathrm{d}q \exp[-\beta H(p, q)], $$

we then recover our dear thermodynamical variables,

$$ \beta = \frac{1}{k_bT},\quad F = -\frac{1}{\beta} \ln(Z), \quad E = -\frac{\partial \ln{Z}}{\partial \beta}, \quad S = -\frac{\partial F(V, T)}{\partial V}, \quad P = -\frac{\partial F(V, T)}{\partial T} $$

My question regareds the last two. Here, we rely on the thermodynamic identity

$$ \mathrm{d}F = -S \mathrm{d}T - p \mathrm{d}V, $$

however to exploit this, we need to express $F$, and therfore $Z$, as a function of of $V$, which I have problems understanding how to do. For ideal gasses, it falls out easily as $Z$ is proportional to $V = \int \mathrm{d}q$. But in genereal, $H$ does not depend on $V$, which makes formulas like

$$ P = - \frac{1}{Z}\sum_r \frac{\partial E_r}{\partial V}, \quad \mathrm{(Statistical\, Mechanics, \,P.\,K.\,Pathria\,(3.3\, eq.11) )} $$ look like nonsens to me: we have nerver told the hamiltonian what the volume does. In his "Elementary Principles of Statistical Mechanics", Gibbs talks about "coordinates $a_1, a_2$ of bodies which we call external, meaning by this simply that they are not to be regarded as forming any part of the system, although their positions afect the forces which act on the system." (p. 47-48 in the project gutenber version). This sounds like a way to tell the hamiltonian, and thus the partition function, about the volume, but I have never seen it anywhere else. How could this be best understood?

$\endgroup$
1
  • $\begingroup$ The volume is in the region of integration for the partition function in the first equation. $\endgroup$ Apr 30, 2020 at 19:05

1 Answer 1

1
$\begingroup$

You either tell the Hamiltonian about the volume, by making the potential energy "outside" the system very large (for example, it could be that your system is in a harmonic potential, and the volume of the system corresponds to a characteristic size of the potential), or you tell the partition function about the volume, by restricting your integration of $dq$ to only refer to particles inside a specified volume. In either case, $Z$ knows about the volume explicitly.

$\endgroup$
2
  • $\begingroup$ How can $\frac{\partial E_r}{\partial V}$ be interpreted if the hamiltonian is not dependent on $V$? $\endgroup$ May 1, 2020 at 12:03
  • $\begingroup$ @martinjohnsrud A formula like that assumes discrete energy levels, so it assumes quantum mechanics, where you explicitly tell the Hamiltonian about the volume. Otherwise, I agree the formula can’t be used. $\endgroup$ May 1, 2020 at 12:59

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.