17
$\begingroup$

In many places in statistical physics we use the partition function. To me, the explanations of their use are clear, but I wonder what their physical significance is. Can anyone please explain with a good example without too many mathematical complications?

$\endgroup$
  • $\begingroup$ Aside from being a normalization factor, many of its significant features for calculations arise from its likeness to Z and Laplace transforms, thanks to the exponential-with-energy Boltzmann distribution, which is kind of a "co-indidence" in that they wouldn't work with a different distribution. $\endgroup$ – WetSavannaAnimal Sep 1 '15 at 10:14
  • 2
    $\begingroup$ Did you read the "meaning" section in the Wikipedia article? If yes, what doesn't satisfy you about "it encodes how the probabilities are partitioned among the different microstates"? $\endgroup$ – ACuriousMind Sep 1 '15 at 10:40
  • 1
    $\begingroup$ Possible duplicate of The unreasonable effectiveness of the partition function $\endgroup$ – tparker Feb 2 '17 at 20:53
11
$\begingroup$

The partition function is a measure of the volume occupied by the system in phase space. Basically, it tells you how many microstates are accessible to your system in a given ensemble. This can be easily seen starting from the microcanonical ensemble.

In the microcanonical ensemble, where every microstate with energy between $E$ and $E+\Delta E$ is equally probable, the partition function is

$$Z_{mc}(N,V,E)= \frac 1 {N! h^{3N}}\int_{E<\mathcal H(\{p,q\})<E+\Delta E} d^{3N}p \ d^{3N} q \tag{1}$$

where the integral is just the hypervolume of the region of phase space where the energy (hamiltonian) $\mathcal H$ of the system is between $E$ and $E+\Delta E$, normalized by $h^{3N}$ to make it dimensionless. The factor $N!^{-1}$ takes into account the fact that by exchanging the "label" on two particles the microstate does not change.

The Boltzmann equation

$$S=k_B \log(Z_{mc})\tag{2}$$

tells you that the entropy is proportional to the logarithm of the total number of microstates corresponding to the macrostate of your system, and this number is just $Z_{mc}$.

In the canonical and grand-canonical ensembles the meaning of the partition function remains the same, but since energy is not anymore fixed the expression is going to change.

The canonical partition function is

$$Z_c(N,V,T)= \frac 1 {N! h^{3N}}\int e^{-\beta \mathcal H(\{p,q\})} d^{3N}p \ d^{3N} q\tag{3}$$

In this case, we integrate over all the phase space, but we assign to every point $\{p,q\}=(\mathbf p_1, \dots \mathbf p_N, \mathbf q_1, \dots \mathbf q_N)$ a weight $\exp(-\beta \mathcal H)$, where $\beta=(k_B T)^{-1}$, so that those states with energy much higher than $k_B T$ are less probable. In this case, the connection with thermodynamics is given by

$$-\frac{F}{T}=k_B \log(Z_c)\tag{4}$$

where $F$ is the Helmholtz free energy.

The grand canonical partition function is

$$Z_{gc}(\mu,V,T)=\sum_{N=0}^\infty e^{\beta \mu N} Z_c(N,V,T)\tag{5}$$

where this time we are also summing over all the possible values of the number of particles $N$, weighting each term by $\exp(\beta \mu N)$, where $\mu$ is the chemical potential.

The connection with thermodynamics is given by

$$\frac{PV}{ T} = k_B \log (Z_{gc}\tag{6})$$

$\endgroup$
5
$\begingroup$

It's $e^{-F/T}$, where $F/T$ is the free energy normalized by the relevant thermodynamic energy scale, the temperature. The exponential is just a monotonic reparameterization, so morally speaking, the partition function is just the free energy that's available to do useful work.

Another interpretation: if you normalize it so that $E = 0$ is the ground state, then roughly speaking, it's the reciprocal of the "fraction of the system that's in the ground state." Extremely heuristically, let $g$ be the total amount of the system that's in the ground state, $e$ be the total amount of the system that's in an exited state, and $s = g + e$ be the total amount of the system. Then $g/s$ is the fraction of the system that's in the ground state, and its reciprocal is $s/g = (g + e)/g = 1 + e/g$. The Boltzmann weight gives that the relative weight (or "amount") of each excited state $i$ with energy $E_i$ relative to the weight of the ground state is $e^{-\beta E_i}$. Summing over all the excited states $i$, we get the partition function $s/g = 1 + e^{-\beta E_1} + e^{-\beta E_2} + \dots$.

$\endgroup$
2
$\begingroup$

Partition function physical meaning is the following: It expresses the number of thermally accesible states that a system provides to carriers (e.g. electrons).

$\endgroup$

protected by Qmechanic Feb 5 '17 at 13:51

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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