From what I understand, in a system $S$ described by a canonical ensemble, the probability that $S$ has energy $E$ is equal to $\frac{1}{Z}e^{-E/kT}$, where $T$ is the "temperature", $k$ the Boltzmann constant, and $Z$ the partition function. I have two questions:

1) Is it obvious that $Z = kT$, since $\int_0^{\infty}e^{-E/kT}dE = kT$?

2) I'm failing to understand where the size of $S$ comes into play. Is this energy distribution true whether $S$ is a system containing 1 or $10^{23}$ atoms? I understand that the heat bath should be much larger than either of these, but don't understand how the size of the system doesn't play a role.

I think I'm missing something...


1 Answer 1


Yes, what you're missing is that the probability is defined over distinct states, not just energies. So, what $Z$ is will depend on how that is defined. For a single spin $1/2$ particle in a magnetic field there are exactly two states with energies $E_+$ and $E_-$, leading to the partition function $Z$ being $$Z = \mathrm{e}^{-E_+/kT} + \mathrm{e}^{-E_-/kT}.$$

If you're talking about a classical particle in a box with volume $V$, then distinct states are those that have distinct position $\mathbf{x}$ and momentum $p$, producing the partition function \begin{align} Z & = \int \mathrm{e}^{-p^2/(2mkT)} \operatorname{d}^3x\operatorname{d}^3p\\ & = V 4\pi \int_0^\infty \mathrm{e}^{-p^2/(2mkT)} p^2 \operatorname{d}p. \end{align}

Where the size comes into play is in both the volume, $V$, and the number of particles involved, $N$. When all of the individual particles are independent you can just raise $Z$ to the power $N$. When the particles aren't independent, you need to do more work to integrate/sum over the degrees of freedom available.

  • $\begingroup$ Even if the particle is a classical, factor $1/h^3$ must be present ( David Tong compared this situation with the vestigial effect, like the male nipple:) $\endgroup$ Mar 29, 2018 at 7:09
  • $\begingroup$ Thanks! My understanding is that the canonical ensemble is derived by considering a small subsystem $S$ (with fixed $N$ and $V$) of a much larger one described by a microcanonical ensemble. The phase space volume in which $S$ has energy $E_s$ is proportional to the phase space volume in which the remainder of the system has energy $E-E_s$, where $E$ is the total system energy. Since phase space volume roughly increases exponentially with energy, a Taylor series expansion gives us the canonical distribution. Doesn't this allow us to consider the probability distribution of energy? $\endgroup$
    – Menachem
    Mar 29, 2018 at 14:29
  • $\begingroup$ @Menachem You still need the degeneracy of each energy level, often called $g$. $\endgroup$ Mar 29, 2018 at 17:02
  • $\begingroup$ Wait, isn't the phase space volume of a given energy its degeneracy? And doesn't the argument depend on the degeneracy increasing roughly exponentially with energy? $\endgroup$
    – Menachem
    Mar 29, 2018 at 17:15
  • $\begingroup$ @Menachem I'm talking about the phase space volume of individual particles, and that grows like $4\pi p^2$. You're talking about the phase space volume of a system of particles, which grows combinatorically with $N$. Either way, the probability won't just be $Z^{-1} \mathrm{e}^{-E/kT}$, it's $Z^{-1} g(E) \mathrm{e}^{-E/kT}$. $\endgroup$ Mar 29, 2018 at 17:28

Your Answer

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

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