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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...

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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.

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  • $\begingroup$ 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:) $\endgroup$ – Aleksey Druggist Mar 29 '18 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 '18 at 14:29
  • $\begingroup$ @Menachem You still need the degeneracy of each energy level, often called $g$. $\endgroup$ – Sean E. Lake Mar 29 '18 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 '18 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 $Z^{-1} g(E) \mathrm{e}^{-E/kT}$. $\endgroup$ – Sean E. Lake Mar 29 '18 at 17:28

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