I've been trying to re-derive the Boltzmann Distribution, and I'm encountering some ideas that are counter-intuitive.

Here is my naive re-derivation: Suppose you have a system with two energy levels $E_1$ and $E_2$, with multiplicities (i.e. degeneracies) $\Omega(E_1)$ and $\Omega(E_2)$ respectively, and with equilibrium observation probabilities of $P(E_1)$ and $P(E_2)$ respectively. You would expect that if you let the system mix for a while and then observed it at equilibrium, that \begin{align} \frac{P(E_1)}{P(E_2)} = \frac{\Omega(E_1)}{\Omega(E_2)} \end{align} So, for example, if $E_2$ had twice the degeneracy of $E_1$, then you'd be twice as likely to observe $E_2$ than $E_1$ when you measured the system at equilibrium.

Now, the entropy is defined to be $S = k\ln \Omega$, and temperature is defined to be \begin{align} \frac{1}{T} = \frac{\partial S}{\partial E} \approx \frac{S(E_1) - S(E_2)}{E_1 - E_2} \end{align} So we should have: \begin{align} \frac{P(E_1)}{P(E_2)} = \frac{\exp(S(E_1)/k)}{\exp(S(E_2)/k)}= \exp\left(\frac{S(E_1) - S(E_2)}{k}\right) = \exp\left(\frac{E_1 - E_2}{kT}\right) = \frac{e^{E_1/kT}}{e^{E_2/kT}} \end{align}

But I'm missing a minus sign in the final exponands, so this is wrong. If I redid the derivation but instead had \begin{align} \frac{P(E_1)}{P(E_2)} = \frac{\Omega(E_2)}{\Omega(E_1)} \end{align} the rest of the maths works out to give me the correct Boltzmann probabilities. But to me this is just so counter-intuitive. It would mean that if $E_2$ had twice the degeneracy of $E_1$, I would actually observe $E_2$ only half as much as $E_1$. How can I reconcile this with my every-day experience of probability?

  • $\begingroup$ You should consider two states with no degeneracy, and consider the degeneracy of the bath instead. $\endgroup$
    – velut luna
    Apr 2, 2017 at 12:58
  • 1
    $\begingroup$ After thinking about it some more, maybe I misunderstood some basic things about the Boltzmann distribution: so... it's meant to be a probability density function for the microstates, not the energy levels, right? I can follow the maths but I just want to make sure my intuition is right, namely: If $E_2$ had twice the degeneracy of $E_1$, it should just be $P(E_1)/P(E_2) = \frac{\Omega(E_1)e^{-E_1/kT}}{\Omega(E_2) e^{-E_2/kT}}$ instead of what I wrote, right? That would gel with my intuition that a more degenerate thing is more likely to be observed. $\endgroup$
    – Paradox
    Apr 2, 2017 at 16:36

2 Answers 2


If E2 E 2 had twice the degeneracy of E1 E 1 , I would actually observe E2 E 2 only half as much as E1 E 1 . How can I reconcile this with my every-day experience of probability?

This statement is incorrect. What the boltzmann distribution really says is if states you have labelled by $E_1$ and $E_2$ are Non degenerate then, $$\frac{P(E_1)}{P(E_2)}=\frac{e^{-\frac{E_1}{k_BT}}}{e^{-\frac{E_2}{k_BT}}}$$

Meaning that the Boltzmann equation talks about micro states, not about degenerate states.

If however, they were degenerate, then:

$$\frac{P(E_1)}{P(E_2)}=\frac{\Omega (E_1) e^{-\frac{E_1}{k_BT}}}{\Omega (E_2) e^{-\frac{E_2}{k_BT}}}$$

Where the symbols have their usual meanings. It would probably serve you best to go through the standard derivation of the results in any statistical mechanics text, like say Rief, and figure out why these statements are indeed true.


While what Anonjohn says is true, even if you started with the right assumption, the rest of your derivation would lead you to $\frac{P(E_1)}{P(E_2)}=1$, which is also wrong.

The reason is that the equation $S(E) = k\ln \Omega(E)$ only applies for the full thermodynamic entropy if you are in the microcanonical ensemble, where $E$ is then the total energy, which is fixed. You need $S$ to refer to the full thermodynamic entropy to use $\frac{1}{T} = \frac{\partial S}{\partial E}$. Note also that we evaluate this derivative at $E$, the total energy, not any particular energy, like $E_1$ or $E_2$ could be.

EDIT (technical comment, which you can ignore): Actually $S(E) = k\ln \Omega(E)$ holds in the canonical ensemble in the thermodynamic limit (large system) and $E_1$ and $E_2$ could be close to $E$. In that case your derivation should work, and the $\frac{P(E_1)}{P(E_2)}=1$ is right at first order.. But you should find the next order, which I believe would give you a Gaussian, which is just a result of the central limit theorem (mathematical version of thermodynamic limit). You will find that the variance of the Gaussian is related to the heat capacity, which goes like the number of particles $N$ (like CTL).

Anyway, here is the right derivation of the Botlzmann distribution, that is closest to what you were attempting.

Derivation of Boltzmann distribution in microcanonical ensemble

The good news is that in the microcanonical ensemble, your intuition that probabilities are proportional to the number of microstates exactly holds. The thing you missed then is that we fix the energy of the total system (you can consider this to be an isolated system, so energy conservation implies this). Now if the energy is fixed, to $E$ say, it makes no sense to ask about the probabilities of other values of energy for the total system. But we can ask about the probability that some part of the system has a certain value energy. Then, we will find the Boltzmann distribution!

So let's find the probability of a part/subsystem having energy $E_1$. The crucial insight is that the probability is proportional to the number of microstates of the total system. We can write this as as the number of microstates of our subsystem (which remember, we will assume is very small compared to the total system), times the number of microstates of the rest of the system. The rest of the system has energy $E-E_1$ to keep the total fixed, giving:

\begin{align} P(E_1) \propto \Omega_S(E_1)\Omega_R(E-E_1) \end{align}

where $\Omega_S$ refers to the number microstates of the subsytem for energy $E_1$, and $\Omega_R$ for the rest of the system. Note that they are different functions, because they refer to different systems.

Now, we can manipulate $\Omega_R(E-E_1) = \exp(S_R(E-E_1)/k)$ as you did, and Taylor exand:

$$\Omega_R(E-E_1) \approx \exp(\frac{S_R(E)-\frac{\partial S_R}{\partial E}E_1}{k})$$

Now, remember we said that the subsystem $S$ is small, so the rest $R$ is very close to the total system, and indeed it's entropy as a function of energy $S_R$ must be very close to that of the total system, $S$. So we can say $\frac{\partial S_R}{\partial E} \approx \frac{\partial S}{\partial E} = \frac{1}{T}$, so $\Omega_R(E-E_1) \approx \exp\left(S(E)/k\right)\exp\left(-E_1/kT\right)$, giving

$$P(E_1) \propto \Omega_S(E_1)\Omega(E)\exp\left(-E_1/kT\right) \propto \Omega_S(E_1)\exp\left(-E_1/kT\right)$$

which is the Boltzmann distribution. Note this limit is exact only as the subsytem becomes an infinitesimal part of the total system (which can be done by considering the "rest of the system" as a heat bath, and making it infinitely large). In that limit, the microcanonical and canonical ensemble become equivalent.

You should probably read about the derivation in the canonical ensemble, to complete your understanding :)


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