I understand that the usual way of deriving the Boltzmann distribution involves considering a small system of energy $\epsilon$ embedded in a much larger heat bath of energy $E - \epsilon$ and the total system energy is $E$.

Given that the bath has accessible microstates $\Omega_b(E - \epsilon)$ and the system has microstates $\Omega_s(\epsilon)$ and the total system + bath has microstates $\Omega_t(E)$.

We now state that the probability of finding the system in energy $\epsilon$ is simply \begin{equation} P(\epsilon) = \frac{\Omega_s(\epsilon)\Omega_b(E - \epsilon)}{\Omega_t(E)} \end{equation}

$\textbf{Question 1}$: This is usually replaced by \begin{equation} P(\epsilon) = \frac{\Omega_b(E - \epsilon)}{\Omega_t(E)} \end{equation}

Why should this be so? Why can we ignore the fact that the number of accessible microstates in the system that should vary with $\epsilon$?

Moving on, the usual trick is to use logarithms, since the $\Omega$ terms are all very large. Thus, \begin{equation} \ln P(\epsilon) = \ln\Omega_b(E - \epsilon) -\ln \Omega_t(E) \end{equation}

A Taylor expansion of the first term about $E$ gives us \begin{equation} \ln P(\epsilon) = c - \epsilon\frac{\partial\ln\Omega_b(E)}{\partial E} + O(\epsilon^2), \end{equation}

where c is just a constant that depends on the bath and will be fixed when we normalize $P(\epsilon)$. Ignorning the higher order terms, we get $P(\epsilon) \propto e^{-\beta\epsilon}$ where we identify $\beta = \frac{\partial\ln\Omega(E)}{\partial E}$.

$\textbf{Question 2}$: Why can we ignore the higher order Taylor terms? I understand that $\epsilon$ is small compared to the heat bath energy $E$ but why am I comparing it to $E$ to get the right measure of the error in probability?


2 Answers 2


I think the confusion here has to do with what the Boltzmann distribution describes. It does not give you the probability of finding your small system with a particular energy. Instead, it tells you the probability of finding it in a particular microstate.

If you want to know the probability of getting a particular energy, you have to sum the Boltzmann probability over the degenerate microstates. This is how you get the Maxwell-Boltzmann distribution.

So, a more correct way to write down what you have above is $$ P_s \propto \Omega_b\left(E - \epsilon\right)\,, $$ where $s$ is some microstate of the small system. This leads you to your formula.

Concerning the second order of $\epsilon$. As you say, in equilibrium, we define $\beta = \frac{\partial \ln\Omega\left(E\right)}{\partial E}$. Since $\beta = \frac{1}{k_B T}$ (a constant), the second derivative of the logarithm vanishes, taking away $\epsilon^2$.

  • $\begingroup$ If I had a system which had many more accessible microstates for energy $\epsilon$ compared to the number of accessible microstates for energy $\epsilon'$, is there not a higher probability of finding my system in the first state compared to the second? Your answer says no i.e. the only thing that matters is the states available to the bath but I don't see why. Maybe that's the intuition that I'm missing? $\endgroup$ Dec 6, 2017 at 5:44
  • $\begingroup$ Almost correct. Think of it this way. The total multiplicity of the composite system is $\Omega(\epsilon,E) = \Omega_s(\epsilon) \Omega_b(E-\epsilon)$. If you fix the microstate in your system, the multiplicity of that is $\Omega_b(E-\epsilon)$ since you choose a single realization from $\Omega_s(\epsilon)$. According to the derivation above, the unnormalized probability of getting one of these microstates is $e^{-\beta\epsilon}$. The probability of finding the system in ANY of the microstates with this energy is $\Omega_s(\epsilon)e^{-\beta\epsilon}$. So, if you have two energies... $\endgroup$
    – IcyOtter
    Dec 6, 2017 at 9:56
  • $\begingroup$ ... $\epsilon$ and $\epsilon'$, the probabilities will be $P(\epsilon)\propto\Omega_s(\epsilon)e^{-\beta\epsilon}$ and $P(\epsilon')\propto\Omega_s(\epsilon')e^{-\beta\epsilon'}$. So to understand which energy state is more likely, we look at both the Boltzmann factor and the multiplicity. $\endgroup$
    – IcyOtter
    Dec 6, 2017 at 9:57
  • $\begingroup$ I see - yeah I understand what you meant in the first paragraph of your answer now. Thank you. $\endgroup$ Dec 6, 2017 at 10:23

Question 1 The quantity $$P(\epsilon)={\Omega_s(\varepsilon)\Omega_b(E-\varepsilon)\over\Omega_t(\varepsilon)}$$ should be interpreted as the probability of the macroscopic state for which the total energy $E$ is divided in $\varepsilon$ for the system and $E-\varepsilon$ for the bath. Performing a Taylor expansion of the logarithm of the last term, one gets $$P(\epsilon)={\rm Cst}\times \Omega_s(\varepsilon)e^{-\beta \epsilon}$$ where $\Omega_s(\varepsilon)$ is the number of microscopic states of the system of energy $\epsilon$. This probability is also the sum of the probabilities of each individual microscopic states whose energies $\epsilon_i$ are equal to $\epsilon$: $$P(\epsilon)=\sum_{i/\epsilon_i=\epsilon} p_i$$ Since these microscopic states are equiprobable, $p_i=p(\epsilon)$ for all $i=1,\ldots,\Omega_s(\varepsilon)$ and $P(\epsilon)=\Omega_s(\epsilon)p(\epsilon)$. The probability to find the system in the $i$-th microscopic state is $$p_i=p(\epsilon_i)={P(\epsilon)\over\Omega_s(\epsilon)} ={\rm Cst}\times e^{-\beta \epsilon_i}$$ I could have used this at the beginning, before performing the Taylor expansion $$p_i={\Omega_b(E-\varepsilon_i)\over\Omega_t(\varepsilon_i)}$$

  • $\begingroup$ Sorry, how did the $\varepsilon$ dependence get absorbed into Cst? The microstates are equiprobable but the number of them depends still on $\varepsilon$, no? $\endgroup$ Dec 4, 2017 at 15:14
  • $\begingroup$ $P(\epsilon)$ is the probability that the system be in one of the $\Omega(\epsilon)$ microstates whose energy is $\epsilon$. Therefore, the probability of the $i$-th microstate is $p_i=P(\epsilon)/\Omega(\epsilon)$. $\endgroup$
    – Christophe
    Dec 4, 2017 at 17:33
  • $\begingroup$ I am not sure to understand your second question. Could you expand it? $\endgroup$
    – Christophe
    Dec 4, 2017 at 17:41
  • $\begingroup$ My point is that although $p_i$ is the same for a given $\varepsilon$, it still is actually $p_i(\varepsilon)$ and cannot be a constant. Otherwise, what you are saying is that regardless of energy, the number of microstates of the system is constant. The question regarding the Taylor expansion is how to decide the scale of the errors - I am computing a probability and therefore I should have an argument that connects the error in my probability estimate with $O(\varepsilon^2)$. How does one argue that $O(\varepsilon^2)$ and higher orders only give small errors in $P(\varepsilon)$? $\endgroup$ Dec 5, 2017 at 8:12
  • $\begingroup$ You are right: $p_i$ depends on $\varepsilon$. It is obvious from the last equation: $p_i=\Omega_b(E-\varepsilon_i)/\Omega_t(\varepsilon_i)$. What I meant is that the $\Omega_s(\varepsilon)$ microstates whose energy is $\varepsilon$ are still equiprobable. $\endgroup$
    – Christophe
    Dec 5, 2017 at 9:31

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