How should we interpret Boltzmann's distribution when the set $\{ E_1, E_2, \cdots, E_k \}$ (in increasing order) of energy levels is a finite set? In that case, the expected energy cannot exceed $E_k$, and in any case the infinite temperature distribution would be uniform over the $k$ microstates, and have expected energy $E = \dfrac{1}{k} \sum_i{E_i}$, wouldn't it?

My understanding is as follows:

Boltzmann's distribution arises from maximizing $-\sum_i{p_i \log p_i}$ subject to $\sum_i{p_i} = 1$ and $\sum_i{p_i E_i} = E$. From this, we get:

$p_i = \dfrac{1}{Z} e^{- \beta E_i}$

$Z = \sum_i{e^{-\beta E_i}}$

where $\beta$ is the Lagrange multipler of the constraint $\beta (\sum_i{p_i E_i} - E)$

So, the case of $\beta = 0$ should arise if we set $E = \infty$, but that is unattainable.

EDIT: Digging a little further, I realized that if we use:

$E(\beta) = - \dfrac{ \partial \log Z(\beta) }{ \partial \beta } = \dfrac{1}{Z}\sum_i{E_i e^{-\beta E_i}}$


$E(0) = \dfrac{1}{k} \sum_i{E_i}$

So, this is a counterintuitive result, that at infinite temperature, this system will have finite average energy. I suppose this means that no physical system has only a finite number of quantum states? That is, there is always a higher energy state for it if you want to add energy?

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    $\begingroup$ Temperature and energy are not the same thing, and not quite so closely related as is sometimes supposed. A system with upper-bounded energy has a vanishing heat capacity at high temperature. $\endgroup$ Sep 29, 2020 at 21:59

3 Answers 3


If a system has a bounded set of accessible energy states, then higher temperature should not be interpreted as more energy. As you have found, such systems admit perfectly well-defined and physical $\beta=0 \iff T=\infty$ configurations. They also admit negative temperature $(\beta<0)$ configurations, where the entropy decreases with increasing system energy. As always, if two systems come into contact then heat will flow in the direction of increasing $\beta$ - this implies that negative temperature systems are "hotter" than positive temperature systems, but this is an artifact of choosing to think about $T$ instead of $\beta$.

I suppose this means that no physical system has only a finite number of quantum states? That is, there is always a higher energy state for it if you want to add energy?

Not at all. Systems like the one you describe are perfectly physical - see e.g. this APS review article from 1956. There's no reason in principle that $\beta$ cannot be zero or negative, but both happen only when the total energy of the system has an upper bound.

  • $\begingroup$ Weird. So, what would happen if you took this system (with finite number of microstates, each with finite energy level), and placed it in a heat bath, and then cranked up the temperature of the heat bath without bound? Would the distribution of the system's microstates be driven just to the uniform distribution? (Even though the skewed distribution of 100% occupancy at the highest energy level would give higher average energy?) $\endgroup$ Sep 29, 2020 at 22:12
  • $\begingroup$ Oh, that is so weird - I didn't realize that Beta going towards -infinity causes the Boltzmann distribution to concentrate 100% occupancy at the highest energy level. Cool! $\endgroup$ Sep 29, 2020 at 22:27
  • $\begingroup$ @HenryBigelow Yes, that's what would happen. $\endgroup$
    – J. Murray
    Sep 29, 2020 at 23:07
  • $\begingroup$ Nice, thanks for the answer and explanation @J. Murray. $\endgroup$ Sep 30, 2020 at 4:14

By requiring that a system has a finite number of energy macrostates, you immediately and necessarily impose that the system has a finite maximum energy, regardless of temperature (which is, of course, the state in which it is in the highest-energy macrostate with probability 1; this maximum-energy state is not a Boltzmann distribution, but all possible Boltzmann distributions will be lower in energy). This is essentially "by construction", so it should not be surprising.

There are certainly many systems that are well modeled as having a finite maximum energy under certain conditions (for example, an ensemble of stationary two-level systems). Temperatures can be infinite at finite energy in these systems, and in fact they can even be negative (for example, a population inversion in a lasing medium is an example of negative-temperature conditions).

As to whether a real-life system ever really has a finite maximum energy, the answer depends on what you're willing to do to it. If you're willing to put in enough energy to destroy your system, decomposing it into a cloud of free elementary particles, then I suppose you can always reach the continuum, but this is sort of a moot point.

  • $\begingroup$ Just one clarification - the system I describe above, I am postulating that it has k microstates, each with energy E_i, i in 1..k. Not macrostates as you had supposed. Not sure if that makes a difference to the analysis. $\endgroup$ Sep 29, 2020 at 22:15
  • $\begingroup$ @HenryBigelow It doesn't actually matter for this answer. Six microstates of different energies implies six macrostates of different energies (since macrostates are merely sets of microstates that share the same thermodynamic variables; in this case, they're all singletons). This answer also works, though, if there are many microstates per macrostate - the only thing that affects is the degeneracy factors on each probability. $\endgroup$ Sep 29, 2020 at 22:22
  • $\begingroup$ Ahh I see. Thanks for the clarification. Yes, I was thinking of index i, for E_i, as indexing microstates, where the E_i were in ascending order, but not necessarily unique values. So I suppose we are talking about the same thing. $\endgroup$ Sep 30, 2020 at 4:13

The parameter $\beta$ is at first only a Lagrangian multiplier. I don't see any contradition if it is zero or negative. While in this case $T = \beta^{-1}$ seems strange.

About a system with a maximum energy, I can imagine the set of the cars of the world, and the energy they spend in a interval of 1s around a given moment. Probably most will have zero energy, but certainly the maximum is bounded.

Another example, that I did as an exercise to understand Boltzmann distribution is income inequality. It is only necessary to replace energy levels by income range, and particles by people. One of the examples was if there was for any reason a maximum income.

The 2 graphics in annex represent maximum income = 1000. The first is for average income of 150, resulting in positive $\beta$. The second for average of 750, resulting in a negative $\beta$. The vertical axis is the % of the total income earned by that range.

In the case of $\beta = 0$ all income levels have the same number of people.

enter image description here


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