Assume the set $Q:=\{1\}$. We define a function $E:Q\to \mathbb{R}$, implemented as $E(q)=q$. Assume a statistical prior defined by the average value $\overline{E}$, then the probability distribution which maximizes the entropy under the statistical prior is the Gibbs ensemble over set $Q$, defined as:

$$ Z=\sum_{q\in Q}e^{-\beta E(q)} = e^{-\beta} $$

Its equation of state is $dS=k_B \beta d\overline{E}$.


  1. I am trying to understand the set up as generally as I can, therefore I will resist the temptation to simplify the treatment by appealing to the symmetries or simplicity of the example.
  2. Clearly $\overline{E}=1$. We can show this explicitly using $\overline{E}:=-\frac{\partial \ln{Z}}{\partial \beta}=\frac{1}{e^{-\beta}} \sum_{q\in Q}E(q)e^{-\beta E(q)}=1$

  3. If we try to find $S$ exactly, I would start from $S=k_B(\ln{Z}+\beta\overline{E})$ then proceed as follows: $S=k_B(\ln{e^{-\beta}}+\beta \overline{E})\implies S=- k_B \beta + k_B \beta \overline{E}=0$. This makes sense since there is only one state for the system to be in, and therefore the entropy should be $0$.

My problem is with the equation of state $dS=k_B \beta d\overline{E}$ and how to interpret in its general form.

  1. Why is the equation of state "suggesting" that $d\overline{E}$ could be varied - by virtue of being a differential? What information is erased during the construction of the equation of state such that the knowledge that Q contains a single element is ignored, leading to a differential? As we recall the equation of state is a constructed directly from the thermodynamics relation $\frac{\partial S}{\partial \overline{E}}=k_b \beta \implies dS=k_B \beta d\overline{E}$.

  2. Looking at the equation of state I would think that the system could varie $\overline{E}$ at the cost of increasing the entropy. But for this example, $\overline{E}$ cannot vary! I do not understand how the equation of state, a general mathematical definition, can be contradicted by a specific case (unless the so-called general definition does not apply generally, or more likely I am assuming something wrong somewhere). Can anyone shed some light?


1 Answer 1


I think you're overthinking this particular case, which is very common to do with such degenerate cases. In this case the equations are 100% correct in a very degenerate way, namely your expression $S=0$ guarantees that $dS = 0$ while your expression that $E=1$ similarly guarantees that $dE=0$, which means that we can trivially state that $dS = \alpha~dE$ for any $\alpha$, of which one happens to be $\alpha = k_\text{B}\beta.$

If one wants a bit more insight—and I am not sure there is more to be had!—a trick a topologist taught me is very helpful. He insisted that he stole one really useful trick from physics and that was to almost always cut out serious reasoning about degenerate cases, by always adding some noise. So if two lines are parallel he expands out to one order, "what if they're not quite parallel, just as a physicist never knows anything absolutely but only within some noise, what if I assume there's a tiny little bit of noise that makes these two things non-parallel and they intersect at some great distance?" and then you can very meaningfully understand that there are no "repeated roots" in polynomials, just roots that are very close together within error thresholds, or there are no "parallel lines that do not intersect", they just intersect very far away.

In this case you would want to state that maybe this particle is not exactly stuck at $Q=\{0\}$ but maybe it's stuck on $Q=\mathbb R$ but some very stiff spring keeps it very close to $q\approx 0$, in which case we can have $E(q) = \frac12 a q^2$ for some big $a$ and then instead of getting your result of $Z=1$ (you made the math a little complicated for yourself by choosing $\{1\}$ instead of $\{0\},$ it happens...) you get a result of $Z=\sqrt{2\pi/a\beta},$ thus $\bar E = 1/(2\beta) = \tau/2,$ as it must be by the equipartition theorem, and then one gets an interesting result that the entropy changes themselves do not depend on $a$, but that just comes out into this additive constant that one gets from integrating $\int d\tau/(2\tau) = \frac12 \ln \tau + C.$ So one ultimately gets some $\frac12 \ln(\tau/a)$ type of expression... so if you try to realize the system in practice it will no longer be "this thing always has zero entropy;" one instead gets a physical model where "this thing is confined by a rigid potential between $-\epsilon$ and $+\epsilon$ but as you dump more and more energy into the thing with a higher and higher temperature, the underlying Gaussian distribution does broaden imperceptibly and raises the entropy by the relation $d\sigma = d(S/k_\text B) = d\tau/(2\tau)$ as the equipartition theorem would demand." (Actually we should probably also add a momentum coordinate as well, so maybe it's $d\tau/\tau$ for one dimension, but meh.)

  • $\begingroup$ Could we then say that drawing thermodynamic cycles on a cartesian graph (say the x axis is the average energy, and the y axis is the entropy) would fail with the system I presented --- in this case the state variables would occupy a point. I might have been under the mistaken assumption that one can use the general equation of state to draw arbitrary cycles, but it appears that the "size/shape" of the cycles is restricted by the microscopic description. I'd appreciate if you could confirm. Also, thanks for your second example, I will try to reproduce it on my own to help my understanding. $\endgroup$
    – Anon21
    Commented Mar 1, 2019 at 18:06
  • $\begingroup$ @AlexandreH.Tremblay I mean yeah if there's a discrete set of states one would rather describe it with a Markov matrix -- a transition matrix to "hop" from one state to another, either in the continuous-time case in a time $dt$ or in the discrete-time case in one timestep... And then one has some nice properties about such matrices, for instance if $\underline 1$ is a row-vector of all 1s then $\underline 1\cdot M=\underline 1$ so $\lambda=1$ is an eigenvalue and all eigenvalues are bound to be on $[-1, 1].$ $\endgroup$
    – CR Drost
    Commented Mar 1, 2019 at 19:55

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