# Equation of state for a set of one element and interpretation

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 $$dS=k_B \beta d\overline{E}$$.

Questions:

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?

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 $$\alpha = k_\text{B}\beta.$$
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.)
• @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 $[-1, 1].$ – CR Drost Mar 1 '19 at 19:55