# Does entropy $S(E)$ as a function of energy $E$ have a negative second order derivative?

From David Tong's lecture, I know that the temperature $$T$$ is defined as a function of energy $$E$$, as $$T(E)=\frac1{S'(E)},$$ where $$S(E)$$ denotes the entropy of an isoenergetic surface of energy $$E\sim E+\delta E$$. Meanwhile, the heat capacity is $$C(T)=\frac{\partial E}{\partial T}=\frac1{T'(E)}=-\frac{1}{T^2S''(E)}.$$ Then we have an immediate corollary (also pointed out by the author) that:

As long as $$C>0$$, we have $$S''(E)<0$$.

However, counter-examples can be found, such as:

• A 1-dimensional harmonic oscillator has $$C>0$$, but its density of states $$D(E)$$ is constant since the energy levels are equally spaced. Therefore, $$S(E)=k_\text{B}\ln [D(E)\delta E]$$ is constant and $$S''(E)=0$$.

• A free particle in a 2-dimensional box with periodic boundary conditions has $$E=p^2/2m$$. Therefore, the isoenergetic surface has volume $$2\pi p\mathrm dp\propto D(E)\mathrm dE$$ and hence $$D(E)$$ is constant. We still have $$C>0$$ and $$S''(E)=0$$.

• A free particle in a 1-dimensional box with periodic boundary condition has $$2\mathrm dp\propto D(E)\mathrm dE$$. We have $$D(E)\propto\frac1{\sqrt{E}}$$ and $$S(E)=-\frac12k_\text B\ln E+\text{const.}$$, which even leads to $$S''(E)>0$$.

Why the paradox? Or what's wrong with the above calculations?

• All the examples given have few degrees of freedom, and thus have nothing to do with thermodynamics. Once they are coupled to a bath, the situation changes, since we have to consider the degrees of freedom and the spectrum of the system + bath. Commented Jul 24 at 10:55
• Just to go on with Roger V. answer. All your systems are integrable, if they were let alone they would not reach equilibrium. Thus, you really need an external coupling as Roger V. was saying to make sense of the concept of entropy, etc... Not that it cannot be defined at the microscopic level (see stochastic thermodynamics), just that in this case it is not useful. In the first case, you explore correctly the isosurface of energy E (funnily enough, the harmonic oscillator is ergodic) but you do not thermalize your velocity degrees of freedom. Same thing for the second and third examples. Commented Jul 25 at 15:06
• For the second and third examples, you do not even explore totally the phase space region with E = cst since your momentum is constant (the energy of a particle is $E = p^2/(2m)$ and it always goes at a velocity $v = p/m$, but the velocity $-v$ would give the particle the same energy and hence must be taken into account in the counting of microstates in the isosurface. But it is not accessible by your particle that always go at $v$ and never at $-v$ Commented Jul 25 at 15:07
• Finally, as Roger V. was saying, if you couple your system to an external source of energy, it is fine. But in this case you are not anymore described by a microcanonical ensemble but by a canonical one. The ensemble would lead to the same results only for non-integrable systems, with short range interactions in the infinite system size limit! Commented Jul 25 at 15:11