# Meaning of zero temperature in terms of $1/T=\partial S/\partial E$

Suppose we have a $$NVE$$ system of a pure crystal. From the definition of the temperature in macroscopic thermodynamics,

$$\frac{1}{T}=\left. \frac{\partial S}{\partial E}\right |_{V,N}$$

. Then suppose the temperature of the system approaches zero. From the equation above, we can deduce that (1) any infinitesimal change of energy results in infinite change of entropy at zero temperature. Alternatively, (2) any infinitesimal change of entropy cannot change energy at zero temperature. I cannot understand why (1) and (2) are true.

Using Boltzmann's definition of entropy in microcanonical ensemble, we get $$\frac{1}{T}= \left. k_{B}\frac{1}{\Omega}\frac{\partial \Omega}{\partial E} \right |_{V,N}$$ i.e. intensive variables are related to density of states of the system. In both classical and quantum mechanics, we know that $$1 \leq\Omega < \infty$$. Thus, at zero temperature, $$\left. \frac{\partial \Omega}{\partial E} \right |_{V,N}\rightarrow \infty$$.

Suppose our system consists of classical particles. Then $$\Omega=\Omega(E,N,V)$$ is a continuous function of $$E$$; thus, if we infinitesimally increase energy of the system,

$$\frac{\partial \Omega}{\partial E}=\frac{\Omega(E_0+\Delta E,N,V)-\Omega(E_0,N,V)}{\Delta E}<\infty$$

where $$E_0$$ is a ground state energy of the system**.

Suppose our system consists of quantum mechanical particles. Then, if the energy change $$\Delta E$$ is so small that such $$\Delta E$$ cannot excite energy of any particle (the energy is discretized), $$\Omega(E_0+\Delta E,N,V)$$ will be zero. In this way, I may argue that

$$\frac{\partial \Omega}{\partial E}=\frac{0-\Omega(E_0,N,V)}{\Delta E} \,\,\text{(not well defined!)}$$

However, in thermodynamics, we regard $$\frac{\partial \Omega}{\partial E}$$ as a well-defined value; thus, I would claim that $$\Delta E$$ is larger than the smallest energy gap between the states of the system but is a sufficiently small value. Then, again,

$$\frac{\partial \Omega}{\partial E}=\frac{\Omega(E_0+\Delta E,N,V)-\Omega(E_0,N,V)}{\Delta E}<\infty$$ .

My reasoning above says that regardless of classical or quantum treatment, any infinitesimal change of energy cannot result in infinite change of entropy at or near to zero temperature, which violates the statement (1). Justifying the statement (2) is even more difficult; any increase of entropy at or near to zero temperature means that some of particles in the system are excited, which will increase the energy of the system from its ground value $$E_0$$.

How can I justify the statements (1) and (2) and understand the meaning of zero temperature in thermodynamics?

** From the Planck's definition of the third law of thermodynamics, $$S \rightarrow 0$$ as $$T \rightarrow 0$$. From the convexity of $$E=E(S)$$ in equilibrium, $$E$$ approaches its ground value as $$S \rightarrow 0$$.

• It is more practical to think about a state close to 0 degrees K, but not at zero degree. The state is not stable and will spontaneously change to a large entropy state with whatever small energy input according to this formula $dS = dE/T$. – user115350 Mar 22 at 22:13

For any finite quantum system, everything must be finite, but in the limit $$N\to \infty$$ we can see this divergence.

Let's consider a simple model - $$N$$ noninteracting two-state systems, with energy levels $$0$$ and $$\epsilon$$. The ground state energy is $$0$$. I will treat the "continuum limit", where we pretend $$E$$ is continuous, first (this should become an acceptable approximation in the limit $$N \to \infty$$)

The entropy $$S{\left(E\right)}$$ is

$$S{\left(E\right)} = \log{\binom{N}{E/\epsilon}} \approx \frac{E}{\epsilon} \left( \log{N} - \log(E/\epsilon)\right)$$,

with the approximation holding for $$N >> E/\epsilon$$.

If we plug in $$E = 0$$, we find $$S = 0$$. We also find that $$\partial{S}/\partial{E} \to \infty$$, giving us $$T=0$$! The divergence of the derivative does not contradict your observation that $$Ω(𝐸_0+Δ𝐸,𝑁,𝑉)−Ω(𝐸_0,𝑁,𝑉) < \infty$$, in the same way that the derivative of $$\sqrt{x}$$ diverges at $$x=0$$, but $$\sqrt{x}$$ is still perfectly finite.

Let's try a discrete analysis now. Note that the entropy of the first excited state is

$$S{\left(\epsilon\right)} = \log{\binom{N}{1}} = \log{N}$$.

So the finite-difference analogue to the inverse temperature is

$$\frac{\Delta S}{\Delta E} = \frac{S{\left(\epsilon\right)}}{\epsilon} = \frac{\log{N}}{\epsilon}$$.

While this is of course finite in a finite system, it diverges in the thermodynamic limit!

• Yes, $\sqrt{x}$ is stable while its derivative is not. Thanks. – Discovery Jun 5 at 17:40