# Question about the proof that heat capacity goes to zero if temperature approaches $0K$

I don't completely understand the proof that is given for the claim that the heat capacity goes to zero, if the temperature approaches $0K$.

They do it as follows, if $C_x$ is the heat capacity where the parameter $x$ is taken constant, and we heat the system from $0K$ to $T_0K$, then :

$$S(T_0) = \int_0^{T_0}{dS} = \int_0^{T_0}{\frac{dQ}{T}} = \int_0^{T_0}{\frac{C_xdT}{T}}$$ (where the differential in the second step is not a real differential)

The only way for this integral to converge, is when $C_x$ is zero in the limit of $T\to 0K$. But I have 2 problems with this. First they claim also that it is impossible to approach zero kelvin, so in fact there doesn't exist a path from $0K$ to $T_0 K$, so it should be impossible to heat the system from $0K$ to $T_0K$. This is worrying me, because in the proof of the claim that it is impossible to reach $0K$, they use a very similar reasoning showing that the integral can not exist.

Secondly the path must be reversible, so even if there exists a path from $0K$ to $T_0K$, how do we know there also exists a reversible one?

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1 - yes the zero temperature limit is not reachable, so you can't measure the heat capacity at zero temperature, what this calculation tell you is that if you measure at smaller and smaller temperatures you will see that C converges towards zero

2- No the reversibility of the path is not important as the entropy is an exact differential

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Thank you, the first point solved my first question. I don't understand the second one completely though: that the entropy is an exact differential implies that if we can calculate the change in entropy along one path, we have calculated the change along every path. But because we try to calculate the change of entropy here with $dS = dQ/T$, and we use $C_x$, it seems to me that there has to be a reversible path with constant $x$, otherwise we couldn't calculate the entropy like that. I don't know though, why there has to be such a reversible path. I hope you can tell me what I'm missing. –  yarnamc May 20 '13 at 21:08
It can't be a reversible process by definition, reversible processes don't change entropy!!! If you want to measure $C_V$ for instance, you have to choose a process in which volume don't change, this is the only constraint, $C_V$ is a state function (consider for example the classical value $3/2 N k$, that you proven wrong in this post), if it wasn't you wouldn't be talking about his limit at zero temperature. –  Ikiperu May 21 '13 at 17:05
The integral must indeed be evaluated along a reversible path. However I do not see why such a path could not be devised. –  jac Jul 28 at 16:07

The best way to see this is to realize that the zero heat capacity is a quantum effect. Classically, the heat capacity does not go to zero.

Quantally what happens is that at low enough temperatures all the particles are in their lowest possible energy states. To get even one particle into a higher energy state requires a small but finite energy $\epsilon$. If the gas is at a finite temperature such that the total energy of the gas is less than $\epsilon$, there can be no movement to a higher energy state, the energy is then constant, and the heat capacity is then zero.

In the equation for entropy in terms of heat capacity, note that the heat capacity has to go to zero faster than $T$. For example $C_V = a\ln T$ won't work.

But in general your technique is not the way to prove that the heat capacity is zero. The actual calculation is statistical in nature and has two cases, one for bosons such as $^4He$ and another for fermions such as electrons.

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I don't get it...why would asking that $\int_0^{T_0}dT\:C(T)/T \:< \infty$ be less general than the statistical approaches you mentioned? It may be less "fundamental" but that's another issue don't you think? –  gatsu May 21 '13 at 8:50
It is maybe not a proof, but rather a good reason to expect a specific heat approaching zero for zero temperature., what is confirmed by quantum statistical mechanics. –  jac Jul 28 at 16:12

Integrability of the inexact differential $\delta Q$ is a law of nature. Although in general Pfaffian differential forms like $\delta Q$ are not integrable, second law of thermodynamics guarantees that an integrating factor always exists and it is $1/T$ in all cases, $T$ being the absolute temperature.

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