# What's the value for $\lim_{T\rightarrow 0}C_V$?

$$C_V=\frac{1}{kT^2}[\overline{E^2} -\overline E ^2]$$ where $$\overline{E}^2=-\frac{1}{Z}\frac{\partial Z}{\partial \beta}$$ and $$\overline{E^2}=-\frac{1}{Z}\frac{\partial^2 Z}{\partial^2 \beta}$$.

$$Z=\sum _ie^{-\beta\cdot E_i}$$ of finite sum is the partition funciton and $$\beta=\frac{1}{kT}$$.

The question is what is $$\lim_{T\rightarrow0^+}C_V$$.

According to my calculation, as $$T\rightarrow 0^+$$, $$[\overline{E^2} -\overline E ^2]$$ approach some finite number $$K$$. Thus $$C_V$$ approach $$\infty$$.

I didn't believe in the begining, so I tried to plot the numerical solution, I also put the simplified function into Wolfram.

However, all the result I got was $$\lim_{T\rightarrow 0^+}C_V=\infty$$.

And it did not make too much sense,

I found two reference

1. http://stp.clarku.edu/notes/chap6.pdf 6.17 seem to support my argument.

2. http://www-personal.umich.edu/~lsander/ESP/chap4.pdf page 5 otherwise.

and my professor's answer sheet says it's suppose to be $$0$$.

• @K_inverse Yeah, but obviously there is a thermal stastical treatment, I couldn't figure out A. how to show $C_V=0$ with thermal statistics derivation. B. how to proof the above calculation for $C_V$ fails. – user9976437 Nov 13 '18 at 8:08
At very low temperature, the states with the highest probability are those at very low energy, close to the ground state. In the limit $$T \rightarrow 0^+$$ it is only the ground state which contributes to the canonical sum over the states. If the ground state is non-degenerate, the distribution probability must have zero-variance.