# Heat capacity $C$ at low temperature

The internal energy, $U = Nk_bT$ where $N$ is particle number, $k_b$ is Boltzmann constant and $T$ is temperature.

Therefore, the heat capacity $C$ is given by $C=\frac{dU}{dT}=k_b$.

However, in the real world $C$ approaches zero as $T$ approaches zero. How do I show this?

One possible solution is, from Einstein's assumption, in the case of 1D harmonic oscillator, I know that

$$U = \hbar \omega \left(n+\frac{1}{2}\right)$$

where

$$n = \frac{1}{e^{\hbar \omega / k_b T}} \, .$$

So I know its slope ($C$) approaches zero as beta increases. I need to use another assumption from Einstein's to show this. How can I do this? Is there any ideas? I think I need to make an assumption for energy level in order to get partition function of it.

• heat capacity is not defined as $\frac{dU}{dT}$ Commented Jan 28, 2015 at 1:22
• Where did you get that $n=1/\exp(\hbar\omega/k_BT)$? The 1D harmonic oscillator requires $n$ to be an integer, whereas your formula gives it a real value. Commented Jan 28, 2015 at 1:32
• sorry for late, thanks, I should use partial derivative form. /// n is integer, right, that value is from calculating partition functions.
– eric
Commented Jan 28, 2015 at 2:57
• dear kyle, we can compare U which you know it contains integer n and U you got it from the partition function. the form is exactly same and we can guess that n is the value I wrote above
– eric
Commented Jan 28, 2015 at 3:18
• The formula you have for $n$ is wrong, I think. Look up the Bose-Einstein distribution. Commented Jan 28, 2015 at 7:08

The heat capacity at constant is defined as the partial derivative of the internal energy with respect to temperature and at a constant volume: $$C_V=\left(\frac{\partial U}{\partial T}\right)_V$$ which is a little different from $dU/dT$ as you've written.

# $C_V$ Via the Microcanonical Ensemble

For the low-temperature solid, you can approximate the solid as 3-independently moving harmonic oscillators, each with energy $$E_n=\hbar\omega\left(n+\frac12\right)$$ In order to distribute $q$ quanta of energy among $3N$ oscillators, the number of possible states (i.e., the multiplicity) turns out to be $$\Omega=\frac{(q+3N-1)!}{q!(3N-1)!}$$ From this you can find the entropy, which leads, with some mathematical work (e.g., finding out what $q$ is in terms of known terms like $N$ and $k_B$), to the energy being $$U=\frac32N\hbar\omega+\frac{3N\hbar\omega}{\exp[\hbar\omega/k_BT]-1}$$ The heat capacity is then $$C_V=3Nk_B\left(\frac{\hbar\omega}{k_BT}\right)^2\frac{\exp[\hbar\omega/k_BT]}{\left(\exp[\hbar\omega/k_BT]-1\right)^2}$$ Which, as $T\to0$, you can show that $C_V\to0$ as well.

# $C_V$ Via the Canonical Ensemble

Here, the partition function is $$Z=\sum_{n=0}^\infty\exp[-E_n/k_BT]$$ from which the internal energy is easily computed and the heat capacity ends up as $$C_V=3Nk_B\left(\frac{\hbar\omega}{2k_BT}\right)^2\frac{1}{\sinh^2[\hbar\omega/2k_BT]}$$ which also goes to zero as $T\to0$.

Note that there are other methods, such as the Debye model, in which you can get the heat capacity.

• Thanks a lot I owe you. could you explain little bit how did you get the number of possible states in the case of microcanonical ensemble?
– eric
Commented Jan 28, 2015 at 3:05
• the internal energy in your upper case is the same with my internal energy. your value is 3 times larger than mine and it is because of the equipartition theorem. I got that from the partition function but wonder how did you get omega (number of possible states). Is there any other way to show C goes to zero at 0 T?
– eric
Commented Jan 28, 2015 at 3:15
• I did a 3D solid, rather than a 1D solid, hence that factor of 3. $\Omega$ is found via a counting argument: how many ways can $q$ quanta of energy be put into $3N-1$ oscillators. You could make an argument based on intuition (e.g., less heat at lower temperatures so less heat capacity), but formally showing an explicit dependence of $C_V$ on $T$ will show the necessary condition. Commented Jan 28, 2015 at 3:31