# Why use dimensionless heat capacity?

Perhaps this is blindly obvious, but in typical discussions of statistical mechanics (with, say, constant volume) one often finds that, rather than using the heat capacity

$$C_V = \frac{\partial E}{\partial T}\bigg|_V = T\frac{\partial S}{\partial T}\bigg|_V,$$

(where the 2nd equality comes from the 1st Law of Thermodynamics), the dimensionless heat capacity

$$\hat{C}_V = \frac{C_V}{Nk}$$

(where $N$ is the # of particles in the system, and $k$ is the Boltzmann constant) is introduced and used instead. So.... why? How is the dimensionless version more useful than the original?

• in physics, it's cool to get rid of the pesky units! It helps in making nice graphical plots, among other things...
– Cham
Aug 30, 2019 at 16:15

One benefit of scaling the heat capacity with another extensive variable is that you end up with an intensive property – heat capacity per number of particles. Similarly, specific heat refers to the heat capacity per mass so that the value of the intensive property can be compared between samples of the same material but with different sizes or geometries for example.

• This explains the choice to use the specific heat $c_v = C_V/N$, but doesn't really explain why Boltzmann's constant is thrown in there. Feb 22, 2017 at 17:26

I think this also comes from the fact that using equipartition theorem, each quadratic degree of freedom in global hamiltonian contributes for $$\frac{1}{2}k_BT$$ in energy and so for $$\frac{1}{2}k_B$$ in $$C_V$$, so basically $$\hat{C_V}=\frac{3}{2}$$ means 3 quadratic degree of freedom (monoatomic ideal gas).

This is a relatively common statistical mechanics convention, for instance in diatomic ideal gases, you have And for solids you get 6 degree of freedom (Note that equipartition theorem is valid only when quantization of energy level can be neglicted in front of $$k_BT$$)