# Energy of a system with quadratic degrees of freedom

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.6.

The definition of temperature in terms of entropy is

$\displaystyle \frac{1}{T}\equiv\frac{\partial S}{\partial U} \ \ \ \ \ (1)$

Schroeder quotes a theorem that states that for any system with only quadratic degrees of freedom in the high temperature limit, the multiplicity ${\Omega}$ is proportional to ${U^{Nf/2}}$ where ${U}$ is the energy, ${N}$ is the number of molecules and ${f}$ is the number of degrees of freedom per molecule. For an ideal gas, this is given by Schroeder’s equation 2.40:

$\displaystyle \Omega\approx\frac{V^{N}\left(2\pi mU\right)^{3N/2}}{h^{3N}N!\left(3N/2\right)!} \ \ \ \ \ (2)$

Since ${f=3}$ for a monatomic ideal gas (3 translational degrees of freedom only), the theorem is valid here. Similarly, for an Einstein solid in the high temperature limit

$\displaystyle \Omega\approx\left(\frac{qe}{N}\right)^{N} \ \ \ \ \ (3)$

and since ${q}$, the number of energy quanta, is proportional to ${U}$. Each oscillator in an Einstein solid is equivalent to a one-dimensional harmonic oscillator, which has two degrees of freedom. [Rather, it has two quadratic terms in its energy: ${\frac{1}{2}kx^{2}}$ for potential energy and ${\frac{1}{2}mv^{2}}$ for its kinetic energy. Each of these is interpreted as a ‘degree of freedom’.] Thus again, ${\Omega\propto U^{Nf/2}=U^{N}}$.

In the general case, we have

$\displaystyle \Omega=AU^{Nf/2} \ \ \ \ \ (4)$

for some constant ${A}$ (that is, ${A}$ doesn’t depend on ${U}$). The entropy is therefore

$\displaystyle S=k\ln\Omega=k\ln A+\frac{kNf}{2}\ln U \ \ \ \ \ (5)$

and the temperature is given by

 $\displaystyle \frac{1}{T}$ $\displaystyle =$ $\displaystyle \frac{kNf}{2U}\ \ \ \ \ (6)$ $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{1}{2}NfkT \ \ \ \ \ (7)$

which is just what the equipartition theorem predicts (${\frac{1}{2}kT}$ energy per degree of freedom).

The formula is valid only for large energies, since if ${U}$ is small enough, ${\Omega}$ eventually becomes less than 1 and the entropy 5 can become negative, which isn’t physically possible.