Equipartition Theorem

The internal energy of a monatomic ideal gas containing $N$ atoms is $(3/2)\,N \,k_B\,T$. (See Section 5.2.3.) This implies that each atom possess, on average, $(3/2)\,k_B\,T$ units of energy. Monatomic particles have only three translational degrees of freedom, corresponding to their motion in three dimensions. They possess no internal rotational or vibrational degrees of freedom. Thus, the mean energy per degree of freedom in a monatomic ideal gas is $(1/2)\,k_B\,T$. In fact, this is a special case of a more general result. Let us now try to prove this result.

Suppose that the energy of a system is determined by $f$ coordinates, $q_k$, and $f$ corresponding momenta, $p_k$, so that

$$E = E(q_1, \cdots, q_f, p_1,\cdots, p_f).$$ (5.366)

Suppose further that:

1. The total energy splits additively into the form

   $$E = \epsilon_i(p_i) + E'(q_1,\cdots, p_f),$$ (5.367)

   where $\epsilon_i$ involves only one variable, $p_i$, and the remaining part, $E'$, does not depend on $p_i$.
2. The function $\epsilon_i$ is quadratic in $p_i$, so that

   $$\epsilon_i(p_i) = b\,p_i^{\,2},$$ (5.368)

   where $b$ is a constant.

The most common situation in which the previous assumptions are valid is where $p_i$ is a momentum. This is because the kinetic energy is usually a quadratic function of each momentum component, whereas the potential energy does not involve the momenta at all. However, if a coordinate, $q_i$, were to satisfy assumptions 1 and 2 then the theorem that we are about to establish would hold just as well.

What is the mean value of $\epsilon_i$ in thermal equilibrium if conditions 1 and 2 are satisfied? If the system is in equilibrium at absolute temperature $T$ then it is distributed according to the Boltzmann probability distribution. (See Section 5.4.7.) In the classical approximation, the mean value of $\epsilon_i$ is expressed in terms of integrals over all phase-space (see Section 5.4.4):

$$\langle\epsilon_i\rangle = \frac{ \int_{-\infty}^{\infty} \exp[-E... ... {\int_{-\infty}^{\infty} \exp[- E(q_1,\cdots, p_f)/k_B\,T]\, dq_1\cdots dp_f}.$$ (5.369)

Condition 1 gives

$$\langle\epsilon_i\rangle =\frac{ \int_{-\infty}^{\infty} \exp[-(\epsilon_i + E')/k_B\,T]\,... ...p_f} {\int_{-\infty}^{\infty} \exp[-(\epsilon_i + E')/k_B\,T]\,dq_1\cdots dp_f}$$

$$= \frac{\int_{-\infty}^{\infty} \exp(-\epsilon_i/k_B\,T)\,\epsilo... ..._i/k_B\,T)\, dp_i \,\int_{-\infty}^{\infty} \exp(-E'/k_B\,T)\,dq_1\cdots dp_f},$$ (5.370)

where use has been made of the multiplicative property of the exponential function, and where the final integrals in both the numerator and denominator extend over all variables, $q_k$ and $p_k$, except for $p_i$. These integrals are equal and, thus, cancel. Hence,

$$\langle\epsilon_i\rangle = \frac{\int_{-\infty}^{\infty} \exp(-\e... ...)\,\epsilon_i\, dp_i} {\int_{-\infty}^{\infty}\exp(-\epsilon_i/k_B\,T)\, dp_i}.$$ (5.371)

This expression can be simplified further because, writing $\beta=1/k_B\,T$,

$$\int_{-\infty}^\infty \exp(-\beta \,\epsilon_i)\,\epsilon_i\, dp_... ...ial \beta} \left[\int_{-\infty}^\infty \exp(-\beta \,\epsilon_i)\, dp_i\right],$$ (5.372)

so

$$\langle\epsilon_i\rangle = - \frac{\partial}{\partial\beta} \ln\left[\int_{-\infty}^\infty \exp(-\beta \,\epsilon_i)\, dp_i\right].$$ (5.373)

According to condition 2,

$$\int_{-\infty}^{\infty} \exp(-\beta\, \epsilon_i)\, dp_i = \int_{... ...\frac{1}{\sqrt{\beta}} \int_{-\infty}^{\infty} \exp\left(- b\, y^{2}\right) dy,$$ (5.374)

where $y = \!\sqrt{\beta} \,p_i$. Thus,

$$\ln \int_{-\infty}^{\infty} \exp(-\beta\, \epsilon_i)\, dp_i = - ... ...{1}{2} \ln \beta + \ln \int_{-\infty}^{\infty} \exp\left(- b \,y^{2}\right) dy.$$ (5.375)

Note that the integral on the right-hand side is independent of $\beta$. It follows from Equation (5.373) that

$$\langle\epsilon_i\rangle = - \frac{\partial}{\partial \beta}\! \left( - \frac{1}{2} \ln \beta\right) = \frac{1}{2\,\beta},$$ (5.376)

giving

$$\langle\epsilon_i\rangle = \frac{1}{2}\, k_B\, T.$$ (5.377)

This result is known as the equipartition theorem. It states that the mean value of every independent quadratic term in the energy is equal to $(1/2)\,k_B\,T$. If all terms in the energy are quadratic then the mean energy is spread equally over all degrees of freedom. (Hence, the name “equipartition.”)