# How does the equipartition theorem apply to vibrational modes?

Specifically, how, based on the idea that energy will be shared equally among all degrees of freedom, does one come to the conclusion that each vibrational mode of a molecule will occupy $kT$ units of energy?

When learning statistical thermodynamics, classes often spend a lot of time talking about the Maxwell-Boltzmann distribution and even deriving that the expectation value for the energy of a monoatomic ideal gas is $\frac{3}{2}kT$ where $\frac12 kT$ comes from energy distributed to each cartesian coordinate plane.

Now, I find that I learn a whole lot from mathematical derivations of these types of things, and from simply working through the math and understanding its physical meaning.

So, I want to know, both on a conceptual level and through some sort of mathematical explanation, why does each vibrational mode of a molecule contribute $kT$ to the overall energy of the molecule when those modes are occupied?

My only guess is that one degree of freedom is kinetic energy (associated with the moving atoms) and the other is potential energy (analogy with a spring works here) but, if that is true, much clarity would be gained from a mathematical demonstration of this.

## 1 Answer

As you probably already know, the equipartition theorem states that each quadratic degree of freedom will, on average, possess an energy $\frac{1}{2}kT$, where a ‘quadratic degree of freedom’ is one for which the energy depends on the square of some property.

Now lets consider the kinetic and potential energies associated with translational, rotational and vibrational energy:

• Translational degrees of freedom: $~E = \frac{1}{2}mv^2$

• Rotational degrees of freedom: $~E = \frac{1}{2}Iω^2$

• Vibrational degrees of freedom: $~E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

As you can see that for translational and rotational degrees of freedom, there is only one quadratic. Hence each translational and rotational state will contain $\frac{1}{2}kT$ energy.

However for vibrational states, the energy contains two quadratic contributions. Therefore each rotational state will contain $kT$ energy.

Now lets consider the reasoning behind this. You are right when you say that the vibrational degree of freedom depends on kinetic and potential energy and hence has 2 quadratic contributions. In the equation for the energy of the vibrational state, $\frac{1}{2}mv^2$ represents the kinetic energy and $\frac{1}{2}kx^2$ represents the potential energy.

A molecule in the vibrational state acts like a harmonic oscillator. So let consider $H_2$. You can imagine that the bond between the 2 hydrogen atoms is like a spring which contracts and expands in length.

The average bond length can be thought as the equilibrium length of the bond. When the bond the contracts, the displacement from this equilibrium length is known as 'x'. Now using Hooke's Law, the potential energy of the system is $\frac{1}{2}kx^2$ where k is the spring constant while the kinetic energy is $\frac{1}{2}mv^2$. Hence the total energy of the system is given by: $~E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$