Average Energy and the Equipartition Theorem

[Plane Wave]

The Equipartition Theorem states that each quadratic degree of freedom contributes 1/2 kT of energy. This can be derived for the translational degrees by integrating the average kinetic energy multiplied by the Maxwell velocity distribution:

$\int_{-\infty }^{\infty } \frac{m v^2}{2} \sqrt{\frac{m}{2 \pi k T}} e^{-\frac{m v^2}{2 k T}}dv=\frac{1}{2}k T$

Doing this integral in 3-dimensions gives $\frac{3}{2}k T$

However the average energy from the Boltzmann energy distribution, which is used in the derivation for energy density per wavelength in blackbody radiation, is

$\int_{0 }^{\infty } E \sqrt{\frac{1}{k T}} e^{-\frac{E}{k T}}dE=k T$

I'm having difficulty reconciling these two results. I understand we are integrating over 2 different variables (velocity for the first and energy for the second), but both results are average energy.

The 1st is a very specific statement on quadratic degrees of freedom. The 2nd is a general result from the Boltzmann energy distribution, which is sometimes smaller than the 1st.

Any thoughts?

 

[DrClaude]

When considering blackbody radiation, you are dealing with oscillatory degrees of freedom. Each corresponds to two quadratic degrees of freedom (kinetic and potential energy, think of the harmonic oscillator), hence $2 \times \frac{1}{2}kT = k T$.

 

[Plane Wave]

Thank you for the response.

Isn't that integral far more general though? I guess what I'm asking is, where in that integral do we specify 2 quadratic degrees of freedom?

 

[jasonRF]

You are on the right track, in that the mean energy is simply
$$\bar{E} = \int E\, p(E) \, dE$$
where p(E) is the probability the system (single particle in this case) has energy between E and E+dE. However, your second equation is assuming 2 degrees of freedom. Here is why. The probability that a system has energy between E and E+dE is
$$p(E) = \sum_n P_n$$
where $P_n$ is the probability of state n, and the sum is only over those states with energy between E and E+dE. Assume dE is small, so $P_n \propto e^{-E/kT}$ for all n. Then
$$p(E) \propto \Omega(E) e^{-E/kT}$$
where $\Omega(E)$ is the number of states with energy between E and E+dE. $\Omega(E)$ isn't so easy to calculute - it is usually easier to calculate $\Phi(E)$, the number of states with energy less than E, and then find
$$\Omega(E) = \frac{d \Phi(E)}{dE}$$

For a particle with energy E, the classical approach tells us
$$\Phi(E) \propto \int dx\, \int dp \,$$
where the integral is over the portion of position-momentum space corresponding to energies less than E.

For a particle in a mono atomic gas, $E = p^2/2m$, which does not depend upon x. So the x integral is just a constant. Hence
$$\Phi(E) \propto \int_0^{\sqrt{2mE}} dp \propto \sqrt{E}.$$
Hence $\Omega(E) \propto E^{-1/2}$ and
$$P(E) = C_1 E^{-1/2} e^{-E/kT}.$$
If you find the normalization constant $C_1$, you should find that the mean energy is kT/2.

For a harmonic oscillator, $E = k x^2 /2 + p^2 / 2m$, so the integral for $\Phi(E)$ is over an ellipse in position-momentum space. The semi-major and semi-minor axes of this ellipse are $\sqrt{2E/k}$ and $\sqrt{2Em}$. Hence the integral, which is just the area of the ellipse, is proportional to E. That is, $\Phi(E) \propto E$ so $\Omega(E)\propto 1$. So for this two degree of freedome harmonic oscillator we find
$$P(E) = C_2 e^{-E/kT}.$$
This is the expression you used in your second case (although you should check your normalization factor - it has the wrong units!). Anyway, find the normalization constant $C_2$ and you should find that the mean energy is kT, just as you found before.

jason

 

[jasonRF]

Just thought I would point out a couple more things

1) If what I wrote is too complicated, a simpler thing you can do is take your velocity space integral and change variables to $E=\frac{1}{2} m v^2$. Note that you can use the fact that the integrand is even, so the integral from -infinity to infinity can be replace by twice the integral from 0 to infinity. Anyway, you should find a square root of energy shows up just as it did in my calculation for a single degree of freedom; you have already done the integral so you know it is 1/2 kT. Since your second integral doesn't have the square root, it cannot be for a single degree of freedom.

2) The calculation I did for one and two quadratic degrees of freedom can easily be generalized to n quadratic degrees of freedom. In that case $\Phi(E)$ is a volume in n-dimensional space, where the size in each dimension is proportional to $\sqrt{E}$. Hence $\Phi(E) \propto E^{n/2}$ and we find
$$p(E) = C_n E^{n/2-1} e^{-E/kT}$$

jason

 

[Plane Wave]

Wow, thank you very much Jason. That was incredibly helpful. I very much appreciate the time and effort it took to write that out.

Thanks again!