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
<br /> \bar{E} = \int E\, p(E) \, dE<br />
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
<br /> p(E) = \sum_n P_n<br />
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
<br /> p(E) \propto \Omega(E) e^{-E/kT}<br />
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
<br /> \Omega(E) = \frac{d \Phi(E)}{dE}<br />

For a particle with energy E, the classical approach tells us
<br /> \Phi(E) \propto \int dx\, \int dp \, <br />
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
<br /> \Phi(E) \propto \int_0^{\sqrt{2mE}} dp \propto \sqrt{E}.<br />
Hence \Omega(E) \propto E^{-1/2} and
<br /> P(E) = C_1 E^{-1/2} e^{-E/kT}.<br />
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
<br /> P(E) = C_2 e^{-E/kT}.<br />
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
<br /> p(E) = C_n E^{n/2-1} e^{-E/kT}<br />

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!