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## Main Question or Discussion Point

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:

[itex]\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[/itex]

Doing this integral in 3-dimensions gives [itex]\frac{3}{2}k T[/itex]

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

[itex]\int_{0 }^{\infty } E \sqrt{\frac{1}{k T}} e^{-\frac{E}{k T}}dE=k T[/itex]

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?

[itex]\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[/itex]

Doing this integral in 3-dimensions gives [itex]\frac{3}{2}k T[/itex]

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

[itex]\int_{0 }^{\infty } E \sqrt{\frac{1}{k T}} e^{-\frac{E}{k T}}dE=k T[/itex]

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?