thermodynamics - Kinetic energy as $pi k_B T$

In the derivation of the Thermal De Broglie Wavelength on Wikipedia, I come across the following:

"In the nonrelativistic case the effective kinetic energy of free particles is $E_K=\pi k_B T$

https://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength

Provided this is correct, in what instances is the kinetic energy of a free particle $\pi k_B T$ and not $\frac 3 2 k_B T$? Thanks for any comments on this.

Answer

...in what instances is the kinetic energy of a free particle...

Remark: The temperature dependent expression for the kinetic energy is not a property of a single particle, but an ensemble.

The $\frac{3}{2}$ comes from a statistical consideration of degrees of freedom of a mechanical particle scheme. The $\pi$ comes form the wave picture.

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Using a characteristic mass and energy, $m$ resp. $E'$, the quantity $p'=\sqrt{m\,E'}$ has the units of momentum. In the realm of thermodynamics, multiples of $k_BT$ are candidates for such a characteristic energy.

$\pi$'s will enter your theory once you compute expectation values assuming a classical dispersion/energy-momentum relation $E(p)=\frac{p^2}{2m}$, resp. $p(E)=\sqrt{2m\,E}$. In a canonical ensemble, a characteristic momentum is given by

$p''=\int_{-\infty}^{\infty}{\mathrm {exp}}\left({-\frac{E(p)}{k_BT}}\right){\mathrm d}p=\sqrt{2m\,(\pi\,k_BT)}$.

Or, before normalization, the computation of the variance $\sim\int_{-\infty}^{\infty}p^2\,{\mathrm {exp}}({-\tfrac{E(p)}{k_BT}})\,{\mathrm d}p$ will generate a factor $\sqrt{\pi}$. I'm not sure about the merit of adopting $\pi$ as the sole proportional constant before I haven't seen what the energy in your application is used for / what's it converted into. As far as I can see you could also absorb them into your temperature scale.