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 $E_K=\pi k_B T$


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.

  • 2
    $\begingroup$ There seems to be some contradictions about this point that I haven't been able to understand. Please take a look at this: i.imgur.com/8IAEQH2.jpg and this: i.imgur.com/Y7waIgP.jpg. I can refer you to the two different paper I got this from if you want, but it is obvious that the average kinetic energy of a particle is different in the three cases. I will wait for someone to explain that. $\endgroup$ – Abanob Ebrahim Aug 19 '15 at 10:11

...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.

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.

  • 1
    $\begingroup$ Apologies, after going through your answer, the factor looks as wrong as it did before. There is no freedom in the definition of the momentum, energy, or absolute temperature. The purely numerical factors must be determined and everything I see indicates that de Broglie has simply determined them incorrectly. An average spherically symmetric particle of an ideal gas simply carries $E=3kT/2$ in average and the fact that it may also be described as a wave can't contradict this fact. $\endgroup$ – Luboš Motl Aug 19 '15 at 9:56
  • $\begingroup$ @LubošMotl: You don't need to violate any exact entropy expressions or the absolute temperature scale. Just consider $k_B^*\equiv \pi\,k_B$ the auxiliary constant. Some like $k_B=1$ and electron volts, and some might find $\approx 10^{-23}J/K$ and kelvins or "pi-kelvins" useful. That's why I said I don't know what would be the merit of adopting another $3.14$ to get energy out of temperature. I showed OP where $\sqrt{\pi}$ pops up. In expectation values, it will pop out anyway, because the density $\rho\propto{\mathrm e}^{-H\,/\,k_BT}$ gets normalized. $\endgroup$ – Nikolaj-K Aug 19 '15 at 10:59
  • $\begingroup$ Sorry, no one else has used $k^*_B$ before, so this is clearly a new invention that can in no way fix the errors in anything that was written before. This frantic redefinition only reinforces the point that the original authors were simply sloppy about all coefficients of order one. One can write formulae that are fine up to numbers of order one. But if one writes actual numerical constants everywhere, they should be correct and only one numerical constant in each equation is correct. $\endgroup$ – Luboš Motl Aug 19 '15 at 11:03
  • $\begingroup$ @LubošMotl: The question is if it matters as long as you keep on using a temperature-unit that's formally distinct from an energy unit. Sure, you may take the reductionist approach and fix all factors using the deeper theory, but "I haven't seen what the energy (...) is converted into" and so if all quantities in the theories account for this strange $\pi$, it may end up not ruining any results. I agree that it's odd that a particle should carry an irrational energy-quantity - I'm not advocating it. But apparently people have used this quantity for some time and it worked for their use. $\endgroup$ – Nikolaj-K Aug 19 '15 at 11:20
  • $\begingroup$ General physical quantities are almost always irrational - and almost never rational (or rational multiples of pi, for that matter). The rationality, like the value, depends on units. None of these things are an issue. The issue is that energy, momentum, as well as temperature are quantities that are well-defined including the numerical factors. There aren't any ambiguities in the values. There aren't any ambiguities about the value of the Boltzmann constant, as long as one keeps its definition, and so on and so on. You're just trying to cover a wrong numerical factor by layers of fog. $\endgroup$ – Luboš Motl Aug 19 '15 at 13:32

The reason for using $E_K=πk_BT$ is to make the "quantum concentration" exactly equal to $\lambda_{th}^{-3}$ (see Kittel and Kroemer, Thermal Physics p. 73). The quantum concentration is $\int \frac{d^3 {\bf k}}{(2\pi)^3} \, e^{-E_K({\bf k})/k_BT}$. In semiconductor theory we multiply this quantity by a factor of 2 from the spin degeneracy and apply it to electron and hole quasiparticles with effective masses determined by the material to obtain the "thermal effective density of states" $N_C$ and $N_V$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.