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The Maxwell-Boltzmann velocity distribution in 3D space is $$ f(v)dv = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 \exp\left(-\frac{m v^2}{2k_B T}\right)dv$$ It gives the probability for a single particle to have a speed in the intervall $[v,v+dv]$. But this probability is not zero for speeds $v > c $ in conflict with special relativity.

Is there a generalization of the Maxwell-Boltzmann velocity distribution which is valid also in the relativistic regime so that $f(v) = 0$ for $v>c$ ? And how can it be derived? Or can a single particle distribution simply not exist for relativistic speeds, because for high energies, we always have pair-production meaning the particle number is not conserved and a single particle distribution can not be defined in a consistent way?

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FYI, the relativistic generalization is called the Maxwell-Juttner distribution. – user1631 Jul 18 '11 at 17:18

The point of a Boltzmann distribution is that it maximizes entropy given a fixed energy. The concept applies to systems with other degrees of freedom besides translational kinetic energy. The general distribution, from Wikipedia is

enter image description here

Thus, the simple adjustment to the Maxwell-Boltzmann distribution you cited is to replace the Newtonian kinetic energy $\frac{mv^2}{2}$ with the relativistic kinetic energy $(\gamma - 1)mc^2$ everywhere it appears in the distribution.

Pair creation is a separate issue that I'll leave to someone else.

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$\gamma-1$ has an imaginary component for $v>c$, which seems to preclude a probability interpretation. Do you have a reference where the consequences of the model you suggest above are developed? Does one have to insert a Heaviside function restriction to $v<c$? That would seem moderately ad-hoc, although perhaps not irredeemably. Pair production is a quantum field theoretic concept that, as you somewhat imply, puts us into a wholly different and barely understood arena. – Peter Morgan Jul 17 '11 at 3:01
@Peter I would think of it as saying that the energies can go from 0 to infinity, and then you get a probability distribution for energies that doesn't seem artificial. From there, you can transform it into a distribution for velocities if you so desire. Doing the transformation you'll naturally stop at v = c. I suppose you are right that you have to insert a Heaviside function. No, I don't have a reference; it was just an off-the-cuff response. – Mark Eichenlaub Jul 17 '11 at 3:09
I believe this ansatz has a problem even for speeds below c. If $E_{kin} = (\gamma-1)m c^2 > 2 m c^2 $ that is for $v > \frac{2}{3} \sqrt{2} c = 0.94c$ the energy is enough for pair production. I guess at this point a distribution based on a constant number of particles cannot be correct, am I right? – asmaier Jul 17 '11 at 11:39
Well, nice try but this is indeed very naive attempt. My feeling is that there is no use in trying to create classical (i.e. non-quantum) relativistic statistical mechanics similarly to trying to create relativistic quantum mechanics. In both cases one needs to introduce possibility of creation/annihilation for the theory to be consistent (or equivalently, pass to the language of fields). And I am not even talking about elementary stuff @asmaier mentions that for relativistic particles concept of kinetic energy is unnatural at best... – Marek Jul 17 '11 at 20:28
@Marek Well, you would know better than I. I'd be interested if you have the time to write up an answer. – Mark Eichenlaub Jul 17 '11 at 20:37

The usual, or text book generalisation, which was first derived in 1911 is not covariant. Presumably a fully covariant distribution would cover all these details assuming that it exists in the first place. The most recent attempt I am aware of is by Ewald Lehmann, "Covariant equilibrium statistical mechanics", Journa of Mathematical Physics 47, 023303,2006.

David Sher

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Per Mark's request, I'll provide an answer.

First, I think it's not possible to obtain such a formula. One can of course naively try to extrapolate various classical formulae but all of these attempts are bound to fail. Here's why: even when constructing relativistic quantum mechanics one finds that the theory is not consistent. For example, in non-relativistic quantum mechanics one has a position operator that can be used to get information on precise position of the particle (up to the uncertainty given by Heisenberg's uncertainty principle). But when one includes relativity into the picture the theory stops being consistent. This reflects the fact that to localize the particle with great precision one has to make experiments with increasingly high energy and at certain point the energy is enough for the appearance of new particles. In fact, creation and annihilation of particles is inevitable in relativistic regime (and in some systems it's not even clear what particles should be and one has to talk about fields instead). The situation is even more pronounced in statistical physics where there are huge number of particles present.

More importantly, there's no need to get such a formula. Consider systems for which it would be useful. Such systems would have to be extremely non-classical (like neutron stars, black holes interior, quark-gluon plasma, etc.) and the concept of velocity would have no meaning as there's no way to observe individual particles of these systems; which is in contrast to the classical case where you can test Maxwell distribution by letting the particles out of the box one by one and seeing how fast they are (the actual experiment is lot more sophisticated of course, but that's unimportant here).

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Do you think the Maxwell-Juttner formula is correct? – John McVirgo Aug 7 '11 at 18:14

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