Misinterpretation of the Maxwell velocity distribution?

[vanhees71]

Again, again and again, I've told you that you have to take into account the JACOBIAN when discussing distributions. Since $E=m v^2/2$ the distribution function for $E$ and $v$ differ by a non-constant Jacobian of the transformation, and that's why the maximum of the distribution function for speed is different from the maximum of the energy-distribution function.

The same holds for averages. Despite the fact that $E=m v^2/2$ the average is $\langle E \rangle=m/2 \langle v^2 \rangle \neq m/2 \langle v \rangle^2$.

 

[Kairos]

It is normal, thanks. Sorry

 

[vanhees71]

Nothing to be sorry for! It's a very important general issue you can easily get confused about, and it's good to really fully understand it. That's the more important, because probabilities are not always very intuitive but need careful mathematical analysis.

 

[Philip Koeck]

I suddenly have a problem with the famous Maxwell velocity distribution. The maximum of this bell-shaped curve is commonly interpreted as the most probable velocity for a particle (see for instance https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution: "The most probable speed, vp, is the speed most likely to be possessed by any molecule..")
But according to Boltzmann's distribution the most probable energy for a particle is zero (decreasing exponential: P(E=Ei) = Exp(-Ei/kT)/Σ), so we should expect that the most probable velocity for a randomly picked particle is zero as well? but the probability of velocity zero is zero in the Maxwell distribution!
thank you in advance if you can enlighten me

Maybe I can help a bit: If you think of discrete energy levels with a certain number of states per level then the Boltzmann factor is proportional to the probability of finding a particle in a given state on a certain level. The MB-distribution for energies, however, gives the probability of finding a particle in any state on a given level. The difference arises because the number of states per level varies. I think this agrees with many of the other replies you got. Obviously for an ideal gas the energy levels have to be replaced by a continuum of energies, but that can be handled easily. The states for the ideal gas are simply all the ways of realizing a certain kinetic energy (coordinates in phase space). The lower the energy the fewer states you have in an ideal gas. That's what's meant by the density of states. Check the derivations, then it should all become clear.