- #1

- 27

- 3

1. If we wish to compute the speed distribution from the velocity distribution we work with, say, [tex]p(v) dv \propto v^2e^{-av^2} dv[/tex] where the ##v^2 dv## comes from considering the volume of velocity space available to the states with speeds in ##[v, v+dv]##. This makes sense, at least mathematically. However ..

2. This distribution seems to say that, if we consider the states close to ##v=0##, we will find no particles with speed 0 (and not many close by), as there is no volume space available to them there (and this appears to be independent of the underlying Boltzmann distribution itself, since ##v^2 f(v) \to 0## as ##v \to 0## for any physically realistic ##f(v)##)

3. But this seems to clash with what the Boltzmann distribution predicts: the probability of finding a particle in a microstate of 0 energy (hence 0 velocity) is greater than finding it with any other energy since the probability of it being in a microstate ##r##, energy ##E_r## is ##p(r) \propto e^{E_r/kT}##.

So I'm confused. The Boltzmann distribution seems to predict that we will often find gas particles with 0 velocity. But the Maxwell-Boltzmann speed distribution seems to say that we will find *no* particles with 0 speed (##\Leftrightarrow## 0 velocity), since there is no velocity space available to them.

Can anyone clear up my confusion?