Distribution of Single Particle Energy and the Maxwell-Boltzmann Gas

In this note, we argue that it is energy and not momentum which is distributed in a Maxwell-Boltzmann equilibrium situation. In the simplest case, one has two particle elastic scattering such that: e1+e2 = e3 + e4 and P(e1)P(e2) = P(e3)P(e4). Taking ln of the second equation and equating to the first, multiplied by a constant -1/T, yields: P(e)= C exp(-e/T). Momentum does not enter into this derivation. One may later introduce momentum degeneracy linked with e through pp dp 4*3.14 (in 3D or dp in 1D) when calculating C Integral pp dp 4*3.14 exp(-e/T) = 1 and C Integral pp dp 4*3.14 exp(-e/T) e = eave. Thus, momentum is used to describe the degeneracy of e, but it is not used to calculate the Maxwell-Botlzman factor, we argue. This argument has implications because one may consider n particles (moving in 1-dimension) with total energy E. One may ask (as in (1)): What is the probability to have one particle with energy e? For this case, (1) writes: Sum over i=1 to n-1 p(i)p(i) = RR-ee where E=RR/2m and p(i) is the one-dimensional momentum of the ith particle. This defines an n-1 sphere which is taken to be proportional to the probability. The issue we have with the assumption is that it is an argument based on momentum, not energy distribution. In other words, it is a distribution based on components of a vector. The approach of (1) does apply to a distribution of vector components. For example, for the case n=2, it yields the probability distribution for a 2-vector (px,py) given in (3). We argue, however, that the Maxwell-Boltzmann equilibrium is associated with a distribution of e (single particle energy) and not momentum and so one should not use the n-1 sphere associated with p(i) (momentum) components in this case