Distribution of Single Particle Energy and the Maxwell-Boltzmann Gas Part II

In Part I, we tried to argue that the Maxwell-Boltzmann distribution is associated with an energy, and not momentum, distribution. We suggested that the phase space associated with momentum is used to calculate allowable momentum states, but not yield the probability distribution.

    In (1), a gas of n  1-dimensional momentum particles is considered by considering the equation:  Sum over i=1,n  p(i)p(i) = RR where E=RR/2m and p(i) is the single component of momentum of the ith particle. If the nth particle has energy e, then Sum over i=1, n-1 p(i)p(i) = RR-p(n)p(n) where p(n)p(n)=e2m. (1) suggests that the probability P(e) is proportional to the surface area of an n-1 sphere, which in turn is proportional to sqrt(E-e)) power(n-1)/2.

   In (2), we suggested that one use nested energy to find the probability. For n=4 particles, the fourth particle has energy e, and the first particle has whatever energy is left over after particles 2 and 3 have been assigned energy. Thus, probability is proportional to:  Integral (0, E-e)  de2  Integral  (0, E-e-e2) de3. We found that in general  P(e) is proportional to (E-e) power (n-2). 

   For high n,  the approach of (1) yields P(e) proportional to (E-e) power(n/2) while the approach of (2) yields P(e) proportional to (E-e) power (n).

   Here we consider the n=2 case to show why there is a difference and argue in favour of the approach of (2). The key difference, we argue, is that for the n=2 case, (1) is equivalent to dividing a line in equal dp pieces from 0 to sqrt(E) C, where C is a constant. Given a p, one immediately knows the value of p2 (the second particle) from e1+e2= E. The probability is then proportional to the length of the line, namely sqrt(2mE).

  The energy approach, on the other hand, does not consider equi-weights, dp, as yielding probability. If one considers the equation  p1p1+ p2p2 = E2m and takes the differential,:  p1 dp1 + p2 dp2 =0. Noting that a given p1 yields the value of p2, the probability weights are p1dp1 and not dp1 (with p1 ranging from 0 to C sqrt(E). In other words, counting p vectors points in a one dimensional line with constant dp does not yield the corresponding probability given the constraint p1p1 + p2p2= 2mE.