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Derivation of the MaxwellBoltzmann Velocity Distribution
We want to derive the MaxwellBoltzmann velocity distribution such that

(9.5) 
We make two assumptions. The first is that the individual
components are statistically independent, i.e. , and
are not correlated with each other. This means that
will only depend on the magnitude of the velocity space vector,
and thus
with
.
Furthermore this means we can write as

(9.6) 
with and being the distribution functions for
and , respectively.
The second assumption is that of isotropy; the distribution should
be spherically symmetric and, thus, it should not matter what direction
we're looking in. This implies that and all have the same
form. We then have

(9.7) 
The derivation now proceeds by taking the derivative of
equation (9.17) with respect to .
Though not immediately obvious the solution to equation (9.20)
is

(9.8) 
The constant has been written as for convenience, and is
implicitly understood to be positive. We now have to work out all
the constants and signs. The first thing to note is that we need to
choose the sign in the argument of the exponential as the
sign leads to unacceptable results at infinity. A heuristic
argument for getting is as follows; note first that
must have units of , e.g. one over velocity
squared, in order to make the argument of the exponential
dimensionless. We add to this the equipartition theorem

(9.9) 
From which we guess that

(9.10) 
The last item to work out is the value of the constant . We can
get that by putting everything into equation (9.15)
and solving. We get that

(9.11) 
Putting this all together gives us the result

(9.12) 
which is the MaxwellBoltzmann velocity distribution. It should be noted
that some of this is not very rigorous. Born (BAB: cite)
provides another derivation.
Next: Bibliography
Up: Miscellaneous Calculations
Previous: Evaluation of Gaussian Integrals
Contents
Ben Breech
20030114