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Derivation of the Maxwell-Boltzmann Velocity Distribution

We want to derive the Maxwell-Boltzmann velocity distribution such that

\begin{displaymath}
dN = \Phi(v_x, v_y, v_z)  dv_x dv_y dv_z.
\end{displaymath} (9.5)

We make two assumptions. The first is that the individual components are statistically independent, i.e. $v_x$, $v_y$ and $v_z$ are not correlated with each other. This means that $\Phi$ will only depend on the magnitude of the velocity space vector, and thus $\Phi = \Phi (v)$ with $v = \sqrt{v_x^2 + v_y^2 + v_z^2}$. Furthermore this means we can write $\Phi$ as

\begin{displaymath}
\Phi(v) = f (v_x) g(v_y) h(v_z)
\end{displaymath} (9.6)


with $f, g$ and $h$ being the distribution functions for $v_x, v_y$ and $v_z$, 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 $f, g$ and $h$ all have the same form. We then have

\begin{displaymath}
\Phi(v) = f(v_x) f(v_y) f(v_z)
\end{displaymath} (9.7)

The derivation now proceeds by taking the derivative of equation (9.17) with respect to $v_x$.

\begin{eqnarray}
\frac{d\Phi}{dv_x} \frac{v_x}{\sqrt{(v_x^2 + v_y^2 + v_z^2)}}
...
...\
\frac{1}{v}\frac{d\Phi}{\Phi} & = & \frac{1}{v_x}\frac{df}{f}
\end{eqnarray}



Though not immediately obvious the solution to equation (9.20) is
\begin{displaymath}
f(v_x) = A \exp(\pm \frac{\lambda}{2} v_x^2)       \
\Phi(v) = A^3 \exp(\pm\frac{\lambda}{2} v^2 / 2).
\end{displaymath} (9.8)


The constant has been written as $\lambda/2$ 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 $\lambda$ is as follows; note first that $\lambda$ must have units of $s^2/m^2$, e.g. one over velocity squared, in order to make the argument of the exponential dimensionless. We add to this the equipartition theorem
\begin{displaymath}
\frac{1}{2}m v_x^2 = \frac{1}{2}k T.
\end{displaymath} (9.9)


From which we guess that
\begin{displaymath}
\lambda = \frac{m}{k T}.
\end{displaymath} (9.10)

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

\begin{displaymath}
A^3 = N \left(\frac{m}{2\pi k T}\right)^{3/2}.
\end{displaymath} (9.11)

Putting this all together gives us the result

\begin{displaymath}
\Phi(v) = N\left(\frac{m}{2\pi k T}\right)^{3/2}\exp\left(-
\frac{m v^2}{2 k T}\right),
\end{displaymath} (9.12)


which is the Maxwell-Boltzmann velocity distribution. It should be noted that some of this is not very rigorous. Born (BAB: cite) provides another derivation.


next up previous contents
Next: Bibliography Up: Miscellaneous Calculations Previous: Evaluation of Gaussian Integrals   Contents
Ben Breech 2003-01-14