4.10 The Maxwell-Boltzmann Distribution

Take-home message: The density of states can be used to derive the Maxwell-Boltzmann distribution of molecular speeds in a gas.

The speed $v$ of a particle is related to the wavenumber $k$ by $mv=\hbar k$. We already know the probability of a particle having $k$ in the range $k\to k+{\rm d}k$, and so we can immediately write down the corresponding probability of the speed being in the range $v\to {\rm d}v$:

$$\begin{eqnarray*} P(k\to k+{\rm d}k)\!\!\!&=&\!\!\!{D(k) e^{-\varepsilon(k)\be... ...style B}T}\right)^{3/2} v^2 e^{-mv^2/2k_{\scriptscriptstyle B}T} \end{eqnarray*}$$

This is called the Maxwell-Boltzmann distribution, and it is plotted below.

We can find the most probable speed (from ${\rm d}P(v)/ {\rm d}v=0$), as well as the mean speed and the rms speed:

$$\begin{eqnarray*} v_p\!\!\!&=&\!\!\!\sqrt{{2k_{\scriptscriptstyle B}T\over m}}\a... ...T\over m}}\approx 1.73\sqrt{{k_{\scriptscriptstyle B}T\over m}} \end{eqnarray*}$$

These are marked on the graph above.

Note that $\hbar$ has disappeared from $P(v)$, which can be derived from the Boltzmann distribution in a purely classical theory provided the normalisation is obtained from requiring the integral of $P(v)$ to be one.

References

• Mandl 7.7
• Bowley and Sánchez 7.4
• Kittel and Kroemer 14