Speed distribution

How many particles are going how fast?

If we knew the precise statistical distribution of the speeds, we could calculate detailed material properties, such as viscosities, mean free paths, etc.

It turns out that with only a handful of plausible assumptions, we can come up with an exact equation for the speed distribution (using classical, rather than quantum mechanical assumptions): The Maxwell-Boltzmann distribution.

Binary collisions

If the density of molecules is low enough then we only need to worry about collisions involving two molecules at a time.

If we fix the initial vector velocities $\myv{v_1}$ and $\myv{v_2}$, are the final velocities $\myv{v_1}'$ and $\myv{v_2}'$ determined or not?

The answer turns out to be kind of "yes"...

Central forces

Assume that the forces (responsible for the change of direction during the collision) between the two particles only ever act along the line that connects them.

$\Rightarrow$ All the velocities are in the same plane.

In order to specify the velocities of both particles, both before and after the collision we need 2 components for 4 vectors = 8 numbers.

Specifying the 2 initial velocity vectors requires 4 constraints, leaving another 4 numbers free.

Momentum conservation

Momentum is conserved in the collision. The constraint that the total initial momentum is equal to the total final momentum can be written as: $$\myv{v_1} + \myv{v_2} = \myv{v_1}'+\myv{v_2}'.$$

Since these vectors all lie in the same plane, this amounts to 2 constraints: one for each vector component in the plane of scattering. We still have 8-4-2=2 numbers free.

Energy conservation

Assume that the collisions are elastic. The total kinetic energy before and after the collision must be the same: $$v_1^2+v_2^2=v_1'^2+v_2'^2.$$

This is only 1 constraint. So, we have 8-4-2-1=1 degree of freedom left: If the final velocities are determined, this last degree of freedom corresponds to a freedom of choice about which particle gets which final velocity.

  • The final velocities *are* fully determined by the initial velocities,
  • However, which particle ends up with which final velocity is *not* determined.

From this point on, we should think of the 1 and 2 subscripts as just indicating the two initial (or two final) velocity vectors to distinguish them from each other: They're no longer going to be associated with one particle or the other.

Velocity distribution function

The velocity distribution function is such that, $F(\myv{v})\,d\myv{v}$, is the number of molecules with velocities in the range $\myv{v}$ to $\myv{v}+d\myv{v}$. So...

...is the number of collisions like the ones shown above in a certain time period. The number of such collisions has to be proportional to the number of molecules with the right initial conditions.

Time reversal symmetry

Because the final velocities are uniquely determined by the initial velocities, there must be just as many collisions with final velocities $\myv{v_1}'$, $\myv{v_2}'$ as have initial velocities $\myv{v_1}$, $\myv{v_2}$.

The laws of mechanics are symmetric with respect to time. A movie of a collision, run backwards, would not violate mechanics. As long as equilibrium has been reached--and therefore entropy is not changing with time--the distribution of velocities should be the same whether the system is run forward or backward in time.

If we run time backwards, the collisions we've been talking about look like this and now have initial velocities $- \myv{v_1}'$, and $- \myv{v_2}'$. Since the number of collisions per unit time has not changed: $$aF(\myv{v_1})F(\myv{v_2})=aF(-\myv{v_1}')F(-\myv{v_2}').$$

The $a$'s cancel.

Moreover, it seems reasonable to assume that all directions of $\myv{v}$ (with the same magnitude $v$) are equally likely. So, we'd expect that $F(\myv{v})=F(-\myv{v})$, and this means... $$F(\myv{v_1})F(\myv{v_2}) =F(\myv{v_1}')F(\myv{v_2}').$$

Taking natural logs of both sides... $$\ln F(\myv{v_1}) + \ln F(\myv{v_2})= \ln F(\myv{v_1}')+ \ln F(\myv{v_2}').$$

Conservation of energy required that: $$v_1^2 + v_2^2=v_1'^2+v_2'^2.$$

This suggests that the functional form of the velocity distribution function should fulfill the condition: $$\ln F(\myv{v}) \propto v^2.$$

With $\myv{v}\cdot\myv{v}=v^2$, this could be fulfilled by: $$F(\myv{v}) = Ae^{-\alpha \myv{v}\cdot\myv{v}}.$$

$\alpha$ *could* be either positive or negative. But we'd like, for a probability density, that $\int_0^\infty F(\myv{v})d\myv{v} = N$ (finite value), and if $(-\alpha)$ is a positive number, the integral is unbounded. So, we will choose $\alpha$ to be a positive quantity.

This is the Maxwell-Boltzmann distribution function

Speed distribution function

Assuming the velocities are uniformly distributed in direction, then the number of velocity vectors with speeds in the range $v$ to $v+dv$ is $N(v)\,dv$, is the number of velocity vectors with tips in a spherical shell of thickness $dv$ at a distance $v$ from the origin: $$N(v) = 4\pi v^2 F(v) dv = 4\pi v^2 Ae^{-\alpha \myv{v}\cdot\myv{v}}.$$

The total number of particles is: $$N=\int_0^\infty N(v)\,dv.$$

The total energy is: $$\frac{3}{2}NkT = \frac{1}{2}m\int_0^\infty v^2 N(v) dv.$$

The constant $\alpha$ can be found from the ratio of these two equations. Then using this in either one, $A$ can be found. [Screenshot of Mathematica solution.] The result is:

$$N(v)dv = 4\pi \(\frac{m}{2\pi k T} \)^{3/2}Nv^2e^{-mv^2/2kT} dv.$$

Using Mathematica, graph 3 plots of $N(v)$ in the same figure: using the values $N=1000$ and $2kT/m =$

  • 0.3
  • 1.0
  • 3.0

Label each plot so we can see what temperature it corresponds to.

As you raise the temperature, you ought to have a greater number of faster molecules, and the peak of the distribution (which ought to track with the average speed) should also be moving to higher $v$-values.

Here are my graphs of $N(v)$ for $N=1000$: (Red is the lowest temperature..)

Optional: You can check your calculations by integrating $\int_0^{v_{max}} N(v')dv'=g(v_{max})$ for one or more values of $2kT/m$ above. The value of $g$ is the number of particles (out of 1000) that have a speed less than or equal to $v_{max}$. If you graph $g(v_{max})$ vs $v_{max}$, the plot ought to start at 0, and then rise and saturate at $N=1000$ as $v_{max}$ gets larger and larger.


Root-mean-square speeds

We used $f(v)$, the fraction of all molecules with speeds between $v$ and $v+dv$, to calculate things like rms speeds. $$f(v) = N(v)/N.$$

So, $$\begin{align}\overline{v^2} & = \int v^2 f(v) dv = \frac{1}{N}\int v^2 N(v) dv =\\ & = 4\pi \(\frac{m}{2\pi k T} \)^{3/2} \int_0^\infty v^4 e^{-mv^2/2kT} dv .\end{align}$$

See Appendix D2 for the integral: $$\begin{align}\overline{v^2} & = 4\pi \(\frac{m}{2\pi k T} \)^{3/2} \[\frac{3}{8 (m/2kT)^2}\( \frac{\pi}{m/2kT}\)^{1/2} \]\\ & = \frac{3kT}{m} .\end{align}$$

The rms speed is $$v_{rms} = \sqrt{\overline{v^2}}=\sqrt{\frac{3kT}{m}} = 1.732\(\frac{kT}{m}\)^{1/2}.$$


Image credits

Mike Brice