# Maxwellian velocity distribution

## Thermal Velocities

### Typical Speed: Back of the Envelope

We can estimate the typical speed of a particle of mass $m$ in thermal equilibrium with its surroundings. In thermal equilibrium, a particle is excited thermally. We can write its kinetic energy as:

${\frac {1}{2}}mv^{2}={\frac {3}{2}}kT\,\!$ Thus, a typical particle has speeds of

$v\approx {\sqrt {\frac {3kT}{m}}}\,\!$ ### Velocity Distribution

Now we will derive $f({\vec {v}})$ , the distribution of velocities of a particle in thermal equilibrium. Since we are in thermal equilibrium, we can use Maxwell-Boltzmann statistics. Recall that for Maxwell-Boltzmann statistics, the probability of an energy state $\epsilon _{i}$ at temperature $T$ is given by:

$P(\epsilon _{i})={\frac {\exp(-\epsilon _{i}/kT)}{\sum _{j}\exp(\epsilon _{j}/kT)}}\,\!$ The denominator is the normalization factor which is calculated by summing over all possible energy states. Now, consider a velocity ${\vec {v}}=(v_{x},v_{y},v_{z})$ . to calculate the probability of a particle of having that particular velocity.

$f({\vec {v}})=C\exp \left(-{\frac {m|{\vec {v}}|^{2}}{2kT}}\right)=C\exp \left(-{\frac {m}{2kT}}(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})\right)\,\!$ To get $C$ , the normalizing factor, we need to sum over all possible velocities.

$1/C=\sum _{i}\exp \left(-{\frac {m{\vec {v_{i}}}^{2}}{2kT}}\right)\,\!$ Because velocities are continuous, the sum then becomes an integral. And since velocity is in three dimensions, we need to integrate over $dv_{x}dv_{y}dv_{z}$ .

$1/C=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\exp \left(-{\frac {m}{2kT}}(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})\right)dv_{x}dv_{y}dv_{z}\,\!$ We note that we can break this up into three Gaussian integrals:

$1/C=\int _{-\infty }^{\infty }\exp \left(-{\frac {mv_{x}^{2}}{2kT}}\right)dv_{x}\int _{-\infty }^{\infty }\exp \left(-{\frac {mv_{y}^{2}}{2kT}}\right)dv_{y}\int _{-\infty }^{\infty }\exp \left(-{\frac {mv_{z}^{2}}{2kT}}\right)dv_{z}\,\!$ We will define ${\tilde {v}}^{2}=2kT/m$ and integrate a Gaussian.

$1/C=({\tilde {v}}{\sqrt {\pi }})^{3}=\left({\frac {2\pi kT}{m}}\right)^{3/2}\,\!$ Plugging it all in, the velocity distribution is simply a Gaussian centered around zero with a typical fluctuation in the velocity of ${\sqrt {2kT/m}}$ :

$f({\vec {v}})=\left({\frac {m}{2\pi kT}}\right)^{3/2}\exp \left(-{\frac {mv^{2}}{2kT}}\right)\,\!$ where $v=|{\vec {v}}|={\sqrt {v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}}$ . Note that $f({\vec {v}})$ has units of $[{\text{cm/s}}]^{-3}$ since to obtain a probability you need to integrate over all three velocity directions.

### Maxwellian Speed Distributions

The Maxwellian Speed Distribution is a bit more interesting than the velocity distribution (and used more often). What we want to find is $f(v)$ which is different than $f({\vec {v}})$ . This is because there are more ways to combine $v_{x}$ , $v_{y}$ , $v_{z}$ to make larger $v$ ’s so there are more available states at higher speeds. For a given speed $v$ , any combination of velocity components that satisfy $v_{x}^{2}+v_{y}^{2}+v_{z}^{2}=v^{2}$ will work. This is analogous to the surface area of a sphere and likewise we find that there are $4\pi v^{2}$ states at any given v. Thus, we can modify the Maxwell velocity distribution with the number of states at a given speed:

$f(v)=4\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}v^{2}\exp \left(-{\frac {mv^{2}}{2kT}}\right)\,\!$ This is the Maxwell speed distribution and note that it has units of $[{\text{cm/s}}]^{-1}$ .

### Typical Speeds Revisited

There are actually three ways to quantify typical speeds of a distribution of particles in thermal equilibrium and our back of the envelope calculation only gives us one of them. The first typical speed is the easiest to calculate: the most probable speed. The most probable speed occurs when $f(v)$ is maximum. To calculate this we set the derivative to zero:

${\frac {df(v)}{dv}}=4\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}\left(2v\exp \left(-{\frac {mv^{2}}{2kT}}\right)-v^{2}{\frac {mv}{kT}}\exp \left(-{\frac {mv^{2}}{2kT}}\right)\right)=0\,\!$ ${\tilde {v}}={\sqrt {2kT/m}}\,\!$ As the tilde suggests, ${\tilde {v}}$ is what we got earlier while solving for the Maxwell velocity distribution. ${\tilde {v}}$ is both the most probable speed and the typical spread in velocities. The next speed we will find is the mean speed, ${\bar {v}}$ . To find this, we find the expectation value of $v$ :

${\bar {v}}=\int _{0}^{\infty }f(v)vdv=\int _{0}^{\infty }4\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}v^{3}\exp \left(-{\frac {mv^{2}}{2kT}}\right)dv\,\!$ Using the result that:

$\int _{0}^{\infty }x^{3}exp(-x^{2}/a^{2})=a^{4}/2\,\!$ we get:

${\bar {v}}=4\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}{\frac {1}{2}}\left({\frac {2kT}{m}}\right)^{2}\,\!$ ${\bar {v}}={\sqrt {\frac {8kT}{\pi m}}}\,\!$ It turns out this is slightly greater than the most probably speed (${\sqrt {8/\pi }}>{\sqrt {2}}$ ). The last speed we will find is the root mean square velocity ($v_{rms}$ ). We define $v_{rms}^{2}\equiv {\overline {v^{2}}}$ . To find ${\overline {v^{2}}}$ :

${\overline {v^{2}}}=\int _{0}^{\infty }f(v)v^{2}dv=\int _{0}^{\infty }4\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}v^{4}\exp \left(-{\frac {mv^{2}}{2kT}}\right)dv\,\!$ Using the result that:

$\int _{0}^{\infty }x^{4}exp(-x^{2}/a^{2})={\frac {3{\sqrt {\pi }}a^{5}}{8}}\,\!$ we get:

${\overline {v^{2}}}=4\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}{\frac {3{\sqrt {\pi }}}{8}}\left({\frac {2kT}{m}}\right)^{5/2}\,\!$ ${\overline {v^{2}}}={\frac {3kT}{m}}\,\!$ Thus, we find that:

$v_{rms}={\sqrt {\frac {3kT}{m}}}\,\!$ We see that the rms speed is the speed we derive from the back of the envelope calculations. We also see that this typical speed is greater than both of the other two speeds. Here is a plot showing the Maxwell speed distribution and the three typical velocities. We can see from this plot that the Maxwell speed distribution is skewed to the right. Plot of the Maxwell Speed Distribution and typical speeds