# Rigorous derivation of the mean free path in a gas

Can anyone supply me with a derivation of the mean free path, of particles in a Maxwell Boltzmann Gas?

Cited in various literature is the formula,

\begin{align} \begin{split} \ell&=\frac{1}{\sqrt{2}n\sigma}, \end{split} \end{align}

which describes the mean free path of an atom/molecule in a Maxwell Boltzmann gas. Within it, $$n$$ is the gas density (assumed to be homogeneous) and $$\sigma$$ is the cross section of colliding particles.

Derivations of the mean free path tend to include a comment about the relative velocity of particles. However, most literature will not treat this as a distribution, and will instead follow a two part approximate method.

First, magnitude of the relative velocity is expressed using equation (1). Within it, $$\theta$$ is the angle between relative velocities $$2$$ and $$1$$.

\begin{align} \tag{1} |\mathbf{v}_{r}|&=\sqrt{\mathbf{v}_{r}\cdot\mathbf{v}_{r}}\\ &=\sqrt{(\mathbf{v_2}-\mathbf{v_1})\cdot(\mathbf{v_2}-\mathbf{v_1})}\\ &=\sqrt{\mathbf{v_1}^2+\mathbf{v_2}^2-2\mathbf{v_1}\cdot\mathbf{v_2}}\\ v_{r}&=\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}\\ \end{align}

Once equation (1) has been found, most authors take an average of both sides. Instead of evaluating this by treating each velocity as a distribution however, they instead take the average under the square root and onto each summed term (equation 2)

\begin{align} \tag{2} \langle v_{r}\rangle&=\big\langle\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}\big\rangle\\ &=\sqrt{\big\langle v_1^2+v_2^2-2v_1v_2cos(\theta)\big\rangle}\\ &=\sqrt{\big\langle v_1^2\big\rangle+\big\langle v_2^2\big\rangle-\big\langle2v_1v_2cos(\theta)\big\rangle}\\ &=\sqrt{\langle v_1\rangle^2+\langle v_2\rangle^2-0}\\ &=\sqrt{\langle v_1\rangle^2+\langle v_2\rangle^2}\\ &=\sqrt{2\langle v_1\rangle^2}\\ &=\sqrt{2}~\langle v_1\rangle\\ \end{align}

A large amount of (2) is wrong. Exceptions are the forth and fifth steps, where an average of the dot product between two arbitrary velocities is zero, justified by the velocity in a Maxwell Boltzmann distribution having no preferred direction. The remaining step involves the average velocity of any particle in a Boltzmann gas being the same ($$\langle v_1\rangle=\langle v_2\rangle$$).

One might think to replace the average of the square of velocity with an rms velocity, however doing so yields the rms relative velocity distribution, which is out by $$8\%$$ from the mean relative velocity.

I would like to see a derivation of the same result without these approximate methods. Failing that, reference to a textbook that provides them would be helpful as well.

References (all use the approximation):

1. Reif, F. (1965) Fundamentals of Statistical and Thermal Physics. McGraw Hill, New York, 273-278.

2. formula for mean free path in two dimensions

3. Why is the mean free path divided by $$\sqrt{2}$$?

4. RMS Free Path vs Mean Free Path

• Commented Dec 17, 2018 at 0:24
• @Thorondor I followed that link, and have been reading through it for the past hour or so. While the original question appears to be on the same topic, there are no answers that provide the derivation. One answer finds the rms relative speed, which is much simpler than the mean relative speed, while another attempts the mean relative speed, but does not complete the derivation (calling it ugly.) Commented Dec 17, 2018 at 4:21

The following is a self answer concerning the rigorous derivation of the relative velocity in a Maxwellian gas. It is here for comparison against other answers.

Assumptions are

• Two particle collisions are the most probable ones, and collisions involving more than this, contribute to the mean relative velocity by a negligible amount.
• The mean relative velocity is strictly positive.

I think that this derivation differs from the one by @Thorondor because within it, $$\mathbf{v_1}^2+\mathbf{v_2}^2\neq\mathbf{v_r}^2$$. Which is to say that the dot product of $$\mathbf{v_1}$$, and $$\mathbf{v_2}$$ is not assumed to be zero. This forces us to integrate over all possible angles between the velocities, which may account for the ~0.07 difference. It is also why the derivation is significantly longer.

The mean free path of an atom/molecule in a Maxwellian gas, depends upon the average relative velocity of each particle to one another. In order to obtain it, we first find the magnitude of velocity $$\mathbf{v}_1$$ relative to all other particles moving with $$\mathbf{v}_2$$. This expression is then averaged for all values of $$\mathbf{v}_2$$, from zero to infinity, producing a mean relative speed given $$\mathbf{v}_1$$ exists.

Following modus ponens, the mean is multiplied by the probability that an atom/molecule has velocity $$\mathbf{v}_1$$, and then averaged for all $$\mathbf{v}_1$$ from zero to infinity.

Consider two velocities $$\mathbf{v}_1$$ and $$\mathbf{v}_2$$ inclined at an angle $$\theta$$, there relative velocity will be described by

\begin{align} (v_r)_{12}&=\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}. \end{align}

All directions are equally probable for $$\mathbf{v}_2$$. To calculate the average of $$(v_r)_{12}$$, we multiply it by the probability that it lies within some solid angle $$d\Omega$$.

\begin{align} \langle(v_r)_{12}\rangle&=\int_{\Omega}\frac{d\Omega}{4\pi}(v_r)_{12}\\ &=\int_{\Omega}\frac{2\pi sin(\theta)d\theta}{4\pi}(v_r)_{12}\\ &=\frac{1}{2}\int_{0}^{\pi}sin(\theta)d\theta(v_1^2+v_2^2-2v_1v_2cos(\theta))^\frac{1}{2} \end{align}

Consider a change of basis, where $$cos(\theta)=x$$, so $$sin(\theta)d\theta=-dx$$. The limits of integration will change to $$cos(0)=1$$ and $$cos(\pi)=-1$$.

\begin{align} \therefore~\langle(v_r)_{12}\rangle&=\frac{1}{2}\int_{1}^{-1}dx(v_1^2+v_2^2-2v_1v_2x)^\frac{1}{2}\\ &=\frac{1}{3}\frac{(v_1^2+v_2^2-2v_1v_2x)^\frac{3}{2}}{2v_1v_2}\Biggr|_{1}^{-1}\\ &=\frac{1}{6v_1v_2}\left((v_1^2+v_2^2+2v_1v_2)^\frac{3}{2}-(v_1^2+v_2^2-2v_1v_2)^\frac{3}{2}\right)\\ &=\frac{1}{6v_1v_2}\left((v_1+v_2)^\frac{3}{2}(v_1+v_2)^\frac{3}{2}+(v_1-v_2)^\frac{3}{2}(v_1-v_2)^\frac{3}{2}\right)\\ &=\frac{1}{6v_1v_2}\left((v_1+v_2)^3+|v_1-v_2|^3\right)\\ \end{align}

We take the magnitude of $$v_1-v_2$$ because $$\langle(v_r)_{12}\rangle$$ should always be positive, and therefore $$\left((v_1-v_2)^2\right)^\frac{3}{2}$$ must also be. This splits the average into two parts.

\begin{align} \langle(v_r)_{12}\rangle&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3-(v_1-v_2)^3\right),~when~v_1\geq v_2,\\ &=\frac{1}{6v_1v_2}\left[2v_2^3+6v_1^2v_2\right]\\ &=\frac{3v_1^2+v_2^2}{3v_1}.\\ \langle(v_r)_{12}\rangle&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3+(v_1-v_2)^3\right),~when~v_2>v_1,\\ &=\frac{3v_2^2+v_1^2}{3v_2}. \end{align}

The average relative velocity of an atom/molecule possessing magnitude and direction $$\mathbf{v_1}$$, with respect to all other particles moving with $$\mathbf{v_2}$$, lying within speeds of zero to infinity is then

\begin{align} \langle(v_r)_1\rangle&=\int_{0}^{\infty}P(v_2)\langle(v_r)_{12}\rangle dv_2,\\ P(v_2)dv_2&=4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}e^{-\frac{mv_2^2}{2k_BT}}v_2^2dv_2. \end{align}

Note that $$\langle(v_r)_1\rangle$$ has two distinct forms depending on the difference between speeds. Because of this we break the integral into two parts.

\begin{align} \langle(v_r)_1\rangle=4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}&\left[\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\ &+\left.\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right] \end{align}

To arrive at the relative velocity of any atom with another, we form a product of the above, and the probability a particle has velocity $$v_1+dv_1$$.

\begin{align} \langle v_r \rangle&=\int_{0}^{\infty}P(v_1)\langle(v_r)_1\rangle dv_1 \end{align}

\begin{align} \langle v_r \rangle=\left(4\pi\left[\frac{m}{2\pi k_BT}\right]^\frac{3}{2}\right)^2\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2&\left[\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\ &+\left.\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right] \end{align}

which may be re-written as

\begin{align} \langle v_r\rangle=\left(\frac{2m}{k_BT}\right)^3&\left[\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\ &+\left.\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right], \end{align}

or

\begin{align} \langle v_r\rangle&=\left(\frac{2m}{k_BT}\right)^3\left[I_1+I_2\right], \end{align}

where $$I_1$$ stands for the first integral in the square brackets of the above, and $$I_2$$ the second one.

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\\ I_2&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2 \end{align}

We wish to show that these two definite integrals are equivalent. To do so, a substitution, then change in the order of integration following Fubini's theorem is employed.

Consider the piecewise function,

\begin{align} \mathbb{I}_{\{v_2\leq v_1\}}&=\begin{cases} 1&v_1\leq v_2\\ 0&v_1>v_2\\ \end{cases} \end{align}

which is zero outside the limit of the second integral of $$I_1$$. Using it, we can rewrite equation $$I_1$$ as

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{\infty}dv_2\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2.\\ \end{align}

We can combine these two integrals with a substitution $$a=v_1$$, allowing us to write the following

\begin{align} I_1&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{ma^2}{2k_BT}}a^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dv_2da.\\ \end{align}

Since the integrals limits are constant, we can apply Fubini's theorem exchange there order.

\begin{align} I_1&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{ma^2}{2k_BT}}a^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dadv_2\\ &=\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{0}^{\infty}dv_1\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2\\ &=\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{v_2}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2\\ \end{align}

Then, because the integral is definite, we can interchange $$v_1$$ with $$v_2$$, and obtain an equivalent volume.

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\\ &=I_2 \end{align}

Hence, the average relative velocity can be rewritten as

\begin{align} \langle v_r\rangle&=2\left(\frac{2m}{k_BT}\right)^3I_2. \end{align}

To evaluate $$I_2$$, first consider the integral

\begin{align} I_0&=\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2. \end{align}

We can break this into parts,

\begin{align} I_0&=\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^3+\frac{v_1^2}{3}\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2, \end{align}

then consider the substitution $$\frac{m}{2k_BT}v_2^2=x$$, so $$dv_2=\frac{k_BT}{mv_2}$$, and the lower limit of integration changes to $$\frac{mv_1^2}{2k_BT}$$.

\begin{align} I_0&=\int_{\frac{mv_1^2}{2k_BT}}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^3+\frac{v_1^2}{3}\int_{\frac{mv_1^2}{2k_BT}}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2\\ &=2\left(\frac{k_BT}{m}\right)^2\int_{\frac{mv_1^2}{2k_BT}}^{\infty}e^{-x}xdx+\frac{v_1^2}{3}\frac{k_BT}{m}\int_{\frac{mv_1^2}{2k_BT}}^{\infty}e^{-x}dx \end{align}

Since

\begin{align} \int_{a}^{b}xe^{-x}dx&=-(x+1)e^{-x}\Biggr|_{a}^{b}, \end{align}

$$I_0$$ simplifies to

\begin{align} I_0&=2\left(\frac{k_BT}{m}\right)^2\left[\frac{m}{2k_BT}v_1^2+1\right]e^{\frac{-mv_1^2}{2k_BT}}\\ &=e^{-\frac{mv_1^2}{2k_BT}}\left[\frac{4}{3}\left(\frac{k_BT}{m}\right)v_1^2+2\left(\frac{k_BT}{m}\right)^2\right] \end{align}

Substituting this into $$I_2$$ yields

\begin{align} I_2=&\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\left[\frac{4}{3}\left(\frac{k_BT}{m}\right)v_1^2+2\left(\frac{k_BT}{m}\right)^2\right] \\=&2\left(\frac{k_BT}{m}\right)^2\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\\ &+\frac{4}{3}\frac{k_BT}{m}\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^4\\ \end{align}

$$I_2$$ is actually a standard integral of the form,

\begin{align} \int_{0}^{\infty}e^{-ax^2}x^n&=\frac{1}{2a^{\frac{n+1}{2}}}\Gamma\left(\frac{n+1}{2}\right), \end{align}

where $$\Gamma$$ is the gamma function, a generalization of the factorial. $$I_2$$ now simplifies to

\begin{align} I_2&=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\left[\Gamma\left(\frac{3}{2}\right)+\frac{2}{3}\Gamma\left(\frac{5}{2}\right)\right]\\ &=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\left[\frac{\sqrt{\pi}}{2}+\frac{2}{3}\frac{3\sqrt{\pi}}{4}\right]\\ &=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\sqrt{\pi} \end{align}

Hence, the average relative speed of any Maxwellian gas particle with respect to any other is

\begin{align} \langle v_r\rangle&=2\left[4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}\right]^2\left(\frac{k_BT}{m}\right)^\frac{7}{2}\sqrt{\pi},\\ &=\frac{2\sqrt{\pi}\cdot16\pi^2}{8\pi^3}\left(\frac{k_BT}{m}\right)^{-3}\left(\frac{k_BT}{m}\right)^\frac{7}{2},\\ &=\frac{4}{\sqrt{\pi}}\sqrt{\frac{k_BT}{m}},\\ &=\sqrt{\frac{16}{\pi}\frac{k_BT}{m}}. \end{align}

And the ratio of mean velocity $$\langle v_r\rangle$$ to mean relative velocity $$\langle v\rangle$$ is

\begin{align} \frac{\langle v\rangle}{\langle v_r\rangle}&=\frac{\sqrt{\frac{8}{\pi}\frac{k_BT}{m}}}{\sqrt{\frac{16}{\pi}\frac{k_BT}{m}}}\\ &=\frac{1}{\sqrt{2}} \end{align}

And thus the mean free path $$\ell$$ is

\begin{align} \ell=\frac{1}{n\sigma\sqrt{2}} \end{align}

• Thanks for taking the time to write this up! I believe this answer is correct. Should I delete mine? As you've shown, my answer was wrong, and I don't want to mislead future readers. Commented Dec 20, 2018 at 7:57
• @Thorondor It is a perfectly valid derivation, under the assumption that all velocities are perpendicular. Just provide an explanation at the start, that it only holds in this case. Commented Dec 21, 2018 at 2:53
• The fact that other directions of velocity, contribute so little to the final result, is actually quite interesting. I think it deserves a spot on the page for this reason. Commented Dec 21, 2018 at 2:53
• a similar proof can be found in the appendix of' Thermal physics by garg, bansal and ghosh'. Commented Sep 11, 2020 at 12:52
• @YasirSadiq That is where I got the majority of the proof from. I added only a few extra steps, it is a great textbook. Commented Sep 11, 2020 at 13:04

Another way:

$$\langle v_r \rangle = \mathcal{N}^2 \iint \vert \vec{v}_1 - \vec{v}_2\vert \ e^{-\frac{m(\vec{v}_1^2+\vec{v}_2^2)}{2k_B T}}\ \mathrm{d}\vec{v}_1 \mathrm{d}\vec{v}_2,$$

wherein the normalized Boltzmann distribution function is $$\mathcal{N} e^{-\frac{m \vec{v}^2}{2k_B T}}$$, $$\ \mathcal{N}$$ being the numerical factor that normalizes the Gaussian distribution.

Rotating velocity coordinates by $$45^{\circ}$$, define new coordinates $$\vec{U}= \frac{\vec{v}_1+\vec{v}_2}{\sqrt{2}},$$ $$\vec{V}= \frac{-\vec{v}_1+\vec{v}_2}{\sqrt{2}}.$$

Since the above transformation is a rotation, the Jacobian is 1.

Rewriting $$\langle v_r \rangle$$ in terms of new coordinates,

\begin{align} \langle v_r \rangle &= \mathcal{N}^2 \iint \vert \sqrt{2} \vec{V}\vert e^{-\frac{m(\vec{V}^2+\vec{U}^2)}{2k_B T}}\ \mathrm{d}\vec{V} \mathrm{d}\vec{U} \\ &= \sqrt{2} \ \underbrace{\int \vert \vec{V}\vert \ \mathcal{N} e^{-\frac{m\vec{V}^2}{2k_B T}}\ \mathrm{d}\vec{V}}_{\text{equals } \langle v \rangle} \ \ \underbrace{\int \mathcal{N} e^{-\frac{m\vec{U}^2}{2k_B T}} \ \mathrm{d}\vec{U}}_{\text{equals }1} \\ & = \sqrt{2} \ \langle v \rangle. \end{align}

The relation, $$\langle v_r \rangle = \sqrt{2} \ \langle v \rangle$$, seems to hold independent of the spatial dimensionality.

• Not sure what you're trying to convey here. $\vec{V}$ is a dummy variable - doesn't matter what you call it. You can also call it $\vec{U}$ or $\vec{z}$ or anything for that matter. Then the formula that you quoted is exactly the formula for relative velocity, no matter what dummy variable you use for integration over the Maxwellian distribution. Commented Apr 25, 2020 at 8:47
• You have defined $\vec{V}$ as $\frac{-\vec{v}_1+\vec{v}_2}{\sqrt{2}}$. Since it has a definition which is dependent on existing variables, it is not a dummy variable. In fact, $\vec{V}=\frac{1}{\sqrt{2}}v_r$. Then, since $\int_{-\infty}^{\infty}f(x)p(x)dx=\langle f(x) \rangle$, where $p(x)$ is a PDF, the part of your 2nd last equation which you state is equal to $\langle v\rangle$, is in fact equal to $\langle |V|\rangle$ instead. Commented Apr 25, 2020 at 10:07
• Nope, any variable that you're integrating over is a dummy variable, irrespective of how you defined it in the first place. If you like, you can do the substitution $\vec{V} = \vec{v}$ (which has Jacobian 1) and then you can see the integral in the form you're familiar with. Commented Apr 25, 2020 at 11:49
• Oops, a small correction, the formula that you quoted is exactly the formula for the "average speed" (not relative velocity). Commented Apr 25, 2020 at 12:23
• $\vec{v}_1$ and $\vec{v}_2$ are not orthogonal or of the same length, so the Jacobian is not 1. Also, it is not possible to ignore the definition given to $\vec{V}$. It cannot be equal to something on one line, then not equal to it on the next, unless you are in contradiction. This means that by substituting $\vec{v}=\vec{V}$, you have also substituted $\vec{v}=\frac{-\vec{v}_1+\vec{v}_2}{\sqrt{2}}$. Commented Apr 26, 2020 at 6:23

Edit: This answer is wrong. I made an error that is equivalent to implicitly assuming that the particles' velocities are always perpendicular (see below). The question author has asked me to leave it up as it is interesting that the final result is so nearly correct despite the mistake.

The Maxwell-Boltzmann distribution is

$$B(\mathbf{v}) = \left( \frac{m}{2\pi k T} \right)^{3/2} e^{-\frac{m|\mathbf{v}|^2}{2kT}}$$

This function has a few nice features that make the following calculations easier. First, it's normalized, so integrals such as $$\int_{\mathbb{R}^3} B(\mathbf{v}) \, d^3\mathbf{v}$$ will always evaluate to 1. Second, since it is a radial function, we'll be able to take advantage of the trick

$$\int_{\mathbb{R}^n} f(|\mathbf{x}|) \, d^n\mathbf{x} = \omega_{n-1} \int_0^{\infty}f(r) r^{n-1} \,dr$$

where $$\omega_{n-1}$$ is the area of an $$n$$-sphere of radius 1.

All right, there's no way to fully answer your question without doing a few ugly integrals, so let's dive in. The mean speed of a particle is

\begin{align} \langle v \rangle &= \frac{\int_{\mathbb{R}^3} v \, B(\mathbf{v}) \, d^3\mathbf{v}}{\int_{\mathbb{R}^3} B(\mathbf{v}) \, d^3\mathbf{v}} \\ &= \int_{\mathbb{R}^3} v \, B(\mathbf{v}) \, d^3\mathbf{v} \\ &= \left( \frac{m}{2\pi k T} \right)^{3/2} \int_{\mathbb{R}^3} v \, e^{-\frac{m|\mathbf{v}|^2}{2kT}} \, d^3\mathbf{v} \\ &= \left( \frac{m}{2\pi k T} \right)^{3/2} \int_0^{\infty} v \, e^{-\frac{mv^2}{2kT}} \, 4\pi v^2 dv \\ &= \left( \frac{m}{2kT} \right)^{3/2} \frac{4}{\sqrt{\pi}} \int_0^{\infty} v^3 \, e^{-\frac{mv^2}{2kT}} \, dv \\ &= \left( \frac{m}{2kT} \right)^{3/2} 2\pi \left( \frac{m}{2kT} \right)^{-2} \\ &= \left( \frac{m}{2kT} \right)^{3/2} \frac{2}{\sqrt{\pi}} \left( \frac{m}{2kT} \right)^{-2} \\ &= \frac{2}{\sqrt{\pi}} \left( \frac{2kT}{m} \right)^{1/2} \end{align}

[Edit: if we make the approximation that the particles' velocities are perpendicular,] the mean relative speed of two particles is

\begin{align} \langle v_r \rangle &\approx \frac{\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} v_r \, B(\mathbf{v})B(\mathbf{v}') \, d^3\mathbf{v} \, d^3\mathbf{v}'}{\int_{\mathbb{R}^3}\int_{\mathbb{R}^3} B(\mathbf{v})B(\mathbf{v}') \, d^3\mathbf{v} \, d^3\mathbf{v}'} \\ &= \int_{\mathbb{R}^3}\int_{\mathbb{R}^3} v_r \, B(\mathbf{v})B(\mathbf{v}') \, d^3\mathbf{v} \, d^3\mathbf{v}' \\ &= \int_{\mathbb{R}^3}\int_{\mathbb{R}^3} v_r \, \left( \frac{m}{2\pi k T} \right)^{3/2} e^{-\frac{m|\mathbf{v}|^2}{2kT}} \left( \frac{m}{2\pi k T} \right)^{3/2} e^{-\frac{m|\mathbf{v'}|^2}{2kT}} \, d^3\mathbf{v} \, d^3\mathbf{v}' \\ &= \left( \frac{m}{2\pi k T} \right)^3 \int_{\mathbb{R}^3}\int_{\mathbb{R}^3} v_r \, e^{-\left(\frac{m|\mathbf{v}|^2}{2kT} + \frac{m|\mathbf{v'}|^2}{2kT}\right)} \, d^3\mathbf{v} \, d^3\mathbf{v}' \\ &= \left( \frac{m}{2\pi k T} \right)^3 \int_{\mathbb{R}^3}\int_{\mathbb{R}^3} v_r \, e^{-\frac{mv_r^2}{2kT}} \, d^3\mathbf{v} \, d^3\mathbf{v}' \\ &= \left( \frac{m}{2\pi k T} \right)^3 \int_0^{\infty} v_r \, e^{-\frac{mv_r^2}{2kT}} \, \pi^3 v_r^5 \, dv_r \\ &= \left( \frac{m}{2 kT} \right)^3 \int_0^{\infty} v_r^6 \, e^{-\frac{mv_r^2}{2kT}} \, dv_r \\ &= \left( \frac{m}{2 kT} \right)^3 \frac{15}{16}\sqrt{\pi} \left( \frac{m}{2kT} \right)^{-7/2} \\ &= \frac{15}{16}\sqrt{\pi} \, \left( \frac{2kT}{m} \right)^{1/2} \end{align}

where on line 6, $$\pi^3 v_r^5$$ is the surface area of a 6-sphere of radius $$v_r$$, and the integral on line 8 was calculated using Wolfram Alpha.

Thus we find that $$\langle v_r \rangle / \langle v \rangle = 15\pi/32 \approx 1.4726$$ which is almost, but not exactly equal to the textbook result $$\sqrt{2} \approx 1.4142$$.

[Edit: the textbook result $$\langle v_r \rangle / \langle v \rangle = \sqrt{2}$$ is correct. For the full, correct proof without any simplifying assumptions, please see the other answer.]

• The funny thing is, that I found a reference to this proof online (google books) but couldn't access the whole proof because I hadn't purchased the book. After two days of searching, I ended up just buying it (an hour ago, because I didn't think I'd receive an answer.) Commented Dec 17, 2018 at 6:43
• If you would like, I could write up my answer (by tomorrow) and the two could sit on the page for comparison. Your answer however, is sufficiently nice to be here on its own (and will also stay accepted), since it doesn't skip steps by saying that $\overline{v_{rel}}$ should be positive. Commented Dec 17, 2018 at 6:45
• There's no gamma function in this proof, though, and the integral on line 8 can be computed exactly. Commented Dec 17, 2018 at 6:47
• Wow that's interesting, maybe there really is a gap in the literature. Thanks for the answer :) Commented Dec 17, 2018 at 6:49
• So I claimed that I would be uploading my own version today yesterday, but it's 3:00 right now and I am only just finishing the on paper proof. It might be another day before I can finish typing it up for SE. Commented Dec 18, 2018 at 4:12