# Maxwell-Boltzmann distribution, average speed in one direction

Consider an ideal gas obeying the Maxwell-Boltzmann distribution i.e

$$f(v) = \bigg(\frac{m}{2 \pi k_{B} T}\bigg)^{3/2} \exp \left(-\frac{m v^{2}}{2 k_{B} T} \right) \, .$$

The probability distribution in 3D velocity space ($v^{2} = v_{x}^2 + v_{y}^2 + v_{z}^2$). How might you determine the average speed the particles are moving at, $\langle |v_{z}| \rangle$, in one direction?

Additionally if my ideal gas is now confined to a hemisphere in velocity space i.e we have the conditions $- \infty \leq v_{x}, v_{y} \leq \infty$ and $0 \leq v_{z} \leq \infty$ but it still has a Maxwell Boltzmann velocity distribution (except I think the normalization factor on $f(v)$ should change) then what is the average speed, or velocity, in the z direction $\langle v_{z} \rangle$, will this be the same as $|\langle v_{z}| \rangle$ from the previous answer?

• What are your ideas as to what would happen? For example, what makes you think that $<|v_{z}|>$ will not change? Mar 22, 2017 at 8:12
• Well I think it shouldn't change because of symmetry, if all the $v_{z} < 0$ have their signs flipped then that should have no effect on the average speed. Also I believe that $|<v_{z}>|$ can be determined by integrating out the x and y dependence of $f(v)$? Mar 22, 2017 at 9:11
• It seems that your distribution should be that of $f(\vec{v})$, but not $f(v)$, which should carry a $v^2$ factor. Sep 10, 2017 at 16:13

The average speed of particles in a particular direction will always be smaller than the average speed of particles.

Imagine that you have three particles with components of their velocity $(1,1,1)\, \rm ms^{-1}$.
Their average speed is $<v>= \sqrt 3\, \rm ms^{-1}$ whilst their average speed in the x-direction is $<v_{\rm x}>=1\, \rm ms^{-1}$

So what you need to do is use the distribution of velocities in the x-direction

$$f(v_\rm{x}) = \left(\frac{m}{2 \pi k_{B} T}\right)^\frac{1}{2} \exp\left ({-\frac{m v_{\rm x}^{2}}{2 k_{B} T}}\right )$$

and do the following integration

$\displaystyle \int_0^\infty v_\rm{x}\, f(v_\rm{x}) \,dv_{\rm x}$ which will give you an answer of $\dfrac {<v>}{4}$ where $<v>$ is the avergae speed of the particles.

• I think this would not be correct as this is not consistent with$<|v_{x}|>$ of the distribution, your integral $\int_{0}^{\infty}v_{x}f(v_{x})dv_{x}$ is not properly weighted by the sum of the probabilities in the $0 \leq v_{x} \leq \infty$ region. Hence it requires a factor of $2$ out the front. I asked this question because I have seen several people stating this result yet I am convinced $<v_{x}> = \frac{<v>}{2}$. where $<v_{x}>$ is the average speed in the x direction and $<v>$ is the average speed in all 3 directions May 1, 2017 at 15:33
• V. Molecular flux lehman.edu/faculty/dgaranin/Statistical_Thermodynamics/… May 1, 2017 at 21:38
• Yes that is for the molecular flux however which only accounts for the particles moving in one specific direction across the barrier, it does not tell you that $<v_{x}>= \frac{1}{4}<v>$. In fact, because only half the density of particles is moving in the correct direction to contribute to this flux I think it is saying $<v_{x}>= \frac{1}{2}<v>$. May 3, 2017 at 5:48

The probability density function (PDF) for a single component (let's take $z$) is

$$f(v_z)=\sqrt{\frac{m}{2\pi k T}} \exp \left(-\frac{m v_z^2}{2kT}\right)$$

Case #1

What you want to calculate is

$$\langle |v_z| \rangle = \int_{-\infty}^{\infty} |v_z| \ f(v_z) \ d v_z = 2 \int_0^\infty v_z \ f(v_z) \ d v_z$$

Using

$$\int_0^{\infty} u e^{-au^2} du = \frac 1 {2a} \tag{1}$$

you get

$$\langle | v_z| \rangle = 2 \cdot \sqrt{\frac{m}{2\pi k T}} \cdot \frac 1 2 \cdot \frac{2kT}{m} = 2 \sqrt{\frac{kT}{2 \pi m}} = \sqrt{\frac{2kT}{\pi m}}$$

Notice that since

$$\langle |v| \rangle = \sqrt{\frac{8kT}{\pi m}}$$

where $|v|= \sqrt{v_x^2+v_y^2+v_z^2}$, you have

$$\langle | v_z| \rangle = \frac 1 2 \langle |v| \rangle$$

Case #2

As you said, in this case the normalization factor must change, since we must require that

$$\int_0^\infty \tilde f(v_z) d v_z = 1$$

The correct PDF is in this case

$$\tilde f(v_z)= \frac 1 2 \sqrt{\frac{m}{2\pi k T}} \exp \left(-\frac{m v_z^2}{2kT}\right)$$

Using again (1), it is straight-forward to obtain from this distribution

$$\langle v_z \rangle_{\tilde f} = \frac 1 2 \sqrt{\frac {kT}{2 \pi m}} = \frac 1 4 \langle |v_z| \rangle = \frac 1 8 \langle |v| \rangle$$