How to work out the relation between the "mean relative speed" and the "mean speed"?

Question

I'm a freshman and am taking the general physics course. I just learned intro thermodynamics. One problem that really puzzles me is the calculation of "collision mean-free path", where calculating the mean relative velocity between gas molecules is needed. Our textbook simply gives a result $$\langle |v_r|\rangle =\sqrt2\langle |v|\rangle $$ without further explanation.
Here I am using angle brackets ($\langle \rangle $) to represent the "mean value" of what's inside. And note that all the velocities here are vectors, so I am using the absolute value symbols to get the "speed".


My professor has provided an explanation as follows:
Suppose that we select an arbitrary molecule A, with the velocity $v$ to the "stationary", as the reference frame. And suppose another arbitrarily selected molecule B has the velocity $v'$ to the "stationary". Therefore, in the reference frame A, B's velocity will be $(v'-v)$, which is just $v_r$, denoting B's "relative velocity" to A.
So we have: $$v_r=v'-v$$ Square both sides, $$|v_r|^2=|v'|^2+|v|^2-2v'\centerdot v$$ Now that we want to obtain the "mean value" of $v_r$ of an immense group of such "molecule B"s in a statistical sense,so my professor tried to work out the "mean value" of both sides: $$\langle |v_r|^2\rangle =\langle |v'|^2\rangle +\langle |v|^2\rangle -2\langle v'\centerdot v\rangle $$ It is plain to see (although there may be a lack of rigorousness) that, statistically $$\langle v'\centerdot v\rangle =0$$ and that $$\langle |v'|^2\rangle =\langle |v|^2\rangle $$ Therefore $$\langle |v_r|^2\rangle =2\langle |v|^2\rangle $$ Here comes the key part.From the above equation my professor concluded that $$\langle |v_r|\rangle =\sqrt2\langle |v|\rangle $$

However, I do not think this plausible step holds water. Because I think that for a statistical variable $x$, $\langle x\rangle ^2$ and $\langle x^2\rangle $ are not necessarily equal. (especially when I later learned something about Maxwell velocity distribution and found that for gas molecules the mean speed $|v|$ is actually smaller than the root mean square speed $\sqrt{\langle |v|^2\rangle }$.) So I think, instead of getting the result we want, the last step in fact gives $$\sqrt{\langle |v_r|^2\rangle }=\sqrt{2}\sqrt{\langle |v|^2\rangle }$$

This problem has been bothering me for several weeks and I want it fully explained, in an explicit and rigor way. I think only by using the knowledge of probability can a mathematically-convincing explanation be achieved. Unluckily I haven't learned much about probability and knows very little about relevant theories. Would anybody help me about this? Merci.

Answer 1

You are right, actually $\frac{\langle v^2 \rangle}{\langle v \rangle^2}=8\pi/3$. For what the professor wrote you do not need to assume that $\langle v^2 \rangle=\langle v \rangle^2 $, but only that $\frac{\langle v^2 \rangle}{\langle v \rangle^2}=\frac{\langle v_r^2 \rangle}{\langle v_r \rangle^2}=C $, with $C$ arbitrary.

Even if this assumption might look more plausible, it is still unjustified, so this is still not a rigorous demonstration. The right way to do it is to use the speed distribution and calculate it by brute force. I tried a couple of times but ended up with terrible integrals. I am not sure now how straightforward it is to do it. So do not be discouraged if you try and fail.

Answer 2

It seems to me that you can use for your gas the Maxwell-Boltzmann distribution of velocities. If my assumption is correct, then the things are as follows.

Let me denote this distribution by $B(v^2)$ because it depends indeed on the square of the velocity. Since your molecules are independent, the distribution describing them is the product of the two independent distributions

$$B(v^2, v'^2) = B(v^2) B(v'^2).\qquad(1)$$

Now, let's calculate the mean value of $v_r^2$ as you requested. As you said,

$$\langle v_r^2\rangle = \int d\vec v \int d\vec v' B(v^2, v'^2) (v^2 + v'^2 -2\vec v\cdot \vec v')$$

By applying (1)

$$\langle v_r^2\rangle = \int d\vec v \,B(v^2)v^2\int d\vec v'B(v'^2) + \int d\vec v \, B(v^2)\int d\vec v' B(v'^2) v'^2 + \int d\vec v\,B(v^2) \vec v \int d\vec v' B(v'^2) \vec v'.$$

Since the distributions $B$ are normalized we get

$$\langle v_r^2\rangle= \int d\vec v\,B(v^2)v^2 + \int d\vec v' B(v'^2) v'^2 + \int d\vec v \, B(v^2) \vec v \int d\vec v' \ B(v'^2) \vec v'.$$

Now, let's concentrate on the last term. Since the integrand under the integral over $\vec v$ is antisymmetrical, this integral is zero. The same with the integral over $\vec v'$. Thus we remain with

$$\langle v_r^2\rangle = \int d\vec v \, B(v^2)v^2 + \int d\vec v' B(v'^2) v'^2 = \langle v^2\rangle + \langle v'^2\rangle = 2\langle v^2\rangle,$$

since it is irrelevant if we name the integration variable $v$ or $v'$.

Therefore, indeed $$\sqrt{\langle v_r^2\rangle} = \sqrt{2\langle v^2\rangle}.$$

About deriving $\langle v_r\rangle$ the calculus is more complicated because

$$|\vec v_r| = \sqrt{v^2 + v'^2 - 2\vec v \cdot\vec v'}.$$

Answer 3

My calculus below is devoted to check if the relation

$$⟨|\vec v_r|⟩= \sqrt {2}⟨|\vec v|⟩\qquad(1)$$

written by the professor, can be true. The calculus is a bit boring – I regret, and the result does not confirm the relation (1).

So, I start from the relation

$$|\vec v_r| = \sqrt {v^2 + v'^2 - 2\vec v \cdot\vec v'}.$$

Now, let's write $\vec v \cdot\vec v'$ in the form $v v'\cos(\theta)$ where $\theta$ is the angle between the two velocity vectors. Reminding that the element of volume in spherical coordinates in the velocity space is $dV = d\vec v = v^2\,dv\sin(\theta)\,d\theta\,d\phi$, let's try to solve the integral

$$\langle|\vec v_r|\rangle = \int d\vec v \,B(v^2) \int v'^2\, dv' B(v'^2)\sin(\theta)\,d\theta\,d\phi \sqrt{v^2 + v'^2 - 2v v'\cos(\theta)},$$

where the second integral is over v from 0 to $\infty$, over $\theta$ from 0 to $\pi$ and over $\phi$ from 0 to $2\pi$. It doesn't look nice, but we see that the integrand doesn't depend on $\phi$, s.t. we can integrate over it

$$\langle|\vec v_r|\rangle = 2\pi \int d\vec v \, B(v^2) \int v'^2\, dv' B(v'^2) \sin(\theta)\,d\theta\sqrt{v^2 + v'^2 - 2v v'\cos(\theta)}.$$

We can also integrate over $\theta$ since $-\sin(\theta)$ is the derivative of $\cos(\theta)$. $$ \begin{aligned} \langle|\vec v_r|\rangle &= -\frac{2\pi}{3} \int \frac{d\vec v}{v} B(v^2) \int v'\,dv'B(v'^2)\bigl[v^2 + v'^2 - 2v v'\cos(\theta)\bigr]^{3/2}|_{\theta = 0}^{\theta = \pi}\\ &=\frac{2\pi}{3} \int \frac{d\vec v}{v}B(v^2) \int v'\,dv'B(v'^2)\bigl[(v^2 + v'^2 + 2v v')^{3/2} - (v^2 + v'^2 - 2v v')^{3/2}\bigr]\\ &=\frac{2\pi}{3} \int \frac{d\vec v}{v}B(v^2) \int v'\,dv'B(v'^2)\bigl[(v + v')^3 - |v - v'|^3\bigr] \end{aligned} $$

Just for the sake of symmetry between $\vec v$ and $\vec v'$, let's notice that no dependence remained here on the angles $\theta$ and $\phi$. If so, let's multiple and divide the integral over $v'$ by $2\pi v'^2$. We get,

$$ \begin{aligned} \langle|\vec v_r|\rangle&= \frac{1}{3} \int \frac{d\vec v}{v}B(v^2) \int \frac{d\vec v'}{v'} B(v'^2)(v + v')^3\\ &- \frac {1}{3} \int \frac{d\vec v}{v}B(v^2) \int \frac{d\vec v'}{v'} B(v'^2)|v - v'|^3 \end{aligned} $$

From now on the calculus is simple but ugly. Because of the absolute value, we will have to split the 2nd integral into two parts: from 0 to $v$, and from $v$ to $\infty$. I will just put it in a slightly simpler form

$$\langle|\vec v_r|\rangle = \frac{1}{3} \int \frac{d\vec v}{v} B(v^2) \int_0^v d\vec v'B(v'^2)(3v^2 + v'^2) + \frac{1}{3} \int d\vec v \ B(v^2) \int_v^{\infty} \frac{d\vec v'}{v'} B(v'^2)(3v'^2+v^2).$$