9
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

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

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

$<v_r^2> = \int d\vec v \int d\vec v' B(v^2, v'^2) (v^2 + v'^2 -2\vec v \vec v')$

By applying (1)

$<v_r^2> = \int d\vec v \ B(v^2)v^2\int d\vec v' B(v'^2) + \int d\vec v \ B(v^2)\int d\vec v' B(v'^2) v'^2 + \int d\vec v \ B(v^2) \vec v \int d\vec v' \ B(v'^2) \vec v'$.

Since the distributions B are normalized we get

$<v_r^2> = \int d\vec v \ B(v^2)v^2 + \int d\vec v' B(v'^2) v'^2 + \int d\vec v \ B(v^2) \vec v \int d\vec v' \ B(v'^2) \vec v'$.

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

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

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

Therefore, indeed $\sqrt {<v_r^2>} = \sqrt {2 <v^2>}$

About deriving $<v_r>$ the calculus is more complicated because

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ @Sofia it is just a reasonable (?) assumption. At least to me is more digestible than than $\langle v^2 \rangle=\langle v \rangle^2$. But the only way to know for sure is to calculate it from the original distribution, as you did. $\endgroup$ – Wolphram jonny Dec 31 '14 at 2:32
  • $\begingroup$ Thank you. I also have considered proving in this way. But as my knowledge is very limited I cannot successfully prove that they share the same C. And I think there is still something to add to Sofia's answer, which I think has so far explained why $<v_r^2>=2<v^2>$ but still has not reached the final result. $\endgroup$ – Vim Dec 31 '14 at 5:07
  • $\begingroup$ @Wolphramjonny : I will also think of it. The truth is that I think that the professor made some mistake. Please look at Vim's words "From the above equation my professor concluded that ...". So, the professor concluded that if $<v_r^2> = 2<v^2>$, then $<|\vec v_r|> = \sqrt {2}<|\vec v|>$. Such an immediate conclusion seems to me strange. For proving the last equality one should start, so I think, from the square root that I wrote on the bottom of my answer. It doesn't seem trivial to integrate it, s.t. the swift conclusion of the professor is quite questionable. $\endgroup$ – Sofia Dec 31 '14 at 23:31
  • $\begingroup$ @sofia I might be wrong, but my intuition tells me that the result should be independent of the particular distribution followed by the molecules. So doing the full calculation could result in a loss of generality. $\endgroup$ – Wolphram jonny Jan 1 '15 at 18:10
  • $\begingroup$ @Wolphramjonny : Jonny you go too far with the speculations. Let's stay closer to our purpose to solve Vim's problem whether the relation written by his professor can be correct. What I got in my calculi is that the professor is wrong. As I suspected, the relation between averages of $v_{rel}$ and $v$ that the professor wrote it's the result of some lack of attention from him, as I told you. I will post an additional answer, dealing with this issue only, and you'll see. But please check my calculus, I am doing this at an hour a bit late. $\endgroup$ – Sofia Jan 1 '15 at 21:47
2
$\begingroup$

My calculus below is devoted to check if the relation

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

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

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

Now, let's write $\vec v \vec v'$ 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

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

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

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

We can also integrate over $\theta$ since $-sin(\theta)$ is the derivative of $cos(\theta)$.

$<|\vec v_r|> = -\frac {2\pi}{3} \int \frac {d\vec v}{v} B(v^2) \ \int v'dv'B(v'^2)\ [v^2 + v'^2 - 2v v'cos(\theta)]^{3/2}|_{\theta = 0}^{\theta = \pi}$

$ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac {2\pi}{3} \int \frac {d\vec v}{v}B(v^2) \ \int v'dv'B(v'^2)[v^2 + v'^2 + 2v v']^{3/2} - [v^2 + v'^2 - 2v v']^{3/2}$

$ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac {2\pi}{3} \int \frac {d\vec v}{v}B(v^2) \ \int v'dv'B(v'^2)[(v + v')^3 - |v - v'|^3]$

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,

$<|\vec v_r|> = \frac {1}{3} \int \frac {d\vec v}{v}B(v^2) \ \int \frac {d\vec v'}{v'} B(v'^2)(v + v')^3$

$ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac {1}{3} \int \frac {d\vec v}{v}B(v^2) \ \int \frac {d\vec v'}{v'} B(v'^2)|v - v'|^3$

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

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

$\endgroup$
  • $\begingroup$ @Sofia I have some trouble comprehending the third displayed equation (the first integral expression) in this post. Why isn't it symmetrical for $v$ and $v'$? I intuit that $v$ and $v'$ should contribute completely equally to $<v_r>$. $\endgroup$ – Vim Jan 3 '15 at 17:08
  • $\begingroup$ @Vim : I introduced a few modifications for stressing the symmetry between $v$ and $v'$. It continues until the before-last equality. In the last equality I break the symmetry "by hand" as people use to say, because I first integrate over $v'$. If I would have integrated first over $v$ I would have obtained the same last equation with $v$ and $v'$ interchanged. $\endgroup$ – Sofia Jan 3 '15 at 18:14
  • $\begingroup$ @Vim : please let me know when you get my answer. This utility makes problems, sometimes I post a comment and it disappears. $\endgroup$ – Sofia Jan 3 '15 at 18:18
  • $\begingroup$ I think I'm a bit stupid...orz...and I still fail to see symmetry in this equation. $$<|\vec v_r|> = \int d\vec v \ B(v^2) \ \int v'^2 dv' B(v'^2)\ sin(\theta)d\theta \ d\phi \ \sqrt {v^2 + v'^2 - 2v v'cos(\theta)}$$, can you give me some hint about that? Thank you! $\endgroup$ – Vim Jan 4 '15 at 17:20
  • $\begingroup$ And another thing.. Are the $v$s and $v'$s in the RHS all vectors? If not, then the LHS is a scalar while RHS appears to be a vector since it contains only one vector term $d\vec{v}$... $\endgroup$ – Vim Jan 4 '15 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.