3
$\begingroup$

In The Feynman Lectures on Physics, Volume I 39-2 The pressure of a gas, the following is presented:

If $v$ is the velocity of an atom, and $v_{x}$ is the $x$-component of $v$, then $mv_{x}$ is the $x$-component of momentum in; but we also have an equal component of momentum 0ut and so the total momentum delivered to the piston by the particle, in one collision, is $2mv_{x}$, because it is "reflected".

Now, we need the number of collisions made by the atoms in a second, or in a certain amount of time $dt;$ then we divide by $dt$. How many atoms are hitting? Let us suppose that there are $N$ atoms in the volume $V$, or $n=N/V$ in each unit volume. To find how many atoms hit the piston, we note that, given a certain amount of time $t$, if a particle has a certain velocity toward the piston it will hit during the time $t$, provided it is close enough. If it is too far away, it goes only part way toward the piston in the time $t$, but does not reach the piston. Therefore it is clear that only those molecules which are within a distance $v_{x}t$ from the piston are going to hit the piston in the time $t$. Thus the number of collisions in a time $t$ is equal to the number of atoms which are in the region within a distance $v_{x}t,$ and since the area of the piston is $A,$ the volume occupied by the atoms which are going to hit the piston is $v_{x}tA$. But the number of atoms that are going to hit the piston is that volume times the number of atoms per unit volume, $nv_{x}tA.$ Of course we do not want the number that hit in a time $t$, we want the number that hit per second, so we divide by the time $t$, to get $nv_{x}A$. (This time $t$ could be made very short; if we feel we want to be more elegant, we call it $dt,$ then differentiate, but it is the same thing.)

So we find that the force is

$$ F=nv_{x}A\times2mv_{x}.\,\,\,(39.3) $$

See, the force is proportional to the area, if we keep the particle density fixed as we change the area! The pressure is then

$$ P=2nmv_{x}^{2}.\,\,\,(39.4) $$

Now we notice a little trouble with this analysis: First, all the molecules do not have the same velocity, and they do not move in the same direction. So, all the $v_{x}^{2}$'s are different! So what we must do, of course, is to take an average of the $v_{x}^{2}$'s, since each one makes its own contribution. What we want is the square of $v_{x}$, averaged over all the molecules:

$$ P=nm\left\langle v_{x}^{2}\right\rangle .\,\,\,(39.5) $$

Did we forget to include the factor 2? No; of all the atoms, only half are headed toward the piston. The other half are headed the other way, so the number of atoms per unit volume that are hitting the piston is only $n/2$.

While I accept the result, I do not understand his development. In particular, what is meant by the "volume" $v_{x}tA$? That volume is introduced with the tacit (and incorrect) assumption that all $v_{x}$'s are equal. An assumption which is subsequently rejected. But the meaning of $v_{x}tA$ in terms of the refined understanding of $v_{x}$ being specific to each atom is never made clear.

Introducing the notation $\Delta V_{x}=v_{x}tA;$ I have found no way to arrive at the advertised result $\left(39.5\right)$ using half the average $x$-component of speed to establish the correct value of $\Delta V_{x}$. For example:

$$ \frac{1}{2}n\left\langle \left|v_{x}\right|\right\rangle A\times2m\left\langle \left|v_{x}\right|\right\rangle =Anm\left\langle \left|v_{x}\right|\right\rangle ^{2}\ne Anm\left\langle v_{x}^{2}\right\rangle . $$

The result $\left(39.5\right)$ can be established by an alternative development which determines the number of collisions per unit time by considering the number of times any specific particle will traverse the $x$-dimension of a box of unit volume to the far side, and then back in a time $t$. But I am interested to know if Feynman's approach can be understood.

Under the assumption that the value $v_{x}$ is specific to each atom, what volume, corresponding to the above, $\Delta V_{x}=v_{x}tA$, should be used to determine the number of collisions per unit time of gas atoms with the piston?

$\endgroup$
2
$\begingroup$

Well, first thing I would like to say is that you are expecting too much rigor from a semi-handwaving demonstration such as this one.

Anyway, what you can argue is that you expect the distribution of $|v_x|$ to have a small (relative) variance, i.e.

$$\frac{\langle |v_x|^2 \rangle - \langle |v_x| \rangle^2}{\langle |v_x| \rangle^2} = \frac{\langle |v_x|^2 \rangle}{\langle |v_x| \rangle^2} -1 \ll 1 $$

Now of course in principle you don't know if this is reasonable. If you actually use the Maxwell-Boltzmann distribution you will discover that

$$\frac{\langle |v_x|^2 \rangle}{\langle |v_x| \rangle^2} -1 = \left(\frac{kT}m\right)\left(\frac{\pi m}{2kT}\right)-1 = \frac \pi 2 -1\approx 0.57 $$

So things are not so good, but not so bad either (we got a number smaller than $1$, but not by much...).

In the end, I would say that you are right in thinking that there is a bit too much handwaving in this proof. Probably, as it often happens in such rough estimates, there is some fortuitous cancellation of errors: you overestimate the value of something, but underestimate the value of something else and in the end you get the correct result. This happens all the time in physics (look at the Drude model for conduction or at the Flory theory of polymers, for example).

Notice that the derivation that you can find for example on Wikipedia is slightly different from Feynman's derivation. The crucial difference is that Feynman estimates the number of collisions per second $f$ as

$$f=\frac{n V(\Delta t)}{\Delta t}$$

The problem is the use of the $V(\Delta t)$, which as you pointed out can only be reasonably estimated using average quantities:

$$V(\Delta t) = A \langle |v_x | \rangle \Delta t$$

In the Wikipedia derivation, a more stringent assumptions are made: the box is a cube of length $L$.

Instead of estimating $f$ using this "average volume" $V(\Delta t)$, it is simply stated that the frequency of the collisions is

$$f=\Delta t^{-1} = \frac{v_x}{2L}$$

where $\Delta t$ is the time it takes a particle to go form one side of the box to the other moving in the $x$ direction.

Then the force is calculated as

$$F= f \Delta p = \frac{v_x}{2L} \cdot 2mv_x = \frac{mv_x^2}{L}$$

Since the two $v_x$ that appears in the previous formula are referred to exactly the same molecule, there is no $\langle |v_x | \rangle$ coming into play and we can safely say, after averaging over all the molecules, that the average force will be

$$\langle F \rangle = \frac{m \langle v_x^2 \rangle}{L}$$

Notice that in this case we don't even have to correct for the factor $2$ which comes out from Feynman's argument. So we get the "right" result without any ambiguity regarding $\langle |v_x|\rangle^2$ vs $\langle v_x^2 \rangle$. The price to pay, however, is that we had to sneak in an additional assumption.

$\endgroup$
  • $\begingroup$ Feynman did say that there were parts of the Lectures that were significantly flawed. My guess is, Vol 1, 39 is an example. He was pole-vaulting over too much detail (IMO). It's not overly difficult to give a more rigorous hand-waving derivation which keeps the math honest. My guess is, he was striving for some intuitive understanding that may not be evident form a more conventional presentation. $\endgroup$ – Steven Thomas Hatton Mar 18 '18 at 22:17
  • $\begingroup$ @StevenHatton I wouldn't call this derivation significantly flawed...It has a very good value as an intuitive explanation, and it gives the correct answer through a reasonably logical argument. Notice that there are slightly different derivations (such as this one) in which the use of $\langle v_x^2\rangle$ is introduced in what to me looks like a clearer way, but with some more assumptions, such as the box being cubic and the particles bouncing perpendicularly on the walls. $\endgroup$ – valerio Mar 18 '18 at 22:58
  • $\begingroup$ @StevenHatton I added some more details to the answer, to show where Feynman's argument diverges from the usual one. $\endgroup$ – valerio Mar 18 '18 at 23:27
  • $\begingroup$ I don't believe we need the tacit assumption that mythical particles move parallel to the edges of the mythical cubical box of volume $V=L^3$. The result stands if we fill the mythical box with real particles, calculate using the velocity components parallel to the edges. Feynman wisely points out that approaching thermal physics axiomatically is futile. The art is to separate primary aspects of the problem from aspects which can be neglected in a course-grained evaluation. OTOH, we need to keep track of which "lies" we told ourselves to obtain usable results. $\endgroup$ – Steven Thomas Hatton Mar 19 '18 at 17:36
  • $\begingroup$ @StevenHatton You're right, they don't have to move parallel to the walls...my bad. I corrected the answer. I still believe that the two derivations are subtly different, though. $\endgroup$ – valerio Mar 19 '18 at 18:01

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.