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I refer to hand-waving a lot in this post. That is not to say that it was in appropriate. Feynman pretty much said at the outset that his treatment of thermal physics was going to be less than rigorous.

I've been rereading Feynman's Lectures on thermal physics with the intent of summarizing the material. I have not been able to sort out actual argument concluding that the molecular speed distribution is homogeneous for an ideal gas in thermal equilibrium. That is, the shape of the distribution curve doesn't change with molecular density.

I accept his argument that the velocity distribution is isotropic at any given location in the body of the gas. That is, statistically, the velocity has no preferred direction.

I accept his hand-wavy argument that the mean kinetic energy per atom is a function of temperature, alone.

But just because the mean velocity is the same everywhere doesn't automatically require the distribution to be the same everywhere.

In Vol I, 40-1 The exponential atmosphere Feynman appears to acknowledge that he has not yet shown that the velocity distribution is necessarily homogeneous:

"So, these are the two questions that we shall try to answer: How are the molecules distributed in space when there are forces acting on them, and how are they distributed in velocity?

"It turns out that the two questions are completely independent, and that the distribution of velocities is always the same. We already received a hint of the latter fact when we found that the average kinetic energy is the same, $\frac{1}{2}kT$ per degree of freedom, no matter what forces are acting on the molecules. The distribution of the velocities of the molecules is independent of the forces, because the collision rates do not depend upon the forces."

However, in section V-I 40-4 on The distribution of molecular speeds, Feynman claims that he has already shown the distribution to be homogeneous.

"We know already that the distribution of velocities is the same, after the argument we made earlier about the temperature being constant all the way through the atmosphere."

I see nothing in the intervening text that implies the velocity distribution is homogeneous. So I am left to wonder if the development of the velocity distribution function is needed to show that the velocity distribution is homogeneous, or if the development depends on the assumption that it is homogeneous.

It seems to me that the homogeneity of velocity distribution follows from the derivation of the velocity distribution formula. But that begs the question of whether a homogeneous velocity distribution is necessary for the assumption that the atoms can be treated as not interacting, for the sake of this development.

This is my method of deriving the one-dimensional velocity distribution formula. It seems to make clearer the connection between gravitational potential energy and the velocity distribution.

If I accept the hand-waving that allows me to treat the molecules as if they are not interacting, and only consider the relationship between upward speed at height $0$ and potential energy at height $h$, then there is an obvious correlation between upward speed distribution and particle density at the corresponding height.

The attached graphic shows how an interval of the velocity distribution maps onto a depiction of the gas in a column. Each point $v$ in the velocity range corresponds to the height at which $\frac{1}{2}mv^{2}=mgh$. In the velocity distribution diagram, the colored regions are of equal width. In the depiction of the column of gas, they are of different widths because $h=\frac{v^{2}}{2g}$.

enter image description here

Ideal gas in thermal equilibrium is confined to a tall cylinder. Assign some arbitrary horizontal plane height $h_{0}=0$. The equation giving density as a function of height is already established: $$n_{h}=n_{0}e^{-h\frac{mg}{kT}}$$

The only component of velocity under discussion is the positive vertical component, denoted $v$. Height is related to velocity by $mgh=\frac{1}{2}mv^{2}$, or

$$h=\frac{v^{2}}{2g}.$$

So

$$n_{h}=n_{v}=n_{0}e^{-v^{2}\frac{m}{2kT}}$$

Assume there is some velocity distribution $\mathscr{P[v]}$ The proportion of atoms with velocity between $v$ and $v+\Delta{v}$ is given by $$\mathscr{P}\left[v,v+\Delta v\right]=\int_{v}^{v+\Delta V}\mathscr{P}\left[v\right]dv.$$

The atoms in that range will come to rest between the heights $h_{v}=\frac{v^{2}}{2g}$ and $h_{v+\Delta v}=\frac{\left(v+\Delta v\right)^{2}}{2g}$. The number of those atoms is equal to the difference in density between those heights.

So the rate at which atoms with velocities between $v$ and $v+\Delta{v}$ pass $h_{0}$ is the rate at which atoms come to rest between $h_{v}$ and $h_{v+\Delta{v}},$ which, assuming $\Delta{v}$ is infinitesimal is $dn_{h}=dn_{v}$. The rate at which such atoms pass $h_{0}$ is proportional to the number density $n_{0}$, the velocity $v$ and the proportion given by $\mathscr{P}\left[v\right]dv$.

$$v\alpha n_{0}\mathscr{P}\left[v\right]dv=dn_{v}=-v\frac{m}{kT}n_{0}e^{-v^{2}\frac{m}{2kT}}dv$$

Canceling out common factors gives:

$$\mathscr{P}\left[v\right]dv=-\frac{m}{\alpha kT}e^{-v^{2}\frac{m}{2kT}}dv$$

The value of $\alpha$ is given by integrating both sides and setting the result equal to unity. The expression obtained using Feynman's development is

$$\mathscr{P}\left[v\right]dv=-\sqrt{\frac{m}{2\pi kT}}e^{-v^{2}\frac{m}{2kT}}dv.$$

Comparing these results gives:

$$\sqrt{\frac{m}{2\pi kT}}=\frac{m}{\alpha kT}.$$

Solving for $\alpha$ gives

$$\alpha = \sqrt{\frac{2\pi m}{kT}}$$.

I did not explicitly assume the velocity distribution is homogeneous in establishing the distribution formula. But the fact that the derivation gives the same result regardless of the arbitrary choice of $h_{0}$ shows the distribution to homogeneous.

So the question remains. Is the assumption that the atoms can be treated as non-interacting for purposes of this derivation implicitly assuming the velocity distribution to be homogeneous?

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  • $\begingroup$ If the system is in thermodynamic equilibrium, why would you expect the velocity distribution to differ? In fact, that the system is in thermodynamic equilibrium (not the same as thermal equilibrium) it implies that the distribution is homogenous. Were it not, then the system would not be in thermodynamic equilibrium because there would be free energy present (probably Gibbs free energy). $\endgroup$ – honeste_vivere Apr 18 '18 at 13:59
  • $\begingroup$ That's begging the question. All I know is that there are a bunch of ideal atoms banging into each other in such a way that something called temperature is constant throughout the gas. Even if I "arbitrarily" assume that the molecular speed distribution curve is the same everywhere that the temperature is the same, I should state it as a "reasonable guess". BTW, my reasoning is off in my derivation in the original post. My factor of proportionality $\alpha$ is mathematically correct, but my physical argument is incorrect. I have been trying to fix that for the past week. $\endgroup$ – Steven Thomas Hatton Apr 23 '18 at 14:44
  • $\begingroup$ The temperature is derived from the second velocity moment of the particle velocity distribution. So yes, if the distributions are the same everywhere, they will share the same temperatures (e.g., see physics.stackexchange.com/a/218643/59023). $\endgroup$ – honeste_vivere Apr 26 '18 at 13:53
  • $\begingroup$ @honeste_vivere I'm not asking if the distributions are the same everywhere. I'm asking where in Feynman's development is that established. Feynman never discusses second moments of velocity distributions. Consider this from VI-46-1 "Now, there are complicated mathematical demonstrations which follow from Newton’s laws to demonstrate that we can get only a certain amount of work out when heat flows from one place to another, but there is great difficulty in converting this into an elementary demonstration. In short, we do not understand it, although we can follow the mathematics." $\endgroup$ – Steven Thomas Hatton Apr 26 '18 at 21:19
  • $\begingroup$ So now I am not entirely sure I know what you are asking. $\endgroup$ – honeste_vivere Apr 27 '18 at 13:44

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