The distribution of molecular speeds
Now we go on to discuss the distribution of velocities, because sometimes it is interesting or useful to know how many of them are moving at different speeds. In order to do that, we may make use of the facts which we discovered with regard to the gas in the atmosphere. We take it to be a perfect gas, as we have already assumed in writing the potential energy, disregarding the energy of mutual attraction of the atoms. The only potential energy that we included in our first example was gravity. We would, of course, have something more complicated if there were forces between the atoms. Thus we assume that there are no forces between the atoms and, for a moment, disregard collisions also, returning later to the justification of this. Now we saw that there are fewer molecules at the height $h$ than there are at the height $0$; according to formula (40.1), they decrease exponentially with height. How can there be fewer at greater heights? After all, do not all the molecules which are moving up at height $0$ arrive at $h$? No!, because some of those which are moving up at $0$ are going too slowly, and cannot climb the potential hill to $h$. With that clue, we can calculate how many must be moving at various speeds, because from (40.1) we know how many are moving with less than enough speed to climb a given distance $h$. Those are just the ones that account for the fact that the density at $h$ is lower than at $0$. Now let us put that idea a little more precisely: let us count how many molecules are passing from below to above the plane $h = 0$ (by calling it height${} = 0$, we do not mean that there is a floor there; it is just a convenient label, and there is gas at negative $h$). These gas molecules are moving around in every direction, but some of them are moving through the plane, and at any moment a certain number per second of them are passing through the plane from below to above with different velocities. Now we note the following: if we call $u$ the velocity which is just needed to get up to the height $h$ (kinetic energy $mu^2/2 = mgh$), then the number of molecules per second which are passing upward through the lower plane in a vertical direction with velocity component greater than $u$ is exactly the same as the number which pass through the upper plane with any upward velocity. Those molecules whose vertical velocity does not exceed $u$ cannot get through the upper plane. So therefore we see that  \begin{equation*} \text{Number passing }h = 0 \text{ with }v_z > u = \text{number passing }h = h \text{ with }v_z > 0. \end{equation*} 
 \begin{equation*} \begin{pmatrix} \text{Number passing }h = 0\\ \text{ with }v_z > u \end{pmatrix} = \begin{pmatrix} \text{number passing }h = h\\ \text{ with }v_z > 0 \end{pmatrix}. \end{equation*}  But the number which pass through $h$ with any velocity greater than $0$ is less than the number which pass through the lower height with any velocity greater than $0$, because the number of atoms is greater; that is all we need. 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. So, since the velocity distributions are the same, and it is just that there are more atoms lower down, clearly the number $n_{> 0}(h)$, passing with positive velocity at height $h$, and the number $n_{> 0}(0)$, passing with positive velocity at height $0$, are in the same ratio as the densities at the two heights, which is $e^{-mgh/kT}$. But $n_{> 0}(h) = n_{> u}(0)$, and therefore we find that \begin{equation*} \frac{n_{> u}(0)}{n_{> 0}(0)} = e^{-mgh/kT} = e^{-mu^2/2kT}, \end{equation*} since $\tfrac{1}{2}mu^2 = mgh$. Thus, in words, the number of molecules per unit area per second passing the height $0$ with a $z$-component of velocity greater than $u$ is $e^{-mu^2/2kT}$ times the total number that are passing through the plane with velocity greater than zero. Now this is not only true at the arbitrarily chosen height $0$, but of course it is true at any other height, and thus the distributions of velocities are all the same! (The final statement does not involve the height $h$, which appeared only in the intermediate argument.) The result is a general proposition that gives us the distribution of velocities. It tells us that if we drill a little hole in the side of a gas pipe, a very tiny hole, so that the collisions are few and far between, i.e., are farther apart than the diameter of the hole, then the particles which are coming out will have different velocities, but the fraction of particles which come out at a velocity greater than $u$ is $e^{-mu^2/2kT}$. Now we return to the question about the neglect of collisions: Why does it not make any difference? We could have pursued the same argument, not with a finite height $h$, but with an infinitesimal height $h$, which is so small that there would be no room for collisions between $0$ and $h$. But that was not necessary: the argument is evidently based on an analysis of the energies involved, the conservation of energy, and in the collisions that occur there is an exchange of energies among the molecules. However, we do not really care whether we follow the same molecule if energy is merely exchanged with another molecule. So it turns out that even if the problem is analyzed more carefully (and it is more difficult, naturally, to do a rigorous job), it still makes no difference in the result. It is interesting that the velocity distribution we have found is just \begin{equation} \label{Eq:I:40:4} n_{> u} \propto e^{-\text{kinetic energy}/kT}. \end{equation} This way of describing the distribution of velocities, by giving the number of molecules that pass a given area with a certain minimum $z$-component, is not the most convenient way of giving the velocity distribution. For instance, inside the gas, one more often wants to know how many molecules are moving with a $z$-component of velocity between two given values, and that, of course, is not directly given by Eq. (40.4). We would like to state our result in the more conventional form, even though what we already have written is quite general. Note that it is not possible to say that any molecule has exactly some stated velocity; none of them has a velocity exactly equal to $1.7962899173$ meters per second.  So in order to make a meaningful statement, we have to ask how many are to be found in some range of velocities. We have to say how many have velocities between $1.796$ and $1.797$, and so on. On mathematical terms, let $f(u)\,du$ be the fraction of all the molecules which have velocities between $u$ and $u + du$ or, what is the same thing (if $du$ is infinitesimal), all that have a velocity $u$ with a range $du$. Figure 40–5 shows a possible form for the function $f(u)$, and the shaded part, of width $du$ and mean height $f(u)$, represents this fraction $f(u)\,du$. That is, the ratio of the shaded area to the total area of the curve is the relative proportion of molecules with velocity $u$ within $du$. If we define $f(u)$ so that the fraction having a velocity in this range is given directly by the shaded area, then the total area must be $100$ percent of them, that is, \begin{equation} \label{Eq:I:40:5} \int_{-\infty}^\infty f(u)\,du = 1. \end{equation} Now we have only to get this distribution by comparing it with the theorem we derived before. First we ask, what is the number of molecules passing through an area per second with a velocity greater than $u$, expressed in terms of $f(u)$? At first we might think it is merely the integral of $\int_u^\infty f(u)\,du$, but it is not, because we want the number that are passing the area per second. The faster ones pass more often, so to speak, than the slower ones, and in order to express how many pass, you have to multiply by the velocity. (We discussed that in the previous chapter when we talked about the number of collisions.)  In a given time $t$ the total number which pass through the surface is all of those which have been able to arrive at the surface, and the number which arrive come from a distance $ut$. So the number of molecules which arrive is not simply the number which are there, but the number that are there per unit volume, multiplied by the distance that they sweep through in racing for the area through which they are supposed to go, and that distance is proportional to $u$. Thus we need the integral of $u$ times $f(u)\,du$, an infinite integral with a lower limit $u$, and this must be the same as we found before, namely $e^{-mu^2/2kT}$, with a proportionality constant which we will get later: \begin{equation} \label{Eq:I:40:6} \int_u^\infty uf(u)\,du = \text{const}\cdot e^{-mu^2/2kT}. \end{equation} Now if we differentiate the integral with respect to $u$, we get the thing that is inside the integral, i.e., the integrand (with a minus sign, since $u$ is the lower limit), and if we differentiate the other side, we get $u$ times the same exponential (and some constants). The $u$’s cancel and we find \begin{equation} \label{Eq:I:40:7} f(u)\,du = Ce^{-mu^2/2kT}\,du. \end{equation} We retain the $du$ on both sides as a reminder that it is a distribution, and it tells what the proportion is for velocity between $u$ and $u + du$. The constant $C$ must be so determined that the integral is unity, according to Eq. (40.5). Now we can  prove1 that \begin{equation*} \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}. \end{equation*} Using this fact, it is easy to find that $C = \sqrt{m/2\pi kT}$. Since velocity and momentum are proportional, we may say that the distribution of momenta is also proportional to $e^{-\text{K.E.}/kT}$ per unit momentum range.  It turns out that this theorem is true in relativity too, if it is in terms of momentum, while if it is in velocity it is not, so it is best to learn it in momentum instead of in velocity: \begin{equation} \label{Eq:I:40:8} f(p)\,dp = Ce^{-\text{K.E.}/kT}\,dp. \end{equation} So we find that the probabilities of different conditions of energy, kinetic and potential, are both given by $e^{-\text{energy}/kT}$, a very easy thing to remember and a rather beautiful proposition. So far we have, of course, only the distribution of the velocities “vertically.” We might want to ask, what is the probability that a molecule is moving in another direction? Of course these distributions are connected, and one can obtain the complete distribution from the one we have, because the complete distribution depends only on the square of the magnitude of the velocity, not upon the $z$-component. It must be something that is independent of direction, and there is only one function involved, the probability of different magnitudes. We have the distribution of the $z$-component, and therefore we can get the distribution of the other components from it. The result is that the probability is still proportional to $e^{-\text{K.E.}/kT}$, but now the kinetic energy involves three parts, $mv_x^2/2$, $mv_y^2/2$, and $mv_z^2/2$, summed in the exponent. Or we can write it as a product: \begin{equation} \begin{aligned} f(v_x,v_y,v_z)&\;dv_x dv_y dv_z\\[.5ex] &\kern{-3.5em}\propto e^{-mv_x^2/2kT}\!\!\cdot e^{-mv_y^2/2kT}\!\!\cdot e^{-mv_z^2/2kT}\,dv_x dv_y dv_z. \end{aligned} \label{Eq:I:40:9} \end{equation} You can see that this formula must be right because, first, it is a function only of $v^2$, as required, and second, the probabilities of various values of $v_z$ obtained by integrating over all $v_x$ and $v_y$ is just (40.7). But this one function (40.9) can do both those things!