The drift speed
We want to describe what happens to a molecule, or several molecules, which are different in some way from the large majority of the molecules in a gas. We shall refer to the “majority” molecules as the “background” molecules, and we shall call the molecules which are different from the background molecules “special” molecules or, for short, the $S$-molecules. A molecule could be special for any number of reasons: It might be heavier than the background molecules. It might be a different chemical. It might have an electric charge—i.e., be an ion in a background of uncharged molecules. Because of their different masses or charges the $S$-molecules may have forces on them which are different from the forces on the background molecules. By considering what happens to these $S$-molecules we can understand the basic effects which come into play in a similar way in many different phenomena. To list a few: the diffusion of gases, electric currents in batteries, sedimentation, centrifugal separation, etc.  We begin by concentrating on the basic process: an $S$-molecule in a background gas is acted on by some specific force $\FLPF$ (which might be, e.g., gravitational or electrical) and in addition by the not-so-specific forces due to collisions with the background molecules. We would like to describe the general behavior of the $S$-molecule. What happens to it, in detail, is that it darts around hither and yon as it collides over and over again with other molecules. But if we watch it carefully we see that it does make some net progress in the direction of the force $\FLPF$. We say that there is a drift, superposed on its random motion. We would like to know what the speed of its drift is—its drift velocity—due to the force $\FLPF$. If we start to observe an $S$-molecule at some instant we may expect that it is somewhere between two collisions. In addition to the velocity it was left with after its last collision it is picking up some velocity component due to the force $\FLPF$. In a short time (on the average, in a time $\tau$) it will experience a collision and start out on a new piece of its trajectory. It will have a new starting velocity, but the same acceleration from $\FLPF$. To keep things simple for the moment, we shall suppose that after each collision our $S$-molecule gets a completely “fresh” start. That is, that it keeps no remembrance of its past acceleration by $\FLPF$. This might be a reasonable assumption if our $S$-molecule were much lighter than the background molecules, but it is certainly not valid in general. We shall discuss later an improved assumption. For the moment, then, our assumption is that the $S$-molecule leaves each collision with a velocity which may be in any direction with equal likelihood. The starting velocity will take it equally in all directions and will not contribute to any net motion, so we shall not worry further about its initial velocity after a collision. In addition to its random motion, each $S$-molecule will have, at any moment, an additional velocity in the direction of the force $\FLPF$, which it has picked up since its last collision. What is the average value of this part of the velocity? It is just the acceleration $\FLPF/m$ (where $m$ is the mass of the $S$-molecule) times the average time since the last collision. Now the average time since the last collision must be the same as the average time until the next collision, which we have called $\tau$, above. The average velocity from $\FLPF$, of course, is just what is called the drift velocity, so we have the relation \begin{equation} \label{Eq:I:43:13} v_{\text{drift}} = \frac{F\tau}{m}. \end{equation} This basic relation is the heart of our subject. There may be some complication in determining what $\tau$ is, but the basic process is defined by Eq. (43.13). You will notice that the drift velocity is proportional to the force. There is, unfortunately, no generally used name for the constant of proportionality. Different names have been used for each different kind of force. If in an electrical problem the force is written as the charge times the electric field, $\FLPF = q\FLPE$, then the constant of proportionality between the velocity and the electric field $\FLPE$ is usually called the “mobility.” In spite of the possibility of some confusion, we shall use the term mobility for the ratio of the drift velocity to the force for any force. We write \begin{equation} \label{Eq:I:43:14} v_{\text{drift}} = \mu F \end{equation} in general, and we shall call $\mu$ the mobility. We have from Eq. (43.13) that \begin{equation} \label{Eq:I:43:15} \mu = \tau/m. \end{equation} The mobility is proportional to the mean time between collisions (there are fewer collisions to slow it down) and inversely proportional to the mass (more inertia means less speed picked up between collisions). To get the correct numerical coefficient in Eq. (43.13), which is correct as given, takes some care. Without intending to confuse, we should still point out that the arguments have a subtlety which can be appreciated only by a careful and detailed study. To illustrate that there are difficulties, in spite of appearances, we shall make over again the argument which led to Eq. (43.13) in a reasonable but erroneous way (and the way one will find in many textbooks!). We might have said: The mean time between collisions is $\tau$. After a collision the particle starts out with a random velocity, but it picks up an additional velocity between collisions, which is equal to the acceleration times the time. Since it takes the time $\tau$ to arrive at the next collision it gets there with the velocity $(F/m)\tau$. At the beginning of the collision it had zero velocity. So between the two collisions it has, on the average, a velocity one-half of the final velocity, so the mean drift velocity is $\tfrac{1}{2}F\tau/m$. (Wrong!)  This result is wrong and the result in Eq. (43.13) is right, although the arguments may sound equally satisfactory. The reason the second result is wrong is somewhat subtle, and has to do with the following: The argument is made as though all collisions were separated by the mean time $\tau$. The fact is that some times are shorter and others are longer than the mean. Short times occur more often but make less contribution to the drift velocity because they have less chance “to really get going.” If one takes proper account of the distribution of free times between collisions, one can show that there should not be the factor $\tfrac{1}{2}$ that was obtained from the second argument. The error was made in trying to relate by a simple argument the average final velocity to the average velocity itself. This relationship is not simple, so it is best to concentrate on what is wanted: the average velocity itself. The first argument we gave determines the average velocity directly—and correctly! But we can perhaps see now why we shall not in general try to get all of the correct numerical coefficients in our elementary derivations! We return now to our simplifying assumption that each collision knocks out all memory of the past motion—that a fresh start is made after each collision. Suppose our $S$-molecule is a heavy object in a background of lighter molecules. Then our $S$-molecule will not lose its “forward” momentum in each collision. It would take several collisions before its motion was “randomized” again. We should assume, instead, that at each collision—in each time $\tau$ on the average—it loses a certain fraction of its momentum. We shall not work out the details, but just state that the result is equivalent to replacing $\tau$, the average collision time, by a new—and longer—$\tau$ which corresponds to the average “forgetting time,” i.e., the average time to forget its forward momentum. With such an interpretation of $\tau$ we can use our formula (43.15) for situations which are not quite as simple as we first assumed.