# How did Feynman derive this formula for the current of particles in a gas?

Feynman Lectures. Lecture 43. Lecture Summary. The setting is that we are considering molecular motion in a gas near equilibrium.

$\text{The current of particles / area sec.} = - \frac{1}{3}lv\frac{dn_a}{dx} = - \mu kT\frac{dn_a}{dx} = - D\frac{d}{dx}(\frac{n_a}{n_0}).$

Here, $D$ is the diffusion constant, $\mu$ the mobility, $k$ the Boltzmann constant, $T$ the temperature,$l$ the mean free path, $v$ the actual molecular velocity (mean), and $n_a$ as a function $n_a(x,y,z)$ the number density of special molecules in a small volume element centered on $(x,y,z)$. $n_0$ is an unknown constant, variable or function, probably the total number density (special + background molecules in unit volume) where a special molecule responds to the field and a background molecule doesn't respond on the field.

Can you show how he got $\frac{n_a}{n_0}$? In Lecture 43 he wrote $D = \tfrac{1}{3}lv$, $D = \mu kT$, but never $D = \mu kT \cdot n_0$.

At the beginning of the Lecture 43 there is the photo of a blackboard, on which Feynman was writing the summary (http://www.feynmanlectures.caltech.edu/img/FLP_I/f43-00/f43-00.jpg but you can browse the image directly only after going to feynmanlectures.caltech.edu/I_43.html). On the blackboard in 6th line (also in Ch. 43-2) Feynman writes that $n_0$ means the total number of particles per unit volume. But in Ch. 43-5 there is formula $n_a = n_0e^{-U/kT}$. This is the exponential probability multiplicator and $n_0$ is the density at zero energy when the field is present (according to Lecture 40, ch. 40-1). So in the context of Ch. 43-5 $n_0$ means the number of special particles per unit volume at zero energy. I assume that there must be absolutely different number density of "background" particles (Feynman's term). Actually I am not sure what Feynman means by $n_0$ in different cases. Besides, the formula in the 9th line looks strange, because there was no $n_0$ at first.

You are probably overthinking this and it is probably an error. Note that there is a very generic "transport equation" $$\frac{\partial\rho}{\partial t} + (\vec v\cdot\nabla)\rho = -\nabla\cdot J + \Phi,$$ which says, if you've never seen it before, "a box of volume $dV$ flows downstream in a fluid moving with velocity field $\vec v$, the time rate of change of stuff $\rho~dV$ in the box is equal to the divergence of the flow of the stuff in general $J$ plus the rate $\Phi$ at which stuff is magically entering the box out of nowhere".
The diffusion constants $D$ occur in this expression usually due to something like $J = -D~\nabla \rho$ and hence for constant $D$ they have the rough pattern $\dot \rho = D~\nabla^2 \rho,$ meaning that they have a sort of universal pattern of units, $[[D]] = \text{m}^2/\text s.$ While Feynman makes things confusing with his "one unit of area" this is probably also what he intends for the ultimate units of $D.$ Whenever you see this pattern of units you should wonder at whether it has some relationship to transport; the most famous case is that the kinematic viscosity $\nu/\rho$ happens to have these units and turns out to be a diffusion constant $D$ for momentum, which of course is conserved and therefore qualifies as a "stuff" for the transport equation to work on.
Now: given $n_a = n_0 e^{-U/kT}$ we have that $e^\text{anything}$ is dimensionless and hence $n_a/n_0$ is dimensionless, hence the SI units of the right-hand side are $(\text{m}^2/\text s)/\text m = \text m / \text s.$ This is inconsistent with his claim of it having units $\text{something}/(\text m^2 ~\text s)$ and is therefore dimensionally incorrect. Probably the $n_0$ is in error and he just meant to reprise the equation $-D~\frac{dn_a}{dx}$ or some such.
• Thank you. I saw some errors in summaries of previous lectures. But here he wrote $\tfrac{d}{dx}$, so he remembered about $dx$. It's not like a mistake made through lack of attention. – Username160611000000 Jan 12 '17 at 17:55