Ohm's law for stones

Question

Suppose I drop stones from a height. Now, we define quantity analogous to electric current called mass current. Now, to get an analogue to Ohm's law, we need drift velocity for stones. For electrons, it's the effective evlocity with which they move forward. For stones, terminal velocity isn't exactly that but it'll do what's required. We already have an analogue to electric potential. Now, we have everything required to get another Ohm's law as: $$V=R_0I_m$$

$I_m=nmAv$, where $m$ is the mass of one stone, $v$ is terminal velocity, $A$ is the area of our imaginarg wire, $n$ is the number of stones per unit volume. It completely works fine. Just like all the potential drop of electrons is converted into vibrations of nuclei because the effective velocity of electrons can't go beyound $v_d$ so that no increase in KE is allowed, similary for stones, all the gravitational potential drop must go to atmospheric heat because no increase in KE is allowed beyound termial velocity.

My question is: What the constant of proportionality $R_0$ here could mean? I know it should be some measure of atmospheric friction, that's why I expected it to be of the dimensions of coefficient of viscocity, but it isn't. How can two quantities of different dimensions be the measure of the same thing? How can it be related to the coefficient of viscocity? EDIT: Oh, I'm sorry. In fact, $\rho_0=R_0\frac{A}{l}$ should be a property of the atmosphere. So, how does it relate to coefficient of viscocity?

Answer 1

The thought-experiment setup dictates a constant flux of objects, which is named in the equation as $I_m$. But there's no rationale on why it should be a constant or how it's determined. So the answer is that $R_0$ is equally arbitrary. It signifies whatever physical forces bring the system to an equilibrium.

Note that the relation is trivial and $R_0$ is unconstrained unless the stones on the bottom can somehow push upward on those being released at the top. Current and flux are conserved even with no resistance, whether it's stones being dropped on the Moon or electrons being accelerated in a cathode ray tube.

So, for the given setup, $R_0$ is merely the gravitational potential (height) from which you're dropping the stones. The atmosphere doesn't enter into it. The terminal velocity term $v$ is cancelled when it's multiplied by $n$, since the density is inversely proportional to speed when quantity is conserved (think of road traffic).

That's not to say that you can't write $R_0$ and possibly use it for reasoning. Resistance as a derivative of energy-level with respect to current is often used in circuit theory, even negative resistance.