Explaining Ohm's Law and Conductivity's constance at particle level

Question

I always wondered since my late high school years, how increasing the voltage can simply be proportional to the flowing current in a resistance.

Given $$\ dU = Eds = \frac{F}{q} ds \implies U \propto v^2 \ ,$$ also the electrical current should be proportional to the speed $v$ of the electrical charges. But this would result in $R$ not being constant but proportional to $v$.

My first instinct was that with growing voltage, electrons would not get faster (because they are always stopped) but the amount of travelling electrons would grow. But I do not think this would be true, since at some moment all (movable) electrons would have to be moved, resulting in no higher current.

Answer 1

First of all, the $E$ one uses in these formulas are macroscopic , and describe accurately only the forces on macroscopically small, yet microscopically large bodies, i.e, a large collection of charge carriers. $J=\sigma E$ is a macroscopic formula (contrary to what high school textbooks like Haliday will tell you).

Next, you are right to expect some slowing down of these electrons, and there certainly are electron-electron collisions (in a rudimentary, classical picture) occuring, and this is accounted for in the derivation of ohm's law, which is not fundamental anyways. Ohm's Law is more of Ohm's specific relation for specific material with some specific considerations. Also keep in mind, this electric field is pretty small (we are inside a conductor, remember?)

In reality, the motion of these electrons are fantastically complicated, with interaction happening with the field, with other electrons, and with other charge carriers.

The electrons have a velocity due to random thermal motion, which is significantly large, but all in all, random. On average, they cancel our to give $\approx 0$. What remains in this average is the small velocity gain in the direction of the field which we call drift velocity.

You wrote down the electrostatic potential, but the velocity we consider here is not the true velocity of each of these particles, this is simply the average velocity we pretend these large collection of charges move with. And also, you are forgetting the important ingredient which will not get accounted for, dissipation. What you claim is only true for a point charge in vaccum in presence of an electric field $E$

Answer 2

Your analysis is based on the false premise that the mobile electrons which are being accelerated by an electric field do not interact with anything.

On a simple model can can think of a copper $(Z=29)$ wire consisting of copper ions with 28 bound electrons vibrating about a fixed position in a lattice and the twenty ninth electron from each copper atom being total free of the atom and moving around the lattice just like gas molecules.
On average the average velocity of the mobile/free elections is zero.

The imposition of an electric field makes those free elections accelerate for a period of time after which they interact (collide) with the lattice ions there is a transfer of kinetic energy (and momentum) between them.
So you now have an assembly of free elections with have a directed velocity due to the electric field superimposed on a random "thermal" velocity.
Thus on average the free electrons move in a direction opposite to the electric field direction at what is called the drift velocity.

An important result is that the resulting drift velocity is proportional to the electric field which in turn means that the voltage (proportional to the electric field) is proportional to the current (proportional to the drift velocity), which is Ohm's law.

The proportionality between the drift velocity and the electric field is explained in an answer to the post Drift velocity in Drude model.

Answer 3

In the semiclassical picture, you are right that the particles would continue to accelerate under a constant potential alone. It is scattering from impurities and lattice motion that makes the particles reach an average velocity.

You added the band theory tag, so I will mention the following. The Fermi surface would translate (and deform if it was allowed to get to extremes) under a voltage. Impurities and phonons (lattice motion) cause the electron states to scatter in the opposite direction. Unbelievably quickly, a steady state is achieve and the Fermi surface is stationary in k-space. Its asymmetric position means there is more net flow of charge in one direction.

Answer 4

I have wondered about this too for years and I take refuge in the knowledge that the laws governing the behavior of the electrons in question are constituitively linear. This means that if we have a solution to an equation, then if we add some constant to the solution we find that it too qualifies as a solution to the equation. This yields a linear relation between the effort and flow variables for a constant value of the resistance variable.

Note that Ohm's law has strict analogs in other systems that have no electrons in them at all. For example, pipes carrying hydraulic fluid moved by pumps obey Ohm's law (as long as the flow speeds are subsonic and the fluid itself is Newtonian).