condensed matter - Do electrons move around a circuit?

We could imagine a simple electronic circuit composed by a power source and a resistor.

It is usual find descriptions as "The moving charged particles in an electric current are called charge carriers. In metals, one or more electrons from each atom are loosely bound to the atom, and can move freely about within the metal. These conduction electrons are the charge carriers in metal conductors." (wikipedia "electric current"). That made me imagine electrons going from power -, running by the wires, crossing the resistor and returning to power +, where they suffer a new kick to power -. All that at near the speed of light.

In this model, it is easy to define "electrical current intensity" as proportional to the number of electrons per second that cross one specific point of the circuit.

However, looking for the physics of resistivity, I can read the phrase "The actual drift velocity of electrons is very small, in the order of magnitude of a meter per hour." (wikipedia "Electrical resistivity and conductivity"). This is a complete different point of view, where electrons are more or less fixed around a fixed position (or moving very slowly), an electricity is a wave of "collisions" or "jumps" of electrons, like in a Newton's cradle (or "do the wave" as public in a football match). Obviously, we do not need severals hours from the switch on to the light on.

So, the question is, do electrons move around a circuit ? References to band theory are welcome.

A related question is: if electrons doesn't moves along the circuit, what is "electrical current intensity"?

Note: I known there are similar questions already posted, but answers of the ones I've read are not very detailed (could be I've not found the best ones). Please, be a few patient before "close vote".

Answer

Your confusion stems from a fundamental misunderstanding about drift velocity. Drift velocity is not the average speed of electron motion, but instead is the average velocity vector. The average speed of free electron motion in a metal can be approximated to be the Fermi speed

$$v_F = \sqrt{\frac{2E_F}{m_e}}$$

where $E_F$ is the Fermi energy. This is incredibly fast - inserting $E_F=10$ eV gives a result that is well over $1000$ km/s.

These electrons are traveling in a solid, though, which is rife with objects to collide with, including other electrons. Therefore, the mean free path of electrons in a metal (i.e. the distance an electron travels until it collides) is typically less than $1$ nm. Therefore, these electrons almost instantaneously collide with something else. A large number of these collisions would serve to essentially randomize the direction of travel of any given electron. When you add a bunch of uniformly-randomly-distributed vectors of roughly equal length together, the resultant is essentially zero, regardless of the actual length of the vectors you added. Therefore, the average velocity vector of an electron should be close to zero, and certainly should be much smaller than its average speed, since its velocity is pointed in an essentially random direction.

When an electric field is applied to a metal, it accelerates electrons in a certain direction, and therefore alters the probability distribution of electron velocity. Velocities in the direction of the field become less probable, and velocities against the direction of the field become more probable. The longer the electric field is allowed to act on a freely-moving electron, the more this probability distribution is distorted. But, as was previously discussed, the time between collisions is quite small due to the density of metals. This means that the electric field can only alter the velocity distribution slightly, which shifts the average velocity vector (i.e. the drift velocity) slightly away from zero.

Another misunderstanding arises from the false assumption that the speed of an electrical signal in a metal is equal to the either the Fermi speed or the drift velocity. In reality, it is unrelated to either of those things. Instead, for conductors the speed of an electrical signal is given by the group velocity of an electromagnetic wave:

$$v_g=\frac{d\omega}{dk}$$

where $\omega(k)$ is the dispersion relation, and is in general derived from the band structure of the material in question. For a good (i.e. close-to-ideal) conductor, the dispersion relation is

$$\omega(k)=\frac{2k^2}{\mu\sigma}$$

for a material with conductivity $\sigma$ and permeability $\mu$. Then the group velocity is

$$v_g=\sqrt{\frac{8\omega}{\mu\sigma}}$$

which, for copper, with $\sigma= 5.96\times 10^7$ S/m and $\mu\approx\mu_0=4\pi\times 10^{-7}$ H/m, and for a plane wave with frequency $1$ GHz, the group velocity is roughly $25$ km/s, and increases with increasing frequency.

EDIT:

The conductivity of a material $\sigma$ is defined by

$$\mathbf{J}=\sigma \mathbf{E}$$

for current density $\mathbf{J}$ and applied electric field $\mathbf{E}$. This essentially means that it's the average number of electrons passing through a unit area per unit time, per unit applied electric field. The higher the conductivity, the less electric field it takes to get electrons to flow. One simple model (specifically, the Drude model) based on similar arguments as above finds that for a material with electron density $n$ and mean time between collisions $\tau$, for DC currents one has

$$\sigma = \frac{ne^2\tau}{m_e}$$

Resistivity $(\rho)$ is defined as the inverse of conductivity. Therefore, again from the Drude model for DC currents, one has

$$\rho = \frac{m_e}{ne^2\tau}.$$