Why is $\vec j\cdot \vec e$ the joule dissipation?

Question

I always see $\vec j\cdot \vec e$ as Joule's dissipation and I don't understand why. For example, if we have a uniform electric field $\vec e=e_o\vec u_x$ and we release an electron in it, it will start moving, accelerating in the direction of $-\vec u_x$, so the we will have a current $\vec j$, and this product $\vec j\cdot\vec e$ will exist, but I don't see that any energy is being dissipated in the form of heat there. What is going on?

I've seen this essentially in Optics texts, when introducing Poynting's theorem.

Answer 1

Joule dissipation (equivalently, Ohmic heating) is a statistical process. It doesn't occur at the microscopic scale, and a single travelling electron in an electrical field is certainly microscopic. Resistivity depends on particle collisions, which turns translational kinetic energy into thermal kinetic energy.

Ohm's law relates the current $\vec{j}$ in a medium to the electric field $\vec{e}$ via the conductivity $\sigma$, $$ \vec{j}=\sigma \vec{e} $$

Another way of looking at this is that dissipation $P$ is given by $$P=\frac{j^2}{\sigma}$$

Written this way, we have an expression very similar to that for Ohmic heating in a resistor, namely that

$$P=I^2 R$$ Since the conductivity is the inverse of the resistivity, this is an intuitive result.

The advantage of Joule's expression is that it allows for the determination of Joule heating at a particular point in space, rather than over the entire resistor (whatever that may be). This is useful in plasma physics, for example, where it may be the case that Joule heating is localized, rather than uniform throughout the plasma. Nonetheless, it is predicated on the medium being strongly collisional, and still must refer to a macroscopic, rather than microscopic effect.

Answer 2

Imagine a cross sectional area of the conductor that carries the current density $\vec{J}$, and take a thin slice, $\Delta x$, of the conductor. The electric field is along the conductor and at 90$^o$ to its cross sectional area, A. Then $\vec{J}$ represents the number of electrons, per second and per unit area, crossing the cross-section of the conductor. The number of electrons crossing per unit area and per second is numerically equal to $n_e=|\vec{J}|/e$. These electrons are “pushed” by the field through the slice $\Delta x$, while the field is

$E=\frac{\Delta U}{\Delta x}$

(dropped the vector notation since we are talking about a 1-D motion in the conductor). Then equation $\vec{J}.\vec{E}$ can be written (without the vector notation) as

$J\times E=\frac{I}{A}\frac{\Delta U}{\Delta x}$

or

$J\times E=\frac{I\Delta U}{A\Delta x}$

The last part of the equation shows that $\vec{J}.\vec{E}$ gives the amount of energy spent by the field (the battery), per second and per unit volume of conductor, to push all these electrons through the slice of area A and thickness $\Delta x$, being at the potential difference $\Delta U$.