# Energy dissipation in current flow

In section 4.8, Energy dissipation in current flow, of Purcell and Morin's Electricity in Magnetism, the expression for the power expended by a resistor is derived. The sections includes the following discussion:

The energy thus expended shows up eventually as heat. In our model of ionic conduction, the way this comes about is quite clear. The ion acquires some extra kinetic energy, as well as momentum, between collisions. A collision, or at most a few collisions, redirects its momentum at random but does not necessarily restore the kinetic energy to normal. For that to happen the ion has to transfer kinetic energy to the obstacle that deflects it. Suppose the charge carrier has a considerably smaller mass than the neutral atom it collides with. The average transfer of kinetic energy is small when a billiard ball collides with a bowling ball. Therefore the ion (billiard ball) will continue to accumulate extra energy until its average kinetic energy is so high that its average loss of energy in a collision equals the amount gained between collisions. In this way, by first “heating up” the charge carriers themselves, the work done by the electrical force driving the charge carriers is eventually passed on to the rest of the medium as random kinetic energy, or heat.

I may be missing out on something elementary but I'm unable to understand the concepts in the paragraph related to energy transfer in the collisions.

• Which line(s) in the excerpt did specifically bother you? – user36790 Jul 12 '16 at 6:30
• "A collision, or at most a few collisions, redirects its momentum at random but does not necessarily restore the kinetic energy to normal. For that to happen the ion has to transfer kinetic energy to the obstacle that deflects it," and "herefore the ion (billiard ball) will continue to accumulate extra energy until its average kinetic energy is so high that its average loss of energy in a collision equals the amount gained between collisions. In this way, by first “heating up” the charge carriers themselves," till the end. Running out of space to copy and paste. – Junaid Aftab Jul 12 '16 at 6:38
• I would say it's pretty poorly written. The electric field doesn't heat the charge carriers. They pick up kinetic energy as they lose potential energy while moving in the field. That is a dynamic and not a thermodynamic process. The thermalization due to collisions with the lattice, on the other hand, is a thermodynamic process. – CuriousOne Jul 12 '16 at 6:45
• Could you shed some light on the dynamical argument the textbook is making, with regards to (in our setup) how does the kinetic energy transfer in between collisions? – Junaid Aftab Jul 12 '16 at 6:51
• @CuriousOne I disagree with you on this one. "heating up" is in quotes. It's a decent description of the microscopic process, which I admit gets a bit frayed when addressing the approach to equilibrium. In any event, the OP is concerned about energy transfer in collisions. I suggest he work out the one-dimensional collision problem and study the extreme cases ... one object much more massive than the other, vice versa, and equal masses. – garyp Jul 12 '16 at 13:38

I think the point made is that the kinetic energy of the electrons (or other charge carrier) will normally be far higher than $kT$. That's because although collisions with the lattice are frequent the electron loses very little energy with each collision.
The point being made is that in a collision between a light object and a heavy one very little of the kinetic energy is transferred to the heavy object. If you bounce a ping pong ball off a battleship the ping pong ball bounces back with almost the same energy it had before the collision. That means the kinetic energy of the electrons keeps growing and growing until it gets big enough for the KE lost in each collision to balance the KE gained from the electric potential. At this point the KE of the electrons is a lot greater than $kT$ so the energy transferred to the lattice is greater than $kT$ and the lattice heats up.