# When the energy in a conductor is not carried by the electrons, how resistances warm up?

The energy flux in a conductor paints the picture that the "collisions" of the electrons with atoms (Drude Model) are not the reason for an resistance warming up in the presence of an electric current, but rather through properties of the surface charge distribution. Although of course any magnetic field (in this case) is influenced by the amount of electrons flowing.

Is this the case?

• This question, casually formulated, perhaps even accidentally, hits the nail right on the head. Commented Feb 29 at 20:22
• Thank you very much. I am just a Software Engineer in an Electrical Engineering Company trying to learn more about electromagnetism, which makes my fundamental knowledge weak at best. Thats why my questions need to be light in math and I try to be concise. Commented Feb 29 at 20:29
• You should read Feynmans discussion of the charging of a capacitor in FLII, Ch 27.4. feynmanlectures.caltech.edu/II_27.html. He reaches the conclusion that 'our “crazy” theory' states that the energy comes from space, not through the wires. This is the rather absurd consequence of gauge invariance. Commented Feb 29 at 20:39
• I will do, but this read will really stretch my understanding. Maybe someone can wrap up his/her understanding for me. Commented Feb 29 at 20:44
• Feynman's lectures are entry level, yet deep and fundamental. So he's one of the people that you are looking for. Commented Feb 29 at 21:11

You need more than just the Poynting vector, you need all of Poynting’s theorem$$\frac{du}{dt}+\nabla\cdot \vec S + \vec J\cdot \vec E =0$$

The Poynting vector describes the flow of energy from one location to another. But energy that simply flows through does not produce any heat. In order for the energy to heat the resistor, it must flow in and then dissipate. The term $$\vec J\cdot \vec E$$ describes that dissipation.

Taken together, Poynting’s theorem says that the fields which are produced by the surface charges result in a flux of energy through space, and then the current dissipates that energy from the fields into the matter.

• But this would mean that you can get rid of the mass of the charges in this model? Is it still the case that kintetic energy of the impacting charge is used for the energy warming up the resistance? This would also mean that the formula above would not be sufficient to calculate all energy flux vectors which result by the collisions? Which would lead to a new question for another time: How does this poynting vector if it exists looka for collision of just one particle with another... Commented Feb 29 at 21:59
• @Niclas indeed, the mass of the charge carrier is not relevant. This is a classical equation and does not involve electrons or any other quantum mechanical entity. The charge carriers could be electrons in a wire, positive ions in an electrolyte, or even holes in a semiconductor. Regardless of the nature of the charge carrier the energy transferred to the matter is $\vec E \cdot \vec J$
– Dale
Commented Feb 29 at 22:07
• OK, but what is the microscopic explanation for "and then the current dissipates that energy from the fields into the matter". Commented Mar 1 at 7:40
• @VladimirFГероямслава In the Drude model, it's collisions of the electrons with obstacles. Not a bad model. "All models are wrong, but some are useful". For a more sophisticated model, as I said in a comment on my answer, the quantum condensed matter rabbit hole is very deep. Commented Mar 1 at 12:13
• @VladimirFГероямслава I agree with JohnDoty. Your preferred QM model explains the $\vec E \cdot \vec J$ term to whatever degree of sophistication and pain you tolerate. The whole point of my answer is that it does not need to “reconcile” with the Poynting vector in any way
– Dale
Commented Mar 1 at 12:45

In the Poynting model, the current produces a magnetic field that, in part, guides the energy flow. In the Drude model, the current is composed of electrons, whose collisions with something (lattice defects, phonons, ...) dissipate energy. The potential that drives that dissipation gives rise to an electric field, which produces a Poynting flux toward the place the dissipation is occurring.

• @Niclas That's the model. Don't take it too seriously. The quantum condensed matter rabbit hole is very deep... Commented Feb 29 at 21:37
• So basically and aware of all simplifications by not applying QM the electrons do not carry the energy all the time but only in the few moments when they get accelerated to their final impact speed? Commented Feb 29 at 21:48
• @Niclas What problem are you trying to solve? It's rarely useful to think about electrons in conductive wires at all. In quantum models, electrons in condensed matter are not the same thing as electrons in vacuum. Normally, we just figure this stuff out with current and resistance, agnostic of the microscopic physics. "All models are wrong, but some are useful." -- George Box Commented Feb 29 at 21:56
• There is no problem to solve. I am just trying to stretch my understanding of how the world works. It is pure interest. Commented Feb 29 at 22:01
• Well, I would argue most physicist have not studied physics because they had a problem to solve. I guess you wonder about useless things too from time to time :) Commented Feb 29 at 22:12