# Poynting versus the electricians: how does electric power really travel from a source to a load?

Suppose I connect a battery to a lamp in the usual way. Obviously, electric power will go from the battery to the lamp, causing the lamp to light up. But exactly what path does the power take to get there? Does it go through the wires? Does it go through the space around the wires?

I know of two different answers to this question, and they seem to contradict each other. Let me explain...

### The electrician's model

There seems to be a standard way of thinking about how electric power works, and I'm going to refer to this as "the electrician's model."

We pick a particular potential (usually ground potential, but we can pick anything) and use this as our reference potential. The amount of power going through any given wire is determined by the formula $$P = IV$$, where $$I$$ is the current going through the wire, and $$V$$ is the potential of the wire minus the reference potential. If the wire's potential is higher than the reference potential, then the wire is carrying power in the same direction as conventional current; if lower, it's carrying power the opposite way. If the wire's potential is equal to the reference potential, it's not carrying power at all, regardless of how much current it's carrying.

Theoretically, you can choose the reference potential arbitrarily. In the scenario with the battery and the lamp, if you take the negative wire as reference, then all the power goes through the positive wire, and if you take the positive wire as reference, then all the power goes through the negative wire. Or you could even choose a completely different potential, and you'll end up with yet another answer. Fortunately, the amount of power that a wire is carrying doesn't actually have any physical significance, so we're free to choose whatever reference potential makes our job the easiest.

To anyone who's worked with electrical wiring, this model makes perfect sense. It's obviously correct... or is it?

### The Poynting vector model

I recently learned that there's another, completely different answer to the question. Simply take the cross product of the electric field and the magnetic field, and the resulting vector field—the Poynting vector—tells you the amount and direction of power flow at every point in space.

However, this model gives us a seemingly bizarre answer to the question of how power gets from the battery to the lamp. According to this model, most of the power is actually going through the space around the wires, since that's where the electric and magnetic fields are. In fact, if the wires are perfect conductors, then the electric field inside them is zero, so none of the power is going through the wires. I can stick a wooden spoon in the middle of the circuit, and power will travel through the spoon!

This model sounds very weird, but I admit that being weird doesn't mean it's actually wrong.

Here's the problem, though. Let's take the positive wire of the circuit as our reference potential. Then the electrons in the negative wire have a positive amount of energy due to where they're located in space. Since the electrons have energy, and they're moving, they must be carrying power. This seems to prove that the power really is going through the wire. No matter what reference potential we choose, we come to the conclusion that the power is going through the wires, not the space around them. That seems like a contradiction.

So what's going on here? Is one model right and the other one wrong? Are they both simply equally valid ways of accounting for the same power transfer?

(I first thought of this question a while ago, and I've learned more about the topic since then. I'm writing the question from the perspective of myself before I learned about it, and I'll write an answer describing what I know now.)

• What you're calling the electrician's model is really just part of what we usually call the lumped element model. Jun 3, 2023 at 2:27
• There are many simplified models in physics that work very well for limited applications. The charge transfer model works quite well for electricians. For electronics designers... not so much. Jun 3, 2023 at 4:42
• Have you been watching Veritasium? Jun 4, 2023 at 11:51
• @RodrigodeAzevedo I sure have! The video that prompted this question is probably the very one you're thinking of. Jun 4, 2023 at 12:00
• @TannerSwett Please consider appending a note mentioning the motivation for this question. Other questions were motivated by Veritasium. Jun 4, 2023 at 12:08

In short, the Poynting vector model is more accurate. The electrician's model is a useful tool for electrical work, but it has shortcomings. The Poynting vector model is fully compatible with Maxwell's equations and works to explain pretty much every electromagnetic phenomenon.

One problem with the electrician's model is that it fails to explain why it's possible for radio waves to carry energy. Radio waves don't have any current, and even if we treat the displacement current as a real current, it's flowing in the wrong direction! (The displacement current is parallel to the electric field, which is perpendicular to the direction of propagation of the wave.) Besides that, there's no consistent way to assign different electric potentials to different points in a radio wave. So this makes the formula $$P = IV$$ impossible to apply here.

The electrician's model also fails to explain how power can get from one winding of a transformer to the other one, since no current travels between the two windings.

The Poynting vector, on the other hand, explains all of this easily: there's an electric field and a magnetic field, and they're perpendicular to each other, so power flows perpendicular to both of them.

As for this argument:

Here's the problem, though. Let's take the positive wire of the circuit as our reference potential. Then the electrons in the negative wire have a positive amount of energy due to where they're located in space. Since the electrons have energy, and they're moving, they must be carrying power. This seems to prove that the power really is going through the wire. No matter what reference potential we choose, we come to the conclusion that the power is going through the wires, not the space around them. That seems like a contradiction.

This argument doesn't quite hold up. Electrons have electric fields around them, and electric fields contain energy, and so we can say that electrons have energy (in aggregate, not individually). Electrons don't have energy from the electric field and also energy due to where they're located in space; the energy in the electric field is the energy electrons have due to where they're located.

If you want to figure out where the power is flowing, you should look at the electric field and how the electric field is "moving," so to speak. Ultimately, you'll end up with... the Poynting vector.

By the way, there is a way to derive the electrician's model from the Poynting vector model! Simply assume that the potential equals your chosen reference potential at all points in space, except for points that are inside of, or immediately adjacent to, a wire or electrical device. The result will be that, outside of the devices, all nonzero electric fields are concentrated at the surfaces of the wires, and the formulas $$P = IV$$ and $$\mathbf{S} = \mathbf{E} \times \mathbf{H}$$ both give the same answer.

The assumption that the electric field is zero outside of wires is, of course, false. But in a lot of electrical work, the effects of that electric field are negligible, and so it's safe to assume that it doesn't exist.

(To be clear, it's sometimes safe to assume that the electric field doesn't exist. Please don't stand in a lightning storm and assume that its electric field doesn't exist.)

• The E field cannot be zero outside the wires because the voltage source still has to set up a voltage difference for the space around, and you also need the E fields so that, together with the B fields, energy can be transported from the outside space into the wires. There is actually a lot of choices and nuances involved in Poynting's argument. Jun 2, 2023 at 18:28
• @naturallyInconsistent That's true. I've updated this answer to try to clarify what I had in mind; what do you think of it now? Jun 2, 2023 at 19:39
• @naturallyInconsistent In case of DC transmission our engineers are trying to make the E-field along a wire as small as possible since it doesn't serve any purpose. All it does is to cause losses. Wires that have a longitudinal field at high frequencies are usually called "antennas" and they serve many different purposes, including cooking. So it's not an either-or, it's both! Jun 3, 2023 at 4:48

Then the electrons in the negative wire have a positive amount of energy due to where they're located in space. Since the electrons have energy, and they're moving, they must be carrying power.

This isn't quite right. If you want to be more careful about your model, the electrostatic potential energy isn't the energy of one electron or one proton, it's a configuration energy. It's the energy that the whole system has due to the arrangement of all the negative and positive charges in the system.

In a simple system with, say, one fixed positive charge and a nearby mobile negative charge, we do often say that the mobile charge "has" some potential energy, but this is a simplification. Really we should say that the system has energy because of the configuration of the two charges relative to each other. (Similarly we can say that a ball raised above the ground has gravitational potential energy, but really g.p.e. is also a configuration energy that depends on all the massive bodies in the system, not a property of one body alone)

Considering the electrostactic potential energy as a configuration energy you can either calculate it from Coulomb's law, or from integration of the electric field, and you should get the same answer (at least for differences between any two configurations)

Suppose I connect a battery to a lamp in the usual way. Obviously, electric power will go from the battery to the lamp, causing the lamp to light up.

This can be described more precisely. A battery as well as a working generator or a solar cell are source and sink at the same time. If you connect the two poles of such a current source with a metallic conductor, electrons flow between the potential difference at the poles.
The lamp (wanted) as well as the conductor (unwanted) do not let the electrons move unhindered. Everything in our usual environment has an electrical resistance and the power loss is calculated as described by you with P=IV.

But exactly what path does the power take to get there? Does it go through the wires? Does it go through the space around the wires?

The ohmic resistance and the number of moving electrons determine the power loss between source and sink. Obviously, the nature of the material of the conductor plays a role. For an incandescent lamp we use high ohmic material or a small conductor cross-section, for the conductors it is the other way around.
But in both cases the power dissipation consists of electromagnetic radiation. This is generated every time the electrons are accelerated. And - evident from the extremely low drift velocity of the electrons in the conductor - this happens incessantly due to the constant collision or change of direction of the moving electrons. Each time EM quanta are emitted, recognizable by the light emission of the bulb or even only by the heating of the conductor.

In short, the power loss takes place in the conductor. The radiation of the power, however, takes place in an EM spectrum with the following extremes:

1. visible light, which is emitted by the surface electrons.
2. thermal radiation, where the emitted quanta are immediately reabsorbed within the material, dispersing to increasingly long wavelength radiation and emerging as IR radiation.

Point two always occurs (hence the law of thermodynamics that no process is lossless), point 1 requires engineering design (or lightning or some other neat electrostatic charge).

• There is a lot that is wrong here. The electrons in a metal are not in constant collision or changing direction. The quantum description is a lot more crazy; what you are thinking is a classical picture that is doomed to lead nowhere because the real motion of the electron waves do not collide with a perfect lattice at all. Jun 3, 2023 at 5:41

Let us consider a very thin wire into which some extra electrons are being pushed at one end. Electrons move through the wire, and there is some over-pressure of electrons. So this is quite a lot like hydraulics.

So the energy transfer happens by an electron pushing the electron ahead of it, by its electric field. Now, how does the energy go through the field of the electron? I don't know. But energy goes through the field through the space where the field is. So it goes mostly through the space outside of the very thin wire.

Let us now consider a thick wire into which some extra electrons are pushed at one end. The situation on the surface of the wire is the same as in the very thin wire, there is an over pressure of electrons. An electron outside the wire does not want to move to the surface, because of the over pressure. But an electron inside the wire has no problem moving to the surface. So the pressure is the same inside the wire as at the surface of the wire. That is the reason the electrons inside the wire contribute to the energy transport.

Now let us stick some static negative charges on the surface of that same thick wire. That increases the aforementioned over-pressure. So more energy flows now, if the current is unchanged.

The reason the energy flows only outside of the wire is that the effects of the extra electrons exist only outside of the wire. An electron moving inside the wire contributes to the electric field and the energy flow outside the wire.