Veritasium Electricity videos: where does the majority of energy really flow? [duplicate]

Question

After watching Veritasium second video on electricity (references at the end), I have some doubts about where the majority of the energy flow actually happens. The reference experiment is the simple circuit made of a battery, switch and load (resistor).

Feynman Lectures: In Volume II, 27-4, Feynman remarks that:

we must say that we do not know for certain what is the actual location in space of the electromagnetic field energy

which puts into question if the common interpretation that the Poynting vector defines the direction of energy flux (power flow) is indeed correct. It is well-known that Poynting's Theorem is valid only on a closed surface, but Feynman remarks seem to point out that, even in that case, the energy of the EM field cannot be localized. Was Feynman too cautious in his statements?

Initial EM field: Looking at Veritasium thought (and also real, albeit scaled-down) experiment, after the switch is closed, an EM field radiates outward (at the speed of light) from the switch towards the load. This field, while generating a current which is orders of magnitude above leakage current, still generates only a transient (weak) current, which is greatly surpassed by the current in the stationary regime. Thus, it seems safe to say that this transient does not carry the majority of the energy.

Energy flux: Considering real-world conductors (perhaps excluding super-conductors), my question boils down to the following: is the majority of the energy carried by the outward EM field located at the surface (or, to be more precise, extremely near the surface, but still outside) of the conductor? Or is it, instead, being carried by the EM field inside the conductor?

Skin Depth: Since our experiment is in DC steady-state, skin depth should be negligible, but I would be really surprised if it turns out not to be so.

Simple "everyday intuition" seems to suggest that the majority of the energy flow should be inside the conductor (with bigger conductors able to carry more energy), but that is generally not a reliable indicator of scientific accuracy.

P.S. I do not know if the answer to this question is greatly affected by framing it as a classical electrodynamics problem, or as a quantum electrodynamics (QED) problem. Although I studied the basics of quantum mechanics, I didn't study QED, so I would really appreciate if someone with that kind of background can shed light also on this.

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Other questions on Veritasium videos, with explanation as to why they are different:

1. Speed of Light - independent issue, already clarified that there is no violation of causality;
2. 1m/c seconds - irrelevant in my question, which is not about time, but about energy;
3. Alternate explanation for energy flow - related, but doesn't answer my question, which is where the majority of the energy is located (spatial distribution of EM energy), not if (or why) there is EM energy near a wire;
4. Insulator, Switch location, larger gap - variants of Veritasium experiment are proposed and questions about time are asked, again, no energy flux is discussed;
5. Loop of wire - variant of Veritasium experiment, again about time and not about energy;
6. Veritasium vs. Science Asylum - related, but focuses on what happens during the transient, and (again) does not clarify the spatial distribution of EM energy near a wire;
7. Question on electrical currents - again about time;
8. EM field around a conductor with current - question about why we don't get electrocuted by the EM field around a wire;
9. Simple circuit - again about time.

Veritasium videos:

• Veritasium first video: Link

• Veritasium second video: Link

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Answer 1

Most of the energy is carried by the fields outside of the conductors. The fields don't have to be "extremely close" to the condcutors. 50/60 Hz high voltage transmission lines have their wires separted by multiple meters and the field exists in the space between.

When transmitting electric/electromagnetic energy using conductors to guide the fields you find that there is a current in the wires. For high power transmission the fields are large and the corresponding currents are large. Ohmic resistance in the wires converts this current into heat. For high power you need large cables to reduce the ohmic resistance and to handle the heat load.

Feynman's remark doesn't make any sense to me. We can calculate the energy density of the electromagnetic field at every point in space. It has a component proportional to $E^2$ and a component proportional to $B^2$. Where the fields are larger the energy density is larger. Pretty straightforward to identify where the energy is located...

This question has nothing at all to do with quantum mechanics or quantum electrodynamics.

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edit addressing points in comments below:

Feynman's equations in Feynman Lectures Vol II 27-3 read

\begin{align} u =& \frac{\epsilon_0}{2} |\mathbf{E}|^2 + \frac{\epsilon_0 c^2}{2}|\mathbf{B}|^2\tag{27.14}\\ \textbf{S} =& \epsilon_0 c^2 \mathbf{E}\times\mathbf{B} \tag{27.15} \end{align}

I take $u$ as defined here to represent the electromagnetic energy. In 27-4 Feynman explains that this definition of $u$ and $\mathbf{S}$ is a sufficienty definition for the electromagnetic energy and energy flow (Poynting vector) because it satisfies

$$ -\frac{du}{dt} = \mathbf{\nabla}\cdot\mathbf{S} + \mathbf{E}\cdot\mathbf{j} \tag{27.5} $$

meaning essentially that the sum of electromagnetic energy $u$, and "matter" energy are conserved together. In Sec 27-4 Feynman is also pointing out that it is possible that there are other definitions for electromagnetic energy and energy flow such as $\tilde{u}$ and $\tilde{\mathbf{S}}$ that would also satify 27.5 and therefore also be valid candidates for what we call energy and energy flow.

That said, because 27.14 and 27.15 are sufficient definitions for electromagnetic energy and energy flow I will simply takes those as THE definitions for the rest of this post. And of course these are the definitions taken anywhere in any related literature so there shouldn't be much issue here.

In the comments the OP clarifies that the question is largely about where (in 3D space) is the electromagnetic energy located for a DC circuit in DC steady state. I want to clarify that the answer here is DIFFERENT than the answer for the transient behavior that occurs immediately after the switch is turned on. This clarification is important because the Veritasium video and the controversy surrounding it are largely focused on this transient time period.

For a circuit in DC steady state there will be electric field lines beginning on the positive voltage side of the circuit (i.e. the positive terminal of the battery and all of the conductor between the positive battery terminal and the bulb positive terminal) and ending on the negative side of the circuit (i.e. the negative terminal of the battery and all of the conductor between the negative battery terminal and the bulb negative terminal). These field lines span from the positive side to the negative side, spanning the space between the two sides.

Note that in DC steady state there is no electric field inside the conductor. These means the field lines terminate on the surface of the conducting wire. Gauss's law then tells us there must be some surface charge on the wires. This is explained some in the Veritasium video.

In addition, the current flowing through the meat of the conducting wires will create magnetic fields curling around the wires.

Where then is the energy? According to 27.14 the energy is located anywhere where the $\mathbf{E}$ and $\mathbf{B}$ fields have non-zero support. The energy density will be greatest closest to the wires because that's where the fields are the biggest. In addition, some thought reveals that there is a non-zero Poynting vector in the space outside the wires which carries energy from the battery to the bulb. The Poynting vector will also be biggest where the fields are biggest, so close to the conductors.

All of that said, there seems to be discussion in the OP and the comments below about "the majority of the energy being close to the wires". Whether this is true or not depends on the geometry of the circuit. If circuit is a big looping circle or square then yes, the fields in the "center" of the circuit will be small compared to the fields at the wires and would be correct to say most of the energy is flowing along the wires. But if the circuit is long and skinny with the postive and negative voltage wires running parallel and very close to each other the field might be just as large in between the wires as it is very close to the wire surfaces. In this case it might be more appropriate to say the energy is flowing in the space between the wires, but "guided" by the wires nonetheless.

This latter case, when the fields are just as large in between the wires as they are near the wires is essentially very similar to an electromagnetic waveguide.

Answer 2

I’m going to carefully choose just one of the issues you raise, which I think is relevant to your entire question:

Initial EM field: Looking at Veritasium thought (and also real, albeit scaled-down) experiment, after the switch is closed, an EM field radiates outward (at the speed of light) from the switch towards the load. This field, while generating a current which is orders of magnitude above leakage current, still generates only a transient (weak) current, which is greatly surpassed by the current in the stationary regime. Thus, it seems safe to say that this transient does not carry the majority of the energy.

The initial (“transient”) current is only weak because of the impedance mismatch at the end of the transmission line. You can get a hint of this if you watch starting at 12:00, where a simulation of the electric and magnetic fields in the circuit shows the switch-is-closed signal propagating down the transmission line, reflecting off of the ends, and then meeting again in the middle. If you watch for two minutes, you’ll see a simulation showing the “circuit,” where the two long wires are connected at the far ends, next to an “antenna,” which is exactly the same as the other long wire except its distant ends are free. The responses are almost exactly the same before the signal reflects from the ends:

A few minutes later, the video derives the “transmission line” model, and then discusses matching the impedance in order to produce the most power during the transient. The transmission line in either direction is measured to have an impedance of about 550Ω, so a load of 1100Ω is used.

Let’s think about why we get a reflection off of the ends. If the circuit is closed, like the loop in the screencapture above, the potential difference at the end (“terminus”) of the transmission line is constrained to be exactly zero. The electromagnetic field conspires to maintain this zero voltage by producing a reflected wave whose electric field is the opposite of the electric field of the incoming wave. Once the reflections have met in the middle, then all of the voltage drop is across the central load, whose current then increases.

Essentially, the circuit has a different set of impedances during the lightspeed travel time to the ends:

 /-----------------(1100Ω)-------------------\
 |                                           | 
 (550Ω)                                  (550Ω) (dynamic impedance)
 |                                           |
 \ -----------------(switch) ----------------/

versus after all of the reflections have died down:

 /-----------------(1100Ω)-------------------\
 |                                           | 
 (0Ω)                                     (0Ω) (steady state)
 |                                           |
 \ -----------------(switch) ----------------/

In the open-circuit case, the “antenna” in the screencapture above, the condition at the end of the transmission line is that the current through the terminus is exactly zero. For the open-ended transmission line, the electric field in the reflected wave has the same sign as the incoming wave, because no charge flows across the gap. The steady-state equivalent circuit is

 /-----------------(1100Ω)-------------------\
 |                                           | 
 (∞ Ω)                                    (∞ Ω) (steady state)
 |                                           |
 \ -----------------(switch) ----------------/

which makes it clear why no current flows in the long-time limit.

You might ask whether there is some special way to build your transmission line, intermediate between the shorted terminus and the open terminus, so that the opposite-sign and same-sign reflected waves cancel out. The answer is that you add a “terminating resistor,” whose resistance is the same as the impedance of the transmission line, so that

 /-----------------(1100Ω)-------------------\
 |                                           | 
 (550Ω)                                  (550Ω) (dynamic AND static impedance)
 |                                           |
 \ -----------------(switch) ----------------/

is the equivalent circuit diagram the entire time.

That is to say, if the transmission line in the video had been properly terminated, you would have seen only the “transient” current, no matter how long the cables in either direction had been. Your statement that

Thus, it seems safe to say that this transient does not carry the majority of the energy.

is correct in this special case, because the transmission line has been designed to have its impedance get smaller over time.

Note that the quantitative results in the video, with a 4V prompt voltage evolving into a 18V steady-state voltage, aren’t consistent with the impedance ratios I’ve quoted in this answer. There may be a reason for that which I’ll suddenly think of in the shower tomorrow. Or, it may be because measuring the inductance and capacitance of this clothesline of a circuit is a mess, and the setup isn’t impedance-matched as well as they had hoped. I did a number of experiments very similar to this one, with industry-standard 50Ω cables, when I was learning how to operate fast electronics, and got gorgeous results. If you have coaxial cable in your house connected to your television or internet, it’s probably a 600Ω cable.

Answer 3

In the case of a steady DC circuit, with a battery and a resistor, there is a difference of potential between the wires from batt+ to resistor and batt- to resistor. As for static conditions, the electric field is minus the gradient of the potential, there is an electric field between those 2 wires. And of course magnetic field due to the current.

It is easy to check using the right hand rule that the Poynting vector is directed from battery to the resistor around both wires. The geometry is too complicated here to perform a calculation, but theoretically we must get P = VI by integrating the Poynting vector crossing an infinite transverse area separating the battery from the resistor. $S = E \times B$ is greater close to the wires because $B$ is more stronger there. The Wikipedia has a much better example for the simpler geometry of a coaxial cable, where the issue is treated quantitatively.

It is misleading however to interpret this flow of energy as something travelling by the air (or vacuum, or any dieletric). The electric and the magnetic field are static, so nothing is travelling out of the wires, (no waves to be specific).