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Consider the following setup.

enter image description here

The answer is $(3)$ - spoiler alert - and it's explained here, using Poynting vector. The thing is though

How can the electromagnetic energy flux even reach the light bulb?

In this picture it's perfectly clear how the energy flux (blue arrows) flows enter image description here

Around the resistor there are both an electric field and a magnetic field so the energy flux, given by all contributions $$ {\bf S} = {1 \over \mu_0} \bf E \times B$$

is radially inward.

But in the first image there aren't electric and magnetic fields around the light bulb (within the first second), so what makes it light up almost immediately?

At first glance, I thought it was the energy flux coming out from the battery that finally reaches the light bulb, but it can't be! Otherwise every source of electromagnetic energy reaching the bulb would make it light up.

If I put it next to an alternating dipole nothing happens (and it should, if all that matters were an energy flux).

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  • $\begingroup$ The video explicitly refers to a more detailed analysis in its description, which is perhaps not ideal but at least explains the setup with the two parallel wires is a transmission line. Putting a bulb next to a dipole is not. Can you be more specific what you want to know that is not already explained there? $\endgroup$
    – ACuriousMind
    Nov 20, 2021 at 18:47
  • $\begingroup$ What I don’t understand is that the shape of the energy flux, pointing radially outward from the battery and radially inward to the light bulb, hardly depends on the current circulating through the transmission line. But how can the flux reach the light bulb before the first second, when the current and the charges have not already reached all the wire? $\endgroup$
    – ric.san
    Nov 21, 2021 at 7:28
  • $\begingroup$ Minor comment to the post (v6): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Dec 31, 2021 at 19:41

3 Answers 3

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You are correct.

The video is only partly right. It is true that EM energy flows along the Poynting vector. This means that energy in a circuit is not transported by the electrons but rather by the fields outside the wire. But despite that, it is not correct that the bulb will light after 1 m/c, it will take the the full second.

There are a couple of important issues that were neglected in the video. The first is the amount of energy flux and its direction. The vast majority of the energy flux is in the space just immediately outside the wires and the energy flows parallel to the wires. There is thus very little energy* that goes across the 1 m gap between the wires. The bulb needs a rather large amount of power to light up, so the small amount that goes across the gap will not do it.

The other issue is the one you identified. It is not enough merely for the energy to reach the bulb. It must flow into the bulb and remain instead of just flowing in and out through it. According to Poynting’s theorem that requires a current in the bulb. The bulb is not an antenna, so it is not designed to produce large currents from small passing EM waves. That means that out of the small amount of power that reaches the bulb an even smaller amount remains in it.

These two issues make the conclusion wrong, even though the discussion of Poynting’s theorem and the idea that energy flows outside the wire is correct. It takes the full second for those fields, traveling just immediately outside the wire to carry the energy the long way around. It actually takes a little longer even, because the speed of this signal around the wire is less than c.

*The “very little energy” comment is assuming a typical setup with a normal lightbulb and a normal battery as shown in the video. It is possible to special build an apparatus that would transfer more energy, but such is not shown. I believe that any apparatus that would cause the bulb to light at 1 m/c would cause it to melt at 1 s.

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    $\begingroup$ Haha. I was thinking exactly the same. When the video of Veritasium popped up on Youtube, I was really happy that someone made a video about the poynting vector. I gave several lectures to students and even professional engineers about RF design and was often shocked that the concept of the poynting vector or the field propagation in general is totally overlooked despite it being one of the most important key ideas for understanding the whole RF area. Sadly the video of Veritasium went completely overboard and created a somehow "warped" image of the actual matter. $\endgroup$
    – GNA
    Nov 23, 2021 at 13:04
  • $\begingroup$ It is not accurate to say that the bulb lights only after 1s. It depends on the resistance of the bulb and the resistance of the wire. During the first second the power supply will be in series with a rather large wave impedance of the transmission line formed by the two wires and increasingly more of the cable DC resistance. After one second, the power supply will be in series with the full DC resistance of the cable, but no longer in series with the wave impedance. $\endgroup$
    – tobalt
    Dec 28, 2021 at 13:58
  • $\begingroup$ The video is not wrong. Both of the “issues” you raise mean that the ratio of the power absorbed by the bulb and the power emitted by the batter/switch at turn on is very very small. BUT if the power emitted by the battery is high enough even the small fraction absorbed by the bulb will be enough to turn it on. This can in principle always be accomplished by increasing the battery voltage. Veritasium specifically says this in the video. In practice the voltage levels may be unfeasible without great difficulty. But no problem in principle. $\endgroup$
    – Jagerber48
    Dec 28, 2021 at 14:48
  • $\begingroup$ @Dale LEDs can turn on at ~100 mW. The power emitted by the battery/switch can always be increased by increasing the voltage. With his words, Veritasium never implied a 12 V battery would turn on a 60 W bulb. If he did that would be incorrdct in my opinion. Perhaps his physical setup was overly misleading to imply that this was the case. But again, listen to what he says. He never implied a 12 V battery would turn on the 60 W bulb. He says you may need to increase voltage like crazy to actually make it light up. $\endgroup$
    – Jagerber48
    Dec 29, 2021 at 3:28
  • $\begingroup$ @Dale fair point about the physical setup. I agree that it is misleading and that Veritasium didn’t do a great job clarifying that the bulb in his physical setup would not light, despite the fact that the theoretical on could light. Could you modify your answer to explicitly indicate you’re specifically talking about a realistic bulb (e.g. 60 W) and realistic battery (e.g. 12V)? I think this would be helpful since there seems to be “in principle” vs “in practice” contention on this topic. $\endgroup$
    – Jagerber48
    Dec 29, 2021 at 14:11
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An oscillating dipole with a large enough dipole moment would cause the bulb to light up. If the resistor is just a resistor with short conductive leads on each end then what happens is the original dipole induces an induced oscillating dipole moment in the resistor. This induce dipole moment results in an electric field pointing into the resistor, a voltage across the resistor, B field lines around the resistor and current through the resistor. All of the above results in the resistor lighting up if it’s a bulb.

The oscillating dipole acts like an emitting antenna, and the resistor a receiving antenna. Obviously the power transfer coefficient will be super super low and geometry dependent. But, for any geometry the transfer coefficient will technically be non zero. So if you use insane amounts of energy radiated by the dipole it will light up the bulb.

The key insight of the Veritasium video that no one seems to explicitly be pointing out is as follows: Under certain conditions, the lumped circuit model for circuits breaks down. If you look at Veritasium's circuit then it is clear the lumped elements are:

  • The battery
  • The bulb
  • The switch
  • The wire between the battery and switch, the wire between the battery and bulb and the wire between the switch and the bulb.

If you just look at this circuit on paper you would expect the bulb to light up instantaneously, no matter the size of the wires. The circuit schematic only encodes information about what lumped elements appear in the circuit and their topology. It does not encode any geometric information about the circuit. The point of the Veritasium video is to show that, under certain conditions, geometry is important to accurately predict circuit behavior.

The example in the Veritasium video is contrived, but these exact same physical principles become critically important at higher frequencies and for more sensitive circuits. For example, it a physics lab, you could imagine a 10 A current supply switching on and inducing a 10 mV or more noise voltage on a nearby sensitive photodetector detection line. This level of noise could be problematic for certain applications. The physical principles are the exact same for the two scenarios.

The effect can be explained either by (1) saying the lumped circuit model fails for certain geometric circuits OR (2) saying that a full model of the circuit requires the inclusion of additional, non-obvious, lumped elements such as distributed capacitors and inductors between and along wires.

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I don't think there is a static electric field between the wires. When the circuit is first closed, there may be some small amount of energy transport via electromagnetic radiation outside the wires due to the acceleration of electrons in the wire. But each wire has no net charge at any time (before the switch is closed, immediately after the switch is closed and the current is increasing, or when the current is established and steady). When a steady current is reached, only the magnetic field exists outside the wires and battery and lamp, so there will be no Poynting vector outside the circuit (outside the wires, battery and lamp). The energy is transported by the electrons.

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