The premise

In a recent video by the pop-sci channel Veritasium, the concept of the flow of electricity and energy transmission in a circuit was discussed. In that video a thought experiment is presented, as seen below. The question is: after the circuit is complete, how long will it take for the lamp to turn on?

Thought experiment

The video concludes that, after the switch is flipped, the lightbulb will turn on after 1/c seconds (1 meter divided by the speed of light), since the lamp and the battery are 1 meter apart. This is explained by showing that, in an electrical circuit, energy is transmitted through an EM field, which propagates at the speed of light, and not through movement of particles in a conductor.

This explanation did not sit right with me, for mainly two reasons:

  1. Clearly, the conductor plays a role in the transmission of energy. This is actually pointed out in the video as well, by saying that after a connection in a circuit is made, the EM field will propagate along the conductor at the speed of light. To me, this would mean that the field would take 1s to propagate through the cables in the thought experiment, and the lamp will turn on after 1s.
  2. For the lamp to turn on, current must flow through it, heating the filament. Simply being in an EM field generated by the battery would not work, there needs to be some potential difference at the lamp's terminals. The signal doesn't "know" what happens at the load until it reaches it. This is a big part of reflection in signal and transmission lines, yet seems to be ignored here. Therefore, the voltage will take 1s to arrive at the lamp in this case.

The question

Is the answer of 1 meter/c seconds correct, or should the answer be something else? If the answer is not 1 meter/c, where does the mistake in the video's reasoning lie?

  • 3
    \$\begingroup\$ Thought experiment : cut the ends off. You now have a step function in a dipole antenna, 1m away from a receiving dipole antenna. When does the latter receive the EM field from the step in current racing out from the switch to both ends of the former? \$\endgroup\$
    – user16324
    Commented Nov 22, 2021 at 20:02
  • \$\begingroup\$ It's something else and it's more complex than what is shown. \$\endgroup\$
    – Andy aka
    Commented Nov 22, 2021 at 21:06
  • \$\begingroup\$ If you're using an incandescent bulb, there's also the fact that they don't turn on instantly, and need time to heat up. LEDs are faster, but they still have a turn-on time. (measured in the picoseconds, but it is there. Diodes don't turn on instantaneously.) \$\endgroup\$
    – Hearth
    Commented Nov 22, 2021 at 21:10
  • 1
    \$\begingroup\$ Mu3, there are complicating factors that haven't been mentioned, yet. Partly, because electric charges must re-arrange themselves on the surfaces of conductors and these changes do not take place at the speed of light. They happen fast, but not that fast. And they are required for current to flow, at all. And that's not a sudden event but one which takes place as things re-assemble due to changes in applied EMF. Mechanisms are being missed, ones that cannot be ignored, in your example. \$\endgroup\$
    – jonk
    Commented Nov 22, 2021 at 21:31
  • 3
    \$\begingroup\$ Dave Jones covers this in a review video he just posted. Rather interesting: youtu.be/VQsoG45Y_00 \$\endgroup\$
    – JYelton
    Commented Nov 24, 2021 at 22:15

4 Answers 4


Veritasium hasn't gone into the vital detail. I'm not sure whether that was because he didn't want to complicate the answer, or because he didn't do enough research.

The important part is that the two 300 km loops of wire are transmission lines.

For the first second after closing the switch, before the reflection gets back from the far end, you could replace those lines in the circuit by resistors equal to their characteristic impedance.

The impedance of a line with two thin conductors 1 m apart will be quite high, many hundreds of ohms, perhaps 1 kohm. The two lines would therefore limit the current into the lamp to a few mA for the first second. If the lamp could operate with that low current, then it would turn on after 1 m/c following the switch closure.

Things get more interesting, and complicated, when a reflection arrives back from the short circuit at the far end. The detail of what happens rather depends on the impedance the lamp presents to the returning reflections. In the simplest case with the lamp matched to the line impedance, the reflection will be absorbed, and the lamp will now operate as if the lines are short-circuited. If the lamp isn't matched, then there will be a further reflection, and further changes of lamp brightness, until the waves propagating on these lines eventually die away.

The many problems this type of video has include:

  • One thing he says is true, energy 'flows' around the wires, not through them
  • but his demonstration doesn't show that
  • and his demonstration is not a real experiment, it's faked
  • He makes an assumption about zero resistance, which is not in itself a problem, but it does mean that any attempt to reproduce what he has done without superconductors is impossible
  • and it confounds the intuition of most non-technical viewers who insist the waves would die out on the long line, which they would on a resistive one
  • Why the lamp would light, a bit, immediately, is just straightforward transmission line theory
  • but he has the orders of magnitude all wrong
  • and the impedance to ground would be just as significant as the lines' impedance to each other, so trying to simulate exactly what he has there would be a) complicated and b) would not show what he's trying to demonstrate.
  • and without a definition of the lamp impedance, what happens when the reflection returns would be uncertain
  • The transmission lines he shows in the video demo area are better coupled to the ground than each other, which will further confuse how that area would react
  • but the lines in the diagram going out into space are clearly isolated from ground
  • Which assumption to make? I have chosen that the lines are isolated from ground, which makes for a simpler thought experiment, without changing the fundamentals. MathKeepsMeBusy makes the non-isolated assumption, which is why he gets such a different answer to me.

In the words of Wolfgang Pauli, Verisatium's demo is not even wrong, it's worse than that.

An experiment, a real physical demonstration, to show the delay of a transmission line would be better done with some well-defined commercial transmission line, ideally 50Ω coax, but some 300Ω open wire antenna wire would also serve and be far closer to the open wires he has used. With the length cut to a few metres or 10s of metres, the resistance would be small, small enough to ignore, but an oscilloscope would be needed to show the microseconds delay. With the line lofted to several metres, it would be sufficiently isolated from ground for my assumptions on isolation to hold and be shown.

I'm not sure I could think of an experiment to 'demonstrate' that energy flows outside of the wires, but it's fairly straightforward to argue that it either does, or that an assumption that it does is consistent with what we see. The Poynting Vector requires electric and magnetic field to create the energy flow. Considering two wires between a battery and a lamp, the battery provides the electric field between the conductors. Only when a current flows do we get the magnetic field between them. I referred to this phenomenon in another answer as a 'power guide' (cf wave guide), as the wires guide the current, and the magnetic field they produce cross the electric field transmits the energy.

  • 1
    \$\begingroup\$ @Mu3 the answer is it depends on your definition of "turns on", as well as some things like the diameter of the wire (thicker wire = more capacitive coupling = lower characteristic impedance = more current flowing "immediately"). \$\endgroup\$
    – hobbs
    Commented Nov 22, 2021 at 20:54
  • 1
    \$\begingroup\$ For 22AWG hookup wire (d=0.64mm) I get Z0=966 ohm; for 12AWG wire (d=2mm), Z0=829 ohm. You have to push the boundaries of the definition of "wire" (e.g. use wide flat copper strap) to get much below 400 ohm with 1 meter spacing. \$\endgroup\$
    – hobbs
    Commented Nov 22, 2021 at 21:00
  • 1
    \$\begingroup\$ @JYelton - Wire resistance has no affect on how fast the signl/voltage moves along the wire. It would affect how bright the light is, but not when it starts to turn on. \$\endgroup\$
    – SteveSh
    Commented Nov 22, 2021 at 22:23
  • 2
    \$\begingroup\$ @JYelton Have you seen his Kelvin Waterdrop HV generator video? He completely skates over the important bit about the placement of the height of the nozzle in the field so the drops break continutity at the right time to enable electrophorous charging. But you're right, I'll edit that down to 'this video is poorly explained'. \$\endgroup\$
    – Neil_UK
    Commented Nov 22, 2021 at 22:24
  • 1
    \$\begingroup\$ @jpp1 added a few notes to my answer. \$\endgroup\$
    – Neil_UK
    Commented Nov 28, 2021 at 13:47

A very, very, very small amount of energy will reach the lamp at the speed of light travelling a direct path between the battery/switch and lamp (1 meter). When the switch is closed, current will begin to flow through the side of the wire with the switch, changing it's voltage relative to the other wire. The two parallel legs of the wire form a capacitor and a transformer, so current will begin to flow in the "bulb" side of the wire as well.

But it will be an extremely small current at first. The full length of the wire should not be taken into account when calculating the size of the capacitor and transformer because the step signal created by closing the switch will not have had time to propagate along the wire. For example, after 3 m/c seconds, an EM disturbance could have propagated 1 meter out from the switch, 1 meter across the space between legs of the wire, and 1 meter back. (This will happen in both directions, so x2). The capacitance formed by two 1 meter wires 1 meter apart will be extremely small. Hence the charge needed to raise the voltage across this capacitor will be extremely small. Hence the current through the lamp will be extremely small. (The transformer effect between the two legs of the wire will likewise be extremely small).

Will this extremely small power turn the lamp "on"? That depends upon what threshold one takes as "on". By most standards, the lamp will be considered off at 1 m/c seconds, and for a considerable period of time after. By those standards, Veritasium is wrong. However, Veritasium is correct insofar as some energy will be delivered to the lamp at the speed of light.

Assuming, as he does, that the wires have zero resistance, the bulk of the power will not arrive until much later, which depends upon characteristics of the transmission lines consisting of the pairs of wires. Once again, there is fuzziness involved in how fast a step signal passes along a transmission line. "When" the signal arrives depends upon the threshold for detection. However, for a "reasonable" sized threshold, the velocity of a step signal through a transmission line is noticeably less than the speed of light, typically 0.5-0.7 times the speed of light. And the length one uses in calculating the time delay is the length of the transmission line, not the shortest path between the wires.

The crux of a disagreement?

In a comment, I wrote in part:

...the current on the side with the battery and switch is not immediately balanced with the current on the side of the lamp. Assume battery is 12V and that it "sees" a 1k impedance, then 12 mA will flow on that side of the transmission line. But it does not follow that 12 mA will be immediately flowing on the other side, where the lamp is. It is only the current flowing through the lamp side lights the lamp.

To which @Neil_UK in part responded:

...My assumption is that the lines have no capacitance to ground, the video demo shows them all floating, so current round the battery-switch-txline-lamp-othertxline-battery obeys current continuity, equal in all parts of the circuit, as it would be for isolated floating lines. Introducing the large capacitances to ground that would be required to get significantly unequal currents is an unneccessary complication in my view

I am willing to be corrected, but I believe that "current continuity" in this case is a poor assumption, and the truth is very far from it. Let me explain.

By an assumption of "current continuity", I understand the assumption that the (conduction) current through the battery and switch is at all times identical to the (conduction) current through the lamp. (Or at least the difference is negligible). I am qualifying the term "current" with the term "conduction" because there are two kinds of current that need to be considered. Conduction current, which is the flow of charges, and displacement current, which is a constant times the rate of change of the electric field. This latter form of current is commonly ignored, but is essential for there to be current continuity through capacitors. It is the sum of conduction current and displacement current that is "conserved" in a circuit.

I believe that conduction current is NOT equal between the switch/battery side of the transmission line and the lamp side, nor even close to equal, in the early period after the switch has been closed, but before the EM signal created by closing that switch reaches the far end of the transmission line. To illustrate why I believe this, consider a one light-second long dipole antenna center-fed by a battery and a switch.


simulate this circuit – Schematic created using CircuitLab

I have used a DPDT switch in this example, whereas Veritasium used a SPST switch. The purpose of the DPDT switch is to allow the two sides of the antenna to be easily brought to the same potential.

If the two sides of the antenna are originally at the same potential, and then the switch is flipped to bring the battery into the circuit, current will flow through the battery, and in each side of the antenna. Assume that this antenna is the only thing in the universe. The conduction current in the antenna is not balanced with any conduction current in "the" return path, because there is no return path! (Or perhaps more correctly, there is no return path for the _conduction current. The return path for the displacement current is the electric field lines between corresponding parts of the two halves of the dipole). Nevertheless, the total current, which includes both the conduction current and the displacement current is the same throughout the length of the antenna. If the conduction current at the feedpoint is I, but at some place x it is Ic, then electric field at x is changing such that the displacement current Id plus the conduction current Ic sums to I.

$$I = I_d + I_c$$

Now consider a second dipole antenna separated by 1 meter from the first, with a lamp across it's center-feed lines.


simulate this circuit

If the two halves of the upper dipole are originally at equal potentials, and the switch is flipped so that the battery is brought into the circuit, current will flow in the upper dipole. This will cause current to flow in the lower dipole as well. But will the currents in the two dipoles be equal? Only if the coupling between the dipoles is 100%. If the dipoles are 1 meter apart, and the wires are thin, then initially, we can expect the coupled current to be small. However, as the current spreads towards the ends, the the coupling may increase, (even though the conduction current will decrease as we approach the ends of the wires).

I assert that in Veritasium's thought experiment, it doesn't matter whether then ends are connected (as in his description) or not (as in the example immediately above) until enough time has elapsed for the EM signal generated by the flipping the switch to reach the ends of the transmission line. At least 1/2 second.

I therefore assert that for the first 1/2 second, the current in battery and switch side will NOT match the current in the bulb side.

  • \$\begingroup\$ I dont think the current will be soo low during the first second. The wave impedance is maybe a few 100 Ohm. a small bulb can have a similar resistance. \$\endgroup\$
    – tobalt
    Commented Nov 28, 2021 at 18:12
  • \$\begingroup\$ @tobalt just added further explanation to my answser \$\endgroup\$ Commented Nov 28, 2021 at 18:32
  • \$\begingroup\$ it all depends on the assumptions. Two poorly coupled dipoles is one such, in which case your explanation is better. Two isolated transmission lines is another, in which case mine is more approrpaite. Mr V has introduced a spherical cow (superconducting lines), but has not specified whether it's in a vacuum (which aspects of the arrangement of those lines is important). We've opted for the opposite assumptions about the vacuum, so have very different answers. Our true argument is with Mr. V, not each other. \$\endgroup\$
    – Neil_UK
    Commented Nov 28, 2021 at 20:28
  • \$\begingroup\$ I don't see how the presence or absence of a vacuum has any effect on whether or not the parallel wires have balanced current prior to the EM signal reaching the ends. I can't think of any good reason why they should be balanced during that time. Continuity of conduction current in a circuit comes about in time due to the re-arrangement of charge density along the wires, and the consequent rearrangement of the E field in the wires. It doesn't come about instantaneously. \$\endgroup\$ Commented Nov 28, 2021 at 20:37
  • 1
    \$\begingroup\$ @MathKeepsMeBusy I'm pointing at the moon, and you're looking at my finger. You're still arguing about what should be, and I'm saying that the problem is too ill defined, that we must state assumptions before attempting to answer it. I'll update my answer to include my assumptions. Once assumptions about the line configuration are stated, we then won't need to have an argument about the lines. Vacuum only has relevance to the spherical cow, not to Mr V's chaos. \$\endgroup\$
    – Neil_UK
    Commented Nov 28, 2021 at 20:46

I don't think Muller (Veritasium) adequately explained why that's the correct answer, though I may have missed it. Here's my own step-by-step reasoning.

In the short moments after the switch is made, an E-field propagates at the speed of light (which is a misnomer, since it's really the "speed of causality") from the E-field around the battery to the E-field that is consequently set up along the two wires connected to the battery.

(Almost) immediately accompanying that E-field surrounding the two wires, is the motion of electrons in the wires. This motion (almost) immediately sets up an M-field around the two wires, which now means that we have an EM-field surrounding the two wires a short distance away from the battery. This EM-field contains a flow of energy, as Poynting shows, which is directed away from the battery in the space surrounding the battery, and towards the two connecting wires in the space surrounding them. Energy is not propagated inside the wires, nor in the direction of the wires. The wire direction is the direction of current, and current is the flow of charge, not the flow of energy.

This EM field propagates in all spatial directions, as with an antenna, not just in the direction of the wires. And in a time 1m/c seconds, that field strikes the space around the bulb and the two wires connected to it. The EM-field now surrounding the bulb and its two wires can be shown by Poynting to direct EM energy into the bulb and into its two connecting wires.

The reader may ask: why do we need wires connected to the bulb, in this explanation? I don't think we do, in the very first moments of time less than 1m/c seconds. Even without wires connected to the bulb, the same thing would happen to the bulb. There will be a flow of energy into the bulb filament (which is really only a wire), but without connecting wires that go back to the battery, that flow won't be maintained for any appreciable time, because the electrons that flowed inside the filament, without connecting wires, have nowhere to go. Their motion thus stops after a very short time, say x/c, where x is about the length of the filament. So without wires, a teeny weensy amount of energy will flow into the filament, not enough energy to significantly increase its temperature.

Thus, the wires are necessary to provide an approach to steady state. Of course, that steady state, as Muller explains, happens after a complicated transient interaction between the establishing EM-field, the battery, wires, and bulb. During this transient, the wires act as a transmission line; there is reflection when the establishing EM-field hits the far end where the wires turn 180 degrees, for instance, and so much more in this very complicated transient. But after a couple seconds, we see steady light from the bulb.

So now having reasoned out these details carefully, I think that's getting to a more full answer to Muller's quiz. Interesting how individual humans can reason, step by step, but in large groups, we often fall on our faces and eventually destroy most everything such careful reasoning has produced.

  • \$\begingroup\$ EEVblog provided an explanation that is similar to yours in his video. It seems that the consensus on the question is that there will be a transient at the bulb at 1m/c seconds, but steady state will be reached later. \$\endgroup\$
    – Mu3
    Commented Nov 30, 2021 at 10:01

If Veritasium assumes that any small current flow in the light bulb or LED is sufficient to turn on the load then turn on time 1{m}/c{m/s} is about 3.33{ns}.

However this does not "prove" that energy does not "flow in" wires. Most of the energy flow follows the wires at or near the surface. No energy would be radiated across the transmission line and/or antenna array unless there is electrical power developed in the length of wire connected to the power source. The development of power is voltage times current in the source and the load. If no power is developed in the wires then there is only an electrostatic field in the battery, no magnetic field, and no transfer of energy from the source to the load.

Edit: This reference describes the rules that apply for electric field boundary conditions at the surface boundary of a lossless conductor:


And this reference describes magnetic field boundary conditions:


This reference is a 35 page pdf including illustrations of poynting vector analysis surrounding a wire:


E (blue), B (black loops), and S (red) Fields for current flow (black arrow) in a perfect conductor wire (uniform wire potential is positive). Note electrical power must be developed to send energy along the wire but the flow of energy along a perfect conductor occurs without loss and so does not require sustained power.

enter image description here

Poynting vector (Red) points radial inward for the loss in a conductor:

enter image description here

Most of the energy flows close to the wire:

enter image description here

This assumes an infinite wire length to give B. Also I am not clear on the meaning of "C" and "s" since there is no definition of "C" and no illustration of the geometry defining "s".

If one assumes, say, a 12 volt light bulb rated 55 mA and 12 volt battery coupled by 1 mm radius ideal wires (18 gauge) then the bulb resistance is about 22 ohm. According to my observation this bulb is barely visible in otherwise complete darkness when powered by a 1.5 volt AA cell. Then about 68 mW must transfer to the bulb in about 3.3 nanoseconds where the full power of the bulb is about 6 watt.

An 18 gauge wire has radius about 1 mm:


This calculator is used to compute characteristic impedence Zo of a copper wire transmission line made of 0.001 meter wire spaced 1 meter apart:


Zo = 828.36 ohm

Let t = 3.3 nanoseconds after the switch closes then the 12 volt source would see a series resistance 2Zo = 1656 ohm in series with the 22 ohm light bulb if the transmission line model is an accurate simulation. This would produce 7 mA in the bulb but only 0.154 volts across the 22 ohm resistor for a total power of 1.1 mW. This is not enough to turn on a typical low power light source.

Also it is a far cry from the 1.21 gigawatts necessary to energize the flux capacitor in the Back to the Future time machine!

  • \$\begingroup\$ The E-field direction is defined as from the + charge to the - charge. That means, whether the wire is superconducting or not, that the E-field direction is along the wire. I thus don't understand why your figure shows an E-field direction radially outward from the wire. This is wrong, and of course, in the real world, wires do have non-zero resistance. Bottom line, the E-field direction is along the wire, which means that the Poynting vector points radially inward towards the wire. Read Feynman's Lectures in Physics, Sec. 27-5. \$\endgroup\$
    – ttonon
    Commented Dec 9, 2021 at 22:39
  • \$\begingroup\$ A perfect conductor, by definition, has infinite conductivity so there is no thermal loss inside or at the surface of the conductor when current flows. When a power source does work to force the flow of current the power developed goes into increasing the energy state of the system and to overcome heat due to finite conductivity. But if the power source is then replaced with short circuit energy will flow perpetually in the perfect conductor. I think the Poynting vector thus points out for a power source; it points in for a power sink; and it runs parallel to wire for conserved flow of energy. \$\endgroup\$ Commented Dec 10, 2021 at 0:40
  • \$\begingroup\$ System Theory, Do you agree that the E-field direction is by convention from the + charge to the - charge? This direction is independent whether there is a conductor of either finite resistance or zero resistance within that field. Do you agree? If so, I don't understand why you keep asserting that if the conductor has zero resistance the E-field direction then magically becomes radially outward. It doesn't make sense. If you don't agree with the convention, you are not talking conventional science. \$\endgroup\$
    – ttonon
    Commented Dec 11, 2021 at 23:37
  • \$\begingroup\$ @ttonon By definition the quantum of positive charge is given by q = e+. Also by definition the electric field exerts a force F = qE on a test charge q = e+. The E field points in the same direction as the force that would apply to q = e+. This means the E field is always normal to an equipotential surface. The E field is normal away from such a surface with positive charge and toward such a surface with negative charge. In a section of lossless conductor where no energy is being stored and no heat is dissipated there is no voltage drop as the current flows so the E field is radial as shown. \$\endgroup\$ Commented Dec 12, 2021 at 0:29

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