How is the answer to this question $1/c$ seconds?

Question

This is a question from the YouTube science content channel Veritasium. The below image shows the question.

The question is about when will the bulb begin to glow after the switch is turned on. The presenter argues that it will take $1/c$ seconds for the bulb to light up, since the separation between load and source is only 1 m, and by Poynting's theorem, the energy will flow from source to load in the direction given by the Poynting vector, and since there is only 1 m separation, it will ideally take $1/c$ seconds for energy to reach the bulb and the bulb will light up. This kind of makes sense, and there are professors of EE and Physics appearing on the video, supporting the argument.

Now, what from my understanding, the Poynting vector only gives a direction in which the energy is flowing. It does not say anything about the time taken for it to each the destination. Also for the bulb to light up, the filaments have to be heated, so there must by a current flow in the filament, which will not happen until an electric field is present inside the filament. Since the wires are a light year long the electric field from the battery can only travel at a speed less than that of light (ideally $c$), so it must take at least a year for the electric field to reach the filament and for it to glow, am I right?

So how is the argument that the bulb will light up in $1/c$ seconds true?

If it were true, then would the light still take only $1/c$ seconds if it was still 1 m away from the battery, but connected using a 100 light year long irregularly coiled wire?

Note: I believe it is assumed that the light will glow even if it carries a minuscule amount of current.

_{I believe this is not a duplicate of this as my question is different. This is is also different from this, as the configuration of the circuit is different.}

Answer 1

I was wondering about this as well. This is not an answer, but reasons why I am confused about the video.

Veritasium made the point that electrons in a wire do not carry energy around the circuit. The energy is carried by the electric and magnetic fields generated by currents in the wire. So when these fields reach the light bulb, energy has been transported and the light bulb will light.

At first glance, this sounds wrong. But Veritasium sometimes picks counter-intuitive situations to explain. But at second glance, it still sounds wrong. I have some thoughts on mechanisms. But I don't know what is correct.

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Suppose you have two parallel infinite wires that don't connect at all. You might expect the wire connected to the battery to generate a current. You might expect fields from that wire to induce a current in the light bulb wire. This would begin as soon as the fields arrive as soon as the fields arrive at the light bulb wire.

But I would expect this to be weaker if the wires are well separated. I would not expect distance to affect the brightness of the bulb when current was flowing in a (resistance free) wire.

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There is another way of thinking about the energy carried by the current. It isn't stored in individual electrons. It is stored as potential energy when electrons are crowded together more densely than the positive charges in the nuclei of the wire. They repel each other. Also when they are spread out less densely than the nuclei, and the nuclei attract them.

You can explain energy transport this way. The battery generates free electrons from chemicals at the negative terminal, and absorbs electrons into chemicals at the positive terminal. These concentrations of charge repel/attract nearby electrons, which starts a current flowing.

In a resistance free wire, you would expect the electrons to easily spread out when pushed. They would be obstructed when the current reaches the bulb. So when everything reaches steady state, you would expect one half the circuit to be negatively charged and the other half positively charged.

This explanation runs into trouble because you don't measure an electric field around a DC current in a wire. You measure a magnetic field.

So is it possible that the electron density change is large enough to explain energy transport, but small enough that the electric field is insignificant?

Yes. Imagine a mechanical analog to this circuit, where energy is transported along a very long spring. Pushing on one end compresses the spring. A compression wave travels down it, carrying energy and an increased mass density. Making the spring very stiff and light has two effects. The spring compresses very little, decreasing the mass density change. And the speed of the wave gets very large.

Electrons are analogous to a very light, stiff spring indeed. Compression waves travel at about $1/3$ c in copper.

Also it is possible to explain the magnetic field in terms of electron density changes. See this old Veritasium video - How Special Relativity Makes Magnets Work. See this newer video for more information - Veritasium's 'How Special Relativity Makes Magnets Work' - EXPLAINED (better)

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It seems that these two explanations of energy transport are to a degree equivalent.

In one, a current generates an electromagnetic field. You can calculate the energy transport from the field.

In the other, charge density causes a current, which causes the field. You can calculate energy transport from the potential energy in the charge density.

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So my question is - Is this right? Is Veritasium also right, expect for the timing?

I am not sure how to post this. It seems I am misusing an answer. But posting a separate question seems to invite closing it as a duplicate of this one. Editing the question with my own version seems to step on the OP's toes. It would be better to answer this in separate answers instead of comments.

Answer 2

My initial intuition after watching this video was that Derek's claim was technically true but only for an uninteresting reason. Namely, he stated that the lamp will light up once any current at all flows through it. Well, of course an infinitely sensitive amperemeter will react to the flip of the switch with the speed of light! (It will probably react to your neurons firing just thinking about it as well.) But given a realistic lamp and a realistic battery, wouldn't the immediate current be many orders of magnitude weaker that the operating current?

Turns out my intuition was wrong, so I'd like to give a quantitative answer to this question.

The circuit consists of two stretches of transmission line connected to each other through the lamp and the battery. Each stretch is shorted at the opposite end from the lamp. The change of voltage and current within the lines is described by the telegrapher's equations, with the lamp, battery and shorted ends defining the boundary conditions.

Here is an equivalent circuit where capacitors and inductors represent capacitance and inductance of an infinitesimal fragment of a transmission line:

• $V_{R} (t,x)$ is the potential difference between the top and the bottom wires at distance $x$ to the right of center at the moment $t$
• $V_{L} (t,-x)$ is the potential difference between the top and the bottom wires at distance $x$ to the left of center at the moment $t$
• $I_{R} (t,x)$ is the current at distance $x$ to the right of center at the moment $t$
• $I_{L} (t,-x)$ is the current at distance $x$ to the left of center at the moment $t$
• $L'$ is the transmission line inductance per unit length
• $С'$ is the transmission line capacitance per unit length
• $\mathcal{E}$ is the electromotive force of the battery
• $R$ is the resistance of the lamp

Let's calculate the current $I(t,0)$ flowing through the lamp, and how the immediate current at $t=0$ relates to the operating current $I_{lim}=\frac{\mathcal{E}}{R}$.

Applying the lossless general solution of the telegrapher's equations to each of the two stretches of transmission line:

$$ \begin{array}{ l l r } V_{R} (t,x)&=v_{R}^{+}\left( t-\frac{x}{u}\right) +v_{R}^{-}\left( t+\frac{x}{u}\right) & \qquad\textrm{(A1)}\\[0.3em] I_{R} (t,x)&=\frac{1}{Z_{0}}\left( v_{R}^{+}\left( t-\frac{x}{u}\right) -v_{R}^{-}\left( t+\frac{x}{u}\right)\right) & \qquad\textrm{(A2)} \\[0.3em] V_{L} (t,-x)&=-v_{L}^{+}\left( t-\frac{x}{u}\right) -v_{L}^{-}\left( t+\frac{x}{u}\right) & \qquad\textrm{(A3)}\\[0.3em] I_{L} (t,-x)&=\frac{1}{Z_{0}}\left( v_{L}^{+}\left( t-\frac{x}{u}\right) -v_{L}^{-}\left( t+\frac{x}{u}\right)\right) & \qquad\textrm{(A4)}\\ \end{array} $$

where $Z_{0} =\sqrt{\frac{L'}{C'}}$ is the characteristic impedance and $u=\sqrt{\frac{1}{L'C'}}$ is the propagation velocity.

$v_{R}^{+}$ and $v_{L}^{+}$ here represent the incident waves travelling from the battery to the right and to the left, respectively. $v_{R}^{-}$ and $v_{L}^{-}$ represent the reflected waves travelling back from the shorted ends.

$v_{R}^{+}(\cdot)$, $v_{R}^{-}(\cdot)$, $v_{L}^{+}(\cdot)$ and $v_{L}^{-}(\cdot)$ can be found from the equations above and the boundary conditions.

The boundary conditions are:

• (B1) There is no current in the line at $t = 0$
• (B2) There is no potential difference in the line at $t = 0$
• (B3) The current is equal to the left and to the right of the lamp
• (B4) Potential differences at $x = 0$ are determined by the EMF and voltage drop at the lamp
• (B5) Far ends are shorted, so there is no potential difference at either end

$$ \begin{array}{ l l r } I_{L} (0,-x)=I_{R} (0,x)=0, &0<x<D & \qquad\textrm{(B1)}& \\[0.3em] V_{L} (0,-x)=V_{R} (0,x)=0,\ \ &0<x<D & \qquad\textrm{(B2)}\\[0.3em] I_{L} (t,0)=I_{R} (t,0), &t \geq 0 & \qquad\textrm{(B3)}\\[0.3em] V_{R} (t,0)-V_{L} (t,0)=\mathcal{E} -R\cdot I_{R} (t,0), &t \geq 0 & \qquad\textrm{(B4)}\\[0.3em] V_{L} (t,-D)=V_{R} (t,D)=0, &t \geq 0 & \qquad\textrm{(B5)} \end{array} $$

The solution to this system of 9 equations is:

$$ \begin{array}{ l l l } v_{R}^{+}(t) &=\frac{Z_{0}\mathcal{E}}{2R}\left( 1-\left(\frac{2Z_{0} -R}{2Z_{0} +R}\right)^{n+1}\right), & \frac{2D}{u}\cdot n \leq t \leq \frac{2D}{u}\cdot (n+1) \\[0.6em] v_{R}^{-}(t) &= -v_{R}^{+}\left(t-\frac{2D}{u}\right)\\[0.4em] &=-\frac{Z_{0}\mathcal{E}}{2R}\left( 1-\left(\frac{2Z_{0} -R}{2Z_{0} +R}\right)^{n}\right), & \frac{2D}{u}\cdot n \leq t \leq \frac{2D}{u}\cdot (n+1) \\[0.6em] v_{L}^{+}(t) &= v_{R}^{+}(t) & \\[0.6em] v_{L}^{-}(t) &= v_{R}^{-}(t) & \\ \end{array} $$

These are piecewise-constant functions that change their values at moments $\frac{2D}{u},\ \frac{4D}{u},\ \frac{6D}{u}$ etc.

($\frac{2D}{u}$ is the amount of time time it takes for the signal to travel to the far end of the transmission line and back. $n$ represents the number of times it has travelled there and back.)

Finally, the current flowing through the lamp is:

$$ \begin{array}{ l l } I_{R}(t,0) &= \frac{1}{Z_{0}}\left( v_{R}^{+}\left( t-\frac{x}{u}\right) -v_{R}^{-}\left( t+\frac{x}{u}\right)\right) \\[0.7em] &= \frac{\mathcal{E}}{2R}\left( 2-\left(\frac{2Z_{0} -R}{2Z_{0} +R}\right)^{n} \cdot \frac{4Z_{0}}{2Z_{0} +R}\right), & \frac{2D}{u}\cdot n \leq t \leq \frac{2D}{u}\cdot (n+1) \end{array} $$

Immediately after the switch is flipped:

$I_{R}(0,0) = \frac{\mathcal{E}}{2Z_{0}+R}$

As time passes, it will tend to the operating current:

$I_{lim}=\lim\limits_{t\rightarrow\infty} I_{R}(t,0) = \frac{\mathcal{E}}{R}$

This implies, roughly speaking, that if $R$ is within an order of magnitude of $2Z_0$, then the immediate current will be within an order of magnitude of the operating current.

This is how it the current changes over time depending on the relative values of $R$ and $Z_0$:

When $R < 2 Z_0$, the lamp initially receives a fraction of the operating current $I_{lim}$, which then jumps up every second (every time a reflected wave comes back to the source) asymptotically approaching $I_{lim}$ from below.

When $R > 2 Z_0$, the current jumps above and below $I_{lim}$ with the amplitude of the jumps tending to zero.

$R = 2 Z_0$ is a special case where the current reaches $I_{lim}$ in a second (after a single reflection) and then doesn't change. (This is because the reflected wave is completely absorbed by the lamp and there is no re-reflection.) The initial current in this case is $\frac{1}{2}I_{lim}$.

Now back to the original question, can $R$ really be of a similar magnitude to $2 Z_0$ if we don't allow any other unrealistic assumptions besides the light second-long superconducting wires? Let's put some numbers on it. Suppose the wires are $a=2\ mm$ wide and positioned at distance $d=1\ m$ from each other. And the battery provides $\mathcal{E}=100\ V$ of EMF. What should the lamp's nominal power be in order to satisfy $R = 2 Z_0$?

$$ \begin{array}{ l l l } L' &= \frac{\mu}{\pi} \cosh^{-1}\frac{d}{2a} &= 2.5\ \mu H/m\\[0.5em] C' &= \frac{\pi\epsilon}{\cosh^{-1}\frac{d}{2a}} &= 4.5\ p F/m \\[0.5em] Z_0 &= \sqrt{\frac{L}{C}} &= 745\ Ω \\[0.5em] R &= 2Z_0 &= 1490\ Ω \\[0.5em] P &= \frac{\mathcal{E}^2}{R} &= 6.7\ W \\ \end{array} \\ $$

(Calculated using this page)

This makes one set of circuit parameters where the immediate current will be half the operating current:

• 100 V battery
• 6.7 W lamp
• 2 mm-wide wires spaced 1 meter apart

Answer 3

No one who knows what they're talking about would ever say "1/c seconds". A commenter called this a minor thing, but I would call it a useful shibboleth. It's a sign that the writer of the video isn't exactly playing fourth-dimensional chess, and a sophisticated analysis of their claim probably isn't necessary. Another sign that he may not be not too knowledgeable about the subject he's chosen to teach to millions is that his alternate answers only have a chance of making sense if the speed of electricity is $c$. I get the impression that he doesn't know that it isn't.

Is the current in the filament nonzero shortly after the switch is closed? Yes. Nothing is ever exactly zero. Does the current become nonzero 1/299792458 s after the switch is closed? No, it's already nonzero before the switch is closed. The current across an open switch isn't exactly zero. There is also an internal current in the battery, and the current it will induce in the filament isn't exactly zero. There are stray electric fields in the environment, etc.

It doesn't really even make sense to talk about a time 1/299792458 s after the switch is closed, because switches can't be closed in such a short time. Even the nonequilibrium current that he's talking about will begin long before the wires actually make contact.

Normally, it would be nitpicking to point out these things, but the effect that leads him to say you've been "lied to" and the other answers to his question are "a big misconception" is itself ridiculously nitpicky. The reasonable answer to his question is that the bulb lights up after somewhat more than 1 second. (The length of the circuit in the video is 2 light seconds, not 2 light years.)

It would be easy to turn this claim into a real experiment: just set up two separate circuits, one with a battery and a dummy resistive load, the other with a light bulb and nothing else, and put the battery and bulb 1 meter apart, and see if the bulb lights up. He either didn't think to do that, or he did it and chose not to publish because he didn't like the outcome.

The analogy with the chain at the beginning of the video is a reasonably good one, though it would be better to use something that is strong in compression as well as in tension. He says he used to teach the chain analogy, but it seems like he never understood it very well himself, since he thinks it's wrong because, e.g., "if it's the electrons that are carrying energy [...] then when [they] flow back to the power station, why are they not carrying energy back?" Well, it's the same in the analogy. The flow of energy in the chain isn't in the direction the chain is moving. I'm not sure he understands that.

The chain analogy even has an (iffy) analog of inductance: when the chain moves, it creates eddies in the air, which propagate across the middle of the loop and exert some pressure on the rest of the chain. If you had a long chain wrapped around a pulley at the far end, and pulled it, you'd create a wind which would move the other half of the chain slightly. But the relevant propagation speed for practically all useful purposes is the speed of sound in the chain, not the speed of sound in the air.

Answer 4

There are three stages:

Before the switch is closed, the circuit is inert. Under a charitable interpretation of Veritasium's simplifying assumptions, there's no residual currents or stray electromagnetic fields interfering with the setup, the wire is a superconductor, and the bulb is hyperefficient (i.e. it will "turn on" as soon as any current flows through it whatsoever, even if it's less than a nanoamp). This is an unphysical spherical cows setup, but not materially different from a high-school physics textbook telling us to imagine a frictionless plane.

Immediately when the switch is closed, the electric field begins to propagate along the wire at the speed of light. Because the circuit is so long and narrow, the field takes a long time to propagate around the whole circuit, and the question is specifically asking about this period of time (before the field has fully propagated). During this period, the setup is entirely equivalent to one in which we have two parallel, disconnected wires, because the field has not yet reached the point where the wires are connected. Therefore, I will describe the wires as if they were disconnected at the far ends.

Here's the thing: If the wires are disconnected, then this setup is not materially different from a pair of dipole antennas, except for the use of direct current (DC) instead of alternating current (AC). Antennas normally use AC in order to create a time-varying electric field, which is necessary if you want that field to actually carry some sort of useful information. For that matter, it is also necessary in order to create electromagnetic waves, which have a far greater range than a static electric field. However, if all you want is to induce a current in another antenna, then a static electric field is perfectly adequate for that purpose. We should of course note that the electric force obeys an inverse square law, and so the antennas must be quite close (or the voltage quite high) in order for this to work in practice. But under the spherical cows assumptions established earlier, this "problem" is quietly ignored. The bulb is now receiving (a pathetically small) electric current.

As the field propagates farther along the wire, the length of wire which is acting as an antenna will increase, and so the current will also increase over time, until eventually the field reaches the endpoints and the circuit is operating in a steady state. This is the point at which the bulb finally receives the full current which we would expect under Ohm's law.

If the circuit is broken at any point, then the electric field will begin to collapse, starting from the point of the break, and expanding in both directions along the circuit until the entire field is once again null. This will not be noticed at the bulb until the field around the bulb collapses, so (as we would expect) it is not possible to send information faster than the speed of light simply by opening a large circuit.

The magnetic field also does various things at various points throughout, but I have omitted it because for the most part, it's "just along for the ride," and not terribly interesting to analyze in its own right. However, you do need to include the magnetic field if you want to calculate the Poynting vector, as Veritasium describes in the video. In short, the magnetic field forms loops surrounding any wire in which charges are moving, in accordance with Ampère's law.

A couple of other notes:

• If we assume that the circuit starts out in a steady state, there should already be a dipole between the ends of the switch. This should create a (highly localized) electric field, which should theoretically induce a very small current on its own, depending on the exact positioning of the setup. In practice, these sorts of dipoles are so pathetically small that you can completely ignore them (unless the voltage is huge). However, it ought to be included under our "spherical cows" assumptions.
• If the wire is not a superconductor (and therefore has nonzero resistance), then some component of the Poynting field will be directed towards the wire. The wire will dissipate the resulting energy as heat.
• The illustration at 9:15 in the video (showing an AC circuit) suggests that the entire field configuration statically blinks on and off when the current reverses, but what actually happens is a great deal more complex (i.e. the electric and magnetic fields do invert, but they don't do so instantaneously). Normally, I wouldn't nitpick something like this, but it's a very important distinction if you want to talk about how a (real) antenna works.
• In addition to functioning "like antennas," the wires also function "like a capacitor," but this doesn't really change our analysis all that much since we're not trying to create a time-varying voltage in the circuit.

Answer 5

The result is determined by the speed of light in the metal, about 2c/3. It will take 3/2 of a year before the effect of the closing of the switch reaches the light bulb. However any finite resistance of the wire will cause the voltage across the bulb to be next to zero, so be expect no light.

[edit] When the switch is closed there will be a time dependent current, which will induce a transient current in the other wire, one meter away. If thus surge in the current is large enough, this may cause the bulb to emit visible light temporarily. A stationary current in the bottom wire will not lead to a current.

[edit] The Poynting vector should be interpreted with caution according to Panofsky&Phillips, Classical Electricity and Magnetism, 2nd edition, p180, 2nd paragraph. I subscribe to this observation.