The lamp that lights before the loop is round

Most electrical intuition comes from circuits small enough that they have no timing at all. Close a switch and a lamp comes on. A question about the order of events seems meaningless, and at bench scale it practically is. The distance where a field changes its mind showed what changes once distances become comparable to a wavelength, and when the source is not heard all at once showed how delay reshapes the field of a thing that is not small. A circuit that is deliberately very long brings the question into sharp focus. It sounds like a riddle, and it went around the internet in that form a few years ago, with some of the answers more confident than they ought to have been.

Take a 12 V battery and a switch in one wire, and a lamp in the other wire, a metre away from the battery. Now route the wires the long way. Each of them runs out a light-second to the right, and a light-second to the left, and at the far ends the two wires are joined. Current cannot get from battery to lamp through the metal without going around a light-second out and a light-second back. The lamp is a metre from the battery in the air between them. How soon does anything happen at the lamp?

There are two tempting answers. One says two seconds, since the current must travel around the conductor. The other, more sophisticated, says that the energy flows through the field, not the wire, and the field need only cross a metre, so the lamp lights up immediately. Each answer has something right, and each gets something wrong. Fortunately the geometry of two parallel wires is well enough known that the matter can be calculated, in a few lines.

Two transmission lines in one loop #

Far from the middle, each pair of rails is a transmission line, the object at the heart of every cable. A step of voltage travelling along such a pair is a sheet of field, moving at the speed of light in the space around the wires, with a current in the wires at the front of it. Those two are tied together. The ratio of voltage to current in the moving step is the line’s characteristic impedance, $Z_0$, set by its geometry alone. For two wires of radius $a$ in air, spaced a distance $D$ apart,

$$Z_0 = \frac{\eta_0}{\pi}\,\text{arccosh}\!\left(\frac{D}{2a}\right),$$

with $\eta_0 \approx 377$ ohms the impedance of free space. For wires 1 mm in radius and a metre apart it is 828 ohms. This is not a resistance. No heat is generated by it. A line of this kind that is long enough looks to anything connected to it exactly like a resistor of 828 ohms, because the current drawn from the source is flowing into new line, charging new capacitance and establishing new magnetic field, at a rate fixed by the geometry, and it keeps on doing so until the front gets to the end. A line has no way of telling from its input whether it goes on for ever.

So at the middle of the circuit, the moment after the fronts have left, the source sees two such lines in series, one to the right and one to the left, and the lamp in series with them too. Kirchhoff’s voltage law is then enough:

$$I = \frac{V}{2Z_0 + R}.$$

For a 12 V battery and a lamp of 30 ohms, the current is 7.1 mA. There are no far ends in this equation. Nothing has had time to report whether the rails join up, or whether they go anywhere. The current through the lamp starts as soon as the field has crossed from one rail to the other, in the metre between them, a time of 3.3 nanoseconds. That is the true sense in which the sophisticated answer is correct. The current flowing in the lamp’s rail, at that moment, is driven by a field that crossed the gap and has nothing to do with whatever the far end looks like.

But it is far from being the lamp’s working current. With the circuit complete and in its final state, the current is $V/R$ = 400 mA. The first current is fifty-six times less.

A stair, not a switch #

After one second, each front reaches its far end. A joined pair of rails is a short circuit, so a wave arriving there reflects with its voltage reversed and its current unchanged in sign. The reflection travels back and reaches the middle at two seconds. There the lamp finds the voltage across each line has changed, and the current steps up. The line’s input voltage is the sum of whatever is going out and whatever has come back, and with $V_+$ the outgoing step and $V_-$ the returning one, each side contributes $Z_0 I + 2V_-$. So a new current follows and sends a new step out, and the process repeats every round trip.

The stair’s height reflects the line again: each return adds $V/Z_0$ approximately, 14 mA for 12 V across 828 ohms, as long as the lamp’s resistance is small compared with the rails’. Seen from a long way off, the stair is a smooth exponential rise. It should be, since a pair of wires a light-second long and joined at the far end is a very large inductor, and the rise of current in an inductor has a time constant $L_{\rm loop}/R$. With a total inductance of $2Z_0\tau$ across both sides, where $\tau$ is the 1 s one-way time, the time constant is $2Z_0\tau/R \approx 55$ seconds, which puts the halfway point at the 38 seconds the stair reaches. The stair is what that smooth curve is made of. Lumped-circuit theory’s exponential rise and the transmission line’s bouncing steps are one calculation, seen at two resolutions.

Calculations of this kind need checking. The algebra gives an exact lattice, but the sign of a reflection is easy to get wrong, and two sides in series are easy to double-count. So here the same loop was also integrated as a pair of delay lines stepped in time, with each wave stepped along its line and a reflection at each join, knowing nothing about the lattice formula. The first six steps agree to within a per cent, and the long-run limit is the required $V/R$.

How much the wires’ spacing matters #

If the first current is limited by $Z_0$, and $Z_0$ is a geometric quantity, the wires’ spacing changes it. Spreading the wires raises the impedance, and moving them together lowers it. It does so slowly, as a logarithm.

The lamp will not visibly glow at 7 mA, and a meter would read it. For the first current to be close to the final value, a line’s impedance would have to be small compared with the lamp’s resistance, and that takes a much lower-impedance line than two wires: closely spaced flat plates, say, or a pair of heavy conductors nearly touching. A typical length of mains flex has an impedance of around 100 ohms, and a coaxial cable around 50; a power cable with broad, closely spaced conductors might get under 20. So with a 200-ohm lamp and a long low-impedance line, the lamp would light near full brightness in the first nanoseconds. How much of the final power arrives at once depends on a ratio that the riddle does not specify.

This is the answer to the sophisticated reasoning. The energy flows through the field, yes. How much energy flows through the field at the lamp’s location, in the first moment, is set by the geometry, and a pair of thin, widely spaced wires does not carry much. A pair of wires does not have to be touching anything to have some capacitance between them and some inductance, and those are the same two numbers as $Z_0$.

What the rails look like in the meantime #

It may seem odd that a current can flow into the rails at all before the circuit is known to be complete. The explanation is displayed in the picture of the line itself.

Half a second in, the first half of the rails carries a voltage and a current; the second half is entirely undisturbed. At the moving front, electrons at the wire’s surface are being shifted, in the way the charge that steers a current round a bend described, to give the rails the surface charge that a voltage between them requires. The current does not need a continuous path through metal to flow, because the circuit is being closed, moment by moment, by a displacement current across the space ahead of the front: the term in the term that made light. Between the wires, the changing field is as good as a conductor at carrying the current on, as it is between the plates of a charging capacitor. The rails are a capacitor being charged, one section after another, at the speed of light.

By one and a half seconds the reflection is partway home. Behind it, the voltage is zero, as it has to be near a short circuit; the current has doubled, because the reversed wave adds to it. The middle of the rails has no information, as yet, that there is a join a light-second away. A local measurement there would read exactly what it read a second ago.

There is one more number that brings out how little this has to do with electrons going anywhere.

At the first current, an electron’s drift along the rails is about a sixth of a micrometre per second. The lamp’s own electrons are set moving, within nanoseconds, by the field that has reached them. The ones in the battery will not come near the lamp in any time of interest. Nor do they need to, as electrons are not carrying energy from one place to the other, except in the sense of being the charges on which the field acts. They are a medium. The half a wire always takes found energy in a circuit behaving with a similar disregard for the path through the metal.

Where the energy comes from in the first second #

The argument over this circuit has often been an argument about where the energy flows, and the transmission line settles that as well. A step travelling along the rails carries energy at a rate equal to $V_+ I$ for each side. It delivers that to new line, half stored in the electric field between the rails and half in the magnetic field around them. Nothing is dissipated in the rails, if they have no resistance. So the battery spends most of its first second’s output on charging up the line, far more than it spends on the lamp. With 7.1 mA and 12 V, the battery is producing about 85 mW, of which the lamp’s 30 ohms is taking 1.5 mW. The rest goes down the rails, in a form that can be drawn: the Poynting vector between the wires, as where the energy of a field is defined it, pointing outward along the line and away from the battery.

The lamp is taking its share from the same field. Between the two rails, near the middle of the circuit, a Poynting flow crosses the gap in the first instant from the battery’s rail to the lamp’s rail, and it delivers power that the lamp absorbs, and this is the honest sense in which energy went “through the air”. So far the claim of the second tempting answer holds. Its error is about how much. Most of the energy flowing past the lamp in that moment is on its way somewhere else.

After a very long time, the steady state is simple: an electric field between the rails only near the battery and the lamp, with hardly any voltage on the rails far out, a current everywhere and an enormous magnetic field all along the rails, storing energy of $\frac{1}{2}L_{\rm loop} I^2$. That is the same energy that had to be built up by the steps, about 130 joules in this example, supplied at 85 mW and then at increasing power. Nearly all of what the battery delivers in the first half-minute goes into that stored field and not into light. Then, if the switch were opened, it would come back as a spark. An inductor is a place to keep energy in a field, and the circuit that fights its own change described the same event at bench scale, where the time is milliseconds and no one thinks about the speed of light.

Scaled experiments have been done. Nobody has a light-second of wire. But with a few hundred metres of cable laid out in a field and a fast oscilloscope, the time scale is microseconds and the steps can be seen clearly. Measurements of that kind have been published online and they reproduce the first step, its dependence on the spacing, and its arrival after about the separation’s light-time, with a slow rise that represents the antenna-like opening. Those experiments were the honest end of the online argument.

Where the rails stop being a line #

The treatment uses transmission-line theory, and that is itself an approximation with a domain of validity. It assumes that the field between the rails has the transverse form of a plane wave, that the spacing is small compared with any wavelength of interest and with the line’s length. This fails at the very start, close to the battery and within a metre or so of it, for the first few nanoseconds. There the wires are not parallel lines to the fields that are spreading out; the battery end radiates as a small antenna, and the true early current through the lamp depends on the three-dimensional geometry of what lies between them, including how the switch and the lamp are mounted. The transition between that first few-nanosecond regime and a clean value of $V/(2Z_0+R)$ takes a few metre-light-times. So the number 7.1 mA is what is reached after those, and not a statement about the very first picosecond of the current’s rise.

A real step is also never perfectly sharp. A switch takes time to close, and a fast rise in a metal wire loses its highest frequencies to the skin depth on the way along. Over a light-second of copper wire this would be serious. Ordinary wire of 1 mm radius has a resistance of about 5 milliohms per metre, or a million and a half ohms in total over 300,000 km. That is enormously larger than the lamp and in fact dwarfs everything else in the circuit. The final current in a real light-second loop of such wire would be microamperes, and the picture of steps of 14 mA would be washed away by heating losses along the first line. So the calculation requires superconducting rails, or a thought-experiment rail with no resistance, as the original puzzle quietly did. A long but actual line, such as a submarine cable, shows the stair’s smoothed and rounded version.

The rails are also assumed to be straight, in empty space, far from anything else. Run for a light-second, they would not be, and a practical layout would have different coupling at different points. In particular, an actual layout where the rails run away from each other would have a very different start: if the two wires leave the battery in opposite directions and only come back near each other at the lamp, then there is no pair of rails to carry a transverse wave until the fronts have travelled some distance. The first current in that case is smaller and set by radiation between two isolated conductors. So the geometry of the question decides the answer, and any puzzle that does not specify it has not been fully asked.

Still open: how much of a real grid is this #

The stair is not a curiosity at long lengths. Printed circuits running at gigahertz clock rates have traces that are many wavelengths long, and so does an undersea power cable or a data centre’s network link. Their designers treat every connection as a line of known impedance, which they match at each end to prevent exactly this kind of bouncing. In a large power system the effect appears when a line is switched on. A surge of voltage travels at close to the speed of light down a transmission line of hundreds of kilometres, reflects at any open or different-impedance point and can reach roughly twice the nominal voltage, a fact that governs the insulation of the line. How far these surges interact when thousands of lines and inverters are connected in a modern grid, where the old rule that loads change slowly no longer holds and where fast electronic converters respond faster than the line’s travel time, is an active question for grid planners.

A lamp a metre from its battery on rails a light-second long begins to conduct after the metre’s 3.3 ns, at $V/(2Z_0+R)$, 7.1 mA for the numbers used here, before any signal has gone round; the current then rises by about 14 mA at each two-second round trip, reaching half of its final 400 mA in about 38 seconds, in stairs that a lumped circuit would call an exponential. Neither of the two tempting answers survived the calculation. The habit worth keeping is to ask what a local measurement could know at the moment in question: the lamp knows about the battery a metre away, and has no information yet about anything further.