The half a wire always takes

Where the energy of a field is found that a capacitor’s energy is stored in the electric field between its plates, at a density of $\tfrac12\varepsilon_0E^2$, and that when a current flows the energy moves through the space around the wires rather than along them. That accounting is exact, and it leads straight to a puzzle that appears in introductory textbooks and has generated a small literature of its own, because the obvious answers are all wrong.

Take a capacitor of capacitance $C$, charged to a voltage $V_0$, so that it holds a charge $Q_0 = CV_0$ and an energy $\tfrac12CV_0^2$. Connect it to a second, identical capacitor, initially empty. Charge flows from the first to the second until their voltages are equal. Charge is conserved, so each ends with $Q_0/2$ and a voltage $V_0/2$. Each then holds an energy $\tfrac12C(V_0/2)^2 = \tfrac18CV_0^2$, and the two together hold $\tfrac14CV_0^2$. Half the energy has disappeared.

The half that goes #

The first answer is that the wire has resistance and the lost half is heat. That is right, and it raises an immediate objection: the resistance can be anything, so the heat ought to depend on it.

The circuit is two capacitors in series with a resistance $R$, so the effective capacitance is $C/2$ and the current falls exponentially from $V_0/R$ with a time constant $\tau = RC/2$. The power dissipated is $I^2R$, and its integral over all time is

$$\int_0^\infty \frac{V_0^2}{R^2}e^{-2t/\tau}R\,dt = \frac{V_0^2}{R}\cdot\frac{\tau}{2} = \frac{V_0^2}{R}\cdot\frac{RC}{4} = \tfrac14CV_0^2.$$

The resistance cancels. The heat is exactly the missing half, for every value of $R$. A resistance ten times larger makes the current ten times smaller, so the power at any moment is ten times smaller; but the current then flows for ten times as long, and the two factors cancel. The first capacitor’s energy falls to a quarter and the second’s rises to a quarter, and the stored total, dashed in the figure, drops from one to one-half and never recovers.

Over eight orders of magnitude of resistance the two falling and rising lines mirror each other exactly, and the energy lost does not move. A wire of a hundredth of an ohm delivers a ten-kilowatt pulse lasting a few picoseconds; a megohm resistor dribbles the same energy out over half a millisecond, at a ten-thousandth of a watt. To make the loss smaller, the wire is the wrong place to look.

Where the final state comes from #

The reason the resistance does not matter is that the final state does not depend on it. Charge is conserved absolutely; the voltages must be equal at the end, because any difference would drive more current; and those two facts fix the final charges, and with them the final energy, without reference to anything that happened on the way. Whatever path the system takes between the two states, it ends with half its energy, and the other half must be somewhere. The resistance only decides how fast it gets there.

That kind of argument is familiar from mechanics, and the match is closer than a resemblance. In collisions are easier than forces, two bodies that collide and stick share their momentum, which is conserved, and end at a common velocity fixed by that sharing; the kinetic energy lost does not depend on what the bodies are made of, how hard they are or how long the impact takes. It depends only on the masses. The capacitor problem is the same problem in different variables.

The dictionary is short. Charge $Q$ is conserved, as momentum $p$ is. The energy stored is $Q^2/2C$, as kinetic energy is $p^2/2m$. Capacitance plays mass. The final common voltage $V = Q/(C_1+C_2)$ is the final common velocity $p/(m_1+m_2)$. The fraction of energy lost is $C_2/(C_1+C_2)$ in the circuit and $m_2/(m_1+m_2)$ in the collision, the same function.

The analogy explains the lopsided cases at once. A large capacitor sharing its charge with a small one loses little: almost all the charge stays where it was, as a lorry hardly slows when a fly sticks to its windscreen. A small capacitor sharing with a large one loses almost everything: almost all its charge moves to where a coulomb stores very little energy, as a fly that hits a parked lorry loses nearly all its kinetic energy. It also explains why nothing about the wire enters. A perfectly inelastic collision loses the same energy whether the bodies are clay or steel with a latch; the material decides only where the energy goes and how fast.

The same structure appears whenever a conserved quantity is shared between two stores that hold it quadratically. Two equal tanks of water joined at the bottom end at half the height of the full one, and lose half its potential energy to the turbulence of the flow between them — the waste the work left in two buckets of water found could have been an engine’s output, had something extracted it instead.

When the wire has no resistance #

The puzzle sharpens if the resistance is taken away. A superconducting wire has none, and the integral above, with $R$ set to zero, gives nonsense: an infinite current for zero time.

The flaw is the model, not the physics. Any loop of wire has inductance — current in it makes a magnetic field, whose energy has to be supplied — and a circuit with inductance and capacitance and little resistance does not decay smoothly. It rings, its current lagging the voltage that drives it, the lag the ring that jumps because its current is late turned into a force.

With little resistance the charge rushes across, overshoots, and rushes back; the empty capacitor charges almost to the full voltage before the current reverses, and energy passes back and forth between the capacitors’ electric fields and the loop’s magnetic field, as it does in the three ways of coming to rest for any lightly damped oscillator. The swings die away, and both voltages settle at $V_0/2$ — and the integrated dissipation is, again, half the starting energy. The inductance changes the shape of the transfer completely and the total loss not at all, because it does not change the final state. It only stores energy temporarily and gives it back.

Moving the charge instead of sharing it #

The ringing figure holds a way round the loss, hidden in its first half-cycle. At the moment the current first falls to zero, the empty capacitor has charged to 9.7 volts and the charged one has fallen to a third of a volt: almost all the charge has moved, not been shared. Open the loop at that instant and the transfer is over, with the second capacitor holding $(9.7/10)^2$, or 94 per cent, of the starting energy. The loss in that half-cycle is only what the resistance takes from one swing of a lightly damped oscillator, roughly $\pi/2Q$ of the energy for a quality factor $Q$, and it shrinks as the resistance does. Here, at last, a better wire helps.

The difference is the inductance, which stores the energy in transit as a magnetic field and gives it back instead of letting the voltage difference drive current straight through a resistance. Its magnetic field keeps the current flowing after the voltages have equalised, carrying the charge on past the halfway point, as a mass on a spring carries on through its equilibrium. In the collision picture, the inductor is a spring between the two bodies, and opening the loop at the first current zero is releasing the spring at the instant the first body has stopped: an elastic collision, in which equal masses exchange their velocities completely and nothing is lost.

That is the whole difference between two families of electronic power converters. A switched-capacitor converter moves energy by connecting capacitors at different voltages, and every connection loses the share computed here, set by the voltage mismatch; its efficiency is high only when the capacitors are switched nearly in balance. An inductor-based converter moves energy through a magnetic field and opens its switches when the current is right, and can approach the efficiency of the elastic case. A circuit designer choosing between them is choosing between a collision that sticks and one that bounces.

The lost half of the sudden connection is also an entropy. Dissipated as heat $\tfrac14CV_0^2$ into surroundings at temperature $T$, it raises their entropy by $CV_0^2/4T$, and nothing in the closed circuit can lower it again: no later manipulation of the two capacitors restores the first one’s charge without spending at least that much work from outside. The sharing is irreversible in exactly the sense entropy is a count made precise, while the inductive transfer, opened at the right instant, is reversible up to its small resistive loss. The arithmetic of charges is a statement of thermodynamics in disguise.

The half that leaves as light #

With no resistance at all, and the loop left closed, the ringing would continue for ever and the capacitors would never settle — unless something else removes energy. Something does.

A loop of wire carrying an oscillating current is an antenna. Every charge going round it is accelerating, and a charge that turns must glow. A small loop radiates as a magnetic dipole, and the power it sends away can be written as the current squared times a radiation resistance,

$$R_{\text{rad}} = 320\pi^4\left(\frac{A}{\lambda^2}\right)^2 \text{ ohms},$$

for a loop of area $A$ ringing at wavelength $\lambda$. This resistance appears in the circuit exactly like an ordinary one, except that the energy it takes leaves as electromagnetic waves instead of as heat, and since the lost half does not depend on what the resistance is, it does not depend on what kind it is either. If the wire is superconducting, the radiation resistance is the only resistance, and all of the lost half is radiated.

In ordinary circuits the radiated share is tiny. A loop is a poor antenna unless it is comparable in size to the wavelength, and the radiation resistance of a ten-centimetre loop ringing at a few megahertz is tens of microohms, while the copper’s resistance at that frequency, with the current crowded into a skin a few hundredths of a millimetre deep — the depth how far a field gets into metal computed — is tens of milliohms. Larger loops radiate more, because the radiation resistance rises as the fourth power of the radius over the wavelength and a larger loop’s lower ringing frequency does not cancel that, but even a loop three metres across made of thick rod radiates only a per cent or two of the loss. The rest is heat.

That resolves the version of the puzzle that asks where the energy goes when the wire is ideal. It goes to infinity, as radio waves. The resolution is not a trick of the limit. The radiation is present in every case, with every resistance; a real circuit always splits its lost half between heat and light in the ratio of its two resistances, and only the proportion changes.

Where the energy travels while it is being lost #

The picture from where the energy of a field is adds one more piece. During the transfer, energy does not flow along the wire like water in a pipe. It flows through the space between the capacitors and the wire, carried by the Poynting vector $\mathbf{E}\times\mathbf{H}$ — the cross product of the electric field, which the charges set up, and the magnetic field of the current. Near a resistive wire the Poynting vector points into the wire from the side, and the energy that arrives that way is the energy that heats it. Near an ideal loop, far from it, the fields are those of the radiation, and the Poynting vector points outwards towards infinity.

The two halves of the energy therefore end in different kinds of places: one half remains in the electric fields between the two pairs of plates, at a quarter of the original density in each — because the field in each is half the original — and the other half has crossed the surface of the wire or the surface of a large sphere. The arithmetic of $\tfrac12\varepsilon_0E^2$ is the arithmetic of the charges, written over space, and it gives the same half.

Charging gently #

The same loss appears whenever a capacitor is charged from a voltage source. Connect an empty capacitor to a battery of voltage $V$ through any resistance, and the battery delivers a charge $CV$ at voltage $V$, an energy $CV^2$; the capacitor stores $\tfrac12CV^2$. The other half is lost, whatever the resistance, for the same reason as before.

That half is the energy every logic gate in every digital circuit wastes each time it switches, charging and discharging the capacitance of the wires and transistors it drives, and it is most of the power a processor uses. It is not a fundamental limit. It comes from switching the full voltage on at once.

The loss in each step is set by the voltage difference across the resistance while the charge moves. Divide the charging into $N$ steps of size $V/N$, and each step loses $\tfrac12C(V/N)^2$, so the total loss is $\tfrac12CV^2/N$. In the limit of a slowly rising supply — one that keeps the voltage across the resistance small at every moment — the loss falls in proportion to the ratio of the circuit’s time constant to the ramp time, and can be made as small as the ramp is made slow.

This is adiabatic charging, and it is the principle behind circuits designed to recover the energy of switching instead of burning it: the supply rises and falls with the logic, slowly, and the energy stored in each node is returned to the supply rather than dumped. The price is time. A circuit that charges its nodes slowly computes slowly, and the trade between energy per operation and speed is exactly the trade between the resistance and the ramp time here. Pushed to its limit, the same reasoning reaches the energy that the bit that has to be paid for found irreducible: the $kT\ln 2$ owed only when information is erased. Charging a capacitor need not erase anything, and in principle need cost nothing at all.

Two plates, a wire and the things left out #

The circuits drawn here are lumped: each capacitor a single capacitance, each wire a single resistance and inductance, with no delay along the wire and no field outside the components. That is accurate when the circuit is much smaller than the wavelength of its ringing, and it is exactly the condition under which the small-loop radiation formula holds; a circuit comparable to its wavelength is a transmission line and an antenna at once, and its transfer is a sequence of reflected pulses rather than a smooth exponential. The capacitors are ideal, with no leakage and no dielectric loss; a real dielectric absorbs some energy and gives some back late, a “soakage” that shows up as a capacitor recovering a little voltage after being discharged. The connection is drawn as instantaneous. If it is made by closing a switch in air, the last fraction of a millimetre is crossed by a spark, which is simply a resistance of a particular kind and takes the same half. The skin-depth resistance assumes a smooth, round copper conductor at room temperature. The domain is circuits from centimetres to metres, ringing at kilohertz to tens of megahertz, with charges small enough that the capacitors stay linear.

Still open: how slowly a computer can afford to switch #

The physics of the lost half is closed. What is not closed is how much of it engineering can recover. Adiabatic logic has been built and shown to cut switching losses by large factors, but it needs power supplies that generate slowly ramping waveforms and recover the returned energy efficiently, and the inductors and resonators that do this are large and lossy compared with the transistors they drive. Leakage currents in modern transistors also leak away some of the charge that a slow ramp keeps in place, putting a floor under how slow is useful. Whether reversible, adiabatic designs can beat conventional circuits at energy per useful operation in real products, at speeds anyone would accept, is being tested now; the arithmetic says the waste is not compulsory, and the engineering has not yet said how much of it can be avoided at a price worth paying.

The puzzle itself reduces to one sentence. Joining a capacitor charged to $V_0$ to an empty one of equal size leaves each at $V_0/2$ holding a quarter of the energy, because charge is conserved; the other half is lost whatever joins them — as heat, in a time $RC/2$ that the resistance sets, or as radiation when the resistance is zero — exactly as a collision that sticks loses $m_2/(m_1+m_2)$ of its energy whatever the bodies are made of. The half is the price of sharing suddenly, and the only way to avoid it is to share slowly.