The product a fusion plasma has to beat

The dust a plasma freezes into a crystal ended with a plasma cold enough to be watched through a video camera, with the charges in it arranging themselves into a lattice. The opposite extreme of the subject has always had a blunter question attached to it. Heat a mixture of deuterium and tritium until its nuclei collide hard enough to fuse, and each reaction releases seventeen and a half million electronvolts. How hot does it have to be, and for how long, before the plasma pays its own way?

John Lawson, an engineer at the British atomic establishment at Harwell, answered in a report written in 1955 and classified until 1957. His answer is a piece of arithmetic that would fit on a postcard, and it has governed every fusion experiment built since. It turns out to contain no reference to how the plasma is held. Density and time appear only as a product, which is why two machines as different as a ring of magnets the size of a building and a laser fired at a capsule the size of a peppercorn are competing against the same curve. It also explains the odd fact that the fusion in a star is feeble.

The reaction that sets the terms #

Deuterium and tritium are chosen because no other pair comes close. The reaction $\mathrm{D} + \mathrm{T} \to {}^4\mathrm{He} + n$ releases 17.6 MeV, a fifth of it in the helium nucleus, the alpha particle, at 3.52 MeV, and the rest in the neutron. The neutron is uncharged and leaves the plasma at once, carrying its 14.1 MeV off to whatever surrounds the plasma. The alpha particle is charged. In a magnetic field or a dense enough plasma it stays, and it gives up its energy by colliding with the fuel. Only that fifth is available to keep the plasma hot.

How often the reaction occurs is set by the reactivity $\langle\sigma v\rangle$, the fusion cross-section times the relative speed, averaged over the thermal distribution. It climbs steeply with temperature, because the nuclei have to tunnel through their mutual repulsion, and the wall that is not quite a wall describes how sharply that tunnelling depends on energy. What matters at a given temperature is the fast tail of the distribution. The value used here comes from the fit Henri-Selim Bosch and Gerald Hale published in 1992, which gives $1.136 \times 10^{-22}\ \text{m}^3/\text{s}$ at 10 keV and $4.33 \times 10^{-22}\ \text{m}^3/\text{s}$ at 20 keV. Temperatures in this subject are quoted in energy units; 1 keV is about 11.6 million kelvin.

With equal numbers of the two nuclei and a total ion density $n$, the alpha heating per unit volume is $\tfrac14 n^2 \langle\sigma v\rangle E_\alpha$. The plasma radiates too. Electrons deflected by ions lose energy as bremsstrahlung, the same radiation a charge that turns must glow derives for any acceleration. That loss goes as $n^2\sqrt{T}$, and a plasma hot enough to fuse is far too thin to absorb it, so it leaves.

A floor that density cannot rescue #

The two losses and gains scale identically with density. That makes the crossing of these two curves unusual: it is a condition on temperature alone. Below about 4.3 keV, some fifty million kelvin, a deuterium–tritium plasma radiates more than its alpha particles can return, and compressing it, cooling its surroundings or waiting longer makes no difference. This is what people mean by the ideal ignition temperature.

Above it, heating rises far faster than radiation, and by 14 keV alpha particles are ahead by a factor of seventeen. Radiation stops being the thing to worry about. The serious loss in any real machine is the slower leakage of the plasma’s own heat: conduction along and across magnetic field lines, turbulence, or in an unconfined plasma simple expansion. How that happens is a very large subject. Lawson’s step was to refuse to go into it, and to give it a single number, a time.

The plasma stores thermal energy $3nT$, counting ions and electrons separately at a common temperature. Let that energy leak away over a time $\tau$, the energy confinement time. It says nothing about the mechanism, and it can be measured directly by turning the heating off and timing the fall. The balance then reads

$$\tfrac14 n^2 \langle\sigma v\rangle E_\alpha = C n^2 \sqrt{T} + \frac{3nT}{\tau},$$

and the density in the last term has one power fewer than in the others. Divide through by $n$ and the density and time come together:

$$n\tau = \frac{3T}{\tfrac14\langle\sigma v\rangle E_\alpha - C\sqrt{T}}.$$

Only the product appears #

The right-hand side holds only temperature and constants. So there is one curve, and a plasma is on one side of it or the other. Its lowest point is near 26 keV, at $1.6 \times 10^{20}$ s/m³; at 14 keV the requirement is about $2.1 \times 10^{20}$. Its shape on the right is worth a moment. The reactivity’s rise slows at high temperature, while the stored energy goes on growing with $T$, so a hotter plasma costs more to keep than the extra reactions pay for. There is no benefit in being arbitrarily hot.

This separation is the whole of Lawson’s insight. It takes a question about magnets, turbulence, lasers and shock waves and turns it into a condition on two numbers multiplied together. Nobody needs to agree on how the plasma is to be held before agreeing on what it has to achieve.

Magnetic machines have a further reason to rewrite the result. A magnetic field can hold a certain pressure, and that pressure is $nT$ up to a constant, so what a given field limits is the product of density and temperature, not the density alone. The same force whichever way the surface faces found that ratio of plasma pressure to magnetic pressure, and that it stays at a few per cent in a tokamak. The figure of merit for such a machine is therefore pressure times time, which is the temperature times the curve above.

The triple product has a sharp minimum of about $3 \times 10^{21}$ keV s/m³ near 14 keV, about 160 million kelvin. This is the number on the wall of every fusion laboratory. It is why a machine’s design temperature is not the highest it could reach, and why reported “records” for temperature alone have rarely meant much. A plasma can be made far hotter than the Sun’s core on a laboratory bench; holding one at $10^{20}$ ions per cubic metre for several seconds, meanwhile, took decades.

ITER, under construction in the south of France, aims for a density of about $10^{20}$ per cubic metre, a temperature of 8.9 keV and a confinement time of 3.7 s, which multiplies to $3.3 \times 10^{21}$. That point sits slightly below the ignition curve at its temperature. It is meant to, since the machine is intended to run with external heating still on and to multiply it tenfold, not to burn alone.

One line, with very different places to stand #

Draw the requirement at a fixed temperature, and it is a straight line on logarithmic axes, sloping down at exactly one decade per decade. The line runs across some sixteen orders of magnitude, and two families of machine have settled at its two ends for reasons outside the line itself.

At the low-density end, a tokamak holds about $10^{20}$ ions per cubic metre, a hundred-thousandth of the density of air, for a few seconds. The density cannot be higher, because the pressure at 160 million kelvin has to be contained by a magnetic field, and even very large superconducting magnets hold only a few atmospheres of plasma. At the high-density end, a laser-driven capsule reaches $10^{31}$ ions per cubic metre, a thousand times solid density, for a tenth of a nanosecond, and in that case nothing holds it at all: the time is just how long the fuel takes to fly apart. The fuel held together only by its inertia follows that route.

In between lies a region nobody has found a use for. Air at sea level has about $2.5 \times 10^{25}$ molecules per cubic metre and would need only 80 microseconds, which sounds easy. The pressure of air at 160 million kelvin, however, is half a million atmospheres, which no magnet can contain, and 80 microseconds is many orders longer than it takes for such a gas to expand. Pulsed machines have explored this gap for decades with magnetic pinches and imploding liners, and the current that squeezes what carries it introduced their basic relation.

What a burn does with its freedom #

A curve drawn at a steady balance says nothing about getting there or staying there, which requires integrating the energy balance in time.

The integration uses nothing more than the two powers and the loss term. With $\tau = 2$ s, below the requirement at any reachable temperature, a plasma started at 14 keV cools and dies. With $\tau = 5$ s the same start climbs. Started from 6 keV instead, however, the same plasma with the same confinement goes out, because at that temperature alpha heating has not yet caught up with the losses, and there is a point of no return between. This is ignition in the literal sense, a threshold in a state and not merely in a parameter: a real machine has to be heated across it by something else.

Having crossed it, the burn is not stable. Near 10 keV the reactivity rises about as the square of the temperature, so a small rise in temperature increases the heating, which raises the temperature further. In the figure the climb stops only when the plot runs out. In a real machine, $\tau$ falls as the plasma heats, because turbulence grows with heating power, and that provides a natural brake; the balance may yet need active control by adjusting the fuel density. A petrol engine or a candle has a chemically limited flame whose rate does not run away with its heat in this manner, and a burning plasma’s closest everyday relation is the heap that sets itself alight, which has the same two-fate structure for a similar reason.

Alpha heating also brings a problem particular to machines that succeed. Burned fuel accumulates as helium ash, which dilutes the fuel, radiates and has to be removed. A plasma with 10 per cent helium needs a noticeably higher triple product. No figure here includes this, and neither does the original criterion.

The fusion in a star is feeble #

The surprising comparison is with the object that every popular account compares fusion to. The centre of the Sun releases about 276 W per cubic metre. A compost heap manages hundreds of watts per cubic metre, and a resting person more than a kilowatt. Fusion in the Sun is not a violent process, and if a cubic metre of its core could be brought to the surface and kept there, it would be a disappointing heater.

The reason is that its first reaction, two protons becoming a deuteron, requires one proton to turn into a neutron during the collision. This is a weak interaction, far too improbable to be seen in any laboratory; its cross-section is computed and has never been measured. The protons a star cannot afford deals with why such a star burns as slowly as it does. Deuterium and tritium skip that step entirely and fuse through the strong force alone, with a rate per collision some twenty-five orders of magnitude larger at comparable energies.

What the Sun has in place of reactivity is confinement. Its energy works its way out by the random walk that the light that takes a hundred thousand years to leave describes, and in Lawson’s terms that gives a density of about $6 \times 10^{31}$ per cubic metre for about $3 \times 10^{12}$ s. The product is many orders of magnitude past anything the laboratory can reach, and the temperature is only 1.3 keV. A star wins by holding its fuel with gravity for an astronomical time, and a laboratory cannot, so it has to choose a far faster reaction and run it ten times hotter. Everything about a fusion reactor that looks unlike a star follows from that one substitution. A star’s stability also has the opposite sign, as the ball of gas that heats up as it cools showed: gravity gives it a negative heat capacity, so a rise in reaction rate expands and cools it, and a star cannot run away the way an unchecked laboratory burn can.

What gain costs, in the same units #

Very few machines are designed to ignite outright. What they aim at is gain, $Q$, the fusion power released divided by the heating power supplied from outside. That number is what appears in headlines, and it has a simple relation to the triple product, because the alpha particles carry a fixed fifth of the fusion power.

Write the balance with an external heating term added. Heating from outside and heating from alphas together pay for the losses, and $Q$ fixes the ratio between them: external power is $P_{\text{fus}}/Q$ and alpha power is $P_{\text{fus}}/5$. The losses to be paid are therefore $P_{\text{fus}}(1/5 + 1/Q)$, whereas at ignition they had only $P_{\text{fus}}/5$ to draw on. Since the required product goes inversely as the heating available per unit of fusion power, a plasma with gain $Q$ needs only

$$\frac{(nT\tau)_Q}{(nT\tau)_{\text{ign}}} = \frac{Q}{Q+5}$$

of the ignition value, ignoring radiation. Scientific break-even, $Q = 1$, needs one-sixth of the product. The tenfold gain ITER is designed for needs two-thirds, and that is what its position in the figure shows. Infinite gain is the full curve.

It is a surprisingly lenient relation at the bottom end and a harsh one at the top. The JET machine in England reached a $Q$ of about 0.67 in 1997 with a triple product well below the curve. Going from a gain of 10 to ignition requires only 50 per cent more in the product, and it is a difference nobody yet needs, because a power plant can happily spend a tenth of its output keeping itself hot. The same relation also makes clear why a headline value of $Q$ is no guide to a power plant. A reactor must turn neutrons into heat and heat into electricity at perhaps 40 per cent, drive its own heating systems from that electricity at perhaps 50 per cent efficiency, and run its magnets and pumps. A plasma $Q$ of 1 is far below the point at which the electric plant breaks even, which is more like a $Q$ of 10 or more.

One temperature, one density, one time #

Everything drawn here is what a zero-dimensional balance can say, with one temperature, one density and one time standing in for profiles that vary by large factors across a real plasma. Ions and electrons are assumed to share a temperature, whereas alpha particles heat mainly the electrons and external beams often heat mainly the ions. The bremsstrahlung formula assumes pure hydrogen isotopes; a small fraction of heavy impurities, whose bremsstrahlung goes as the square of their charge and which also radiate line emission, moves the whole curve upward.

The reactivity is the well-measured part. Its uncertainty is less than a per cent at these temperatures, a fact owed to careful low-energy beam experiments and their fit. The confinement time is the part where the criterion refuses to speak. $\tau$ is whatever the machine delivers. Its dependence on size, field, current and heating power is known in tokamaks only from empirical fits to many machines. Those fits predict that a larger machine confines longer; ITER’s size was fixed by them, and they are extrapolated well beyond any device that has yet run. Whether turbulence continues to obey them at ITER’s size is an important question that the machine exists to test.

Still open: whether the product can be reached more cheaply #

All the arguments above apply to deuterium–tritium, and the fuel brings its own problems. Tritium has a 12-year half-life and has to be bred from lithium in the machine’s wall by the neutrons it produces, and those 14 MeV neutrons damage and activate whatever they strike. Other fuels with fewer neutrons exist, and their requirements can be read off the same arithmetic. Deuterium with helium-3, and proton with boron-11, both require reactivities that peak at temperatures of several hundred keV, at which bremsstrahlung in hot electrons takes far more of the output. A proton-boron plasma may be unable to reach the balance at all without keeping its ions and electrons far from equilibrium, and whether it can do so is disputed.

High-temperature superconducting magnets, which became practical in the 2010s, have changed one number. The pressure a magnet can hold goes as the square of its field, and a doubled field permits a plasma of the same performance in a machine a small fraction of the volume. Several private ventures are now building on that reasoning, and whether the empirical confinement laws hold at those high fields, along with whether the magnets and first walls survive, is being tested as this is written.

The ignition condition contains a temperature floor of about 4.3 keV that no density can beat, and above it a product: about $3 \times 10^{21}$ keV s/m³ near 14 keV. Density and time trade one for one, so a thin gas held for seconds and a crushed pellet held for a tenth of a nanosecond are the same answer, and a star, at 276 W per cubic metre, reaches it on time alone.