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The Lawson Criterion & Triple Product

Source lecture(s): PC368 Lec 1

Intuition

A burning plasma is a race between two rates: fusion alphas depositing energy, and the plasma losing it. Fusion power scales as \(n^2\langle\sigma v\rangle\); losses scale as \(3nT/\tau_E\), where \(\tau_E\) is the energy confinement time — how long the plasma would take to cool if you switched the heating off. Setting gain above loss and cancelling one power of \(n\) leaves a single combined figure of merit.

The criterion

Lawson criterion for D–T ignition

\[n\,T\,\tau_E \;\gtrsim\; 3\times10^{21}\ \text{keV}\cdot\text{s}\cdot\text{m}^{-3}\]

Three knobs, one product. You may trade freely among them — and the two mainstream approaches to fusion do exactly that, in opposite directions:

Magnetic (MCF) Inertial (ICF)
Density \(n\) \(10^{20}\) m⁻³ \(10^{30}\) m⁻³
Confinement \(\tau_E\) ~1–10 s ~\(10^{-11}\) s
Temperature \(10^8\) K \(10^8\) K
Strategy hold a thin plasma for a long time crush a dense one and let inertia hold it for an instant

Ten orders of magnitude in density, ten in time, the same temperature, the same product. A tokamak is a bottle; a laser implosion is a hammer.

Where the number comes from

Power balance per unit volume, for a 50:50 D–T mix (\(n_D = n_T = n/2\)):

\[P_\alpha = \frac{n^2}{4}\langle\sigma v\rangle E_\alpha, \qquad P_{\rm loss} = \frac{3nT}{\tau_E}\]

Ignition requires \(P_\alpha \ge P_{\rm loss}\):

\[\frac{n^2}{4}\langle\sigma v\rangle E_\alpha \ \ge\ \frac{3nT}{\tau_E} \quad\Longrightarrow\quad n\tau_E \ \ge\ \frac{12\,T}{\langle\sigma v\rangle E_\alpha}\]

In the 10–20 keV window \(\langle\sigma v\rangle \propto T^2\) to good accuracy, so the right side goes as \(1/T\), and multiplying through by \(T\) leaves the right-hand side roughly constant. That is the small miracle that makes the triple product a meaningful single number rather than a temperature-dependent curve.

Worked example

An ITER-like design point sits at \(n = 10^{20}\) m⁻³ and \(T = 15\) keV. What confinement time does it need?

\[\tau_E \gtrsim \frac{3\times10^{21}}{(10^{20})(15)} = 2\ \text{s}\]

Two seconds — which sounds trivial until you remember it means a plasma at 150 million degrees losing heat no faster than a well-insulated house. Now run the same arithmetic for ICF at \(n = 10^{30}\) m⁻³:

\[\tau_E \gtrsim \frac{3\times10^{21}}{(10^{30})(15)} = 2\times10^{-10}\ \text{s}\]

0.2 nanoseconds. A compressed pellet only has to hold together for the time light takes to cross a dinner plate — which is exactly what "inertial confinement" buys you: no bottle required, just the fuel's own mass failing to get out of the way fast enough.

Progress

Fusion triple product has improved by more than five orders of magnitude since 1960, faster over that period than transistor density. JET reached \(Q \approx 0.67\) in 1997 and 69 MJ of fusion energy in 2023; NIF achieved target gain \(Q > 1\) in December 2022. ITER is designed for \(Q = 10\) at 500 MW — the first burning plasma, where alpha self-heating dominates external heating.

Performance also scales steeply with machine size, roughly \(\tau_E \propto R^{2}\) under standard confinement scalings. That is the awkward reason fusion devices keep getting bigger and more expensive rather than smaller and cheaper — and the reason high-field superconducting magnets, which buy performance without size, attracted so much private capital after 2018.

Common mistakes

  • Treating \(\tau_E\) as a particle confinement time. It is an energy confinement time. Particles often stay much longer than their heat does.
  • Assuming higher \(T\) is always better. Above ~30 keV the D–T cross-section falls and bremsstrahlung losses climb; the triple-product requirement rises again. There is an optimum, and it is around 15 keV.
  • Forgetting that \(n\tau_E\) alone is not enough. The older Lawson form \(n\tau_E\) is only meaningful at a stated temperature; the triple product is what allows machines at different temperatures to be compared on one axis.

Knowledge graph position

Prerequisites: Fusion energy, plasma. Leads to: magnetic confinement, transport and confinement scaling.

Quiz

Q1 (conceptual). Why is the triple product \(nT\tau_E\) a more useful figure of merit than \(n\tau_E\) alone?

Answer

Because \(\langle\sigma v\rangle \propto T^2\) near the operating point, the ignition requirement on \(n\tau_E\) itself depends on temperature (\(\propto 1/T\)). Multiplying by \(T\) absorbs that dependence and gives a threshold that is nearly constant across 10–20 keV, so machines at different temperatures can be plotted on one axis.

Q2 (computational). A stellarator runs at \(n = 3\times10^{19}\) m⁻³, \(T = 10\) keV, \(\tau_E = 0.5\) s. How far is it from ignition?

Answer

\(nT\tau_E = 3\times10^{19}\times10\times0.5 = 1.5\times10^{20}\) keV·s·m⁻³ — a factor of 20 below the \(3\times10^{21}\) threshold.

Q3 (MCQ). ICF and MCF differ by ten orders of magnitude in density but need the same temperature. This is because temperature is set by:

  • (a) the energy confinement time
  • (b) the Coulomb barrier and the fusion cross-section
  • (c) the magnetic field strength
  • (d) the plasma frequency
Answer

(b). Temperature is fixed by nuclear physics — the energy needed to tunnel the Coulomb barrier at a useful rate. Only \(n\) and \(\tau_E\) are engineering choices, which is precisely why the two approaches can trade them off against each other.