Lawson Criterion

The Lawson criterion sets the minimum bar a fusion reactor must clear to “get out more energy than you put in.” It was derived in 1957 by the British physicist J.D. Lawson, and for more than half a century it has served as a yardstick for measuring how far fusion development has come. This page builds up, starting from the simplest ideas, to explain why energy balance gives rise to three quantities—density, temperature, and confinement time—and how they connect to ignition and the Q value.

Start with Intuition (High School)

Think of fusion as a campfire. To keep a fire burning, the rate at which you add wood (the rate at which energy is produced) has to exceed the rate at which heat escapes (the rate at which energy is lost). If heat escapes faster, the fire goes out. It is the same in a fusion reactor: within the plasma (an extremely hot gas in which atomic nuclei and electrons fly around separately), the heat produced by reactions must not lose out to the heat leaking away.

So what helps increase the heat produced and reduce the heat lost? Three things matter.

The first is density. Fusion only happens when nuclei collide, so the more crowded the particles are, the more often collisions occur and the more reactions take place. It is like how a room packed with people makes it easier for people to bump into each other.

The second is temperature. Every nucleus carries a positive electric charge, so as they approach they repel each other through the electric force. To overcome this repulsion and get close enough, they need to collide at tremendous speed. High temperature means particles are flying around fast, so the higher the temperature, the more they can cross the wall of repulsion and fuse. The temperature fusion requires is an astonishing figure of about 100 million degrees.

The third is confinement time. Even if you go to the trouble of heating things up, it is pointless if the heat escapes right away. How long the plasma can hold on to its heat—that staying power—is the confinement time. Just as a thermos keeps water warm for a long time, the ability to trap heat without letting it escape is essential.

Think of the Lawson criterion as a numerical statement of how much of these three—density, temperature, and confinement time—you need for the fire to keep burning on its own. Making just one of them large is not enough; the condition is met only when the product of all three exceeds a certain value.

Understand the Physics (Undergraduate)

Let us write the energy balance out properly. The thermal energy the plasma holds per unit volume comes from each particle carrying on average $\frac{3}{2}k_B T$ of energy. Since ions and electrons are present in equal numbers, using the density $n$ (the electron density) we can write

$$W = 3 n k_B T$$

Here $k_B$ is Boltzmann’s constant and $T$ is the temperature. The ions and electrons each carry $\frac{3}{2}n k_B T$, so together they give $3 n k_B T$.

The rate at which this energy is lost is described by the energy confinement time $\tau_E$. By definition, the loss power per unit volume is

$$P_{\text{loss}} = \frac{W}{\tau_E} = \frac{3 n k_B T}{\tau_E}$$

The longer $\tau_E$ is, the more slowly energy escapes for the same stored energy—in other words, the better the heat is retained.

Now consider the power produced by fusion reactions. In the D-T reaction, which uses deuterium (D) and tritium (T) as fuel, each reaction releases 17.6 MeV of energy. Of this, 3.5 MeV is carried off by an alpha particle (a helium nucleus) and 14.1 MeV by a neutron. The electrically charged alpha particle stays inside the plasma and heats it from within, while the neutron, carrying no charge, slips through the plasma and escapes outward.

When D and T are mixed in equal amounts, each has density $n/2$. The reaction rate per unit volume is given by $\frac{n}{2}\cdot\frac{n}{2}\langle\sigma v\rangle = \frac{n^2}{4}\langle\sigma v\rangle$. Here $\langle\sigma v\rangle$ is the reactivity, the product of the reaction cross section and relative velocity averaged over the temperature distribution—a quantity set by the temperature. Taking the total energy per reaction to be $E_f = 17.6$ MeV, the fusion power density is

$$P_{\text{fus}} = \frac{n^2}{4}\langle\sigma v\rangle E_f$$

The key point is that it scales as the square of the density: doubling the density makes the reactions four times as frequent.

Ignition is the state in which the plasma can maintain its heat from alpha-particle heating alone, with no external heating. Writing the condition where the alpha-particle share of the power $P_\alpha = \frac{n^2}{4}\langle\sigma v\rangle E_\alpha$ (with $E_\alpha = 3.5$ MeV) exactly balances the loss power gives

$$\frac{n^2}{4}\langle\sigma v\rangle E_\alpha = \frac{3 n k_B T}{\tau_E}$$

Solving this for $n\tau_E$ gives

$$n\tau_E = \frac{12 k_B T}{\langle\sigma v\rangle E_\alpha}$$

This is the most basic form of the Lawson criterion. If the product of density and confinement time on the left exceeds the temperature-dependent value on the right, the fire keeps burning on its own. For the D-T reaction, the temperature that minimizes the right-hand side is around 10 to 20 keV (about 100 to 200 million degrees), and the condition there is roughly

$$n\tau_E \gtrsim 1.5 \times 10^{20}\ \text{m}^{-3}\cdot\text{s}$$

Now that we have the formula, in words it says: “if the product of the number of particles per cubic meter and the number of seconds the heat is retained exceeds roughly $1.5\times10^{20}$, you reach ignition.”

Deepen the Theory (Graduate)

In the derivation above we lumped the losses together as $W/\tau_E$, but more precisely the losses have several channels. The main ones are transport loss, in which particles and heat are carried out to the edges of the device, and bremsstrahlung, in which electrons emit electromagnetic radiation and cool as they are deflected near nuclei. $\tau_E$ is an empirical quantity that captures these transport losses collectively, and its physical content is the subject of transport theory. We cover this in detail on the Transport Phenomena page.

The reason the triple product $nT\tau_E$ is used as a performance metric also becomes clear from this derivation. Rewriting the Lawson criterion above in the form of $nT\tau_E$ gives

$$nT\tau_E \gtrsim \frac{12 (k_B T)^2}{\langle\sigma v\rangle E_\alpha}$$

The right-hand side is a function of temperature alone, and for the D-T reaction it forms a nearly flat minimum around $10$ to $20$ keV. In this temperature range $\langle\sigma v\rangle$ increases roughly in proportion to $T^2$, so the temperature dependence on the right-hand side cancels out, and the required value of $nT\tau_E$ becomes a constant that is nearly independent of temperature. That is exactly why the triple product is a convenient yardstick that is little affected by small differences in temperature. The triple product required for ignition is estimated to be roughly

$$nT\tau_E \gtrsim 3 \times 10^{21}\ \text{keV}\cdot\text{m}^{-3}\cdot\text{s}$$

The strength of this metric is that it condenses everything into the product of three independently measurable quantities: density, temperature, and confinement time.

Even without reaching ignition, you can obtain a net gain while supplementing with external heating. This is where the fusion energy gain factor $Q$ enters. $Q$ is defined as the ratio of fusion output power to external heating power.

$$Q = \frac{P_{\text{fus}}}{P_{\text{ext}}}$$

$Q = 1$ is scientific breakeven (the state where the heating power put in equals the fusion output). $Q = 10$ is ITER’s target value, meaning obtaining ten times the external heating as fusion output. $Q \to \infty$ is ignition, corresponding to the state where the reaction sustains itself with zero external heating. The fraction of the plasma’s self-heating is set by the contribution of alpha-particle heating, and in terms of $Q$ the self-heating fraction is given by $\frac{Q}{Q+5}$. This comes from the fact that, in the D-T reaction, the alpha particle carries 3.5 MeV of the total 17.6 MeV—that is, $1/5$. At $Q = 5$ self-heating is half, and at $Q \to \infty$ it is 100 percent self-heating, which is ignition.

Research Frontier (PhD)

The strategy for reaching the Lawson criterion differs greatly depending on the confinement approach. Details of each approach are collected on the Confinement Schemes page.

Magnetic confinement uses strong magnetic fields to confine the plasma for a long time, as in tokamaks and stellarators. The density is relatively low, around $10^{20}\ \text{m}^{-3}$, but the triple product is built up by extending the confinement time to the order of seconds. It is a combination of low density and long time. In tokamaks, the central research topics are the transition to the high-confinement mode (H-mode) and the improvement of confinement by increasing device size. The key to determining $\tau_E$ is how to suppress turbulent transport in the plasma, and the mechanism of transport-barrier formation remains an active research subject today.

Inertial confinement compresses and heats a fuel pellet all at once with powerful lasers, reacting it in the very brief time before the particles fly apart under their own inertia. The density reaches an overwhelmingly high $10^{31}\ \text{m}^{-3}$, but the confinement time is under a nanosecond. It tries to satisfy the Lawson criterion with a combination that is the exact opposite of the magnetic approach: ultra-high density and ultra-short time. In the inertial approach, instead of the product of density and confinement time, the areal density $\rho R$ (the density-radius product) of the compressed fuel is used as the metric, and $\rho R \gtrsim 0.3\ \text{g/cm}^2$ is taken as the rough target for ignition.

When comparing what different devices have achieved, looking at which combination of density, temperature, and time was used to reach the same triple product value reveals the character of each approach. In the magnetic approach, JT-60U reached a triple product on the order of $1.5\times10^{21}\ \text{keV}\cdot\text{m}^{-3}\cdot\text{s}$ in the 1990s, and JET recorded $Q = 0.67$ in 1997. In the inertial approach, the National Ignition Facility (NIF) in the United States reported in December 2022 the scientific achievement of extracting more fusion energy than the laser energy put in. ITER is under construction with the goal of $Q \ge 10$ (500 MW of fusion output). New approaches such as SPARC, which strengthen the magnetic field with high-temperature superconducting magnets to shrink the device, are advancing in parallel, giving rise to a trend of aiming for high triple products in smaller devices. When reading papers, terms such as triple product, Lawson criterion, ignition, energy gain (Q), alpha heating, and density-radius product appear frequently.

It is also worth keeping in mind that the type of reaction itself affects the Lawson criterion. Up to this point we have assumed the D-T reaction, but the choice of fuel changes both the required temperature and the triple product. The properties of each reaction are covered in detail on the Fusion Reactions page.

Check Your Understanding

Q1. Which combination correctly gives the three quantities that matter in the Lawson criterion?

A Mass, charge, and pressure B Density, temperature, and confinement time C Magnetic field strength, device size, and operating time D Current, voltage, and resistance

Q2. Why is the fusion power density proportional to the square of the density n?

A Because the temperature rises as the square of the density B Because reactions occur through the collision of two nuclei, so the frequency is proportional to the product of each reaction partner's density C Because energy is stored as the square of the density D Because the confinement time is proportional to the square of the density

Q3. Which is the correct description of the fusion energy gain factor Q?

A Q = 1 is ignition, and Q = 10 represents scientific breakeven B Q = 1 is scientific breakeven, Q = 10 is ITER's target, and Q of infinity represents ignition C Q always represents the magnitude of the external heating itself D Q = 10 means a fusion output one tenth of the external heating

Q4. Which correctly describes the difference in how magnetic confinement and inertial confinement satisfy the same Lawson criterion?

A Both satisfy the condition with the same density and confinement time B Magnetic confinement uses high density and short time, inertial confinement uses low density and long time C Magnetic confinement extends the confinement time to seconds at low density, while inertial confinement achieves ultra-high density in a short time under a nanosecond D Magnetic confinement raises only the temperature, and inertial confinement raises only the density

Q5. Why is the triple product nTtau_E more convenient than the product of density and confinement time n tau_E for comparing device performance?

A Because the triple product is easier to calculate B Because for the D-T reaction the triple product required for ignition barely depends on temperature and is nearly a constant C Because the triple product is set by density alone, so temperature need not be measured D Because the required value of the triple product keeps falling as the temperature rises

Related Topics

• Fusion Reactions: This is the foundation for understanding how the choice of fuel changes the Lawson criterion.
• Transport Phenomena: This covers in detail the physics of the losses that set the confinement time $\tau_E$.
• Confinement Schemes: This lets you compare the devices and strategies of the magnetic and inertial approaches.