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---
title: "Lawson criterion"
date: 2021-10-06
categories:
- Physics
- Plasma physics
layout: "concept"
---
For sustained nuclear fusion to be possible,
the **Lawson criterion** must be met,
from which some required properties
of the plasma and the reactor chamber can be deduced.
Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion,
but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way.
If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction,
then the following inequality must be satisfied
in order to have harvestable energy:
$$\begin{aligned}
P_\mathrm{loss}
\le P_\mathrm{fus} + P_\mathrm{aux}
\end{aligned}$$
We can rewrite $P_\mathrm{aux}$ using the definition
of the **energy gain factor** $Q$,
which is the ratio of the output and input powers of the fusion reaction:
$$\begin{aligned}
Q
\equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}}
\quad \implies \quad
P_\mathrm{aux}
= \frac{P_\mathrm{fus}}{Q}
\end{aligned}$$
Returning to the inequality, we can thus rearrange its right-hand side as follows:
$$\begin{aligned}
P_\mathrm{loss}
\le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q}
= P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big)
= P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
\end{aligned}$$
We assume that the plasma has equal species densities $n_i = n_e$,
so its total density $n = 2 n_i$.
Then $P_\mathrm{fus}$ is as follows,
where $f_{ii}$ is the frequency
with which a given ion collides with other ions,
and $E_\mathrm{fus}$ is the energy released by a single fusion reaction:
$$\begin{aligned}
P_\mathrm{fus}
= f_{ii} n_i E_\mathrm{fus}
= \big( n_i \Expval{\sigma v} \big) n_i E_\mathrm{fus}
= \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus}
\end{aligned}$$
Where $\Expval{\sigma v}$ is the mean product
of the velocity $v$ and the collision cross-section $\sigma$.
Furthermore, assuming that both species have the same temperature $T_i = T_e = T$,
the total energy density $W$ of the plasma is given by:
$$\begin{aligned}
W
= \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e
= 3 k_B T n
\end{aligned}$$
Where $k_B$ is Boltzmann's constant.
From this, we can define the **confinement time** $\tau_E$
as the characteristic lifetime of energy in the reactor, before leakage.
Therefore:
$$\begin{aligned}
\tau_E
\equiv \frac{W}{P_\mathrm{loss}}
\quad \implies \quad
P_\mathrm{loss}
= \frac{3 n k_B T}{\tau_E}
\end{aligned}$$
Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$
into the inequality, we arrive at:
$$\begin{aligned}
\frac{3 n k_B T}{\tau_E}
\le \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
\end{aligned}$$
This can be rearranged to the form below,
which is the original Lawson criterion:
$$\begin{aligned}
n \tau_E
\ge \frac{Q}{Q + 1} \frac{12 k_B T}{\Expval{\sigma v} E_\mathrm{fus}}
\end{aligned}$$
However, it turns out that the highest fusion power density
is reached when $T$ is at the minimum of $T^2 / \Expval{\sigma v}$.
Therefore, we multiply by $T$ to get the Lawson triple product:
$$\begin{aligned}
\boxed{
n T \tau_E
\ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}}
}
\end{aligned}$$
For some reason,
it is often assumed that the fusion is infinitely profitable $Q \to \infty$,
in which case the criterion reduces to:
$$\begin{aligned}
n T \tau_E
\ge \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}}
\end{aligned}$$
## References
1. M. Salewski, A.H. Nielsen,
*Plasma physics: lecture notes*,
2021, unpublished.
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