From 2a5024543ed99f569fbade8744e6c8001f2edb02 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Tue, 12 Oct 2021 14:29:59 +0200
Subject: Expand knowledge base

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+---
+title: "Lawson criterion"
+firstLetter: "L"
+publishDate: 2021-10-06
+categories:
+- Physics
+- Plasma physics
+
+date: 2021-10-04T14:49:24+02:00
+draft: false
+markup: pandoc
+---
+
+# Lawson criterion
+
+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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