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+---
+title: "Grad-Shafranov equation"
+date: 2022-03-06
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+Nuclear fusion reactors tend to have a torus shape,
+in which the plasma is confined by a **pinch**,
+i.e. by [magnetic fields](/know/concept/magnetic-field/)
+chosen so that the [Lorentz force](/know/concept/lorentz-force/)
+stops particles escaping.
+Effectively, we are taking a cylindrical [screw pinch](/know/concept/screw-pinch/)
+and bending it into a torus.
+
+We would like to find the equilibrium state of the plasma
+in the general case of a reactor with toroidal symmetry.
+Using ideal [magnetohydrodynamics](/know/concept/magnetohydrodynamics/) (MHD),
+we start by assuming that the fluid is stationary,
+and that the confining field $\vb{B}$ is fixed:
+
+$$\begin{aligned}
+    \vb{u}
+    = 0
+    \qquad \qquad
+    \pdv{\vb{u}}{t}
+    = 0
+    \qquad \qquad
+    \pdv{\vb{B}}{t}
+    = 0
+    \qquad \qquad
+    \vb{E}
+    = 0
+\end{aligned}$$
+
+Notice that $\vb{E} = 0$ is a result of the ideal generalized Ohm's law.
+Under these assumptions, the relevant MHD equations to be solved are
+Gauss' law for magnetism, Ampère's law, and the MHD momentum equation, respectively:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \vb{B}
+    \qquad \qquad
+    \mu_0 \vb{J}
+    = \nabla \cross \vb{B}
+    \qquad \qquad
+    \nabla p
+    = \vb{J} \cross \vb{B}
+\end{aligned}$$
+
+The goal is to analyze them in this order,
+exploiting toroidal symmetry along the way,
+to arrive at a general equilibrium condition.
+[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $(r, \theta, z)$
+are a natural choice, with the $z$-axis running through the middle of the torus.
+
+As preparation, it is a good idea to write $\vb{B}$
+as the curl of a magnetic vector potential $\vb{A}$,
+which looks like this in cylindrical polar coordinates:
+
+$$\begin{aligned}
+    \vb{B}
+    = \nabla \cross \vb{A}
+    = \begin{bmatrix}
+        \displaystyle \frac{1}{r} \pdv{A_z}{\theta} - \pdv{A_\theta}{z} \\
+        \displaystyle \pdv{A_r}{z} - \pdv{A_z}{r} \\
+        \displaystyle \frac{1}{r} \Big( \pdv{(r A_\theta)}{r} - \pdv{A_r}{\theta} \Big)
+    \end{bmatrix}
+    = \begin{bmatrix}
+        \displaystyle - \pdv{A_\theta}{z} \\
+        \displaystyle \pdv{A_r}{z} - \pdv{A_z}{r} \\
+        \displaystyle \frac{1}{r} \pdv{(r A_\theta)}{r}
+    \end{bmatrix}
+\end{aligned}$$
+
+Here, it is convenient to define the so-called **stream function** $\psi$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \psi
+        \equiv r A_\theta
+    }
+\end{aligned}$$
+
+Such that $\vb{B}$ can be written as below,
+where we will regard $B_\theta$ as a given quantity:
+
+$$\begin{aligned}
+    \vb{B}
+    = \begin{bmatrix}
+        \displaystyle -\frac{1}{r} \pdv{\psi}{z} \\
+        B_\theta \\
+        \displaystyle \frac{1}{r} \pdv{\psi}{r}
+    \end{bmatrix}
+    \qquad \mathrm{where} \qquad
+    B_\theta
+    = \pdv{A_r}{z} - \pdv{A_z}{r}
+\end{aligned}$$
+
+
+Inserting this into Gauss' law,
+we see that it is trivially satisfied,
+thanks to circular symmetry guaranteeing that $\ipdv{B_\theta}{\theta} = 0$:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \vb{B}
+    &= - \frac{1}{r} \pdv{}{r}\bigg( \frac{r}{r} \pdv{\psi}{z} \bigg)
+    + \frac{1}{r} \pdv{B_\theta}{\theta}
+    + \pdv{}{z}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+    \\
+    &= - \frac{1}{r} \mpdv{\psi}{r}{z} + \frac{1}{r} \mpdv{\psi}{z}{r}
+    = 0
+\end{aligned}$$
+
+What matters is that we have expressions for the components of $\vb{B}$.
+Moving on, to find the current density $\vb{J}$,
+we use Ampère's law and symmetry to get:
+
+
+$$\begin{aligned}
+    \vb{J}
+    = \frac{1}{\mu_0} \nabla \cross \vb{B}
+    = \frac{1}{\mu_0}
+    \begin{bmatrix}
+        \displaystyle \frac{1}{r} \pdv{B_z}{\theta} - \pdv{B_\theta}{z} \\
+        \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\
+        \displaystyle \frac{1}{r} \Big( \pdv{(r B_\theta)}{r} - \pdv{B_r}{\theta} \Big)
+    \end{bmatrix}
+    = \frac{1}{\mu_0}
+    \begin{bmatrix}
+        \displaystyle 0 \\
+        \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\
+        \displaystyle \frac{1}{r} \pdv{(r B_\theta)}{r}
+    \end{bmatrix}
+\end{aligned}$$
+
+Where we have assumed that $B_\theta$ depends only on $r$, not $z$ or $\theta$.
+Substituting this into the MHD momentum equation
+gives the following pressure gradient $\nabla p$:
+
+$$\begin{aligned}
+    \nabla p
+    &= \vb{J} \cross \vb{B}
+    = \begin{bmatrix}
+        J_\theta B_z - J_z B_\theta \\
+        J_z B_r - J_r B_z \\
+        J_r B_\theta - J_\theta B_r
+    \end{bmatrix}
+    = \begin{bmatrix}
+        J_\theta B_z - J_z B_\theta \\
+        J_z B_r \\
+        - J_\theta B_r
+    \end{bmatrix}
+\end{aligned}$$
+
+Now, the idea is to focus on this $r$-component to get an equation for $\psi$,
+whose solution can then be used to calculate the $\theta$ and $z$-components of $\nabla p$.
+Therefore, we evaluate:
+
+$$\begin{aligned}
+    \pdv{p}{r}
+    &= J_\theta B_z - J_z B_\theta
+    \\
+    &= \frac{1}{\mu_0} \bigg( \pdv{B_r}{z} - \pdv{B_z}{r} \bigg) B_z
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+    \\
+    &= - \frac{1}{\mu_0} \bigg( \pdv{}{z}\Big(\frac{1}{r} \pdv{\psi}{z}\Big)
+    + \pdv{}{r}\Big(\frac{1}{r} \pdv{\psi}{r}\Big) \bigg) \frac{1}{r} \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+    \\
+    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdvn{2}{\psi}{z} + \pdv{}{r}\Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+\end{aligned}$$
+
+By using the chain rule to rewrite $\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$,
+we get $\ipdv{\psi}{r}$ in each term:
+
+$$\begin{aligned}
+    \pdv{\psi}{r} \pdv{p}{\psi}
+    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdvn{2}{\psi}{z} + \pdv{}{r}\Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta
+\end{aligned}$$
+
+Dividing out $\ipdv{\psi}{r}$ and multiplying by $\mu_0 r^2$
+leads us to the **Grad-Shafranov equation**,
+which gives the equilibrium condition of a plasma in a toroidal reactor:
+
+$$\begin{aligned}
+    \boxed{
+        \pdvn{2}{\psi}{z} + r \pdv{}{r}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+        = - \mu_0 r^2 \pdv{p}{\psi} - r \pdv{(r B_\theta)}{\psi} B_\theta
+    }
+\end{aligned}$$
+
+Weirdly, $\psi$ appears both as an unknown and as a differentiation variable,
+but this equation can still be solved analytically by
+assuming a certain $\psi$-dependence of $p$ and $r B_\theta$.
+
+Suppose that $B_\theta$ is induced by a poloidal electrical current $I_\mathrm{pol}$,
+i.e. a current around the "tube" of the torus,
+then, assuming $I_\mathrm{pol}$ only depends on $r$, we have:
+
+$$\begin{aligned}
+    B_\theta
+    = \frac{\mu_0 I_\mathrm{pol}(r)}{2 \pi r}
+\end{aligned}$$
+
+Inserting this into the Grad-Shafranov equation yields its following alternative form:
+
+$$\begin{aligned}
+    \boxed{
+        \pdvn{2}{\psi}{z} + r \pdv{}{r}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+        = - \mu_0 r^2 \pdv{p}{\psi} - \frac{\mu_0^2}{8 \pi^2} \pdv{I_\mathrm{pol}^2}{\psi}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
+