path: root/content/know/concept/grad-shafranov-equation/index.pdc

---
title: "Grad-Shafranov equation"
firstLetter: "G"
publishDate: 2022-03-06
categories:
- Physics
- Plasma physics

date: 2022-01-30T19:27:07+01:00
draft: false
markup: pandoc
---

# Grad-Shafranov equation

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 $\pdv*{B_\theta}{\theta} = 0$:

$$\begin{aligned}
    0
    = \nabla \cdot \vb{B}
    &= - \frac{1}{r} \pdv{r} \bigg( \frac{r}{r} \pdv{\psi}{z} \bigg)
    + \frac{1}{r} \pdv{B_\theta}{\theta}
    + \pdv{z} \bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
    \\
    &= - \frac{1}{r} \pdv{\psi}{r}{z} + \frac{1}{r} \pdv{\psi}{z}{r}
    = 0
\end{aligned}$$

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:

$$\begin{aligned}
    \pdv{p}{r}
    &= J_\theta B_z - J_z B_\theta
    \\
    &= \frac{1}{\mu_0} \bigg( \pdv{B_r}{z} - \pdv{B_z}{r} \bigg) B_z
    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
    \\
    &= - \frac{1}{\mu_0} \bigg( \pdv{z}\Big(\frac{1}{r} \pdv{\psi}{z}\Big) + \pdv{r}\Big(\frac{1}{r} \pdv{\psi}{r}\Big) \bigg) \frac{1}{r} \pdv{\psi}{r}
    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
    \\
    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdv[2]{\psi}{z} + \pdv{r} \Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
\end{aligned}$$

By using the chain rule to rewrite $\pdv*{r} = (\pdv*{\psi}{r}) \; \pdv*{\psi}$,
we get $\pdv*{\psi}{r}$ in each term:

$$\begin{aligned}
    \pdv{\psi}{r} \pdv{p}{\psi}
    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdv[2]{\psi}{z} + \pdv{r} \Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
    - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta
\end{aligned}$$

Dividing out $\pdv*{\psi}{r}$ 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:

$$\begin{aligned}
    \boxed{
        \pdv[2]{\psi}{z} + r \pdv{r} \bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
        = - \mu_0 r^2 \pdv{p}{\psi} - r \pdv{(r B_\theta)}{\psi} B_\theta
    }
\end{aligned}$$

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:

$$\begin{aligned}
    \boxed{
        \pdv[2]{\psi}{z} + r \pdv{r} \bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
        = - \mu_0 r^2 \pdv{p}{\psi} - \frac{\mu_0^2}{8 \pi^2} \pdv{I_\mathrm{pol}^2}{\psi}
    }
\end{aligned}$$



## References
1.  M. Salewski, A.H. Nielsen,
    *Plasma physics: lecture notes*,
    2021, unpublished.