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Categories: Physics, Plasma physics.

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 chosen so that the Lorentz force stops particles escaping. Effectively, we are taking a cylindrical 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 (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 $(r, \theta, z)$ 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.

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