Categories: Physics, Plasma physics.

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

\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{\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{\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{\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{\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}

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 M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.