Categories: Physics, Plasma physics.

# Screw pinch

A pinch is a type of plasma confinement, which relies on magnetic fields to squeeze the plasma into the desired area. Examples include tokamaks and stellarators, although the term pinch is typically introduced for simpler 1D confinement.

Suppose that we want to pinch a plasma into a cylindrical shape. The general way of doing this is called a screw pinch. For simplicity, let the cylinder be infinitely long, so that it is natural to work in cylindrical polar coordinates $$(r, \theta, z)$$.

Using the framework of ideal magnetohydrodynamics (MHD), let us start by assuming that the fluid is stationary, and that the confining field $$\vb{B}$$ is fixed. From the (ideal) generalized Ohm’s law, it then follows that the electric field $$\vb{E} = 0$$:

\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}

To get the plasma’s equilibrium state for a given $$\vb{B}$$, we first solve Ampère’s law for the current density $$\vb{J}$$, and then the MHD momentum equation for the pressure $$p$$. Symmetries should be used whenever possible to reduce these equations:

\begin{aligned} \nabla \cross \vb{B} = \mu_0 \vb{J} \qquad \qquad \vb{J} \cross \vb{B} = \nabla p \end{aligned}

Note that the latter implies that $$\nabla p$$ is always orthogonal to $$\vb{J}$$ and $$\vb{B}$$, meaning that the current density and magnetic field must follow surfaces of constant pressure.

## ϴ-pinch

In a so-called ϴ-pinch, the confining field $$\vb{B}$$ is parallel to the $$z$$-axis, and its magntiude $$B_z$$ may only depend on $$r$$. Concretely, we have:

\begin{aligned} \vb{B} = B_z(r) \: \vu{e}_z \end{aligned}

Where $$\vu{e}_z$$ is the basis vector of the $$z$$-axis. This $$\vb{B}$$ confines the plasma thanks to the Lorentz force, which makes charged particles gyrate around magnetic field lines.

Using Ampère’s law, we find that the resulting current density $$\vb{J}$$, expressed in $$(r, \theta, z)$$:

\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} \pdv{B_z}{r} \: \vu{e}_\theta \end{aligned}

Where we have used that only $$B_z$$ is nonzero, and that it only depends on $$r$$. This yields a circular current parallel to $$\vu{e}_\theta$$, hence the name ϴ-pinch.

Next, we use the MHD momentum equation to find the pressure gradient $$\nabla p$$. The cross product is easy to evaluate, since $$\vb{B}$$ is parallel to $$\vu{e}_z$$, and $$\vb{J}$$ is parallel to $$\vu{e}_\theta$$:

\begin{aligned} \nabla p &= \vb{J} \cross \vb{B} = J_\theta \vu{e}_\theta \cross B_z \vu{e}_z = J_\theta B_z \vu{e}_r = - \frac{1}{\mu_0} \pdv{B_z}{r} B_z \: \vu{e}_r \end{aligned}

Consequently, $$\nabla p$$ is parallel to $$\vu{e}_r$$, and only depends on $$r$$ through $$B_z$$. Along the $$r$$-direction, the above equation can be rewritten into the following equilibrium condition:

\begin{aligned} \boxed{ \pdv{r} \bigg( p + \frac{B_z^2}{2 \mu_0} \bigg) = 0 } \end{aligned}

In other words, the parenthesized expression does not depend on $$r$$.

## Z-pinch

Meanwhile, in a so-called Z-pinch, we create an $$r$$-dependent current $$\vb{J}$$ parallel to the $$z$$-axis:

\begin{aligned} \vb{J} = J_z(r) \: \vu{e}_z \end{aligned}

We can then deduce $$\vb{B}$$ from Ampère’s law, using that only $$J_z$$ is nonzero, and that $$\pdv*{B_r}{\theta} = 0$$ due to circular symmetry:

\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 r} \pdv{(r B_\theta)}{r} \: \vu{e}_z \end{aligned}

Therefore, $$\vb{J}$$ induces a circular $$\vb{B} = B_\theta(r) \: \vu{e}_\theta$$, which confines the plasma for the same reason as in the ϴ-pinch: the Lorentz force makes particles gyrate around magnetic field lines.

Next, the resulting pressure gradient $$\nabla p$$ is found from the MHD momentum equation:

\begin{aligned} \nabla p &= \vb{J} \cross \vb{B} = J_z \vb{e}_z \cross B_\theta \vb{e}_\theta = - J_z B_\theta \vu{e}_r = - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta \: \vu{e}_r \end{aligned}

Once again, $$\nabla p$$ is parallel to $$\vu{e}_r$$ and only depends on $$r$$. After rearranging, we thus arrive at the following equilibrium condition in the $$r$$-direction:

\begin{aligned} \boxed{ \pdv{r} \bigg( p + \frac{B_\theta^2}{2 \mu_0} \bigg) + \frac{B_\theta^2}{\mu_0 r} = 0 } \end{aligned}

## Screw pinch

Thanks to the linearity of electromagnetism, a ϴ-pinch and Z-pinch can be combined to create a screw pinch, where $$\vb{J}$$ and $$\vb{B}$$ both have nonzero $$\theta$$ and $$z$$-components. By performing the above procedure again, the following equilibrium condition is obtained:

\begin{aligned} \boxed{ \pdv{r} \bigg( p + \frac{B_z^2}{2 \mu_0} + \frac{B_\theta^2}{2 \mu_0} \bigg) + \frac{B_\theta^2}{\mu_0 r} = 0 } \end{aligned}

Which simply combines the terms of the preceding equations. Indirectly, this result is relevant for certain types of nuclear fusion reactor, e.g. the tokamak, which basically consists of a screw pinch bent into a torus. The resulting equilibrium is given by the Grad-Shafranov equation.

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.