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:

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:

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

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.

## References

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*Plasma physics: lecture notes*, 2021, unpublished.