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,θ,z)(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 B\vb{B} is fixed. From the (ideal) generalized Ohm’s law, it then follows that the electric field E=0\vb{E} = 0:

u=0ut=0Bt=0E=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 B\vb{B}, we first solve Ampère’s law for the current density J\vb{J}, and then the MHD momentum equation for the pressure pp. Symmetries should be used whenever possible to reduce these equations:

×B=μ0JJ×B=p\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 p\nabla p is always orthogonal to J\vb{J} and B\vb{B}, meaning that the current density and magnetic field must follow surfaces of constant pressure.


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

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

Where e^z\vu{e}_z is the basis vector of the zz-axis. This B\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 J\vb{J}, expressed in (r,θ,z)(r, \theta, z):

J=1μ0×B=1μ0[1rBzθBθzBrzBzr1r((rBθ)rBrθ)]=1μ0Bzre^θ\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 BzB_z is nonzero, and that it only depends on rr. This yields a circular current parallel to e^θ\vu{e}_\theta, hence the name ϴ-pinch.

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

p=J×B=Jθe^θ×Bze^z=JθBze^r=1μ0BzrBze^r\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, p\nabla p is parallel to e^r\vu{e}_r, and only depends on rr through BzB_z. Along the rr-direction, the above equation can be rewritten into the following equilibrium condition:

r(p+Bz22μ0)=0\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 rr.


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

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

We can then deduce B\vb{B} from Ampère’s law, using that only JzJ_z is nonzero, and that Br/θ=0\ipdv{B_r}{\theta} = 0 due to circular symmetry:

J=1μ0×B=1μ0[1rBzθBθzBrzBzr1r((rBθ)rBrθ)]=1μ0r(rBθ)re^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 r} \pdv{(r B_\theta)}{r} \: \vu{e}_z \end{aligned}

Therefore, J\vb{J} induces a circular B=Bθ(r)e^θ\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 p\nabla p is found from the MHD momentum equation:

p=J×B=Jzez×Bθeθ=JzBθe^r=1μ0r(rBθ)rBθe^r\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, p\nabla p is parallel to e^r\vu{e}_r and only depends on rr. After rearranging, we thus arrive at the following equilibrium condition in the rr-direction:

r(p+Bθ22μ0)+Bθ2μ0r=0\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 J\vb{J} and B\vb{B} both have nonzero θ\theta and zz-components. By performing the above procedure again, the following equilibrium condition is obtained:

r(p+Bz22μ0+Bθ22μ0)+Bθ2μ0r=0\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.