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+---
+title: "Screw pinch"
+firstLetter: "S"
+publishDate: 2022-03-06
+categories:
+- Physics
+- Plasma physics
+
+date: 2022-01-30T19:27:25+01:00
+draft: false
+markup: pandoc
+---
+
+# Screw pinch
+
+A **pinch** is a type of plasma confinement,
+which relies on [magnetic fields](/know/concept/magnetic-field/)
+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](/know/concept/cylindrical-polar-coordinates/)
+$(r, \theta, z)$.
+
+Using the framework of ideal [magnetohydrodynamics](/know/concept/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](/know/concept/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](/know/concept/maxwells-equations/)
+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](/know/concept/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](/know/concept/grad-shafranov-equation/).
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.