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diff --git a/content/know/concept/screw-pinch/index.pdc b/content/know/concept/screw-pinch/index.pdc new file mode 100644 index 0000000..bb4e0b5 --- /dev/null +++ b/content/know/concept/screw-pinch/index.pdc @@ -0,0 +1,209 @@ +--- +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. |