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diff --git a/source/know/concept/grad-shafranov-equation/index.md b/source/know/concept/grad-shafranov-equation/index.md new file mode 100644 index 0000000..8bef8af --- /dev/null +++ b/source/know/concept/grad-shafranov-equation/index.md @@ -0,0 +1,226 @@ +--- +title: "Grad-Shafranov equation" +date: 2022-03-06 +categories: +- Physics +- Plasma physics +layout: "concept" +--- + +Nuclear fusion reactors tend to have a torus shape, +in which the plasma is confined by a **pinch**, +i.e. by [magnetic fields](/know/concept/magnetic-field/) +chosen so that the [Lorentz force](/know/concept/lorentz-force/) +stops particles escaping. +Effectively, we are taking a cylindrical [screw pinch](/know/concept/screw-pinch/) +and bending it into a torus. + +We would like to find the equilibrium state of the plasma +in the general case of a reactor with toroidal symmetry. +Using ideal [magnetohydrodynamics](/know/concept/magnetohydrodynamics/) (MHD), +we start by assuming that the fluid is stationary, +and that the confining field $\vb{B}$ is fixed: + +$$\begin{aligned} + \vb{u} + = 0 + \qquad \qquad + \pdv{\vb{u}}{t} + = 0 + \qquad \qquad + \pdv{\vb{B}}{t} + = 0 + \qquad \qquad + \vb{E} + = 0 +\end{aligned}$$ + +Notice that $\vb{E} = 0$ is a result of the ideal generalized Ohm's law. +Under these assumptions, the relevant MHD equations to be solved are +Gauss' law for magnetism, Ampère's law, and the MHD momentum equation, respectively: + +$$\begin{aligned} + 0 + = \nabla \cdot \vb{B} + \qquad \qquad + \mu_0 \vb{J} + = \nabla \cross \vb{B} + \qquad \qquad + \nabla p + = \vb{J} \cross \vb{B} +\end{aligned}$$ + +The goal is to analyze them in this order, +exploiting toroidal symmetry along the way, +to arrive at a general equilibrium condition. +[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $(r, \theta, z)$ +are a natural choice, with the $z$-axis running through the middle of the torus. + +As preparation, it is a good idea to write $\vb{B}$ +as the curl of a magnetic vector potential $\vb{A}$, +which looks like this in cylindrical polar coordinates: + +$$\begin{aligned} + \vb{B} + = \nabla \cross \vb{A} + = \begin{bmatrix} + \displaystyle \frac{1}{r} \pdv{A_z}{\theta} - \pdv{A_\theta}{z} \\ + \displaystyle \pdv{A_r}{z} - \pdv{A_z}{r} \\ + \displaystyle \frac{1}{r} \Big( \pdv{(r A_\theta)}{r} - \pdv{A_r}{\theta} \Big) + \end{bmatrix} + = \begin{bmatrix} + \displaystyle - \pdv{A_\theta}{z} \\ + \displaystyle \pdv{A_r}{z} - \pdv{A_z}{r} \\ + \displaystyle \frac{1}{r} \pdv{(r A_\theta)}{r} + \end{bmatrix} +\end{aligned}$$ + +Here, it is convenient to define the so-called **stream function** $\psi$ as follows: + +$$\begin{aligned} + \boxed{ + \psi + \equiv r A_\theta + } +\end{aligned}$$ + +Such that $\vb{B}$ can be written as below, +where we will regard $B_\theta$ as a given quantity: + +$$\begin{aligned} + \vb{B} + = \begin{bmatrix} + \displaystyle -\frac{1}{r} \pdv{\psi}{z} \\ + B_\theta \\ + \displaystyle \frac{1}{r} \pdv{\psi}{r} + \end{bmatrix} + \qquad \mathrm{where} \qquad + B_\theta + = \pdv{A_r}{z} - \pdv{A_z}{r} +\end{aligned}$$ + + +Inserting this into Gauss' law, +we see that it is trivially satisfied, +thanks to circular symmetry guaranteeing that $\ipdv{B_\theta}{\theta} = 0$: + +$$\begin{aligned} + 0 + = \nabla \cdot \vb{B} + &= - \frac{1}{r} \pdv{}{r}\bigg( \frac{r}{r} \pdv{\psi}{z} \bigg) + + \frac{1}{r} \pdv{B_\theta}{\theta} + + \pdv{}{z}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg) + \\ + &= - \frac{1}{r} \mpdv{\psi}{r}{z} + \frac{1}{r} \mpdv{\psi}{z}{r} + = 0 +\end{aligned}$$ + +What matters is that we have expressions for the components of $\vb{B}$. +Moving on, to find the current density $\vb{J}$, +we use Ampère's law and symmetry to get: + + +$$\begin{aligned} + \vb{J} + = \frac{1}{\mu_0} \nabla \cross \vb{B} + = \frac{1}{\mu_0} + \begin{bmatrix} + \displaystyle \frac{1}{r} \pdv{B_z}{\theta} - \pdv{B_\theta}{z} \\ + \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\ + \displaystyle \frac{1}{r} \Big( \pdv{(r B_\theta)}{r} - \pdv{B_r}{\theta} \Big) + \end{bmatrix} + = \frac{1}{\mu_0} + \begin{bmatrix} + \displaystyle 0 \\ + \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\ + \displaystyle \frac{1}{r} \pdv{(r B_\theta)}{r} + \end{bmatrix} +\end{aligned}$$ + +Where we have assumed that $B_\theta$ depends only on $r$, not $z$ or $\theta$. +Substituting this into the MHD momentum equation +gives the following pressure gradient $\nabla p$: + +$$\begin{aligned} + \nabla p + &= \vb{J} \cross \vb{B} + = \begin{bmatrix} + J_\theta B_z - J_z B_\theta \\ + J_z B_r - J_r B_z \\ + J_r B_\theta - J_\theta B_r + \end{bmatrix} + = \begin{bmatrix} + J_\theta B_z - J_z B_\theta \\ + J_z B_r \\ + - J_\theta B_r + \end{bmatrix} +\end{aligned}$$ + +Now, the idea is to focus on this $r$-component to get an equation for $\psi$, +whose solution can then be used to calculate the $\theta$ and $z$-components of $\nabla p$. +Therefore, we evaluate: + +$$\begin{aligned} + \pdv{p}{r} + &= J_\theta B_z - J_z B_\theta + \\ + &= \frac{1}{\mu_0} \bigg( \pdv{B_r}{z} - \pdv{B_z}{r} \bigg) B_z + - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta + \\ + &= - \frac{1}{\mu_0} \bigg( \pdv{}{z}\Big(\frac{1}{r} \pdv{\psi}{z}\Big) + + \pdv{}{r}\Big(\frac{1}{r} \pdv{\psi}{r}\Big) \bigg) \frac{1}{r} \pdv{\psi}{r} + - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta + \\ + &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdvn{2}{\psi}{z} + \pdv{}{r}\Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r} + - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta +\end{aligned}$$ + +By using the chain rule to rewrite $\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$, +we get $\ipdv{\psi}{r}$ in each term: + +$$\begin{aligned} + \pdv{\psi}{r} \pdv{p}{\psi} + &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdvn{2}{\psi}{z} + \pdv{}{r}\Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r} + - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta +\end{aligned}$$ + +Dividing out $\ipdv{\psi}{r}$ and multiplying by $\mu_0 r^2$ +leads us to the **Grad-Shafranov equation**, +which gives the equilibrium condition of a plasma in a toroidal reactor: + +$$\begin{aligned} + \boxed{ + \pdvn{2}{\psi}{z} + r \pdv{}{r}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg) + = - \mu_0 r^2 \pdv{p}{\psi} - r \pdv{(r B_\theta)}{\psi} B_\theta + } +\end{aligned}$$ + +Weirdly, $\psi$ appears both as an unknown and as a differentiation variable, +but this equation can still be solved analytically by +assuming a certain $\psi$-dependence of $p$ and $r B_\theta$. + +Suppose that $B_\theta$ is induced by a poloidal electrical current $I_\mathrm{pol}$, +i.e. a current around the "tube" of the torus, +then, assuming $I_\mathrm{pol}$ only depends on $r$, we have: + +$$\begin{aligned} + B_\theta + = \frac{\mu_0 I_\mathrm{pol}(r)}{2 \pi r} +\end{aligned}$$ + +Inserting this into the Grad-Shafranov equation yields its following alternative form: + +$$\begin{aligned} + \boxed{ + \pdvn{2}{\psi}{z} + r \pdv{}{r}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg) + = - \mu_0 r^2 \pdv{p}{\psi} - \frac{\mu_0^2}{8 \pi^2} \pdv{I_\mathrm{pol}^2}{\psi} + } +\end{aligned}$$ + + + +## References +1. M. Salewski, A.H. Nielsen, + *Plasma physics: lecture notes*, + 2021, unpublished. + |