From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- .../know/concept/grad-shafranov-equation/index.md | 46 +++++++++++----------- 1 file changed, 23 insertions(+), 23 deletions(-) (limited to 'source/know/concept/grad-shafranov-equation') diff --git a/source/know/concept/grad-shafranov-equation/index.md b/source/know/concept/grad-shafranov-equation/index.md index 35d23f4..b86c032 100644 --- a/source/know/concept/grad-shafranov-equation/index.md +++ b/source/know/concept/grad-shafranov-equation/index.md @@ -20,7 +20,7 @@ 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: +and that the confining field $$\vb{B}$$ is fixed: $$\begin{aligned} \vb{u} @@ -36,7 +36,7 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Notice that $\vb{E} = 0$ is a result of the ideal generalized Ohm's law. +Notice that $$\vb{E} = 0$$ is a result of the ideal generalized Ohm's law. Under these assumptions, the relevant MHD equations to be solved are Gauss' law for magnetism, Ampère's law, and the MHD momentum equation, respectively: @@ -54,11 +54,11 @@ $$\begin{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. +[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 $\vb{A}$, +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: $$\begin{aligned} @@ -76,7 +76,7 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -Here, it is convenient to define the so-called **stream function** $\psi$ as follows: +Here, it is convenient to define the so-called **stream function** $$\psi$$ as follows: $$\begin{aligned} \boxed{ @@ -85,8 +85,8 @@ $$\begin{aligned} } \end{aligned}$$ -Such that $\vb{B}$ can be written as below, -where we will regard $B_\theta$ as a given quantity: +Such that $$\vb{B}$$ can be written as below, +where we will regard $$B_\theta$$ as a given quantity: $$\begin{aligned} \vb{B} @@ -103,7 +103,7 @@ $$\begin{aligned} Inserting this into Gauss' law, we see that it is trivially satisfied, -thanks to circular symmetry guaranteeing that $\ipdv{B_\theta}{\theta} = 0$: +thanks to circular symmetry guaranteeing that $$\ipdv{B_\theta}{\theta} = 0$$: $$\begin{aligned} 0 @@ -116,8 +116,8 @@ $$\begin{aligned} = 0 \end{aligned}$$ -What matters is that we have expressions for the components of $\vb{B}$. -Moving on, to find the current density $\vb{J}$, +What matters is that we have expressions for the components of $$\vb{B}$$. +Moving on, to find the current density $$\vb{J}$$, we use Ampère's law and symmetry to get: @@ -138,9 +138,9 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -Where we have assumed that $B_\theta$ depends only on $r$, not $z$ or $\theta$. +Where we have assumed that $$B_\theta$$ depends only on $$r$$, not $$z$$ or $$\theta$$. Substituting this into the MHD momentum equation -gives the following pressure gradient $\nabla p$: +gives the following pressure gradient $$\nabla p$$: $$\begin{aligned} \nabla p @@ -157,8 +157,8 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -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 $\nabla p$. +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: $$\begin{aligned} @@ -176,8 +176,8 @@ $$\begin{aligned} - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta \end{aligned}$$ -By using the chain rule to rewrite $\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$, -we get $\ipdv{\psi}{r}$ in each term: +By using the chain rule to rewrite $$\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$$, +we get $$\ipdv{\psi}{r}$$ in each term: $$\begin{aligned} \pdv{\psi}{r} \pdv{p}{\psi} @@ -185,7 +185,7 @@ $$\begin{aligned} - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta \end{aligned}$$ -Dividing out $\ipdv{\psi}{r}$ and multiplying by $\mu_0 r^2$ +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: @@ -196,13 +196,13 @@ $$\begin{aligned} } \end{aligned}$$ -Weirdly, $\psi$ appears both as an unknown and as a differentiation variable, +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 $r B_\theta$. +assuming a certain $$\psi$$-dependence of $$p$$ and $$r B_\theta$$. -Suppose that $B_\theta$ is induced by a poloidal electrical current $I_\mathrm{pol}$, +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: +then, assuming $$I_\mathrm{pol}$$ only depends on $$r$$, we have: $$\begin{aligned} B_\theta -- cgit v1.2.3