Expand knowledge base

5 files changed, 444 insertions, 4 deletions

diff --git a/content/know/concept/cylindrical-parabolic-coordinates/index.pdc b/content/know/concept/cylindrical-parabolic-coordinates/index.pdc
index 3460de4..e3257b5 100644
--- a/content/know/concept/cylindrical-parabolic-coordinates/index.pdc
+++ b/content/know/concept/cylindrical-parabolic-coordinates/index.pdc
@@ -43,7 +43,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Cylindrical parabolic coordinates form an orthogonal
-[curvilinear](/know/concept/curvilinear-coordinates/) system,
+[curvilinear system](/know/concept/curvilinear-coordinates/),
 so we would like to find its scale factors $h_\sigma$, $h_\tau$ and $h_z$.
 The differentials of the Cartesian coordinates are as follows:
 
diff --git a/content/know/concept/cylindrical-polar-coordinates/index.pdc b/content/know/concept/cylindrical-polar-coordinates/index.pdc
index 6242b9f..0b2acd1 100644
--- a/content/know/concept/cylindrical-polar-coordinates/index.pdc
+++ b/content/know/concept/cylindrical-polar-coordinates/index.pdc
@@ -47,7 +47,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The cylindrical polar coordinates form an orthogonal
-[curvilinear](/know/concept/curvilinear-coordinates/) system,
+[curvilinear system](/know/concept/curvilinear-coordinates/),
 whose scale factors $h_r$, $h_\varphi$ and $h_z$ we want to find.
 To do so, we calculate the differentials of the Cartesian coordinates:
 
@@ -141,7 +141,7 @@ $$\begin{aligned}
             \\
             &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big)
             \\
-            &+ \frac{\vu{e}_\varphi}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big)
+            &+ \frac{\vu{e}_z}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big)
         \end{aligned}
     }
 \end{aligned}$$
diff --git a/content/know/concept/grad-shafranov-equation/index.pdc b/content/know/concept/grad-shafranov-equation/index.pdc
new file mode 100644
index 0000000..afd24b0
--- /dev/null
+++ b/content/know/concept/grad-shafranov-equation/index.pdc
@@ -0,0 +1,231 @@
+---
+title: "Grad-Shafranov equation"
+firstLetter: "G"
+publishDate: 2022-03-06
+categories:
+- Physics
+- Plasma physics
+
+date: 2022-01-30T19:27:07+01:00
+draft: false
+markup: pandoc
+---
+
+# Grad-Shafranov equation
+
+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 $\pdv*{B_\theta}{\theta} = 0$:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \vb{B}
+    &= - \frac{1}{r} \pdv{r} \bigg( \frac{r}{r} \pdv{\psi}{z} \bigg)
+    + \frac{1}{r} \pdv{B_\theta}{\theta}
+    + \pdv{z} \bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+    \\
+    &= - \frac{1}{r} \pdv{\psi}{r}{z} + \frac{1}{r} \pdv{\psi}{z}{r}
+    = 0
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \pdv{p}{r}
+    &= J_\theta B_z - J_z B_\theta
+    \\
+    &= \frac{1}{\mu_0} \bigg( \pdv{B_r}{z} - \pdv{B_z}{r} \bigg) B_z
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+    \\
+    &= - \frac{1}{\mu_0} \bigg( \pdv{z}\Big(\frac{1}{r} \pdv{\psi}{z}\Big) + \pdv{r}\Big(\frac{1}{r} \pdv{\psi}{r}\Big) \bigg) \frac{1}{r} \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+    \\
+    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdv[2]{\psi}{z} + \pdv{r} \Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+\end{aligned}$$
+
+By using the chain rule to rewrite $\pdv*{r} = (\pdv*{\psi}{r}) \; \pdv*{\psi}$,
+we get $\pdv*{\psi}{r}$ in each term:
+
+$$\begin{aligned}
+    \pdv{\psi}{r} \pdv{p}{\psi}
+    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdv[2]{\psi}{z} + \pdv{r} \Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta
+\end{aligned}$$
+
+Dividing out $\pdv*{\psi}{r}$ 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:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv[2]{\psi}{z} + r \pdv{r} \bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+        = - \mu_0 r^2 \pdv{p}{\psi} - r \pdv{(r B_\theta)}{\psi} B_\theta
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv[2]{\psi}{z} + r \pdv{r} \bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+        = - \mu_0 r^2 \pdv{p}{\psi} - \frac{\mu_0^2}{8 \pi^2} \pdv{I_\mathrm{pol}^2}{\psi}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
+
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.
diff --git a/content/know/concept/spherical-coordinates/index.pdc b/content/know/concept/spherical-coordinates/index.pdc
index 4768110..dc8acd2 100644
--- a/content/know/concept/spherical-coordinates/index.pdc
+++ b/content/know/concept/spherical-coordinates/index.pdc
@@ -51,7 +51,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The spherical coordinate system is an orthogonal
-[curvilinear](/know/concept/curvilinear-coordinates/) system,
+[curvilinear system](/know/concept/curvilinear-coordinates/),
 whose scale factors $h_r$, $h_\theta$ and $h_\varphi$ we want to find.
 To do so, we calculate the differentials of the Cartesian coordinates: