User:Bpsullivan/GS

Derivation:
To begin we assume that the system is 2-dimensional with z as the invariant axis, i.e. $\partial /\partial z=0$ for all quantities. Then the magnetic field can be written in cartesian coordinates as

$\mathbf {B} =(\partial A/\partial y,-\partial A/\partial x,B_{z}(x,y))$

or more compactly,

$\mathbf {B} =\nabla A\times {\hat {\mathbf {z} }}+{\hat {\mathbf {z} }}B_{z}$,

where $A(x,y){\hat {\mathbf {z} }}$ is the vector potential for the in-plane (x and y components) magnetic field. Note that based on this form for B we can see that A is constant along any given magnetic field line, since $\nabla A$ is everywhere perpendicular to B.

Two dimensional, stationary, magnetic structures are described by the balance of pressure forces and magnetic forces, i.e.:

$\nabla p=\mathbf {j} \times \mathbf {B}$,

where p is the plasma pressure and j is the electric current. Note from the form of this equation that we also know p is a constant along any field line, (again since $\nabla p$ is everywhere perpendicular to B. Additionally, the two-dimensional assumption ($\partial /\partial z$) means that the z- component of the left hand side must be zero, so the z-component of the magnetic force on the right hand side must also be zero. This means that $\mathbf {j} _{\perp }\times \mathbf {B} _{\perp }=0$, i.e. $\mathbf {j} _{\perp }$ is parallel to $\mathbf {B} _{\perp }$.

We can break the right hand side of the previous equation into two parts:

$\mathbf {j} \times \mathbf {B} =j_{z}({\hat {\mathbf {z} }}\times \mathbf {B_{\perp }} )+\mathbf {j_{\perp }} \times {\hat {\mathbf {z} }}B_{z}$,

where the $\perp$ subscript denotes the component in the plane perpendicular to the $z$-axis. The z component of the current in the above equation can be written in terms of the one dimensional vector potential as $j_{z}=-\nabla ^{2}A/\mu _{0}.$. The in plane field is

$\mathbf {B} _{\perp }=\nabla A\times {\hat {\mathbf {z} }}$,

and using Ampère's Law the in plane current is given by

$\mathbf {j} _{\perp }=(1/\mu _{0})\nabla B_{z}\times {\hat {\mathbf {z} }}$.

In order for this vector to be parallel to $\mathbf {B} _{\perp }$ as required, the vector $\nabla B_{z}$ must be perpendicular to $\mathbf {j} _{\perp }$, and $B_{z}$ must therefore, like $p$ be a field like invariant.

Rearranging the cross products above, we see that that

${\hat {\mathbf {z} }}\times \mathbf {B} _{\perp }=\nabla A$,

and

$\mathbf {j} _{\perp }\times \mathbf {\hat {z}} =-(1/\mu _{0})B_{z}\nabla B_{z}$

These results can be subsituted into the expression for $\nabla p$ to yield:

$\nabla p=-[(1/\mu _{0})\nabla ^{2}A]\nabla A-(1/\mu _{0})B_{z}\nabla B_{z}.$

Now, since $p$ and $B_{\perp }$ are constants along a field line, and functions only of $A$, we note that $\nabla p=(dp/dA)\nabla A$ and $\nabla B_{z}=(dB_{z}/dA)\nabla A$. Thus, factoring out $\nabla A$ and rearraging terms we arrive at the Grad Shafranov equation:

<math>\nabla^2 A = -\mu_0 \frac{d}{dA}(p + \frac{B_z^2}{2\mu_0})