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 $$\nabla p = (d p /dA)\nabla A$$ and $$ \nabla B_z = (d B_z/dA)\nabla A$$. Thus, factoring out $$\nabla A$$ and rearraging terms we arrive at the Grad Shafranov equation:
 * \nabla^2 A = -\mu_0 \frac{d}{dA}(p + \frac{B_z^2}{2\mu_0})