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2002 I 1

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The magnetic field is:

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

The equilibrium momentum equation gives:

$\frac{1}{c}\mathbf{J}\times\mathbf{B}=\nabla p$

Using:

$\mathbf{J}=\frac{c}{4\pi}\nabla\times\mathbf{B}$

Plugging in:

$\mathbf{J}=\frac{c}{4\pi}\nabla\times\left(\nabla\psi\times\hat{z}+B_{z}\hat{z}\right)$

Using vector identities:

$\mathbf{J}=\frac{c}{4\pi}\left(-\hat{z}\left(\nabla^{2}\psi\right)+\frac{\partial B_{z}}{\partial\psi}\nabla\psi\times\hat{z}\right)$

Plugging this in:

$\frac{1}{4\pi}\left(-\hat{z}\left(\nabla^{2}\psi\right)+\frac{\partial B_{z}}{\partial\psi}\nabla\psi\times\hat{z}\right)\times\left(\nabla\psi\times\hat{z}+B_{z}\hat{z}\right)=\nabla p$

Or:

$\frac{1}{4\pi}\left(-\left(\nabla^{2}\psi\right)\hat{z}\times\left(\nabla\psi\times\hat{z}\right)+\frac{\partial B_{z}}{\partial\psi}\left(\nabla\psi\times\hat{z}\right)\times B_{z}\hat{z}\right)=\nabla p$

Expanding:

$\frac{1}{4\pi}\left(-\left(\nabla^{2}\psi\right)\nabla\psi-\frac{\partial B_{z}}{\partial\psi}B_{z}\nabla\psi\right)=\nabla p$

Since the only vector on the left is $\nabla\psi$ , $p$ is a function only of $ψ$ , and we can write $\nabla p=\partial p/\partial\psi\,\nabla\psi$ , and dot with $\nabla\psi$ :

$-\left(\nabla^{2}\psi\right)-\frac{\partial B_{z}}{\partial\psi}B_{z}=4\pi\frac{\partial p}{\partial\psi}$

Or:

$\nabla^{2}\psi=-\frac{\partial B_{z}}{\partial\psi}B_{z}-4\pi\frac{\partial p}{\partial\psi}$

In slab coordinates:

$\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{\partial^{2}\psi}{\partial y^{2}}=-\frac{\partial B_{z}}{\partial\psi}B_{z}-4\pi\frac{\partial p}{\partial\psi}$

We choose the equilibrium:

$\mathbf{B}_{0}\left(x\right)=\nabla\psi_{0}\times\hat{z}+B_{0z}\hat{z}$

We are given:

B 0 y = - b 1 x - b 2 x 3

From the $\mathbf{B}_{0}$ definition:

$B_{0y}=-\frac{\partial\psi_{0}}{\partial x}=-b_{1}x-b_{2}x^{3}$

So:

$\psi_{0}=b_{1}\frac{x^{2}}{2}+b_{2}\frac{x^{4}}{4}$

The current is:

$\mathbf{j}=\frac{c}{4\pi}\nabla\times\mathbf{B}$

So:

$j_{0z}=\frac{c}{4\pi}\frac{\partial B_{0y}}{\partial x}=-b_{1}-3b_{2}x^{2}$

The equilibrium equation becomes for $p = 0$ :

$\nabla^{2}\psi\nabla\psi=-B_{z}\nabla B_{z}$

If we take the curl (to get rid of the unknown $B z$ term):

$\nabla\times\left(\nabla^{2}\psi\nabla\psi\right)=-\nabla\times\left(B_{z}\nabla B_{z}\right)$

Writing this out:

$\nabla^{2}\psi\nabla\times\nabla\psi+\nabla\left(\nabla^{2}\psi\right)\times\nabla\psi=-B_{z}\nabla\times\nabla B_{z}-\nabla B_{z}\times\nabla B_{z}$

$\nabla\times\nabla f=0$ for any scalar function:

$\nabla\left(\nabla^{2}\psi\right)\times\nabla\psi=0$

This gives the first order equation:

$\nabla\left(\nabla^{2}\psi_{0}\right)\times\nabla\left[\psi_{1}\cos ky\right]+\nabla\left[\nabla^{2}\left(\psi_{1}\cos ky\right)\right]\times\nabla\psi_{0}=0$

Since $ψ 1$ and $ψ 0$ are only functions of $x$ :

$\left(\psi_{0}^{\prime\prime\prime}\hat{x}\right)\times\left(\hat{x}\psi_{1}^{\prime}\cos ky-\hat{y}k\psi_{1}\sin ky\right)+\nabla\left[\psi_{1}^{\prime\prime}\cos ky-k^{2}\psi_{1}\cos ky\right]\times\left(\hat{x}\psi_{0}^{\prime}\right)=0$

Dotting with $\hat{z}$ (the only nonzero component):

$-\psi_{0}^{\prime\prime\prime}k\psi_{1}\sin ky+k\left(\psi_{1}^{\prime\prime}\sin ky-k^{2}\psi_{1}\sin ky\right)\psi_{0}^{\prime}=0$

Or:

$\left(\psi_{1}^{\prime\prime}-k^{2}\psi_{1}\right)\psi_{0}^{\prime}=\psi_{0}^{\prime\prime\prime}\psi_{1}$

Using $\psi_{0}=b_{1}\frac{x^{2}}{2}+b_{2}\frac{x^{4}}{4}$ :

$\left(\psi_{1}^{\prime\prime}-k^{2}\psi_{1}\right)\left(b_{1}+b_{2}x^{2}\right)=6b_{2}\psi_{1}$

We write:

$\hat{n}\cdot\mathbf{B}=0$

at the boundary:

$x=a\left(1+\epsilon\cos\left(ky\right)\right)$

So:

$\frac{dx}{dy}=-a\epsilon k\sin\left(ky\right)$

And:

$\hat{n}\propto\hat{x}+\hat{y}a\epsilon k\sin\left(ky\right)$

At the other boundary:

$\hat{n}\propto-\hat{x}+\hat{y}a\epsilon k\sin\left(ky\right)$

So:

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

The boundary condition:

$\mathbf{B}\cdot\hat{n}=0$

At the upper boundary:

$B_{x}+B_{y}a\epsilon k\sin\left(ky\right)=0$

Putting in the definition for the $B$ :

$\frac{\partial\psi}{\partial y}+\frac{\partial\psi}{\partial x}a\epsilon k\sin\left(ky\right)=0$

Using $\psi=\psi_{0}\left(x\right)-\psi_{1}\left(x\right)\cos\left(ky\right)$ :

$\psi_{1}k\sin\left(ky\right)+\frac{\partial\psi_{0}}{\partial x}a\epsilon k\sin\left(ky\right)=0$

Simplifying:

$\psi_{1}=-\frac{\partial\psi_{0}}{\partial x}a\epsilon$

At the lower boundary, we get by analogy:

$\psi_{1}=\frac{\partial\psi_{0}}{\partial x}a\epsilon$

So:

$\psi_{1}\left(x=\pm a\right)=\mp\left.a\epsilon\frac{\partial\psi_{0}}{\partial x}\right|_{x=\pm a}$

For the equation:

$\left(\psi_{1}^{\prime\prime}-k^{2}\psi_{1}\right)\left(b_{1}+b_{2}x^{2}\right)=6b_{2}\psi_{1}$

Near $x = 0$ :

$\left(\psi_{1}^{\prime\prime}-k^{2}\psi_{1}\right)b_{1}\approx6b_{2}\psi_{1}$

Which can be written as:

$\psi_{1}^{\prime\prime}=\left(k^{2}+\frac{6b_{2}}{b_{1}}\right)\psi_{1}$

We get:

$\psi_{1}=ce^{-\left|x\right|\sqrt{k^{2}+6b_{2}/b_{1}}}$

The jump condition across the boundary will be:

$\Delta^{\prime}=\frac{\psi_{1}^{\prime}\left(0_{+}\right)-\psi_{1}^{\prime}\left(0_{-}\right)}{\psi_{1}\left(0\right)}=\frac{-c\sqrt{k^{2}+\frac{6b_{2}}{b_{1}}}-c\sqrt{k^{2}+\frac{6b_{2}}{b_{1}}}}{c}=-2\sqrt{k^{2}+\frac{6b_{2}}{b_{1}}}$

If $b 2 = 0$ :