# Boozer coordinates

Boozer coordinates are a set of magnetic coordinates in which the diamagnetic $\nabla\psi\times\mathbf{B}$ lines are straight besides those of magnetic field $\mathbf{B}$. The periodic part of the stream function of $\mathbf{B}$ and the scalar magnetic potential are flux functions (that can be chosen to be zero without loss of generality) in this coordinate system.

## Form of the Jacobian for Boozer coordinates

Multiplying the covariant representation of the magnetic field by $\mathbf{B}\cdot$ we get

$B^2 = \mathbf{B}\cdot\nabla\chi = \frac{I_{tor}}{2\pi}\mathbf{B}\cdot\nabla\theta + \frac{I_{pol}^d}{2\pi}\mathbf{B}\cdot\nabla\phi + \mathbf{B}\cdot\nabla\tilde\chi~.$

Now, using the known form of the contravariant components of the magnetic field for a magnetic coordinate system we get

$\mathbf{B}\cdot\nabla\tilde\chi = B^2 - \frac{1}{4\pi^2\sqrt{g}}\left(I_{tor}\Psi_{pol}' + I_{pol}^d\Psi_{tor}' \right)~,$

where we note that the term in brackets is a flux function. Taking the flux surface average $\langle\cdot\rangle$ of this equation we find $(I_{tor}\Psi_{pol}' + I_{pol}^d\Psi_{tor}') = 4\pi^2\langle B^2\rangle/\langle(\sqrt{g})^{-1}\rangle = \langle B^2\rangle V'$, so that we have

$\mathbf{B}\cdot\nabla\tilde\chi = B^2 - \frac{1}{4\pi^2\sqrt{g}}\langle B^2\rangle V' ~,$

In Boozer coordinates, the LHS of this equation is zero and therefore we must have

$\sqrt{g_B} = \frac{V'}{4\pi^2}\frac{\langle B^2\rangle} {B^2}$

## Contravariant representation of the magnetic field in Boozer coordinates

Using this Jacobian in the general form of the magnetic field in magnetic coordinates one gets.

$\mathbf{B} = 2\pi\frac{d\Psi_{pol}}{dV}\frac{B^2}{\langle B^2\rangle}\mathbf{e}_\theta + 2\pi\frac{d\Psi_{tor}}{dV}\frac{B^2}{\langle B^2\rangle}\mathbf{e}_\phi$

so, in Boozer coordinates,

$B^\theta = 2\pi\frac{d\Psi_{pol}}{dV}\frac{B^2}{\langle B^2\rangle} \quad \text{and} \quad B^\phi = 2\pi\frac{d\Psi_{tor}}{dV}\frac{B^2}{\langle B^2\rangle}$

## Covariant representation of the magnetic field in Boozer coordinates

The covariant representation of the field is also relatively simple when using Boozer coordinates, since the angular covariant $B$-field components are flux functions in these coordinates

$\mathbf{B} = -\tilde\eta\nabla\psi + \frac{I_{tor}}{2\pi}\nabla\theta + \frac{I_{pol}^d}{2\pi}\nabla\phi~.$

The covariant $B$-field components are explicitly

$B_\psi = -\tilde{\eta} \quad , \quad B_\theta =\frac{I_{tor}}{2\pi} \quad \text{and} \quad B_\phi = \frac{I_{pol}^d}{2\pi}~.$

It then follows that

$\nabla\psi\times\mathbf{B} = \nabla\psi\times\nabla\left(\frac{I_{tor}}{2\pi}\theta + \frac{I_{pol}^d}{2\pi}\phi\right)~,$

and then the 'diamagnetic' lines are straight in Boozer coordinates and given by ${I_{tor}}\theta + {I_{pol}^d}\phi = \mathrm{const.}$.

It is also useful to know the expression of the following object in Boozer coordinates

$\frac{\nabla V\times\mathbf{B}}{B^2} = -\frac{2\pi I_{pol}^d}{\langle B^2\rangle}\mathbf{e}_\theta + \frac{2\pi I_{tor}}{\langle B^2\rangle}\mathbf{e}_\phi~.$

The above expressions adopt very simple forms for the 'vacuum' field, i.e. one with $\nabla\times\mathbf{B} = 0$. In this case $I_{tor} = 0$ and $\tilde{\eta} = 0$ leaving, e.g.

$\mathbf{B} = \frac{I_{pol}^d}{2\pi}\nabla\phi,\quad (\text{for a vacuum field)}$

In a low-$\beta$ stellarator the equilibrium magnetic field is approximatelly given by the vauum value.