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.