Boozer coordinates

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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.