Boozer coordinates: Difference between revisions

Jump to navigation Jump to search
(Created page with 'Boozer coordinates are a set of magnetic coordinates in which the diamagnetic <math>\nabla\psi\times\mathbf{B}</math> lines are straight besides the those of magnetic field <math…')
 
Line 10: Line 10:
  \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)~,
  \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)~,
</math>
</math>
where we note that the term in brackets is a flux function. Taking the [[Flux coordinates#flux surface average|flux surface average]] <math>\langle\cdot\rangle</math> of this equation we find <math>(I_{tor}\Psi_{pol}' + I_{pol}^d\Psi_{tor}') = \langle B^2\rangle/4\pi^2\langle(\sqrt{g})^{-1}\rangle^{-1}</math>, so that we have
where we note that the term in brackets is a flux function. Taking the [[Flux coordinates#flux surface average|flux surface average]] <math>\langle\cdot\rangle</math> of this equation we find <math>(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' </math>, so that we have
:<math>
\mathbf{B}\cdot\nabla\tilde\chi = B^2  - \frac{1}{4\pi^2\sqrt{g}}\langle B^2\rangle V' ~,
</math>


In Boozer coordinates, the LHS of this equation is zero and therefore we must have <math>\sqrt{g_B}^{-1} = f(\psi)B^2</math>
In Boozer coordinates, the LHS of this equation is zero and therefore we must have  
:<math>
\sqrt{g_B} = \frac{V'}{4\pi^2}\frac{\langle B^2\rangle} {B^2}
</math>
204

edits

Navigation menu