Hamada coordinates: Difference between revisions

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== Magnetic field and current density expressions in Hamada vector basis ==
== Magnetic field and current density expressions in Hamada vector basis ==
With the form of the Hamada coordinates' Jacobian we can now write the explicit contravariant form of the magnetic field in terms of the '''Hamada''' basis vectors
With the form of the Hamada coordinates' Jacobian we can now write the explicit [[Flux coordinates#Contravariant Form|contravariant form]] of the magnetic field in terms of the '''Hamada''' basis vectors
:<math>
:<math>
  \mathbf{B} = 2\pi\Psi_{pol}'(V)\mathbf{e}_\theta + 2\pi\Psi_{tor}'(V)\mathbf{e}_\phi~.
  \mathbf{B} = 2\pi\Psi_{pol}'(V)\mathbf{e}_\theta + 2\pi\Psi_{tor}'(V)\mathbf{e}_\phi~.
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This has the nice property of having flux constant contravariant coefficients (functions of the radial coordinate only).
This has the nice property of having flux constant contravariant coefficients (functions of the radial coordinate only).


The covariant expression is less clean
The [[Flux coordinates#Convariant Form |covariant expression]] is less clean
:<math>
:<math>
\mathbf{B} =  \frac{I_{tor}}{2\pi}\nabla\theta + \frac{I_{pol}^d}{2\pi}\nabla\phi + \nabla\tilde\chi~.
\mathbf{B} =  \frac{I_{tor}}{2\pi}\nabla\theta + \frac{I_{pol}^d}{2\pi}\nabla\phi + \nabla\tilde\chi~.
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\langle B_\theta\rangle = \langle\mathbf{B}\cdot\mathbf{e}_\theta\rangle = \left\langle \frac{I_{tor}}{2\pi} + \frac{\partial \tilde\chi}{\partial \theta}\right\rangle = \frac{I_{tor}}{2\pi} + (V')^{-1}\int\partial_\theta\tilde\chi \sqrt{g} d\theta d\phi = \frac{I_{tor}}{2\pi}
\langle B_\theta\rangle = \langle\mathbf{B}\cdot\mathbf{e}_\theta\rangle = \left\langle \frac{I_{tor}}{2\pi} + \frac{\partial \tilde\chi}{\partial \theta}\right\rangle = \frac{I_{tor}}{2\pi} + (V')^{-1}\int\partial_\theta\tilde\chi \sqrt{g} d\theta d\phi = \frac{I_{tor}}{2\pi}
</math>
</math>
where the integral over <math>\theta</math> is zero because the jacobian in Hamda coordinates is not a function of this angle. Similarly
where the integral over <math>\theta</math> is zero because the jacobian in Hamada coordinates is not a function of this angle. Similarly
:<math>
:<math>
\langle B_\phi\rangle = \langle\mathbf{B}\cdot\mathbf{e}_\phi\rangle = \frac{I^d_{pol}}{2\pi}~.
\langle B_\phi\rangle = \langle\mathbf{B}\cdot\mathbf{e}_\phi\rangle = \frac{I^d_{pol}}{2\pi}~.
</math>
</math>
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