Toroidal coordinates: Difference between revisions

No edit summary
Line 193: Line 193:
If we consider an equilibrium magnetic field such that <math> \mathbf{j}\times\mathbf{B} \propto \nabla\psi</math>, then both <math> \mathbf{B}\cdot\nabla\psi = 0</math> and <math> \nabla\times\mathbf{B}\cdot\nabla\psi = 0</math> and the magnetic field can be written as
If we consider an equilibrium magnetic field such that <math> \mathbf{j}\times\mathbf{B} \propto \nabla\psi</math>, then both <math> \mathbf{B}\cdot\nabla\psi = 0</math> and <math> \nabla\times\mathbf{B}\cdot\nabla\psi = 0</math> and the magnetic field can be written as
:<math>
:<math>
\mathbf{B} = \eta\nabla\psi + \nabla\chi
\mathbf{B} = \beta\nabla\psi + \nabla\chi
</math>
</math>
where <math>\chi</math> is identified as the magnetic ''scalar'' potential. Its general form is
where <math>\chi</math> is identified as the magnetic ''scalar'' potential. Its general form is
:<math>
:<math>
\chi(\psi, \theta, \phi) = \frac{I_{tor}}{2\pi}\theta + \frac{I_{pol}^d}{2\pi}\phi + \tilde\chi(\psi, \theta, \phi)
\chi(\psi, \theta, \phi) = \frac{I_{tor}}{2\pi}\theta + \frac{I_{pol}^d}{2\pi}\phi + \tilde\chi(\psi, \theta, \phi)
</math>
In fact, noting that
:<math>
\int_S \mu_0\mathbf{j}\cdot d\mathbf{S}
= \int_{\partial S}\mathbf{B}\cdot d\mathbf{l}
= \oint(\beta\nabla\psi + \nabla\chi)\cdot d\mathbf{l}
= \oint(\beta d\psi + d\chi)
</math>
and choosing an integration circuit contained within a flux surface we get
:<math>
\int_S \mu_0\mathbf{j}\cdot d\mathbf{S}
= \Delta \chi = \frac{I_{tor}}{2\pi}\Delta\theta + \frac{I_{pol}^d}{2\pi}\Delta\phi~.
</math>
If we now chose a ''toroidal'' circuit <math>(\Delta\theta = 0, \Delta\phi = 2\pi)</math> we get
:<math>
I_{pol}^d = \int_S \mu_0\mathbf{j}\cdot d\mathbf{S}\; ; ~\mathrm{with}~ \partial S ~\mathrm{such~that}~ (\Delta\theta = 0, \Delta\phi = 2\pi)~.
</math>
</math>


Anonymous user