Special case: metric tensor for $(\psi , \theta , \phi )$ coordinate system

Suppose that $(\psi , \theta , \phi )$ are arbitrary general coordinates except that $\phi$ is the usual toroidal angle in cylindrical coordinates. Then $\nabla \phi = 1 / R \hat{\ensuremath{\boldsymbol{\phi}}}$ is perpendicular to both $\nabla \psi$ and $\nabla \theta$. Using this, Eq. (152) is simplified to

$$\left(\begin{array}{c} \nabla \psi\\ \nabla \theta\\ \nabla \phi ... ...i \mathcal{J}\\ \nabla \psi \times \nabla \theta \mathcal{J} \end{array}\right)$$ (166)

Similarly, Eq. (165) is simplified to

$$\left(\begin{array}{c} \nabla \theta \times \nabla \phi \mathcal{... ...t(\begin{array}{c} \nabla \psi\\ \nabla \theta\\ \nabla \phi \end{array}\right)$$ (167)

[Note that the matrix in Eqs. (166) and (167) should be the inverse of each other. The product of the two matrix,

$$\left(\begin{array}{ccc} \vert \nabla \theta \vert^2 \mathcal{J}^... ... \theta & \vert \nabla \theta \vert^2 & 0\\ 0 & 0 & 1 / R^2 \end{array}\right),$$ (168)

can be calculated to give

$$\left(\begin{array}{ccc} A & 0 & 0\\ 0 & A & 0\\ 0 & 0 & 1 \end{array}\right),$$

where

$$A = \vert \nabla \theta \vert^2 \vert \nabla \psi \vert^2 \mathcal{J}^2 / R^2 - (\nabla \theta \cdot \nabla \psi)^2 \mathcal{J}^2 .$$

By using the definition of the Jacobian in Eq. (105), it is easy to verify that $A = 1$, i.e.,

$$\vert \nabla \theta \vert^2 \vert \nabla \psi \vert^2 - (\nabla \theta \cdot \nabla \psi)^2 = \frac{R^2}{\mathcal{J}^2},$$ (169)

]