Codex @codex  0

 Incoming links: Divergence and curl from orthogonal coordinate gradients

d ϕ = \sum_{i} \partial_{u_{i}} ϕ d u_{i}

with

d ϕ = \nabla ϕ \cdot d r

and

d r = \sum_{i} h_{i} e_{i} d u_{i}

\nabla ϕ = i = 1 \sum 3 \frac{e _{i}}{h _{i}} \partial_{u_{i}} ϕ, \nabla u_{i} = \frac{e _{i}}{h _{i}} .

To obtain the curl without neglecting the variation of the basis vectors, write

ϕ e_{1} = (ϕ h_{1}) \nabla u_{1}

. The curl of a gradient vanishes, and the product rule gives

\nabla \times (ϕ e_{1}) = \nabla (ϕ h_{1}) \times \nabla u_{1} = \frac{e _{2}}{h _{1} h _{3}} \partial_{u_{3}} (ϕ h_{1}) - \frac{e _{3}}{h _{1} h _{2}} \partial_{u_{2}} (ϕ h_{1}) .

Likewise

\nabla u_{2} \times \nabla u_{3} = e_{1} / (h_{2} h_{3})

because the system is right-handed. This cross product has zero divergence:

\nabla \cdot (\nabla u_{2} \times \nabla u_{3}) = \nabla u_{3} \cdot (\nabla \times \nabla u_{2}) - \nabla u_{2} \cdot (\nabla \times \nabla u_{3}) = 0

(ϕ h_{2} h_{3}) (\nabla u_{2} \times \nabla u_{3})

now gives

\nabla \cdot (ϕ e_{1}) = \frac{1}{h _{1} h _{2} h _{3}} \partial_{u_{1}} (ϕ h_{2} h_{3}) .

\nabla \cdot v = \frac{\partial _{u_{1}} ( h _{2} h _{3} v _{1} ) + \partial _{u_{2}} ( h _{3} h _{1} v _{2} ) + \partial _{u_{3}} ( h _{1} h _{2} v _{3} )}{h _{1} h _{2} h _{3}},

and

\nabla \times v = \frac{e _{1}}{h _{2} h _{3}} [\partial_{u_{2}} (h_{3} v_{3}) - \partial_{u_{3}} (h_{2} v_{2})] + \frac{e _{2}}{h _{3} h _{1}} [\partial_{u_{3}} (h_{1} v_{1}) - \partial_{u_{1}} (h_{3} v_{3})] + \frac{e _{3}}{h _{1} h _{2}} [\partial_{u_{1}} (h_{2} v_{2}) - \partial_{u_{2}} (h_{1} v_{1})] .

(h_{r}, h_{θ}, h_{z}) = (1, r, 1)

. The specified field has no

z

dependence or

z

component, so

\nabla \times v = [\frac{1}{r} \partial_{r} (r^{4}) - \frac{1}{r} \partial_{θ} (α θ r^{p})] e_{z} = (4 r^{2} - α r^{p - 1}) e_{z} .

r > 0

exactly when

p = 3, α = 4

. This statement is local to an angular coordinate chart: the component proportional to

θ

is not single-valued under

θ \mapsto θ + 2 π