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 Incoming links: Curl in spherical coordinates

Del in cylindrical and spherical coordinates Created 2026-10-14 Updated 2026-10-18

This collects the gradient, divergence, curl and Laplacian formulas in cylindrical and spherical coordinates. Coordinate unit vectors vary with position, so taking their derivatives is essential when translating vector calculus operators from Cartesian coordinates. Curl in spherical coordinates is one of these formulas.

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Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 12C / Solution Created 2026-09-24 Updated 2026-10-14

Use suffix notation, with repeated indices summed. Contracting the two Levi-Civita symbols gives

[u \times (\nabla \times A)]_{i} = ϵ_{ijk} u_{j} ϵ_{k l m} \partial_{l} A_{m} = u_{j} (\partial_{i} A_{j} - \partial_{j} A_{i}) .

Applying the product rule to

h = A_{i} B_{i}

and substituting the two evolution equations yields

\partial_{t} h = B_{i} u_{j} \partial_{i} A_{j} - B_{i} u_{j} \partial_{j} A_{i} + B_{i} \partial_{i} ψ + A_{i} B_{j} \partial_{j} u_{i} - A_{i} u_{j} \partial_{j} B_{i} = B_{i} \partial_{i} (u_{j} A_{j} + ψ) - u_{j} \partial_{j} (A_{i} B_{i}) .

In the second line we renamed dummy indices in

A_{i} B_{j} \partial_{j} u_{i}

and then combined product derivatives. Hence

f = u \cdot A + ψ, \partial_{t} h = B \cdot \nabla f - u \cdot \nabla h .

No zero-divergence assumption was used in this local identity.

When both fields are solenoidal vector fields, the product rule for divergence rewrites this as a conservation law:

\partial_{t} h = \nabla \cdot (f B - h u) .

Because the volume is fixed, differentiation under its integral and the divergence theorem give

\frac{d H}{d t} = \int_{\partial V} (f B - h u) \cdot n d S = 0

under the two tangency conditions. Thus the volume integral is conserved. This is helicity conservation by tangent boundary conditions; when

B = \nabla \times A

, the density is the magnetic helicity density.

For the given axisymmetric field, the curl in spherical coordinates gives

B_{r} B_{θ} B_{ϕ} = \frac{1}{r sin θ} \partial_{θ} (sin θ A_{ϕ}) = 2 (a^{2} - r^{2}) cos θ, = - \frac{1}{r} \partial_{r} (r A_{ϕ}) = (4 r^{2} - 2 a^{2}) sin θ, = \frac{1}{r} \partial_{r} (r A_{θ}) = 3 a r sin θ .

There are no

ϕ

derivatives and

A_{r} = 0

. Consequently

h = A_{θ} B_{θ} + A_{ϕ} B_{ϕ} = a r^{2} (4 r^{2} - 2 a^{2}) sin^{2} θ + 3 a r^{2} (a^{2} - r^{2}) sin^{2} θ = a r^{2} (a^{2} + r^{2}) sin^{2} θ .

The volume form in spherical polar coordinates is

r^{2} sin θ d r d θ d ϕ

, so

H = 2 πa \int_{0}^{a} r^{4} (a^{2} + r^{2}) d r \int_{0}^{π} sin^{3} θ d θ = \frac{8 πa}{3} (\frac{a ^{7}}{5} + \frac{a ^{7}}{7}) = \frac{32 π a ^{8}}{35} .

Here

B_{r} = 0

at

r = a

, as required for the magnetic-field tangency condition in the conservation argument; tangency of

u

is a separate hypothesis.

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 3 / 9A / c / Solution Created 2026-09-24 Updated 2026-09-25

Each field has only an azimuthal component and is independent of

r

and

ϕ

. The curl in spherical coordinates therefore reduces to

(\nabla \times A)_{r} = \frac{1}{r s i n θ} \frac{\partial}{\partial θ} (sin θ A_{ϕ}), (\nabla \times A)_{θ} = - \frac{1}{r} \frac{\partial}{\partial r} (r A_{ϕ}),

and its

ϕ

component vanishes. The half-angle identities

sin θ tan \frac{θ}{2} = 1 - cos θ, sin θ (- cot \frac{θ}{2}) = - (1 + cos θ)

show in both cases that

\nabla \times A_{+} = \nabla \times A_{-} = \frac{e _{r}}{r ^{2}} = \frac{x}{r ^{3}} .

Apply part (a) with

g (r) = r^{- 3}

. Then

\nabla \cdot \frac{x}{r ^{3}} = r (- 3 r^{- 4}) + 3 r^{- 3} = 0

away from the origin, explicitly confirming that each resulting curl has zero divergence.

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 1 / Solution Created 2026-10-18 Updated 2026-10-20

Vector identity. Work in Cartesian coordinates and use the Einstein summation convention. Contraction of the two Levi-Civita symbols gives

[u \times (\nabla \times u)]_{i} = ϵ_{ijk} u_{j} ϵ_{k l m} \partial_{l} u_{m} = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) u_{j} \partial_{l} u_{m} = u_{j} \partial_{i} u_{j} - u_{j} \partial_{j} u_{i} = \frac{1}{2} \partial_{i} (u_{j} u_{j}) - [(u \cdot \nabla) u]_{i} .

Thus the convective acceleration identity is

(u \cdot \nabla) u = \frac{1}{2} \nabla∣ u ∣^{2} - u \times (\nabla \times u) .

Each term is a geometrically defined vector field: the directional derivative of a vector along itself, the gradient of a scalar field, and a cross product with the curl. Equality in a Cartesian frame is equality of these vectors, so it holds in every coordinate system. When using a varying coordinate basis, derivatives of the varying basis vectors must be included; differentiating only the component functions would describe a different operation.

Spherical metric. Let

R (r, θ, ϕ) = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ) .

Differentiating gives the coordinate tangent vectors

R_{r} R_{θ} R_{ϕ} = (sin θ cos ϕ, sin θ sin ϕ, cos θ), = r (cos θ cos ϕ, cos θ sin ϕ, - sin θ), = r sin θ (- sin ϕ, cos ϕ, 0) .

Their pairwise dot products vanish, proving that spherical coordinates are orthogonal coordinates. Their lengths are the scale factors of orthogonal coordinates:

h_{r} = 1, h_{θ} = r, h_{ϕ} = r sin θ, d s^{2} = d r^{2} + r^{2} d θ^{2} + r^{2} sin^{2} θ d ϕ^{2} .

These formulas use the usual chart

r > 0, 0 < θ < π

; the origin and polar axis are coordinate singularities.

Azimuthal field. Write the speed as

U (r)

to distinguish it from the velocity field

u = U (r) e_{ϕ}

. Using the curl in spherical coordinates,

\nabla \times u = \frac{U c o t θ}{r} e_{r} - \frac{U + r U ^{'}}{r} e_{θ} .

Also

\frac{1}{2} \nabla∣ u ∣^{2} = U U^{'} e_{r}

. Since

e_{ϕ} \times e_{r} = e_{θ}

and

e_{ϕ} \times e_{θ} = - e_{r}

,

u \times (\nabla \times u) = \frac{U ( U + r U ^{'} )}{r} e_{r} + \frac{U ^{2} c o t θ}{r} e_{θ} .

The

U U^{'}

terms cancel in the convective acceleration identity, giving

[(u \cdot \nabla) u]_{r} = - \frac{U ^{2}}{r}, [(u \cdot \nabla) u]_{θ} = - \frac{U ^{2} cot θ}{r}, [(u \cdot \nabla) u]_{ϕ} = 0.

This centripetal acceleration of an azimuthal flow points toward the polar axis. A streamline is a circle of cylindrical radius

r sin θ

, so its acceleration is

- U^{2} e_{ρ} / (r sin θ)

, with

e_{ρ} = sin θ e_{r} + cos θ e_{θ}

. This independent geometric argument recovers both nonzero components and explains why

U^{'}

cannot appear: the particle travels at fixed

r

.

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 1 / c / Solution Created 2026-10-18 Updated 2026-10-20

For an axisymmetric azimuthal vector field, the curl in spherical coordinates has components

(\nabla \times A)_{r} = \frac{1}{r s i n θ} \partial_{θ} (sin θ A_{ϕ}), (\nabla \times A)_{θ} = - \frac{1}{r} \partial_{r} (r A_{ϕ}), (\nabla \times A)_{ϕ} = 0.

The angular expressions are

sin θ A_{ϕ}^{(+)} = (1 - cos θ) / r

and

sin θ A_{ϕ}^{(-)} = - (1 + cos θ) / r

, each with

θ

sin θ / r

. Also

r A_{ϕ}

is independent of

r

, so the second component vanishes. Consequently, on each potential's regular patch,

\nabla \times A^{(+)} = \nabla \times A^{(-)} = \frac{e _{r}}{r ^{2}} = B .

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