Del in cylindrical and spherical coordinates

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Coordinate conversions

Note that the operation $\arctan (\frac{A}{B})$ must be interpreted as the two-argument inverse tangent, atan2.

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Unit vector conversions

Del formula

^α This page uses $θ$ for the polar angle and $φ$ for the azimuthal angle, which is common notation in physics. The source that is used for these formulae uses $θ$ for the azimuthal angle and $φ$ for the polar angle, which is common mathematical notation. In order to get the mathematics formulae, switch $θ$ and $φ$ in the formulae shown in the table above.
^β Defined in Cartesian coordinates as $\partial_{i} A \otimes e_{i}$ . An alternative definition is $e_{i} \otimes \partial_{i} A$ .
^γ Defined in Cartesian coordinates as $e_{i} \cdot \partial_{i} T$ . An alternative definition is $\partial_{i} T \cdot e_{i}$ .

Differential elements

Calculation rules

$div grad f \equiv \nabla \cdot \nabla f \equiv \nabla^{2} f$
$curl grad f \equiv \nabla \times \nabla f = 0$
$div curl A \equiv \nabla \cdot (\nabla \times A) = 0$
$curl curl A \equiv \nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A$ (Lagrange's formula for del)
$\nabla^{2} (f g) = f \nabla^{2} g + 2 \nabla f \cdot \nabla g + g \nabla^{2} f$
$\nabla^{2} (P \cdot Q) = Q \cdot \nabla^{2} P - P \cdot \nabla^{2} Q + 2 \nabla \cdot [(P \cdot \nabla) Q + P \times \nabla \times Q]$ (From )

Cartesian derivation

$\begin{matrix} div A & = lim_{V \to 0} \frac{\iint_{\partial V} A \cdot d S}{∭_{V} d V} \\ = \frac{[A_{x} (x + d x) - A_{x} (x)] d y d z + [A_{y} (y + d y) - A_{y} (y)] d x d z + [A_{z} (z + d z) - A_{z} (z)] d x d y}{d x d y d z} \\ = \frac{\partial A_{x}}{\partial x} + \frac{\partial A_{y}}{\partial y} + \frac{\partial A_{z}}{\partial z} \end{matrix}$

$\begin{matrix} (curl A)_{x} & = lim_{S^{⊥ \hat{x}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{[A_{z} (y + d y) - A_{z} (y)] d z - [A_{y} (z + d z) - A_{y} (z)] d y}{d y d z} \\ = \frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z} \end{matrix}$

The expressions for $(curl A)_{y}$ and $(curl A)_{z}$ are found in the same way.

Cylindrical derivation

$\begin{matrix} div A & = lim_{V \to 0} \frac{\iint_{\partial V} A \cdot d S}{∭_{V} d V} \\ = \frac{[A_{ρ} (ρ + d ρ) (ρ + d ρ) - A_{ρ} (ρ) ρ] d ϕ d z + [A_{ϕ} (ϕ + d ϕ) - A_{ϕ} (ϕ)] d ρ d z + [A_{z} (z + d z) - A_{z} (z)] d ρ (ρ + d ρ / 2) d ϕ}{ρ d ϕ d ρ d z} \\ = \frac{1}{ρ} \frac{\partial (ρ A_{ρ})}{\partial ρ} + \frac{1}{ρ} \frac{\partial A_{ϕ}}{\partial ϕ} + \frac{\partial A_{z}}{\partial z} \end{matrix}$

$\begin{matrix} (curl A)_{ρ} & = lim_{S^{⊥ \hat{ρ}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{A_{ϕ} (z) (ρ + d ρ) d ϕ - A_{ϕ} (z + d z) (ρ + d ρ) d ϕ + A_{z} (ϕ + d ϕ) d z - A_{z} (ϕ) d z}{(ρ + d ρ) d ϕ d z} \\ = - \frac{\partial A_{ϕ}}{\partial z} + \frac{1}{ρ} \frac{\partial A_{z}}{\partial ϕ} \end{matrix}$

$\begin{matrix} (curl A)_{ϕ} & = lim_{S^{⊥ \hat{ϕ}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{A_{z} (ρ) d z - A_{z} (ρ + d ρ) d z + A_{ρ} (z + d z) d ρ - A_{ρ} (z) d ρ}{d ρ d z} \\ = - \frac{\partial A_{z}}{\partial ρ} + \frac{\partial A_{ρ}}{\partial z} \end{matrix}$

$\begin{matrix} (curl A)_{z} & = lim_{S^{⊥ \hat{z}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{A_{ρ} (ϕ) d ρ - A_{ρ} (ϕ + d ϕ) d ρ + A_{ϕ} (ρ + d ρ) (ρ + d ρ) d ϕ - A_{ϕ} (ρ) ρ d ϕ}{ρ d ρ d ϕ} \\ = - \frac{1}{ρ} \frac{\partial A_{ρ}}{\partial ϕ} + \frac{1}{ρ} \frac{\partial (ρ A_{ϕ})}{\partial ρ} \end{matrix}$

$\begin{matrix} curl A & = (curl A)_{ρ} \hat{ρ} + (curl A)_{ϕ} \hat{ϕ} + (curl A)_{z} \hat{z} \\ = (\frac{1}{ρ} \frac{\partial A_{z}}{\partial ϕ} - \frac{\partial A_{ϕ}}{\partial z}) \hat{ρ} + (\frac{\partial A_{ρ}}{\partial z} - \frac{\partial A_{z}}{\partial ρ}) \hat{ϕ} + \frac{1}{ρ} (\frac{\partial (ρ A_{ϕ})}{\partial ρ} - \frac{\partial A_{ρ}}{\partial ϕ}) \hat{z} \end{matrix}$

Spherical derivation

$\begin{matrix} div A & = lim_{V \to 0} \frac{\iint_{\partial V} A \cdot d S}{∭_{V} d V} \\ = \frac{[A_{r} (r + d r) (r + d r)^{2} - A_{r} (r) r^{2}] \sin θ d θ d ϕ + [A_{θ} (θ + d θ) \sin (θ + d θ) - A_{θ} (θ) \sin θ] r d r d ϕ + [A_{ϕ} (ϕ + d ϕ) - A_{ϕ} (ϕ)] r d r d θ}{d r r d θ r \sin θ d ϕ} \\ = \frac{1}{r^{2}} \frac{\partial (r^{2} A_{r})}{\partial r} + \frac{1}{r \sin θ} \frac{\partial (A_{θ} \sin θ)}{\partial θ} + \frac{1}{r \sin θ} \frac{\partial A_{ϕ}}{\partial ϕ} \end{matrix}$

$\begin{matrix} (curl A)_{r} & = lim_{S^{⊥ \hat{r}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{A_{θ} (ϕ) r d θ + A_{ϕ} (θ + d θ) r \sin (θ + d θ) d ϕ - A_{θ} (ϕ + d ϕ) r d θ - A_{ϕ} (θ) r \sin (θ) d ϕ}{r d θ r \sin θ d ϕ} \\ = \frac{1}{r \sin θ} \frac{\partial (A_{ϕ} \sin θ)}{\partial θ} - \frac{1}{r \sin θ} \frac{\partial A_{θ}}{\partial ϕ} \end{matrix}$

$\begin{matrix} (curl A)_{θ} & = lim_{S^{⊥ \hat{θ}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{A_{ϕ} (r) r \sin θ d ϕ + A_{r} (ϕ + d ϕ) d r - A_{ϕ} (r + d r) (r + d r) \sin θ d ϕ - A_{r} (ϕ) d r}{d r r \sin θ d ϕ} \\ = \frac{1}{r \sin θ} \frac{\partial A_{r}}{\partial ϕ} - \frac{1}{r} \frac{\partial (r A_{ϕ})}{\partial r} \end{matrix}$

$\begin{matrix} (curl A)_{ϕ} & = lim_{S^{⊥ \hat{ϕ}} \to 0} \frac{\int_{\partial S} A \cdot d ℓ}{\iint_{S} d S} \\ = \frac{A_{r} (θ) d r + A_{θ} (r + d r) (r + d r) d θ - A_{r} (θ + d θ) d r - A_{θ} (r) r d θ}{r d r d θ} \\ = \frac{1}{r} \frac{\partial (r A_{θ})}{\partial r} - \frac{1}{r} \frac{\partial A_{r}}{\partial θ} \end{matrix}$

$\begin{matrix} curl A & = (curl A)_{r} \hat{r} + (curl A)_{θ} \hat{θ} + (curl A)_{ϕ} \hat{ϕ} \\ = \frac{1}{r \sin θ} (\frac{\partial (A_{ϕ} \sin θ)}{\partial θ} - \frac{\partial A_{θ}}{\partial ϕ}) \hat{r} + \frac{1}{r} (\frac{1}{\sin θ} \frac{\partial A_{r}}{\partial ϕ} - \frac{\partial (r A_{ϕ})}{\partial r}) \hat{θ} + \frac{1}{r} (\frac{\partial (r A_{θ})}{\partial r} - \frac{\partial A_{r}}{\partial θ}) \hat{ϕ} \end{matrix}$

Unit vector conversion formula

The unit vector of a coordinate parameter u is defined in such a way that a small positive change in u causes the position vector $r$ to change in $u$ direction.

Therefore, $\frac{\partial r}{\partial u} = \frac{\partial s}{\partial u} u$ where s is the arc length parameter.

For two sets of coordinate systems $u_{i}$ and $v_{j}$ , according to chain rule, $\begin{matrix} d r & = \sum_{i} \frac{\partial r}{\partial u_{i}} d u_{i} = \sum_{i} \frac{\partial s}{\partial u_{i}} {\hat{u}}_{i} d u_{i} = \sum_{j} \frac{\partial s}{\partial v_{j}} {\hat{v}}_{j} d v_{j} \\ = \sum_{j} \frac{\partial s}{\partial v_{j}} {\hat{v}}_{j} \sum_{i} \frac{\partial v_{j}}{\partial u_{i}} d u_{i} \\ = \sum_{i} \sum_{j} \frac{\partial s}{\partial v_{j}} \frac{\partial v_{j}}{\partial u_{i}} {\hat{v}}_{j} d u_{i} . \end{matrix}$

Now, we isolate the $i$ ^th component. For $i \neq k$ , let $d u_{k} = 0$ . Then divide on both sides by $d u_{i}$ to get: $\frac{\partial s}{\partial u_{i}} {\hat{u}}_{i} = \sum_{j} \frac{\partial s}{\partial v_{j}} \frac{\partial v_{j}}{\partial u_{i}} {\hat{v}}_{j} .$