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 Incoming links: Dot-product gradient identity

[F \times (\nabla \times G)]_{i} = ϵ_{ijk} F_{j} ϵ_{k ℓ m} \partial_{ℓ} G_{m} = F_{j} \partial_{i} G_{j} - F_{j} \partial_{j} G_{i} .

(F \cdot \nabla) G_{i} = F_{j} \partial_{j} G_{i}

cancels the second term. Interchanging

F, G

gives another pair whose sum is

G_{j} \partial_{i} F_{j}

. The total right-hand side is therefore

F_{j} \partial_{i} G_{j} + G_{j} \partial_{i} F_{j} = \partial_{i} (F_{j} G_{j}),

the

i

th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:

\nabla (F \cdot G) = (F \cdot \nabla) G + (G \cdot \nabla) F + F \times (\nabla \times G) + G \times (\nabla \times F) .

It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

[F \times (\nabla \times G)]_{i} = ϵ_{ijk} F_{j} ϵ_{k ℓ m} \partial_{ℓ} G_{m} = (δ_{i ℓ} δ_{jm} - δ_{im} δ_{j ℓ}) F_{j} \partial_{ℓ} G_{m} = F_{j} \partial_{i} G_{j} - F_{j} \partial_{j} G_{i} .

The last term cancels

(F \cdot \nabla) G_{i}

. Interchanging

F, G

gives the analogous cancellation with

(G \cdot \nabla) F_{i}

. The surviving terms are

F_{j} \partial_{i} G_{j} + G_{j} \partial_{i} F_{j} = \partial_{i} (F_{j} G_{j})

\nabla (F \cdot G) = (F \cdot \nabla) G + (G \cdot \nabla) F + F \times (\nabla \times G) + G \times (\nabla \times F) .

The proof only requires continuously differentiable vector fields; neither field needs to be an irrotational vector field.