Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 3C (Vector Calculus)

Table of contents
- i 3C
  - Solution i
- ii 3C
  - Solution ii

i

Solution

Use the Levi-Civita symbol and sum over repeated indices. The contraction of two Levi-Civita symbols and the product rule give

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

Reading the four terms as divergence and directional derivatives proves the curl of a cross product identity

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

ii

Solution

The contraction of two Levi-Civita symbols gives

[F \times (\nabla \times G)]_{i} = F_{j} \partial_{i} G_{j} - F_{j} \partial_{j} G_{i}

. Similarly

[G \times (\nabla \times F)]_{i} = G_{j} \partial_{i} F_{j} - G_{j} \partial_{j} F_{i}

. Add the two directional-derivative terms to cancel the last terms, leaving

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

The product rule has therefore proved the gradient of a dot product identity

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

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