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Gradient of (a·r)/rⁿ

Short answer

Apply product rule: ∇((a·r) r^-n) = (a·r)(-n r^{-n-2} r) + r^{-n} a = a/r^n - (n/r^{n+2})(a·r) r.

If r be the position vector and a is a constant vector then prove that ∇(a·r / rⁿ) = a/rⁿ − (n/r^(n+2)) (a·r) r.

Let r = xi + yj + zk, r = |r|, and a a constant vector. We use ∇r = r/r (so ∇(rᵐ) = m r^(m−2) r) and ∇(a · r) = a.

Product rule

Write a · r / rⁿ = (a · r) · r⁻ⁿ and use ∇(uv) = u∇v + v∇u:

∇[(a · r) r⁻ⁿ] = (a · r) ∇(r⁻ⁿ) + r⁻ⁿ ∇(a · r)

The two gradients

  • ∇(r⁻ⁿ) = −n r^(−n−1) ∇r = −n r^(−n−1) (r/r) = −n r^(−n−2) r.
  • ∇(a · r) = ∇(a₁x + a₂y + a₃z) = a₁i + a₂j + a₃k = a.

Combine

∇(a·r / rⁿ) = (a · r)(−n r^(−n−2) r) + r⁻ⁿ a
            = a/rⁿ − (n / r^(n+2)) (a · r) r

Hence ∇(a·r / rⁿ) = a/rⁿ − (n/r^(n+2)) (a · r) r.

✓ Answer checked by Claude Opus 5.5 (high effort)

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