Skip to main content

Divergence of (a×r)/r is zero

Short answer

Apply divergence product rule: ∇(1/r) = -r/r³ and ∇·(a×r)=0; thus ∇·((a×r)/r) = (-r/r³)·(a×r) = 0, since r·(a×r)=0.

If r be the position vector and a is constant vector then prove that ∇·((a×r)/r) = 0.

Let r = xi + yj + zk, r = |r|, a a constant vector. Use ∇ · (φF) = ∇φ · F + φ(∇ · F) with φ = 1/r and F = a × r.

∇φ

∇(1/r) = −(1/r²)∇r = −(1/r²)(r/r) = −r/r³.

∇ · (a × r)

Using ∇ · (a × r) = r · (∇ × a) − a · (∇ × r). Here ∇ × a = 0 (a is constant) and ∇ × r = 0, so ∇ · (a × r) = 0.

(Direct check: a × r = (a₂z − a₃y, a₃x − a₁z, a₁y − a₂x); ∂/∂x of the first is 0, ∂/∂y of the second is 0, ∂/∂z of the third is 0.)

Combine

∇ · ((a × r)/r) = (−r/r³) · (a × r) + (1/r)(0)
                = −(1/r³) r · (a × r) + 0
                = −(1/r³) [r a r]  = 0

because r · (a × r) is a scalar triple product with a repeated vector r, so it is zero (r is perpendicular to a × r).

Hence ∇ · ((a × r)/r) = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 5 questions →