- Asked 2/16
- 2068 Bhadra · 5 marks
- 2068 Magh · 5 marks
Curl of a×b
Short answer
For constant a, expand curl(a×b) componentwise, add and subtract a₁∂b₁/∂x, to obtain a div b − (a·∇)b. Hence curl(a×b) = a div b − (a·∇)b.
Prove that curl(a×b) = a div b − (a·∇)b. [Right side as printed; b after (a·∇) is probably intended]
[Reading: a is a constant vector (this is the case in which the printed result is true), and the term (a · ∇)b means the operator a₁∂/∂x + a₂∂/∂y + a₃∂/∂z applied to b. For a variable a the full identity is curl(a×b) = a div b − b div a + (b·∇)a − (a·∇)b, which reduces to the printed one when a is constant.]
Let a = (a₁, a₂, a₃) be constant and b = (b₁, b₂, b₃), functions of x, y, z.
a × b = (a₂b₃ − a₃b₂) i + (a₃b₁ − a₁b₃) j + (a₁b₂ − a₂b₁) k.
x-component of the curl
[curl(a×b)]ₓ = ∂/∂y (a₁b₂ − a₂b₁) − ∂/∂z (a₃b₁ − a₁b₃)
= a₁ ∂b₂/∂y − a₂ ∂b₁/∂y − a₃ ∂b₁/∂z + a₁ ∂b₃/∂z
Add and subtract a₁ ∂b₁/∂x:
= a₁ (∂b₁/∂x + ∂b₂/∂y + ∂b₃/∂z) − (a₁ ∂/∂x + a₂ ∂/∂y + a₃ ∂/∂z) b₁
= a₁ (div b) − (a · ∇) b₁
Other components
The y and z components follow by cyclic change of (x, y, z) and (1, 2, 3):
[curl(a×b)]_y = a₂ div b − (a·∇)b₂
[curl(a×b)]_z = a₃ div b − (a·∇)b₃
Hence curl(a × b) = a div b − (a · ∇)b, for constant a.
✓ Answer checked by Claude Opus 5.5 (high effort)
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