Page 27 - Intro to Tensor Calculus

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EXAMPLE 1.1-19. In Cartesian coordinates prove the vector identity

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curl (fA)= ∇× (fA)= (∇f) × A + f(∇× A).
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Solution: Let B =curl (fA) and write the components as
B i = e ijk (fA k ) ,j
= e ijk [fA k,j + f ,j A k ]
= fe ijk A k,j + e ijk f ,j A k .

This index form can now be expressed in the vector form
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B =curl (fA)= f(∇× A)+ (∇f) × A
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EXAMPLE 1.1-20. Prove the vector identity ∇· (A + B)= ∇· A + ∇· B
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Solution: Let A + B = C and write this vector equation in the index notation as A i + B i = C i .We then
have
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∇· C = C i,i =(A i + B i ) ,i = A i,i + B i,i = ∇· A + ∇· B.
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EXAMPLE 1.1-21. In Cartesian coordinates prove the vector identity (A ·∇)f = A ·∇f
Solution: In the index notation we write
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(A ·∇)f = A i f ,i = A 1 f ,1 + A 2 f ,2 + A 3 f ,3
∂f ∂f ∂f
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= A 1 1 + A 2 2 + A 3 3 = A ·∇f.
∂x ∂x ∂x
EXAMPLE 1.1-22. In Cartesian coordinates prove the vector identity

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∇× (A × B)= A(∇· B) − B(∇· A)+(B ·∇)A − (A ·∇)B
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Solution: The pth component of the vector ∇× (A × B)is
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b e p · [∇× (A × B)] = e pqk [e kji A j B i ] ,q
= e pqk e kji A j B i,q + e pqk e kji A j,q B i
By applying the e − δ identity, the above expression simplifies to the desired result. That is,
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b e p · [∇× (A × B)] = (δ pj δ qi − δ pi δ qj )A j B i,q +(δ pj δ qi − δ pi δ qj )A j,q B i
= A p B i,i − A q B p,q + A p,q B q − A q,q B p
In vector form this is expressed
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∇× (A × B)= A(∇· B) − (A ·∇)B +(B ·∇)A − B(∇· A)