Page 68 - Intro to Tensor Calculus

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I 50. Let a i and b i for i =1,... ,n denote arbitrary vectors and form the dyadic

Φ= a 1 b 1 + a 2 b 2 + ··· + a n b n .
By definition the first scalar invariant of Φ is

φ 1 = a 1 · b 1 + a 2 · b 2 + ··· + a n · b n
where a dot product operator has been placed between the vectors. The first vector invariant of Φ is defined
~
φ = a 1 × b 1 + a 2 × b 2 + ··· + a n × b n
where a vector cross product operator has been placed between the vectors.
(a) Show that the first scalar and vector invariant of

Φ= b e 1 b e 2 + b e 2 b e 3 + b e 3 b e 3
are respectively 1 and b e 1 + b e 3 .
(b) From the vector f = f 1 b e 1 + f 2 b e 2 + f 3 b e 3 one can form the dyadic ∇f having the matrix components
 ∂f 1 ∂f 2 ∂f 3 
∂x ∂x ∂x
∇f =  ∂f 1 ∂f 2 ∂f 3  .
∂y ∂y ∂y
∂f 1 ∂f 2 ∂f 3
∂z ∂z ∂z
Show that this dyadic has the first scalar and vector invariants given by
∂f 1 ∂f 2 ∂f 3
∇· f = + +
∂x ∂y ∂z

∂f 3 ∂f 2 ∂f 1 ∂f 3 ∂f 2 ∂f 1
∇× f = − b e 1 + − b e 2 + − b e 3
∂y ∂z ∂z ∂x ∂x ∂y
I 51. Let Φ denote the dyadic given in problem 50. The dyadic Φ 2 defined by
1 X
Φ 2 = a i × a j b i × b j
2
i,j
is called the Gibbs second dyadic of Φ, where the summation is taken over all permutations of i and j.When
i = j the dyad vanishes. Note that the permutations i, j and j, i give the same dyad and so occurs twice
in the final sum. The factor 1/2 removes this doubling. Associated with the Gibbs dyad Φ 2 are the scalar
invariants
1 X
φ 2 = (a i × a j ) · (b i × b j )
2
i,j
1 X
φ 3 = (a i × a j · a k )(b i × b j · b k )
6
i,j,k
Show that the dyad
Φ= as + tq + cu
has
the first scalar invariant φ 1 = a · s + b · t + c · u
~
the first vector invariant φ = a × s + b × t + c × u
Gibbs second dyad Φ 2 = b × ct × u + c × au × s + a × bs × t
second scalar of Φ φ 2 =(b × c) · (t · u)+(c × a) · (u × s)+(a × b) · (s × t)

third scalar of Φ φ 3 =(a × b · c)(s × t · u)