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1. Show that the decomposition of a tensor into the symmetric and anti-symmetric
parts is unique.
๐ =1
2(๐ + ๐T) +
1
2(๐ โ ๐T) = sym๐ + skw๐
Suppose there is another decomposition into symmetric and antisymmetric parts
similar to the above so that โ๐ such that ๐ =1
2(๐ + ๐T) +
1
2(๐ โ ๐T). Now take the
inner product of the two expressions for the tensor ๐ and a symmetric tensor ๐
๐:๐ = (๐ฌ๐ฒ๐ฆ๐ + ๐ฌ๐ค๐ฐ๐): ๐
= (๐ฌ๐ฒ๐ฆ๐): ๐
= (1
2(๐ + ๐T) +
1
2(๐ โ ๐T)) : ๐
= (1
2(๐ + ๐T)) : ๐
since the inner products of symmetric and anti-symmetric tensors vanish. The
above equation leads to,
(sym ๐): ๐ โ (1
2(๐ + ๐T)) : ๐ = (sym๐ โ
1
2(๐ + ๐T)) : ๐ = ๐
Since ๐ is arbitrary, it is clear that,
sym ๐ โ1
2(๐ + ๐T) = ๐
or, sym๐ โก1
2(๐ + ๐T) =
1
2(๐ + ๐T) showing uniqueness of the symmetrical part. The
uniqueness of the anti-symmetrical part is arrived at by taking the inner product with an
antisymmetric tensor at the beginning.
2. The trilinear mapping, [ , ,] :V ร V ร V โ R from the product set
V ร V ร V to real space is defined by: [๐, ๐, ๐] โก ๐ โ (๐ ร ๐) = (๐ ร ๐) โ ๐. Show
that [ ๐, ๐, ๐] = [๐, ๐, ๐] = [๐, ๐, ๐] = โ[๐, ๐, ๐] = โ[๐, ๐, ๐] = โ[๐, ๐, ๐]
In component form,
[๐, ๐, ๐] = ๐๐๐๐๐๐๐๐๐๐
Cyclic permutations of this, upon remembering that (๐, ๐, ๐) are dummy indices,
yield,
๐๐๐๐๐๐๐๐๐๐ = [๐, ๐, ๐] = ๐๐๐๐๐๐๐๐๐๐
= ๐๐๐๐๐๐๐๐๐๐ = [๐, ๐, ๐] = ๐๐๐๐๐๐๐๐๐๐
The other results follow from antisymmetric arrangements and the nature of ๐๐๐๐.
3. Given that, [๐, ๐, ๐] โก ๐ โ (๐ ร ๐) = (๐ ร ๐) โ ๐. Show that this product vanishes if
the vectors (๐, ๐, ๐) are linearly dependent.
Suppose it is possible to find scalars ๐ผ and ๐ฝ such that, ๐ = ๐ผ๐ + ๐ฝ๐. It therefore
means that,
[๐, ๐, ๐] = ๐๐๐๐๐๐๐๐๐๐ = ๐๐๐๐(๐ผ๐๐ + ๐ฝ๐๐)๐๐๐๐
= ๐ผ๐๐๐๐๐๐๐๐๐๐ + ๐ฝ๐๐๐๐๐๐๐๐๐๐
= 0
Note that ๐๐๐๐๐๐ is symmetric in ๐ and ๐, ๐๐๐๐๐๐ is symmetric in ๐ and ๐ and ๐๐๐๐ is
antisymmetric in ๐, ๐ and ๐ Because each term is the product of a symmetric and an
antisymmetric object which must vanish.
4. Show that the product of a symmetric and an antisymmetric object vanishes.
Consider the product sum, ๐๐๐๐๐๐๐๐๐๐ in which ๐๐๐๐ is symmetric in ๐ and ๐ and ๐๐๐๐ is
antisymmetric in ๐, ๐ and ๐. Only the shared symmetrical and antisymmetrical
indices ๐, ๐ are relevant here.
๐๐๐๐๐๐๐๐๐๐ = โ๐๐๐๐๐๐๐๐๐๐ = โ๐๐๐๐๐๐๐๐๐๐ = โ๐๐๐๐๐๐๐๐๐๐ = 0
The first equality on account of the antisymmetry of ๐๐๐๐ in ๐, ๐; the second on the
symmetry of ๐๐๐๐ in ๐, ๐; the third on the fact that ๐, ๐ are dummy indices. These
vanish because a non-trivial scalar quantity cannot be the negative of itself.
This is a general rule and its observation makes a number of steps easy to see
transparently. Watch out for it.
5. Show that the product ๐๐ = (
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
)(
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
) = ๐ Can be
written in indicial notation as, ๐๐๐๐๐๐ = ๐๐๐.
To show this, apply summation convention and see that,
for ๐ = 1, ๐ = 1, ๐11๐11 + ๐12๐21 + ๐13๐31 = ๐11
for ๐ = 1, ๐ = 2, ๐11๐12 + ๐12๐22 + ๐13๐32 = ๐12
for ๐ = 1, ๐ = 3, ๐11๐13 + ๐12๐23 + ๐13๐33 = ๐13
for ๐ = 2, ๐ = 1, ๐21๐11 + ๐22๐21 + ๐23๐31 = ๐21
for ๐ = 2, ๐ = 2, ๐21๐12 + ๐22๐22 + ๐23๐32 = ๐22
for ๐ = 2, ๐ = 3, ๐21๐13 + ๐22๐23 + ๐23๐33 = ๐23
for ๐ = 3, ๐ = 1, ๐31๐11 + ๐32๐21 + ๐33๐31 = ๐31
for ๐ = 3, ๐ = 2, ๐31๐12 + ๐32๐22 + ๐33๐32 = ๐32
for ๐ = 3, ๐ = 3, ๐31๐13 + ๐32๐23 + ๐33๐33 = ๐33
It is necessary to go through this manual process to gain the experience of seeing
this transparently in future. Similary, ๐๐๐ = ๐ can be written in indicial notation as,
๐๐๐๐๐๐ = ๐๐๐ which again becomes clear after a manual expansion after invoking the
summation convention.
6. Given that vectors ๐ฎ, ๐ฏ and ๐ฐ are linearly independent, and that the tensor ๐ is not
singular, show that the set ๐๐ฎ, ๐๐ฏ and ๐๐ฐ are also linearly independent.
If ๐ is not singular, then its determinant exists and is not equal to zero. Therefore,
det ๐ =[๐๐ฎ, ๐๐ฏ, ๐๐ฐ]
[๐ฎ, ๐ฏ, ๐ฐ]โ 0
Consequently, [๐๐ฎ, ๐๐ฏ, ๐๐ฐ] โ 0. Which shows that ๐๐ฎ, ๐๐ฏ and ๐๐ฐ are also linearly
independent.
7. Given that vectors ๐ฎ, ๐ฏ and ๐ฐ are linearly independent, and that the tensor ๐ is not
singular, show that the set ๐๐ฎ, ๐๐ฏ and ๐๐ฐ are also linearly independent.
If ๐ is not singular, if ๐๐ฎ, ๐๐ฏ and ๐๐ฐ are also linearly dependent, then โ๐ผ, ๐ฝ and ๐พ all
real such that ๐ผ๐๐ฎ + ๐ฝ๐๐ฏ + ๐พ๐๐ฐ = ๐จ. But ๐ฎ, ๐ฏ and ๐ฐ are linearly independent. This
means that ๐ผ๐ฎ + ๐ฝ๐ฏ + ๐พ๐ฐ โ ๐จ.
๐ผ๐๐ฎ + ๐ฝ๐๐ฏ + ๐พ๐๐ฐ = ๐(๐ผ๐ฎ + ๐ฝ๐ฏ + ๐พ๐ฐ) = ๐.
This means that ๐ผ๐ฎ + ๐ฝ๐ฏ + ๐พ๐ฐ = ๐จ. This states that set of linearly independent
vectors is linearly dependent! That is a contradiction!
8. Given that vectors ๐ฎ and ๐ฏ are linearly independent, and that the tensor ๐ is not
singular, show that the set ๐๐ฎ and ๐๐ฏ are also linearly independent.
If ๐ is not singular, if ๐๐ฎ and ๐๐ฏ are also linearly dependent, then โ๐ผ, and ๐ฝ both
real such that ๐ผ๐๐ฎ + ๐ฝ๐๐ฏ = ๐จ. But ๐ฎ, ๐ฏ and ๐ฐ are linearly independent. This means
that ๐ผ๐ฎ + ๐ฝ๐ฏ โ ๐จ.
๐ผ๐๐ฎ + ๐ฝ๐๐ฏ = ๐(๐ผ๐ฎ + ๐ฝ๐ฏ) = ๐.
This means that ๐ผ๐ฎ + ๐ฝ๐ฏ = ๐จ. This states that set of linearly independent vectors is
linearly dependent! That is a contradiction!
.
9. Given that vectors ๐ฎ and ๐ฏ are linearly independent, and that the tensor ๐ is not
singular, show that the set ๐๐ฎ and ๐๐ฏ are also linearly independent.
If ๐ is not singular, then its determinant exists and is not equal to zero. Therefore
the cofactor, ๐c = ๐โT det ๐ โ 0 also exists and is non-zero. The linear
independence of ๐ฎ and ๐ฏ means that the parallelogram formed by them has a non-
trivial area ๐ฎ ร ๐ฏ โ 0. Now, the parallelogram formed by ๐๐ฎ and ๐๐ฏ is also non zero
because,
๐๐ฎ ร ๐๐ฏ = ๐c(๐ฎ ร ๐ฏ) โ 0
Hence ๐๐ฎ and ๐๐ฏ are also linearly independent.
10. Given three vectors ๐ฎ, ๐ฏ and ๐ฐ, show that (๐ฐ ร ๐ฎ) ร (๐ฐ ร ๐ฏ) = (๐ฐ โ ๐ฐ)(๐ฎ ร ๐ฏ)
and that for the unit vector ๐, [๐, ๐ ร ๐ฎ, ๐ ร ๐ฏ] = [๐, ๐ฎ, ๐ฏ]
(๐ฐ ร ๐ฎ) ร (๐ฐ ร ๐ฏ) = [(๐ฐ ร ๐ฎ) โ ๐ฏ]๐ฐ โ [(๐ฐ ร ๐ฎ) โ ๐ฐ]๐ฏ
= [(๐ฐ ร ๐ฎ) โ ๐ฏ]๐ฐ
= [(๐ฎ ร ๐ฏ) โ ๐ฐ]๐ฐ
= (๐ฐ โ ๐ฐ)(๐ฎ ร ๐ฏ)
Consequently,
[๐, ๐ ร ๐ฎ, ๐ ร ๐ฏ] = ๐ โ [(๐ ร ๐ฎ) ร (๐ ร ๐ฏ)]
= ๐ โ [(๐ โ ๐)(๐ฎ ร ๐ฏ)]
= (๐ฎ ร ๐ฏ) โ (๐ โ ๐)๐
= (๐ฎ ร ๐ฏ) โ ๐ = [๐, ๐ฎ, ๐ฏ]
making use of the symmetry of (๐ โ ๐).
11. Given three vectors ๐ฎ, ๐ฏ and ๐ฐ, using the result, (๐ฐ ร ๐ฎ) ร (๐ฐ ร ๐ฏ) =
(๐ฐ โ ๐ฐ)(๐ฎ ร ๐ฏ), show that [(๐ฎ ร ๐ฏ), (๐ฏ ร ๐ฐ), (๐ฐ ร ๐ฎ)] = [๐ฎ, ๐ฏ,๐ฐ]2
From the given result,
[(๐ฎ ร ๐ฏ), (๐ฏ ร ๐ฐ), (๐ฐ ร ๐ฎ)] = โ(๐ฎ ร ๐ฏ) โ (๐ฐ ร ๐ฏ) ร (๐ฐ ร ๐ฎ)
= โ(๐ฎ ร ๐ฏ) โ (๐ฐ โ ๐ฐ)(๐ฏ ร ๐ฎ)
= (๐ฎ ร ๐ฏ)((๐ฐ โ ๐ฎ ร ๐ฏ)๐ฐ)
= [๐ฎ, ๐ฏ,๐ฐ]2
12. For any tensor ๐จ, define (Sym(๐จ))๐๐
=๐
๐(๐จ๐๐ + ๐จ๐๐). Show that Sym(๐จT๐บ๐จ) =
๐จTSym(๐บ)๐จ
Clearly, Sym(๐บ) =๐
๐(๐บ + ๐บT)
It also follows that,
๐จTSym(๐บ)๐จ = 1
2๐จT(๐บ + ๐บT)๐จ
=1
2(๐จT๐บ๐จ + ๐จT๐บT๐จ)
But Sym(๐จT๐บ๐จ) =๐
๐(๐จT๐บ๐จ + ๐จT๐บT๐จ).
Hence Sym(๐จT๐บ๐จ) = ๐จTSym(๐บ)๐จ
13. Given that the trace of a dyad ๐ โ ๐, tr(๐ โ ๐) = ๐ โ ๐. By expressing the tensors
๐ and ๐ in component form, show that tr(๐๐) = tr(๐๐) = tr(๐T๐T) = tr(๐T๐T)
In component form, ๐ = ๐๐๐๐ ๐ โ ๐ ๐, ๐ = ๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ.
๐๐ = ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ
tr(๐๐) = ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ = ๐๐๐๐๐ผ๐ฝ๐๐๐ฝ๐๐๐ผ
= ๐๐๐๐๐๐
๐T๐T = ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ฝ โ ๐ ๐ผ) = ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ผ๐๐๐ฝ
so that
tr(๐T๐T) = ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ผ๐๐๐ฝ = ๐๐๐๐๐ผ๐ฝ๐๐๐ผ๐๐๐ฝ
= ๐๐๐๐๐๐
Similar computations lead to the conclusion that
tr(๐๐) = tr(๐๐) = tr(๐T๐T) = tr(๐T๐T)
14. Given an arbitrary tensor ๐ a skew tensor ๐ and a symmetric tensor ๐. Show that
๐: ๐ = ๐: ๐T = ๐: sym๐
๐:๐ = โ๐:๐T = ๐: skw๐
๐:๐ = 0
Note that ๐ = sym ๐ + skw๐ , and ๐T = sym ๐ โ skw๐ . Also note that the inner
product between a skew and a symmetric tensor vanishes. Consequently,
๐: ๐ = ๐: (sym ๐ + skw๐)
= ๐: sym ๐ + ๐: skw๐
= ๐: sym ๐
= ๐: ๐T
๐:๐ = ๐: (sym๐ + skw๐)
= ๐: sym ๐ + ๐: skw๐
= ๐: skw ๐
= โ๐: ๐T
To show that ๐:๐ = 0. Observe that, in component form, ๐ = ๐๐๐๐ ๐ โ ๐ ๐, ๐ =
๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ .
๐T๐ = ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ
๐:๐ = tr ๐T๐
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ = ๐๐๐๐๐ผ๐ฝ๐๐๐ฝ๐๐๐ผ
= ๐๐๐๐๐๐ = โ๐๐๐๐
๐๐
= โ๐๐๐๐๐๐ = โ๐๐๐๐
๐๐
Which vanishes because it is equal to the negative of itself.
15. Show that if for every skew tensor ๐, the inner product ๐:๐ = 0, it must follow
that ๐ is symmetric.
๐T๐ = ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ
๐:๐ = tr ๐T๐
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ = ๐๐๐๐๐ผ๐ฝ๐๐๐ฝ๐๐๐ผ
= ๐๐๐๐๐๐ = โ๐๐๐๐
๐๐ = 0 = ๐๐๐๐๐๐
Since all inner products ๐:๐ = 0. But,
๐๐๐๐๐๐ = ๐๐๐๐
๐๐ = ๐๐๐๐๐๐
So that (๐๐๐ โ ๐๐๐)๐๐๐ = 0 โ ๐๐๐ = ๐๐๐ Hence, ๐ is symmetric.
16. Show that if for every symmetric tensor ๐, the inner product ๐:๐ = 0, it must
follow that ๐ is anti-symmetric.
๐T๐ = ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ
๐:๐ = tr ๐T๐
= ๐๐๐๐๐ผ๐ฝ๐ ๐ โ ๐ ๐ฝ๐๐๐ผ = ๐๐๐๐๐ผ๐ฝ๐๐๐ฝ๐๐๐ผ
= ๐๐๐๐๐๐ = ๐๐๐๐
๐๐ = 0 = โ๐๐๐๐๐๐
Since all inner products ๐:๐ = 0. But,
๐๐๐๐๐๐ = ๐๐๐๐
๐๐ = โ๐๐๐๐๐๐
So that ๐๐๐(๐๐๐ + ๐๐๐) = 0 โ ๐๐๐ = โ๐๐๐ Hence, ๐ is anti-symmetric
17. If we can find ๐ผ, ๐ฝ and ๐พ unit tensor, ๐ = ๐ผ๐ โ ๐ + ๐ฝ๐ โ ๐ + ๐พ๐ โ ๐, show that
unless ๐ โ ๐ = ๐ผโ1 then ๐, ๐ and ๐ cannot be linearly independent.
In the expression,
๐ = ๐ผ๐ โ ๐ + ๐ฝ๐ โ ๐ + ๐พ๐ โ ๐
Take a product on the right with vector ๐,
๐๐ = ๐ผ๐(๐ โ ๐) + ๐ฝ๐(๐ โ ๐) + ๐พ๐(๐ โ ๐)
โ ๐(1 โ ๐ผ(๐ โ ๐)) = ๐ฝ๐(๐ โ ๐) + ๐พ๐(๐ โ ๐)
๐ = ๐ฝ๐(๐ โ ๐)
1 โ ๐ผ(๐ โ ๐)+
๐พ๐(๐ โ ๐)
(1 โ ๐ผ(๐ โ ๐))
So that this expression enables us to write ๐ in terms of ๐ and ๐.
18. Divergence of a product: Given that ๐ is a scalar field and ๐ฏ a vector field, show
that div(๐๐ฏ) = (grad๐) โ ๐ฏ + ๐ div ๐ฏ
grad(๐๐ฏ) = (๐๐ฃ๐),๐ ๐ ๐ โ ๐ ๐
= ๐,๐ ๐ฃ๐๐ ๐ โ ๐ ๐ + ๐๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐
= ๐ฏ โ (grad ๐) + ๐ grad ๐ฏ
Now, div(๐๐ฏ) = tr(grad(๐๐ฏ)). Taking the trace of the above, we have:
div(๐๐ฏ) = ๐ฏ โ (grad ๐) + ๐ div ๐ฏ
19. Show that grad(๐ฎ ยท ๐ฏ) = (grad ๐ฎ)T๐ฏ + (grad ๐ฏ)T๐ฎ
๐ฎ ยท ๐ฏ = ๐ข๐๐ฃ๐ is a scalar sum of components.
grad(๐ฎ ยท ๐ฏ) = (๐ข๐๐ฃ๐),๐ ๐ ๐
= ๐ข๐ ,๐ ๐ฃ๐๐ ๐ + ๐ข๐๐ฃ๐ ,๐ ๐ ๐
Now grad ๐ฎ = ๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ swapping the bases, we have that,
(grad ๐ฎ)T = ๐ข๐ ,๐ (๐ ๐ โ ๐ ๐).
Writing ๐ฏ = ๐ฃ๐๐ ๐ , we have that, (grad ๐ฎ)T๐ฏ = ๐ข๐ ,๐ ๐ฃ๐(๐ ๐ โ ๐ ๐)๐ ๐ = ๐ข๐ ,๐ ๐ฃ๐๐ ๐๐ฟ๐
๐ =
๐ข๐ ,๐ ๐ฃ๐๐ ๐
It is easy to similarly show that ๐ข๐๐ฃ๐ ,๐ ๐ ๐ = (grad ๐ฏ)T๐ฎ. Clearly,
grad(๐ฎ ยท ๐ฏ) = (๐ข๐๐ฃ๐),๐ ๐ ๐ = ๐ข๐ ,๐ ๐ฃ๐๐ ๐ + ๐ข๐๐ฃ๐ ,๐ ๐ ๐
= (grad ๐ฎ)T๐ฏ + (grad ๐ฏ)T๐ฎ
As required.
20. Show that grad(๐ฎ ร ๐ฏ) = (๐ฎ ร)grad ๐ฏ โ (๐ฏ ร)grad ๐ฎ
๐ฎ ร ๐ฏ = ๐๐๐๐๐ข๐๐ฃ๐๐ ๐
Recall that the gradient of this vector is the tensor,
grad(๐ฎ ร ๐ฏ) = (๐๐๐๐๐ข๐๐ฃ๐),๐ ๐ ๐ โ ๐ ๐
= ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐๐ ๐ โ ๐ ๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐
= โ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐๐ ๐ โ ๐ ๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐
= โ (๐ฏ ร)grad ๐ฎ + (๐ฎ ร)grad ๐ฏ
21. Show that div (๐ฎ ร ๐ฏ) = ๐ฏ โ curl ๐ฎ โ ๐ฎ โ curl ๐ฏ
We already have the expression for grad(๐ฎ ร ๐ฏ) above; remember that
div (๐ฎ ร ๐ฏ) = tr[grad(๐ฎ ร ๐ฏ)]
= โ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐๐ ๐ โ ๐ ๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐
= โ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐๐ฟ๐๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐ ๐ฟ๐
๐
= โ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐ = ๐ฏ โ curl ๐ฎ โ ๐ฎ โ curl ๐ฏ
22. Given a scalar point function ๐ and a vector field ๐ฏ, show that curl (๐๐ฏ) =
๐ curl ๐ฏ + (grad ๐) ร ๐ฏ.
curl (๐๐ฏ) = ๐๐๐๐(๐๐ฃ๐),๐ ๐ ๐
= ๐๐๐๐(๐,๐ ๐ฃ๐ + ๐๐ฃ๐ ,๐ )๐ ๐
= ๐๐๐๐๐,๐ ๐ฃ๐๐ ๐ + ๐๐๐๐๐๐ฃ๐ ,๐ ๐ ๐
= (grad๐) ร ๐ฏ + ๐ curl ๐ฏ
23. Show that div (๐ฎ โ ๐ฏ) = (div ๐ฏ)๐ฎ + (grad ๐ฎ)๐ฏ
๐ฎ โ ๐ฏ is the tensor, ๐ข๐๐ฃ๐๐ ๐ โ ๐ ๐ . The gradient of this is the third order tensor,
grad (๐ฎ โ ๐ฏ) = (๐ข๐๐ฃ๐),๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
And by divergence, we mean the contraction of the last basis vector:
div (๐ฎ โ ๐ฏ) = (๐ข๐๐ฃ๐)๐(๐ ๐ โ ๐ ๐)๐
๐
= (๐ข๐๐ฃ๐)๐๐ ๐๐ฟ๐
๐ = (๐ข๐๐ฃ๐)๐๐ ๐
= ๐ข๐ ,๐ ๐ฃ๐๐ ๐ + ๐ข๐๐ฃ๐ ,๐ ๐ ๐
= (grad ๐ฎ)๐ฏ + (div ๐ฏ)๐ฎ
24. For a scalar field ๐ and a tensor field ๐ show that grad (๐๐) = ๐grad ๐ + ๐ โ
grad๐. Also show that div (๐๐) = ๐ div ๐ + ๐grad๐
grad(๐๐) = (๐๐๐๐),๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
= (๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ )๐ ๐ โ ๐ ๐ โ ๐ ๐
= ๐ โ grad๐ + ๐grad ๐
Furthermore, we can contract the last two bases and obtain,
div(๐๐) = (๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ )๐ ๐ โ ๐ ๐ โ ๐ ๐
= (๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ )๐ ๐๐ฟ๐๐
= ๐๐๐ ๐,๐ ๐ ๐ + ๐๐๐๐ ,๐ ๐ ๐
= ๐grad๐ + ๐ div ๐
25. For two arbitrary tensors ๐ and ๐, show that grad(๐๐) = (grad ๐T)T๐ + ๐grad๐
grad(๐๐) = (๐๐๐๐๐๐),๐ผ ๐ ๐ โ ๐ ๐ โ ๐ ๐ผ
= (๐๐๐,๐ผ๐๐๐ + ๐๐๐๐๐๐ ,๐ผ )๐ ๐ โ ๐ ๐ โ ๐ ๐ผ
= (๐๐๐๐๐๐,๐ผ + ๐๐๐๐๐๐ ,๐ผ )๐ ๐ โ ๐ ๐ โ ๐ ๐ผ
= (grad ๐T)T๐ + ๐ grad๐
26. For two arbitrary tensors ๐ and ๐, show that div(๐๐) = (grad ๐): ๐ + ๐div ๐
grad(๐๐) = (๐๐๐๐๐๐),๐ผ ๐ ๐ โ ๐ ๐ โ ๐ ๐ผ
= (๐๐๐,๐ผ๐๐๐ + ๐๐๐๐๐๐ ,๐ผ )๐ ๐ โ ๐ ๐ โ ๐ ๐ผ
div(๐๐) = (๐๐๐,๐ผ๐๐๐ + ๐๐๐๐๐๐ ,๐ผ )๐ ๐(๐ ๐ โ ๐ ๐ผ)
= (๐๐๐,๐ผ๐๐๐ + ๐๐๐๐๐๐ ,๐ผ )๐ ๐๐ฟ๐
๐ผ
= (๐๐๐,๐๐๐๐ + ๐๐๐๐๐๐ ,๐ )๐ ๐
= (grad ๐): ๐ + ๐div ๐
27. For two arbitrary vectors, ๐ฎ and ๐ฏ, show that grad(๐ฎ ร ๐ฏ) = (๐ฎ ร)grad๐ฏ โ
(๐ฏ ร)grad๐ฎ
grad(๐ฎ ร ๐ฏ) = (๐๐๐๐๐ข๐๐ฃ๐),๐ ๐ ๐ โ ๐ ๐
= (๐๐๐๐๐ข๐ ,๐ ๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐ )๐ ๐ โ ๐ ๐
= (๐ข๐ ,๐ ๐๐๐๐๐ฃ๐ + ๐ฃ๐ ,๐ ๐
๐๐๐๐ข๐)๐ ๐ โ ๐ ๐
= โ(๐ฏ ร)grad๐ฎ + (๐ฎ ร)grad๐ฏ
28. For a vector field ๐ฎ, show that grad(๐ฎ ร) is a third ranked tensor. Hence or
otherwise show that div(๐ฎ ร) = โcurl ๐ฎ.
The secondโorder tensor (๐ฎ ร) is defined as ๐๐๐๐๐ข๐๐ ๐ โ ๐ ๐. Taking the covariant
derivative with an independent base, we have
grad(๐ฎ ร) = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
This gives a third order tensor as we have seen. Contracting on the last two bases,
div(๐ฎ ร) = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
= ๐๐๐๐๐ข๐ ,๐ ๐ ๐๐ฟ๐๐
= ๐๐๐๐๐ข๐ ,๐ ๐ ๐
= โcurl ๐ฎ
29. Show that div (๐๐) = grad ๐
Note that ๐๐ = (๐๐๐ผ๐ฝ)๐ ๐ผ โ ๐ ๐ฝ . Also note that
grad ๐๐ = (๐๐๐ผ๐ฝ),๐ ๐ ๐ผ โ ๐ ๐ฝ โ ๐ ๐
The divergence of this third order tensor is the contraction of the last two bases:
div (๐๐) = tr(grad ๐๐) = (๐๐๐ผ๐ฝ),๐ (๐ ๐ผ โ ๐ ๐ฝ)๐ ๐ = (๐๐๐ผ๐ฝ),๐ ๐
๐ผ๐๐ฝ๐
= ๐,๐ ๐๐ผ๐ฝ๐๐ฝ๐๐ ๐ผ
= ๐,๐ ๐ฟ๐ผ๐ ๐ ๐ผ = ๐,๐ ๐
๐ = grad ๐
30. Show that curl (๐๐) = ( grad ๐) ร
Note that ๐๐ = (๐๐๐ผ๐ฝ)๐ ๐ผ โ ๐ ๐ฝ , and that curl ๐ = ๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ so that,
curl (๐๐) = ๐๐๐๐(๐๐๐ผ๐),๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐(๐,๐ ๐๐ผ๐)๐ ๐ โ ๐ ๐ผ = ๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐
= ( grad ๐) ร
31. Show that the dyad ๐ฎ โ ๐ฏ is NOT, in general symmetric: ๐ฎ โ ๐ฏ = ๐ฏ โ ๐ฎ โ
(๐ฎ ร ๐ฏ) ร
๐ฎ ร ๐ฏ = ๐๐๐๐๐ข๐๐ฃ๐ ๐ ๐
((๐ฎ ร ๐ฏ) ร) = ๐๐ผ๐๐ฝ๐๐๐๐๐ข๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ
= โ(๐ฟ๐ผ๐๐ฟ๐ฝ
๐ โ ๐ฟ๐ผ๐๐ฟ๐ฝ
๐ ) ๐ข๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ
= (โ๐ข๐ผ๐ฃ๐ฝ + ๐ข๐ฝ๐ฃ๐ผ)๐ ๐ผ โ ๐ ๐ฝ
= ๐ฏ โ ๐ฎ โ ๐ฎ โ ๐ฏ
32. Show that curl (๐ฏ ร) = (div ๐ฏ)๐ โ grad ๐ฏ
(๐ฏ ร) = ๐๐ผ๐ฝ๐๐ฃ๐ฝ ๐ ๐ผ โ ๐ ๐
curl ๐ = ๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
so that
curl (๐ฏ ร) = ๐๐๐๐๐๐ผ๐ฝ๐๐ฃ๐ฝ ,๐ ๐ ๐ โ ๐ ๐ผ
= (๐๐๐ผ๐๐๐ฝ โ ๐๐๐ฝ๐๐๐ผ) ๐ฃ๐ฝ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐ฃ๐ ,๐ ๐ ๐ผ โ ๐ ๐ผ โ ๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐
= (div ๐ฏ)๐ โ grad ๐ฏ
33. Show that div (๐ฎ ร ๐ฏ) = ๐ฏ โ curl ๐ฎ โ ๐ฎ โ curl ๐ฏ
div (๐ฎ ร ๐ฏ) = (๐๐๐๐๐ข๐๐ฃ๐),๐
Noting that the tensor ๐๐๐๐ behaves as a constant under a covariant differentiation,
we can write,
div (๐ฎ ร ๐ฏ) = (๐๐๐๐๐ข๐๐ฃ๐),๐
= ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐ ,๐
= ๐ฏ โ curl ๐ฎ โ ๐ฎ โ curl ๐ฏ
34. Given a scalar point function ๐ and a vector field ๐ฏ, show that curl (๐๐ฏ) =
๐ curl ๐ฏ + (grad๐) ร ๐ฏ.
curl (๐๐ฏ) = ๐๐๐๐(๐๐ฃ๐),๐ ๐ ๐
= ๐๐๐๐(๐,๐ ๐ฃ๐ + ๐๐ฃ๐ ,๐ )๐ ๐
= ๐๐๐๐๐,๐ ๐ฃ๐๐ ๐ + ๐๐๐๐๐๐ฃ๐ ,๐ ๐ ๐
= (grad๐) ร ๐ฏ + ๐ curl ๐ฏ
35. Show that curl (grad ๐) = ๐จ
For any tensor ๐ฏ = ๐ฃ๐ผ๐ ๐ผ
curl ๐ฏ = ๐๐๐๐๐ฃ๐ ,๐ ๐ ๐
Let ๐ฏ = grad ๐. Clearly, in this case, ๐ฃ๐ = ๐,๐ so that ๐ฃ๐ ,๐ = ๐,๐๐ . It therefore follows
that,
curl (grad ๐) = ๐๐๐๐๐,๐๐ ๐ ๐ = ๐จ.
The contraction of symmetric tensors with anti-symmetric led to this conclusion.
Note that this presupposes that the order of differentiation in the scalar field is
immaterial. This will be true only if the scalar field is continuous โ a proposition we
have assumed in the above.
36. Show that curl (grad ๐ฏ) = ๐
For any tensor ๐ = T๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
curl ๐ = ๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
Let ๐ = grad ๐ฏ. Clearly, in this case, T๐ผ๐ฝ = ๐ฃ๐ผ ,๐ฝ so that ๐๐ผ๐ ,๐ = ๐ฃ๐ผ ,๐๐. It therefore
follows that,
curl (grad ๐ฏ) = ๐๐๐๐๐ฃ๐ผ ,๐๐ ๐ ๐ โ ๐ ๐ผ = ๐.
The contraction of symmetric tensors with anti-symmetric led to this conclusion.
Note that this presupposes that the order of differentiation in the vector field is
immaterial. This will be true only if the vector field is continuous โ a proposition we
have assumed in the above.
37. Show that curl (grad ๐ฏ)T = grad(curl ๐ฏ)
From previous derivation, we can see that, curl ๐ = ๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ. Clearly,
curl ๐T = ๐๐๐๐๐๐๐ผ,๐ ๐ ๐ โ ๐ ๐ผ
so that curl (grad ๐ฏ)T = ๐๐๐๐๐ฃ๐,๐ผ๐ ๐ ๐ โ ๐ ๐ผ. But curl ๐ฏ = ๐๐๐๐๐ฃ๐,๐ ๐ ๐ . The gradient of
this is,
grad(curl ๐ฏ) = (๐๐๐๐๐ฃ๐,๐ ),๐ผ ๐ ๐ โ ๐ ๐ผ = ๐๐๐๐๐ฃ๐ ,๐๐ผ ๐ ๐ โ ๐ ๐ผ = curl (grad ๐ฏ)T
38. Show that div (grad ๐ ร grad ฮธ) = 0
grad ๐ ร grad ฮธ = ๐๐๐๐๐,๐ ๐,๐ ๐ ๐
The gradient of this vector is the tensor,
grad(grad ๐ ร grad ๐) = (๐๐๐๐๐,๐ ๐,๐ ),๐ ๐ ๐ โ ๐ ๐
= ๐๐๐๐๐,๐๐ ๐,๐ ๐ ๐ โ ๐ ๐ + ๐๐๐๐๐,๐ ๐,๐๐ ๐ ๐ โ ๐ ๐
The trace of the above result is the divergence we are seeking:
div (grad ๐ ร grad ฮธ) = tr[grad(grad ๐ ร grad ๐)]
= ๐๐๐๐๐,๐๐ ๐,๐ ๐ ๐ โ ๐ ๐ + ๐๐๐๐๐,๐ ๐,๐๐ ๐ ๐ โ ๐ ๐
= ๐๐๐๐๐,๐๐ ๐,๐ ๐ฟ๐๐ + ๐๐๐๐๐,๐ ๐,๐๐ ๐ฟ๐
๐
= ๐๐๐๐๐,๐๐ ๐,๐+ ๐๐๐๐๐,๐ ๐,๐๐ = 0
Each term vanishing on account of the contraction of a symmetric tensor with an
antisymmetric.
39. Show that curl curl ๐ฏ = grad(div ๐ฏ) โ grad2๐ฏ
Let ๐ฐ = curl ๐ฏ โก ๐๐๐๐๐ฃ๐,๐ ๐ ๐ . But curl ๐ฐ โก ๐๐ผ๐ฝ๐พ๐ค๐พ ,๐ฝ ๐ ๐ผ . Upon inspection, we find
that ๐ค๐พ = ๐๐พ๐๐๐๐๐๐ฃ๐ ,๐ so that
curl ๐ฐ โก ๐๐ผ๐ฝ๐พ(๐๐พ๐๐๐๐๐๐ฃ๐,๐ ),๐ฝ ๐ ๐ผ = ๐๐พ๐๐
๐ผ๐ฝ๐พ๐๐๐๐๐ฃ๐ ,๐๐ฝ ๐ ๐ผ
Now, it can be shown that ๐๐พ๐๐๐ผ๐ฝ๐พ๐๐๐๐ = ๐๐ผ๐๐๐ฝ๐ โ ๐๐ผ๐๐๐ฝ๐ so that,
curl ๐ฐ = (๐๐ผ๐๐๐ฝ๐ โ ๐๐ผ๐๐๐ฝ๐)๐ฃ๐ ,๐๐ฝ ๐ ๐ผ
= ๐ฃ๐ฝ ,๐๐ฝ ๐ ๐ โ ๐๐ฝ๐๐ฃ๐ผ ,๐๐ฝ ๐ ๐ผ
= grad(div ๐ฏ) โ grad2๐ฏ
Also recall that the Laplacian (grad2) of a scalar field ๐ is, grad2๐ = ๐๐๐๐,๐๐ . In
Cartesian coordinates, this becomes,
grad2๐ = ๐๐๐๐,๐๐ = ๐ฟ๐๐ ๐,๐๐ = ๐,๐๐
as the unit (metric) tensor now degenerates to the Kronecker delta in this special
case. For a vector field, grad2๐ฏ = ๐๐ฝ๐๐ฃ๐ผ ,๐๐ฝ ๐ ๐ผ .
Also note that while grad is a vector operator, the Laplacian (grad2) is a scalar
operator.
40. Given that ๐(๐ก) = |๐(๐ก)|, Show that ๏ฟฝฬ๏ฟฝ(๐ก) =๐
|๐(๐ก)|: ๏ฟฝฬ๏ฟฝ
๐2 โก ๐:๐
Now, ๐
๐๐ก(๐2) = 2๐
๐๐
๐๐ก=
๐๐
๐๐ก: ๐ + ๐:
๐๐
๐๐ก= 2๐:
๐๐
๐๐ก
as inner product is commutative. We can therefore write that
๐๐
๐๐ก=
๐
๐:๐๐
๐๐ก=
๐
|๐(๐ก)|: ๏ฟฝฬ๏ฟฝ
as required.
41. Given a tensor field ๐, obtain the vector ๐ฐ โก ๐T๐ฏ and show that its divergence is
๐: (โ๐ฏ) + ๐ฏ โ div ๐
The gradient of ๐ฐ is the tensor, (๐๐๐๐ฃ๐),๐ ๐ ๐ โ ๐ ๐. Therefore, divergence of ๐ (the
trace of the gradient) is the scalar sum, ๐๐๐๐ฃ๐ ,๐ ๐๐๐ + ๐๐๐ ,๐ ๐ฃ๐๐๐๐ . Expanding, we
obtain,
div (๐T๐ฏ) = ๐๐๐๐ฃ๐ ,๐ ๐๐๐ + ๐๐๐ ,๐ ๐ฃ๐๐๐๐
= ๐๐๐ ,๐ ๐ฃ๐ + ๐๐
๐๐ฃ๐ ,๐
= (div ๐) โ ๐ฏ + tr(๐Tgrad ๐ฏ)
= (div ๐) โ ๐ฏ + ๐: (grad ๐ฏ)
Recall that scalar product of two vectors is commutative so that
div (๐T๐ฏ) = ๐: (grad ๐ฏ) + ๐ฏ โ div ๐
42. For a second-order tensor ๐ define curl ๐ โก ๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ show that for any
constant vector ๐, (curl ๐) ๐ = curl (๐T๐)
Express vector ๐ in the invariant form with covariant components as ๐ = ๐๐ฝ๐ ๐ฝ . It
follows that
(curl ๐) ๐ = ๐๐๐๐๐๐ผ๐,๐ (๐ ๐ โ ๐ ๐ผ)๐
= ๐๐๐๐๐๐ผ๐,๐ ๐๐ฝ(๐ ๐ โ ๐ ๐ผ)๐ ๐ฝ
= ๐๐๐๐๐๐ผ๐,๐ ๐๐ฝ๐ ๐๐ฟ๐ฝ
๐ผ
= ๐๐๐๐(๐๐ผ๐),๐ ๐ ๐๐๐ผ
= ๐๐๐๐(๐๐ผ๐๐๐ผ),๐ ๐ ๐
The last equality resulting from the fact that vector ๐ is a constant vector. Clearly,
(curl ๐) ๐ = curl (๐T๐)
43. For any two vectors ๐ฎ and ๐ฏ, show that curl (๐ฎ โ ๐ฏ) = [(grad ๐ฎ)๐ฏ ร]๐ +
(curl ๐ฏ) โ ๐ where ๐ฏ ร is the skew tensor ๐๐๐๐๐ฃ๐ ๐ ๐ โ ๐ ๐.
Recall that the curl of a tensor ๐ป is defined by curl ๐ป โก ๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ. Clearly
therefore,
curl (๐ฎ โ ๐ฏ) = ๐๐๐๐(๐ข๐ผ๐ฃ๐),๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐(๐ข๐ผ ,๐ ๐ฃ๐ + ๐ข๐ผ๐ฃ๐ ,๐ ) ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐๐ข๐ผ ,๐ ๐ฃ๐ ๐ ๐ โ ๐ ๐ผ + ๐๐๐๐๐ข๐ผ๐ฃ๐,๐ ๐ ๐ โ ๐ ๐ผ
= (๐๐๐๐๐ฃ๐ ๐ ๐) โ (๐ข๐ผ ,๐ ๐ ๐ผ) + (๐๐๐๐๐ฃ๐ ,๐ ๐ ๐) โ (๐ข๐ผ๐ ๐ผ)
= (๐๐๐๐๐ฃ๐ ๐ ๐ โ ๐ ๐)(๐ข๐ผ ,๐ฝ ๐ ๐ฝ โ ๐ ๐ผ) + (๐๐๐๐๐ฃ๐ ,๐ ๐ ๐) โ (๐ข๐ผ๐ ๐ผ)
= โ(๐ฏ ร)(grad ๐ฎ)๐ป + (curl ๐ฏ) โ ๐ฎ
= [(grad ๐ฎ)๐ฏ ร]๐ป + (curl ๐ฏ) โ ๐ฎ
upon noting that the vector cross is a skew tensor.
44. Show that curl (๐ฎ ร ๐ฏ) = div(๐ฎ โ ๐ฏ โ ๐ฏ โ ๐ฎ)
The vector ๐ฐ โก ๐ฎ ร ๐ฏ = ๐ค๐๐ ๐ = ๐๐๐ผ๐ฝ๐ข๐ผ๐ฃ๐ฝ๐ ๐ and curl ๐ฐ = ๐๐๐๐๐ค๐ ,๐ ๐ ๐ . Therefore,
curl (๐ฎ ร ๐ฏ) = ๐๐๐๐๐ค๐,๐ ๐ ๐
= ๐๐๐๐๐๐๐ผ๐ฝ(๐ข๐ผ๐ฃ๐ฝ),๐ ๐ ๐
= (๐ฟ๐ผ๐ ๐ฟ๐ฝ
๐โ ๐ฟ๐ฝ
๐ ๐ฟ๐ผ๐) (๐ข๐ผ๐ฃ๐ฝ),๐ ๐ ๐
= (๐ฟ๐ผ๐ ๐ฟ๐ฝ
๐โ ๐ฟ๐ฝ
๐ ๐ฟ๐ผ๐) (๐ข๐ผ ,๐ ๐ฃ๐ฝ + ๐ข๐ผ๐ฃ๐ฝ ,๐ )๐ ๐
= [๐ข๐ ,๐ ๐ฃ๐ + ๐ข๐๐ฃ๐ ,๐โ (๐ข๐ ,๐ ๐ฃ๐ + ๐ข๐๐ฃ๐ ,๐ )]๐ ๐
= [(๐ข๐๐ฃ๐),๐โ (๐ข๐๐ฃ๐),๐ ]๐ ๐
= div(๐ฎ โ ๐ฏ โ ๐ฏ โ ๐ฎ)
since div(๐ฎ โ ๐ฏ) = (๐ข๐๐ฃ๐),๐ผ ๐ ๐ โ ๐ ๐ โ ๐ ๐ผ = (๐ข๐๐ฃ๐),๐ ๐ ๐ .
45. Given a scalar point function ๐ and a second-order tensor field ๐, show that
curl (๐๐) = ๐ curl ๐ + ((grad๐) ร)๐T where [(grad๐) ร] is the skew tensor
๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐
curl (๐๐) โก ๐๐๐๐(๐๐๐ผ๐),๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐(๐,๐ ๐๐ผ๐ + ๐๐๐ผ๐ ,๐ ) ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐๐,๐ ๐๐ผ๐ ๐ ๐ โ ๐ ๐ผ + ๐๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
= (๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐) (๐๐ผ๐ฝ๐ ๐ฝ โ ๐ ๐ผ) + ๐๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ
= ๐ curl ๐ + ((grad๐) ร)๐T
46. For a second-order tensor field ๐ป, show that div(curl ๐) = curl(div ๐T)
Define the second order tensor ๐ as
curl ๐ โก ๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ = ๐.๐ผ๐ ๐ ๐ โ ๐ ๐ผ
The gradient of ๐บ is ๐.๐ผ๐ ,๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ = ๐๐๐๐๐๐ผ๐,๐๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ
Clearly,
div(curl ๐) = ๐๐๐๐๐๐ผ๐ ,๐๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ = ๐๐๐๐๐๐ผ๐ ,๐๐ฝ ๐ ๐ ๐๐ผ๐ฝ
= ๐๐๐๐๐๐ฝ๐ ,๐๐ฝ ๐ ๐ = curl(div ๐T)
47. The position vector in Cartesian coordinates ๐ซ = ๐ฅ๐๐๐. Show that (a) div ๐ซ = ๐, (b)
div (๐ซ โ ๐ซ) = ๐๐ซ, (c) div ๐ซ = 3, and (d) grad ๐ซ = ๐ and (e) curl (๐ซ โ ๐ซ) = โ๐ซ ร
grad ๐ = ๐ฅ๐ ,๐ ๐๐ โ ๐๐
= ๐ฟ๐๐๐๐ โ ๐๐ = ๐
div ๐ = ๐ฅ๐ ,๐ ๐๐ โ ๐๐
= ๐ฟ๐๐๐ฟ๐๐ = ๐ฟ๐๐ = 3. ๐ โ ๐ = ๐ฅ๐๐๐ โ ๐ฅ๐๐๐ = ๐ฅ๐๐ฅ๐๐๐ โ ๐๐grad(๐ โ ๐)
= (๐ฅ๐๐ฅ๐),๐ ๐๐ โ ๐๐ โ ๐๐ = (๐ฅ๐ ,๐ ๐ฅ๐ + ๐ฅ๐๐ฅ๐ ,๐ )๐๐ โ ๐๐ โ ๐๐
= (๐ฟ๐๐๐๐ + ๐๐๐ฟ๐๐)๐ฟ๐๐๐๐ = (๐ฟ๐๐๐๐ + ๐๐๐ฟ๐๐)๐๐
= 4๐ฅ๐๐๐ = 4๐
curl(๐ โ ๐) = ๐๐ผ๐ฝ๐พ(๐ฅ๐๐ฅ๐พ),๐ฝ ๐๐ผ โ ๐๐
= ๐๐ผ๐ฝ๐พ(๐ฅ๐ ,๐ฝ ๐ฅ๐พ + ๐ฅ๐๐ฅ๐พ,๐ฝ )๐๐ผ โ ๐๐
= ๐๐ผ๐ฝ๐พ(๐ฟ๐๐ฝ๐ฅ๐พ + ๐ฅ๐๐ฟ๐พ๐ฝ)๐๐ผ โ ๐๐
= ๐๐ผ๐๐พ๐ฅ๐พ๐๐ผ โ ๐๐ + ๐๐ผ๐ฝ๐ฝ๐ฅ๐๐๐ผ โ ๐๐ = โ๐๐ผ๐พ๐๐ฅ๐พ๐๐ผ โ ๐๐ = โ๐ ร
48. Define the magnitude of tensor ๐ as, |๐| = โtr(๐๐T) Show that โ|๐|
โ๐=
๐
|๐|
By definition, given a scalar ๐ผ, the derivative of a scalar function of a tensor ๐(๐) is
โ๐(๐)
โ๐:๐ = lim
๐ผโ0
โ
โ๐ผ๐(๐ + ๐ผ๐)
for any arbitrary tensor ๐.
In the case of ๐(๐) = |๐|,
โ|๐|
โ๐: ๐ = lim
๐ผโ0
โ
โ๐ผ|๐ + ๐ผ๐|
|๐ + ๐ผ๐| = โtr(๐ + ๐ผ๐)(๐ + ๐ผ๐)T = โtr(๐๐T + ๐ผ๐๐T + ๐ผ๐๐T + ๐ผ2๐๐T)
Note that everything under the root sign here is scalar and that the trace operation
is linear. Consequently, we can write,
lim๐ผโ0
โ
โ๐ผ|๐ + ๐ผ๐| = lim
๐ผโ0
tr (๐๐T) + tr (๐๐T) + 2๐ผtr (๐๐T)
2โtr(๐๐T + ๐ผ๐๐T + ๐ผ๐๐T + ๐ผ2๐๐T)=
2๐:๐
2โ๐: ๐=
๐
|๐|: ๐
So that,
โ|๐|
โ๐: ๐ =
๐
|๐|: ๐
or,
โ|๐|
โ๐=
๐
|๐|
as required since ๐ is arbitrary.
49. Show that ๐๐ฐ3(๐)
๐๐=
๐det(๐)
๐๐= ๐c the cofactor of ๐.
Clearly ๐c = det(๐) ๐โT = ๐ฐ3(๐) ๐โT. Details of this for the contravariant
components of a tensor is presented below. Let
det(๐) โก |๐| โก ๐ =1
3!๐๐๐๐๐๐๐ ๐ก๐๐๐๐๐๐ ๐๐๐ก
Differentiating wrt ๐๐ผ๐ฝ , we obtain,
๐๐
๐๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ =
1
3!๐๐๐๐๐๐๐ ๐ก [
๐๐๐๐
๐๐๐ผ๐ฝ๐๐๐ ๐๐๐ก + ๐๐๐
๐๐๐๐
๐๐๐ผ๐ฝ๐๐๐ก + ๐๐๐๐๐๐
๐๐๐๐ก
๐๐๐ผ๐ฝ] ๐ ๐ผ โ ๐ ๐ฝ
=1
3!๐๐๐๐๐๐๐ ๐ก [๐ฟ๐
๐ผ๐ฟ๐๐ฝ๐๐๐ ๐๐๐ก + ๐๐๐๐ฟ๐
๐ผ๐ฟ๐ ๐ฝ๐๐๐ก + ๐๐๐๐๐๐ ๐ฟ๐
๐ผ๐ฟ๐ก๐ฝ] ๐ ๐ผ โ ๐ ๐ฝ
=1
3!๐๐ผ๐๐๐๐ฝ๐ ๐ก[๐๐๐ ๐๐๐ก + ๐๐๐ ๐๐๐ก + ๐๐๐ ๐๐๐ก]๐ ๐ผ โ ๐ ๐ฝ
=1
2!๐๐ผ๐๐๐๐ฝ๐ ๐ก๐๐๐ ๐๐๐ก๐ ๐ผ โ ๐ ๐ฝ โก [๐c]๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
Which is the cofactor of [๐๐ผ๐ฝ] or ๐บ
50. For a scalar variable ๐ผ, if the tensor ๐ = ๐(๐ผ) and ๏ฟฝฬ๏ฟฝ โก๐๐
๐๐ผ, Show that
๐
๐๐ผdet(๐) =
det(๐) tr(๏ฟฝฬ๏ฟฝ๐โ๐)
Let ๐จ โก ๏ฟฝฬ๏ฟฝ๐ปโ1 so that, ๏ฟฝฬ๏ฟฝ = ๐จ๐ป. In component form, we have ๏ฟฝฬ๏ฟฝ๐๐ = ๐ด๐
๐ ๐๐๐. Therefore,
๐
๐๐ผdet(๐ป) =
๐
๐๐ผ(๐๐๐๐๐๐
1๐๐2๐๐
3) = ๐๐๐๐(๏ฟฝฬ๏ฟฝ๐1๐๐
2๐๐3 + ๐๐
1๏ฟฝฬ๏ฟฝ๐2๐๐
3 + ๐๐1๐๐
2๏ฟฝฬ๏ฟฝ๐3)
= ๐๐๐๐(๐ด๐1๐๐
๐๐๐2๐๐
3 + ๐๐1๐ด๐
2 ๐๐๐๐๐
3 + ๐๐1๐๐
2๐ด๐3๐๐
๐)
= ๐๐๐๐ [(๐ด11๐๐
1 + ๐ด21๐๐
2 + ๐ด31๐๐
3 ) ๐๐2๐๐
3 + ๐๐1 ( ๐ด1
2๐๐1 + ๐ด2
2๐๐2 + ๐ด3
2๐๐3 ) ๐๐
3
+ ๐๐1๐๐
2 ( ๐ด13๐๐
1 + ๐ด23๐๐
2 + ๐ด33๐๐
3)]
All the boxed terms in the above equation vanish on account of the contraction of a
symmetric tensor with an antisymmetric one.
(For example, the first boxed term yields, ๐๐๐๐๐ด21๐๐
2๐๐2๐๐
3
Which is symmetric as well as antisymmetric in ๐ and ๐. It therefore vanishes. The
same is true for all other such terms.)
โ
โ๐ผdet(๐) = ๐๐๐๐[(๐ด1
1๐๐1)๐๐
2๐๐3 + ๐๐
1(๐ด22๐๐
2)๐๐3 + ๐๐
1๐๐2(๐ด3
3๐๐3)]
= ๐ด๐๐๐๐๐๐๐๐
1๐๐2๐๐
3 = tr(๏ฟฝฬ๏ฟฝ๐โ1) det(๐)
as required.
51. Prove Liouvilleโs Theorem that for a scalar variable ๐ผ, if the tensor ๐ = ๐(๐ผ) and
๏ฟฝฬ๏ฟฝ โก๐๐
๐๐ผ,
๐
๐๐ผdet(๐) = det(๐) tr(๏ฟฝฬ๏ฟฝ๐โ๐) by direct methods.
We choose three constant, linearly independent vectors ๐, ๐ and ๐ so that
[๐, ๐, ๐] det(๐) = [๐๐, ๐๐, ๐๐]
Differentiating both sides, noting that the RHS is a product,
[๐, ๐, ๐]๐
๐๐ผdet(๐)
=๐
๐๐ผ[๐๐, ๐๐, ๐๐]
= [๐๐
๐๐ผ๐, ๐๐, ๐๐] + [๐๐,
๐๐
๐๐ผ๐, ๐๐] + [๐๐, ๐๐,
๐๐
๐๐ผ๐]
= [๐๐
๐๐ผ๐โ1๐๐, ๐๐, ๐๐] + [๐๐,
๐๐
๐๐ผ๐โ1๐๐, ๐๐] + [๐๐, ๐๐,
๐๐
๐๐ผ๐โ1๐๐]
= tr (๐๐
๐๐ผ๐โ1) [๐๐, ๐๐, ๐๐]
Clearly, ๐
๐๐ผdet(๐) = det(๐) tr (
๐๐
๐๐ผ๐โ1)
52. If ๐ is invertible, show that โ
โ๐(log det(๐)) = ๐โ๐
โ
โ๐(log det(๐)) =
โ(log det(๐))
โdet(๐) โdet(๐)
โ๐
=1
det(๐)๐c =
1
det(๐)det(๐) ๐โT
= ๐โT
53. If ๐ is invertible, show that โ
โ๐(log det(๐โ1)) = โ๐โ๐
โ
โ๐(log det(๐โ1)) =
โ(log det(๐โ1))
โdet(๐โ1) โdet(๐โ1)
โ๐โ1
โ๐โ1
โ๐
=1
det(๐โ1)๐โ๐(โ๐โ2)
=1
det(๐โ1)det(๐โ1)๐๐(โ๐โ2)
= โ๐โ๐
54. Given that ๐ is a constant tensor, Show that โ
โ๐tr(๐๐) = ๐T
In invariant components terms, let ๐ = A๐๐๐ ๐ โ ๐ ๐ and let ๐ = S๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ.
๐๐ = A๐๐S๐ผ๐ฝ(๐ ๐ โ ๐ ๐)(๐ ๐ผ โ ๐ ๐ฝ)
= A๐๐S๐ผ๐ฝ(๐ ๐ โ ๐ ๐ฝ)๐ฟ๐๐ผ
= A๐๐S๐๐ฝ(๐ ๐ โ ๐ ๐ฝ)
tr(๐๐) = A๐๐S๐๐ฝ(๐ ๐ โ ๐ ๐ฝ)
= A๐๐S๐๐ฝ๐ฟ๐๐ฝ
= A๐๐S๐๐
โ
โ๐tr(๐๐) =
โ
โS๐ผ๐ฝtr(๐๐)๐ ๐ผ โ ๐ ๐ฝ
=โA๐๐S๐๐
โS๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
= A๐๐๐ฟ๐๐ผ๐ฟ๐
๐ฝ๐ ๐ผ โ ๐ ๐ฝ = A๐๐๐ ๐ โ ๐ ๐ = ๐T =
โ
โ๐(๐T: ๐)
as required.
55. Given that ๐ and ๐ are constant tensors, show that โ
โ๐tr(๐๐๐T) = ๐T๐
First observe that tr(๐๐๐T) = tr(๐T๐๐). If we write, ๐ โก ๐T๐, it is obvious from
the above that โ
โ๐tr(๐๐) = ๐T. Therefore,
โ
โ๐tr(๐๐๐T) = (๐T๐)๐ = ๐T๐
56. Given that ๐ and ๐ are constant tensors, show that โ
โ๐tr(๐๐T๐T) = ๐T๐
Observe that tr(๐๐T๐T) = tr(๐T๐๐T) = tr[๐(๐T๐)T] = tr[(๐T๐)T๐]
[The transposition does not alter trace; neither does a cyclic permutation. Ensure
you understand why each equality here is true.] Consequently,
โ
โ๐tr(๐๐T๐T) =
โ
โ๐ tr[(๐T๐)T๐] = [(๐T๐)T]๐ = ๐T๐
57. Let ๐ be a symmetric and positive definite tensor and let ๐ผ1(๐), ๐ผ2(๐)and๐ผ3(๐) be
the three principal invariants of ๐ show that (a) ๐๐ฐ1(๐)
๐๐= ๐ the identity tensor, (b)
๐๐ฐ2(๐)
๐๐= ๐ผ1(๐)๐ โ ๐ and (c)
๐๐ผ3(๐)
๐๐= ๐ผ3(๐) ๐โ1
๐๐ฐ1(๐บ)
๐๐บ can be written in the invariant component form as,
๐๐ผ1(๐)
๐๐=
๐๐ผ1(๐)
๐๐๐๐
๐ ๐ โ ๐ ๐
Recall that ๐ผ1(๐) = tr(๐) = ๐ฮฑฮฑ hence
๐๐ผ1(๐)
๐๐=
๐๐ผ1(๐)
๐๐๐๐
๐ ๐ โ ๐ ๐ =๐๐ฮฑ
ฮฑ
๐๐๐๐๐ ๐ โ ๐ ๐
= ๐ฟ๐ผ๐ ๐ฟ๐
๐ผ๐ ๐ โ ๐ ๐ = ๐ฟ๐๐๐ ๐ โ ๐ ๐
= ๐
which is the identity tensor as expected. ๐๐ผ2(๐)
๐๐ in a similar way can be written in the invariant component form as,
๐๐ผ2(๐)
๐๐=
1
2
๐๐ผ1(๐)
๐๐๐๐
[๐ฮฑฮฑ๐๐ฝ
๐ฝโ ๐๐ฝ
ฮฑ๐ฮฑ๐ฝ] ๐ ๐ โ ๐ ๐
where we have utilized the fact that ๐ผ2(๐) =1
2[tr2(๐) โ tr(๐2)]. Consequently,
๐๐ผ2(๐)
๐๐=
1
2
๐
๐๐๐๐[๐ฮฑ
ฮฑ๐๐ฝ๐ฝ
โ ๐๐ฝฮฑ๐ฮฑ
๐ฝ] ๐ ๐ โ ๐ ๐
=1
2[๐ฟ๐ผ
๐ ๐ฟ๐๐ผ๐๐ฝ
๐ฝ+ ๐ฟ๐ฝ
๐ ๐ฟ๐๐ฝ๐ฮฑ
ฮฑ โ ๐ฟ๐ฝ๐ ๐ฟ๐
๐ผ๐ฮฑ๐ฝ
โ ๐ฟ๐ผ๐ ๐ฟ๐
๐ฝ๐๐ฝ
ฮฑ] ๐ ๐ โ ๐ ๐
=1
2[๐ฟ๐
๐๐๐ฝ๐ฝ
+ ๐ฟ๐๐๐ฮฑ
ฮฑ โ ๐๐๐โ ๐๐
๐] ๐ ๐ โ ๐ ๐ = (๐ฟ๐
๐๐ฮฑฮฑ โ ๐๐
๐)๐ ๐ โ ๐ ๐
= ๐ผ1(๐)๐ โ ๐
det(๐บ) โก |๐บ| โก ๐ =1
3!๐๐๐๐๐๐๐ ๐ก๐๐๐๐๐๐ ๐๐๐ก
Differentiating wrt ๐๐ผ๐ฝ , we obtain,
๐๐
๐๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ =
1
3!๐๐๐๐๐๐๐ ๐ก [
๐๐๐๐
๐๐๐ผ๐ฝ๐๐๐ ๐๐๐ก + ๐๐๐
๐๐๐๐
๐๐๐ผ๐ฝ๐๐๐ก + ๐๐๐๐๐๐
๐๐๐๐ก
๐๐๐ผ๐ฝ] ๐ ๐ผ โ ๐ ๐ฝ
=1
3!๐๐๐๐๐๐๐ ๐ก [๐ฟ๐
๐ผ๐ฟ๐๐ฝ๐๐๐ ๐๐๐ก + ๐๐๐๐ฟ๐
๐ผ๐ฟ๐ ๐ฝ๐๐๐ก + ๐๐๐๐๐๐ ๐ฟ๐
๐ผ๐ฟ๐ก๐ฝ] ๐ ๐ผ โ ๐ ๐ฝ
=1
3!๐๐ผ๐๐๐๐ฝ๐ ๐ก[๐๐๐ ๐๐๐ก + ๐๐๐ ๐๐๐ก + ๐๐๐ ๐๐๐ก]๐ ๐ผ โ ๐ ๐ฝ
=1
2!๐๐ผ๐๐๐๐ฝ๐ ๐ก๐๐๐ ๐๐๐ก๐ ๐ผ โ ๐ ๐ฝ โก [๐c]๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
Which is the cofactor of [๐๐ผ๐ฝ] or ๐บ
58. For a tensor field ๐ฉ, The volume integral in the region ฮฉ โ โฐ, โซ (grad ๐ฉ)ฮฉ
๐๐ฃ =
โซ ๐ฉ โ ๐โฮฉ
๐๐ where ๐ is the outward drawn normal to ๐ฮฉ โ the boundary of ฮฉ.
Show that for a vector field ๐
โซ (div ๐)ฮฉ
๐๐ฃ = โซ ๐ โ ๐๐ฮฉ
๐๐
Replace ๐ฉ by the vector field ๐ we have,
โซ (grad ๐)ฮฉ
๐๐ฃ = โซ ๐ โ ๐โฮฉ
๐๐
Taking the trace of both sides and noting that both trace and the integral are linear
operations, therefore we have,
โซ (div ๐)ฮฉ
๐๐ฃ = โซ tr(grad ๐)ฮฉ
๐๐ฃ
= โซ tr(๐ โ ๐)โฮฉ
๐๐
= โซ ๐ โ ๐๐ฮฉ
๐๐
59. Show that for a scalar function Hence the divergence theorem
becomes,โซ (grad ๐)ฮฉ
๐๐ฃ = โซ ๐๐๐ฮฉ
๐๐
Recall that for a vector field, that for a vector field ๐
โซ (div ๐)ฮฉ
๐๐ฃ = โซ ๐ โ ๐๐ฮฉ
๐๐
if we write, ๐ = ๐๐ where ๐ is an arbitrary constant vector, we have,
โซ (div[๐๐])ฮฉ
๐๐ฃ = โซ ๐๐ โ ๐ง๐ฮฉ
๐๐ = ๐ โ โซ ๐๐ง๐ฮฉ
๐๐
For the LHS, note that, div[๐๐] = tr(grad[๐๐])
grad[๐๐] = (๐๐๐),๐ ๐ ๐ โ ๐ ๐ = ๐๐๐,๐ ๐ ๐ โ ๐ ๐
The trace of which is,
๐๐๐,๐ ๐ ๐ โ ๐ ๐ = ๐๐๐,๐ ๐ฟ๐๐= ๐๐๐,๐ = ๐ โ grad ๐
For the arbitrary constant vector ๐, we therefore have that,
โซ (div[๐๐])ฮฉ
๐๐ฃ = ๐ โ โซ grad ๐ ฮฉ
๐๐ฃ = ๐ โ โซ ๐๐ง๐ฮฉ
๐๐
โซ grad ๐ ฮฉ
๐๐ฃ = โซ ๐๐ง๐ฮฉ
๐๐
60. For the tensor ๐, given that ๐๐ป = ๐ฐ, show that ๐ป๐ = ๐๐ป = ๐ฐ.
Consider ๐ป๐๐ป๐ where ๐ is a vector. Since ๐๐ป = ๐ฐ, it follows that ๐ป๐๐ป๐ = ๐ป๐ฐ๐ =
๐ป๐ โก ๐ where ๐ is also a vector. Clearly,
๐ป๐๐ป๐ = ๐ป๐๐ = ๐
which immediately shows that ๐ป๐ = ๐ฐ as required to be shown.
61. Given the basis vectors ๐ ๐ , ๐ = 1,2,3 show that for any tensor ๐, the product
๐๐ ๐ = ๐๐ผ๐๐ ๐ผ
In component form, let ๐ = ๐๐ผ๐ฝ(๐ ๐ผ โ ๐ ๐ผ) so that
๐๐ ๐ = ๐๐ผ๐ฝ(๐ ๐ผ โ ๐ ๐ฝ)๐ ๐
= ๐๐ผ๐ฝ(๐ ๐ฝ โ ๐ ๐)๐ ๐ผ = ๐๐ผ๐ฝ๐ฟ๐
๐ฝ๐ ๐ผ
= ๐๐ผ๐๐ ๐ผ
62. Show that the trace of a tensor ๐, in component form, can be written as ๐๐.๐ =
๐๐๐๐๐๐ = ๐๐๐๐๐๐ .
Observe that any three linearly independent vectors can be treated as a basis of a
coordinate system, ๐ ๐ , ๐ = 1,2,3. The existence of the dual of these vectors can be
taken as given. Consequently,
tr (๐) โก ๐ผ1(๐) =[๐๐ 1, ๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
=[(๐๐1๐
๐), ๐ 2, ๐ 3] + [๐ 1, (๐๐2๐ ๐), ๐ 3] + [๐ 1, ๐ 2, (๐๐3๐
๐)]
[๐ 1, ๐ 2, ๐ 3]
=[(๐๐1๐
๐ผ๐๐ ๐ผ), ๐ 2, ๐ 3] + [๐ 1, (๐๐2๐๐ฝ๐๐ ๐ฝ), ๐ 3] + [๐ 1, ๐ 2, (๐๐3๐
๐พ๐๐ ๐พ)]
[๐ 1, ๐ 2, ๐ 3]
=๐1
๐ผ[๐ ๐ผ , ๐ 2, ๐ 3] + ๐2๐ฝ[๐ 1, ๐ ๐ฝ , ๐ 3] + ๐3
๐พ[๐ 1, ๐ 2, ๐ ๐พ]
๐123
=๐1
๐ผ๐๐ผ23 + ๐2๐ฝ๐1๐ฝ3 + ๐3
๐พ๐12๐พ
๐123
=๐1
1๐123 + ๐22๐123 + ๐3
3๐123
๐123 = ๐๐
.๐ = ๐๐๐๐๐๐ = ๐๐๐๐๐๐
63. Show that [๐๐ ๐ , ๐ ๐ , ๐ ๐] + [๐ ๐ , ๐๐ ๐ , ๐ ๐] + [๐ ๐ , ๐ ๐ , ๐๐ ๐] = ๐๐ผ๐ผ[๐ ๐ , ๐ ๐ , ๐ ๐]
Note that,
[๐๐ 1, ๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐ 2, ๐๐ 3]
= [(๐๐1๐ ๐), ๐ 2, ๐ 3] + [๐ 1, (๐๐2๐
๐), ๐ 3] + [๐ 1, ๐ 2, (๐๐3๐ ๐)]
= [(๐๐1๐๐ผ๐๐ ๐ผ), ๐ 2, ๐ 3] + [๐ 1, (๐๐2๐
๐ฝ๐๐ ๐ฝ), ๐ 3] + [๐ 1, ๐ 2, (๐๐3๐๐พ๐๐ ๐พ)]
= ๐1๐ผ[๐ ๐ผ , ๐ 2, ๐ 3] + ๐2
๐ฝ[๐ 1, ๐ ๐ฝ , ๐ 3] + ๐3
๐พ[๐ 1, ๐ 2, ๐ ๐พ]
= ๐1๐ผ๐๐ผ23 + ๐2
๐ฝ๐1๐ฝ3 + ๐3
๐พ๐12๐พ
= ๐11๐123 + ๐2
2๐123 + ๐33๐123
= ๐๐ผ๐ผ๐123 = ๐๐ผ
๐ผ[๐ 1, ๐ 2, ๐ 3]
Swapping ๐ 2 and ๐ 3, it is clear that,
[๐๐ 1, ๐ 3, ๐ 2] + [๐ 1, ๐๐ 3, ๐ 2] + [๐ 1, ๐ 3, ๐๐ 2]
= โ[๐๐ 1, ๐ 2, ๐ 3] โ [๐ 1, ๐๐ 2, ๐ 3] โ [๐ 1, ๐ 2, ๐๐ 3]
= ๐๐ผ๐ผ๐132 = ๐๐ผ
๐ผ[๐ 1, ๐ 3, ๐ 2]
Continuing, we have that
[๐๐ ๐ , ๐ ๐ , ๐ ๐] + [๐ ๐ , ๐๐ ๐ , ๐ ๐] + [๐ ๐ , ๐ ๐ , ๐๐ ๐] = ๐๐ผ๐ผ[๐ ๐ , ๐ ๐ , ๐ ๐] = ๐๐ผ
๐ผ๐๐๐๐
64. Show that ๐ผ1(๐ป) โก tr(๐ป) =[๐ป๐,๐,๐]+[๐,๐ป๐,๐]+[๐,๐,๐ป๐]
[๐,๐,๐] is independent of the choice of
the linearly independent vectors ๐, ๐ and ๐.
Let us refer each vector to a covariant basis so that, ๐ = ๐๐๐ ๐ , ๐ = ๐๐๐ ๐ , and ๐ =
๐๐๐ ๐ . Hence,
๐ผ1(๐ป) โก tr(๐ป) =[๐ป(๐๐๐ ๐), ๐
๐๐ ๐ , ๐๐๐ ๐] + [๐๐๐ ๐ , ๐ป(๐๐๐ ๐), ๐
๐๐ ๐] + [๐๐๐ ๐ , ๐๐๐ ๐ , ๐ป(๐๐๐ ๐)]
[๐, ๐, ๐]
=๐๐๐๐๐๐([๐ป๐ ๐ , ๐ ๐ , ๐ ๐] + [๐ ๐ , ๐ป๐ ๐ , ๐ ๐] + [๐ ๐ , ๐ ๐ , ๐ป๐ ๐])
[๐, ๐, ๐]
But [๐๐ ๐ , ๐ ๐ , ๐ ๐] + [๐ ๐ , ๐๐ ๐ , ๐ ๐] + [๐ ๐ , ๐ ๐ , ๐๐ ๐] = ๐๐ผ๐ผ[๐ ๐ , ๐ ๐ , ๐ ๐]. We have that
๐ผ1(๐ป) =๐๐๐๐๐๐๐๐ผ
๐ผ[๐ ๐ , ๐ ๐ , ๐ ๐]
๐๐๐๐๐๐๐๐๐๐
=๐๐๐๐๐๐๐๐๐๐
๐๐๐๐๐๐๐๐๐๐๐๐ผ
๐ผ = ๐๐ผ๐ผ
Which is obviously independent of the choice of ๐, ๐ and ๐.
65. Write the second tensor invariant in terms of components
As previously observed, any three linearly independent vectors can be treated as the
basis of a coordinate system, ๐ ๐ , ๐ = 1,2,3. The existence of the dual of these vectors
can be taken as given. Consequently,
๐ผ2(๐) =[๐๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐๐ 3] + [๐๐ 1, ๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
The first of the numerator terms can be simplified as,
[๐๐ 1, ๐๐ 2, ๐ 3] = [(๐๐1๐ ๐), (๐๐2๐
๐), ๐ 3]
= [(๐๐1๐๐ผ๐๐ ๐ผ), (๐๐2๐
๐ฝ๐๐ ๐ฝ), ๐ 3] = ๐1๐ผ๐2
๐ฝ[๐ ๐ผ , ๐ ๐ฝ , ๐ 3]
The other terms are similarly simplified. Clearly,
๐ผ2(๐) =๐1
๐ผ๐2๐ฝ[๐ ๐ผ , ๐ ๐ฝ , ๐ 3] + ๐2
๐ฝ๐3
๐พ[๐ 1, ๐ ๐ฝ , ๐ ๐พ] + ๐1
๐ผ๐3๐พ[๐ ๐ผ , ๐ 2, ๐ ๐พ]
[๐ 1, ๐ 2, ๐ 3]
=๐1
๐ผ๐2๐ฝ๐๐ผ๐ฝ3 + ๐2
๐ฝ๐3
๐พ๐1๐ฝ๐พ + ๐1
๐ผ๐3๐พ๐๐ผ2๐พ
๐123
=[(๐1
1๐22 โ ๐1
2๐21) + (๐2
2๐33 โ ๐2
3๐32) + (๐3
3๐11 โ ๐3
1๐13)]๐123
๐123
=1
2(๐๐ผ
๐ผ๐๐ฝ๐ฝ
โ ๐๐ฝ๐ผ๐๐ผ
๐ฝ)
66. Using direct notation, show that the second invariant of a tensor is the trace of its
cofactor.
Given three basis vectors, (๐ 1, ๐ 2, ๐ 3)
๐ผ2(๐) =[๐๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐๐ 3] + [๐๐ 1, ๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
=๐c(๐ 1 ร ๐ 2) โ ๐ 3 + ๐ 1 โ ๐c(๐ 2 ร ๐ 3) + [๐c(๐ 3 ร ๐ 1) โ ๐ 2]
[๐ 1, ๐ 2, ๐ 3]
=(๐ 1 ร ๐ 2) โ ๐cT๐ 3 + (๐ 2 ร ๐ 3) โ ๐cT๐ 1 + (๐ 3 ร ๐ 1) โ ๐cT๐ 2
[๐ 1, ๐ 2, ๐ 3]
=[๐ 1, ๐ 2, ๐
cT๐ 3] + [๐cT๐ 1, ๐ 2, ๐ 3] + [๐ 3, ๐ 1, ๐cT๐ 2]
[๐ 1, ๐ 2, ๐ 3]
= ๐ผ1(๐cT) = ๐ผ1(๐
c)
Which is the trace of its cofactor as required.
67. Express the third tensor invariant in terms of its components.
As previously observed, any three linearly independent vectors can be treated as the
basis of a coordinate system, ๐ ๐ , ๐ = 1,2,3. The existence of the dual of these vectors
can be taken for granted. Consequently, for any tensor ๐,
๐ผ3(๐) =[๐๐ 1, ๐๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]=
[(๐๐1๐ ๐), (๐๐2๐
๐), (๐๐3๐ ๐)]
๐123
=[(๐๐1๐
๐ผ๐๐ ๐ผ), (๐๐2๐๐ฝ๐๐ ๐ฝ), (๐๐3๐
๐พ๐๐ ๐พ)]
๐123=
๐1๐ผ๐2
๐ฝ๐3
๐พ[๐ ๐ผ , ๐ ๐ฝ , ๐ ๐พ]
๐123
=๐1
๐ผ๐2๐ฝ๐3
๐พ๐๐ผ๐ฝ๐พ
๐123= ๐๐ผ๐ฝ๐พ๐1
๐ผ๐2๐ฝ๐3
๐พ= det ๐
68. For the tensor ๐, the cofactor is defined as ๐c โก ๐โT det ๐. Using direct notation
and the property of the cofactor that ๐๐ ร ๐๐ = ๐c(๐ ร ๐) for any two vectors ๐, ๐,
show that the third invariant of a tensor, defined as ๐ผ3(๐) =[๐๐ 1,๐๐ 2,๐๐ 3]
[๐ 1,๐ 2,๐ 3] for any set
of independent vectors, (๐ 1, ๐ 2, ๐ 3), is its determinant.
[๐๐ 1, ๐๐ 2, ๐๐ 3] = ๐๐ 1 โ (๐๐ 2 ร ๐๐ 3)
= ๐๐ 1 โ ๐c(๐ 2 ร ๐ 3)
= (๐ 2 ร ๐ 3) โ (๐c)T๐๐ 1
= (๐ 2 ร ๐ 3) โ ๐cT๐๐ 1
= (๐ 2 ร ๐ 3) โ (det ๐)๐ 1
so that
๐ผ3(๐) =[๐๐ 1, ๐๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
= det ๐(๐ 2 ร ๐ 3) โ ๐ 1
[๐ 1, ๐ 2, ๐ 3]
= det ๐
69. In component form, the third tensor invariant of a tensor ๐, ๐ผ3(๐) =
๐๐ผ๐ฝ๐พ๐1๐ผ๐2
๐ฝ๐3
๐พ= det ๐. Show that ๐๐๐๐๐๐ผ
๐๐๐ฝ๐๐๐พ
๐ = ๐๐ผ๐ฝ๐พ det ๐.
We do this by first establishing the fact that the LHS is completely antisymmetric in
๐ผ, ๐ฝ and ๐พ. We note that the indices ๐, ๐ and ๐ are dummy and therefore,
๐๐๐๐๐๐ผ๐๐๐ฝ
๐๐๐พ
๐ = ๐๐๐๐๐๐ผ๐๐๐ฝ
๐๐๐พ
๐ = ๐๐๐๐๐๐พ๐๐๐ผ
๐๐๐ฝ๐= โ๐๐๐๐๐๐พ
๐๐๐ฝ๐๐๐ผ
๐
Showing that a simple swap of ๐ผ and ๐พ changes the sign. This is similarly true for the
other pairs in the lower symbols. Thus we establish anti-symmetry in ๐ผ, ๐ฝ and ๐พ.
Noting that both sides take the same values when ๐ผ, ๐ฝ and ๐พ are equal to 1, 2 and 3
respectively. The arrangement of the indices makes this value positive or negative in
the same antisymmetric way. This completes the proof. Similarly we can write,
๐๐๐๐๐๐๐ผ๐๐
๐ฝ๐๐
๐พ= ๐๐๐๐๐๐
1๐๐2๐๐
3๐๐ผ๐ฝ๐พ = ๐๐ผ๐ฝ๐พdet ๐
70. Given the basis vectors, ๐ 1, ๐ 2, ๐ 3 and their dual, ๐ 1, ๐ 2, ๐ 3, Show that for any
other basis pair, ๐1, ๐2, ๐3 and their dual, ๐1, ๐2, ๐3, the relationship,
(๐๐ โ ๐ ๐ผ)(๐๐ โ ๐ ๐ฝ) = ๐ฟ๐ผ๐ฝ
holds. By simply reversing the step, it is immediately obvious that,
(๐๐ โ ๐ ๐ผ)(๐๐ โ ๐ ๐ฝ) = [(๐๐ โ ๐๐)๐ ๐ผ] โ ๐ ๐ฝ
Observe that the expression in the parentheses is the unit tensor โ having no effect
on a vector; it follows that,
(๐๐ โ ๐ ๐ผ)(๐๐ โ ๐ ๐ฝ) = [(๐๐ โ ๐๐)๐ ๐ผ] โ ๐ ๐ฝ = ๐ ๐ผ โ ๐ ๐ฝ = ๐ฟ๐ผ๐ฝ
71. Show that the first invariant has the same value in every coordinate system.
We have shown elsewhere that for any tensor ๐ป the first invariant, ๐ผ1(๐ป)=
tr(๐ป) โก[{๐๐ 1}, ๐ 2, ๐ 3] + [๐ 1, {๐๐ 2}, ๐ 3] + [๐ 1, ๐ 2, {๐๐ 3}]
[๐ 1, ๐ 2, ๐ 3]= ๐๐๐๐
๐๐ = ๐๐.๐
We proceed to show that this quantity has the same value in any other independent
set of basis vectors. Let (๐ธ1, ๐ธ2, ๐ธ3) be another arbitrary set of linearly independent
vectors. They therefore form a basis of a coordinate system. Let (๐ธ1, ๐ธ2, ๐ธ3) be the
dual to the set. It is easy to establish that the new set will be related to the old in;
๐๐ = (๐๐ โ ๐ ๐ผ)๐ ๐ผ , and ๐๐ = (๐๐ โ ๐ ๐ฝ)๐ ๐ฝ
Let us write the (๐, ๐)๐กโ components in the ๐ธ โ bases as ๏ฟฝฬ๏ฟฝ๐.๐.
Clearly,
๏ฟฝฬ๏ฟฝ๐.๐
โก ๐๐ โ ๐๐๐ = (๐๐ โ ๐ ๐ผ)๐ ๐ผ โ [๐(๐๐ โ ๐ ๐ฝ)๐ ๐ฝ]
= (๐๐ โ ๐ ๐ผ)(๐๐ โ ๐ ๐ฝ)[๐ ๐ผ โ ๐๐ ๐ฝ]
= [(๐๐ โ ๐๐)๐ ๐ผ] โ ๐ ๐ฝ [๐๐ผ
.๐ฝ]
Contracting ๐ with ๐, we obtain the sum,
๏ฟฝฬ๏ฟฝ๐.๐ = ๐๐ โ ๐๐๐ = [(๐๐ โ ๐๐)๐
๐ผ] โ ๐ ๐ฝ [๐๐ผ.๐ฝ
]
= ๐ ๐ถ โ ๐ ๐ฝ๐๐ผ.๐ฝ
= ๐ฟ๐ฝ๐ผ๐๐ผ
.๐ฝ= ๐๐ผ
.๐ผ
So that the trace has the same value in any arbitrarily chosen coordinate system
including curvilinear ones whether orthogonal or not.
72. Express the second invariant of a tensor in terms of traces only. Show that the
second invariant has the same value in all coordinate systems.
For arbitrary vectors bases, (๐ 1, ๐ 2, ๐ 3) the second principal invariant of a tensor ๐
๐ผ2(๐) =[๐๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐๐ 3] + [๐๐ 1, ๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3], or
๐ผ2(๐) =1
2(๐๐ผ
๐ผ๐๐ฝ๐ฝ
โ ๐๐ฝ๐ผ๐๐ผ
๐ฝ) =
1
2(๐ผ1
2(๐) โ ๐ผ1(๐2))
That is, half of the square of the trace minus the trace of the square. These therefore
are unchanged by coordinate transformation because traces are unchanged bt
coordinate transformation.
73. Show that the third tensor invariant is unaffected by a coordinate transformation.
We begin by showing that we can express the third invariant in terms of traces only.
Given a tensor ๐, by Cayley Hamilton theorem,
๐3 โ ๐ผ1๐2 + ๐ผ2๐ โ ๐ผ3 = 0
Note that ๐ผ1 = ๐ผ1(๐), ๐ผ2 = ๐ผ2(๐) and ๐ผ3 = ๐ผ3(๐) the first, second and third invariants
of ๐ are all scalar functions of the tensor ๐. Taking the trace of the above equation,
๐ผ1(๐3) = ๐ผ1(๐)๐ผ1(๐
2) โ ๐ผ2(๐)๐ผ1(๐) + 3๐ผ3(๐)
= ๐ผ1(๐บ)(๐ผ12(๐) โ 2๐ผ2(๐)) โ ๐ผ1(๐)๐ผ2(๐) + 3๐ผ3(๐)
= ๐ผ13(๐) โ 3๐ผ1(๐)๐ผ2(๐) + 3๐ผ3(๐)
Or, ๐ผ3(๐) =1
3(๐ผ1(๐
3) โ ๐ผ13(๐) + 3๐ผ1(๐)๐ผ2(๐))
which show that the third invariant is itself expressible in terms of traces only. It is
therefore invariant in value as a result of coordinate transformation.
74. For any tensor ๐, the arbitrary vector ๐ฎ and the scalar ๐, show that the eigenvalue
problem, ๐๐ฎ = ๐๐ฎ leads to the characteristic equation, ๐3 โ ๐ผ1๐2 + ๐ผ2๐ โ ๐ผ3 = 0,
where ๐ผ1 = ๐ผ1(๐), ๐ผ2 = ๐ผ2(๐) and ๐ผ3 = ๐ผ3(๐) the first, second and third invariants of
๐.
Writing the tensor and vector in component forms, we have
๐๐ฎ = ๐๐.๐(๐ ๐ โ ๐ ๐)๐ข๐๐ ๐ = ๐๐ฎ = ๐๐ข๐๐
๐
So that,
๐๐ฎ โ ๐๐ฎ = ๐๐.๐๐ ๐(๐ ๐ โ ๐ ๐)๐ข๐ โ ๐๐ข๐๐
๐ = ๐๐.๐๐ ๐๐ข๐ โ ๐๐ข๐๐
๐
= ๐๐.๐๐ ๐๐ข๐ โ ๐๐ข๐๐ฟ๐
๐๐ ๐ = (๐๐
.๐โ ๐๐ฟ๐
๐)๐ข๐๐
๐ = ๐
Which is possible only if the coefficient determinant, |๐๐.๐
โ ๐๐ฟ๐๐| vanishes.
Expanding, we find that,
โ๐31๐2
2๐13 + ๐2
1๐32๐1
3 + ๐31๐1
2๐23 โ ๐1
1๐32๐2
3 โ ๐21๐1
2๐33 + ๐1
1๐22๐3
3 + ๐21๐1
2๐ โ ๐11๐2
2๐
+ ๐31๐1
3๐ + ๐32๐2
3๐ โ ๐11๐3
3๐ โ ๐22๐3
3๐ + ๐11๐2 + ๐2
2๐2 + ๐33๐2 โ ๐3
= โ๐31๐2
2๐13 + ๐2
1๐32๐1
3 + ๐31๐1
2๐23 โ ๐1
1๐32๐2
3 โ ๐21๐1
2๐33 + ๐1
1๐22๐3
3
+ (๐21๐1
2 โ ๐11๐2
2 + ๐31๐1
3 + ๐32๐2
3 โ ๐11๐3
3 โ ๐22๐3
3)๐ + (๐11 + ๐2
2 + ๐33)๐2
โ ๐3 = 0
Or, ๐3 โ ๐ผ1๐2 + ๐ผ2๐ โ ๐ผ3 = 0, as required.
75. For arbitrary vectors ๐, ๐ and ๐, and a tensor ๐ use the relationship,
[๐๐ โ ๐๐, ๐๐ โ ๐๐, ๐๐ โ ๐๐ ] = det(๐๐ โ ๐)[๐, ๐, ๐] and the characteristic equation
๐3 โ ๐ผ1(๐)๐2 + ๐ผ2(๐)๐ โ ๐ผ3(๐) = 0 to find expressions for the for the invariants of
๐. The characteristic equation, det(๐๐ โ ๐) = 0 immediately implies that
[๐๐ โ ๐๐, ๐๐ โ ๐๐, ๐๐ โ ๐๐ ] = 0.
Expanding the scalar triple product, we have
[๐, ๐, ๐]๐3 โ ([๐๐, ๐, ๐] + [๐, ๐๐, ๐] + [๐, ๐, ๐๐])๐2
+ ([๐๐, ๐๐, ๐] + [๐, ๐๐, ๐๐] + [๐๐, ๐, ๐๐])๐ โ [๐๐, ๐๐, ๐๐] = 0
From which we can see that,
๐ผ1(๐) =[๐๐, ๐, ๐] + [๐, ๐๐, ๐] + [๐, ๐, ๐๐]
[๐, ๐, ๐]
๐ผ2(๐) =[๐๐, ๐๐, ๐] + [๐, ๐๐, ๐๐] + [๐๐, ๐, ๐๐]
[๐, ๐, ๐], and
๐ผ3(๐) =[๐๐, ๐๐, ๐๐]
[๐, ๐, ๐]
Assuming we have carefully chosen [๐, ๐, ๐] โ 0.
76. For any two vectors ๐ฎ and ๐ฏ, find the principal invariants of the dyad ๐ฎ โ ๐ฏ
The first principal invariant ๐ผ1(๐ฎ โ ๐ฏ) is the trace, defined as,
tr(๐ฎ โ ๐ฏ ) =[{(๐ฎ โ ๐ฏ) ๐ 1}, ๐ 2, ๐ 3] + [๐ 1, {(๐ฎ โ ๐ฏ) ๐ 2}, ๐ 3] + [๐ 1, ๐ 2, {(๐ฎ โ ๐ฏ)๐ 3}]
[๐ 1, ๐ 2, ๐ 3]
=1
๐123
{[๐ฃ1๐ฎ, ๐ 2, ๐ 3] + [๐ 1, ๐ฃ2๐ฎ, ๐ 3] + [๐ 1, ๐ 2, ๐ฃ3๐ฎ]}
=1
๐123{(๐ฃ1๐ฎ) โ (๐23๐๐
๐) + (๐31๐๐ ๐) โ (๐ฃ2๐ฎ) + (๐12๐๐
๐) โ (๐ฃ3๐ฎ)}
=1
๐123
{(๐ฃ1๐ฎ) โ (๐231๐ 1) + (๐312๐
2) โ (๐ฃ2๐ฎ) + (๐123๐ 3) โ (๐ฃ3๐ฎ)} = ๐ฃ๐๐ข
๐ .
The second principal invariant ๐ผ2(๐ฎ โ ๐ฏ) =
=[{(๐ฎ โ ๐ฏ) ๐ 1}, (๐ฎ โ ๐ฏ)๐ 2, ๐ 3] + [๐ 1, {(๐ฎ โ ๐ฏ) ๐ 2}, (๐ฎ โ ๐ฏ)๐ 3] + [(๐ฎ โ ๐ฏ)๐ 1, ๐ 2, {(๐ฎ โ ๐ฏ)๐ 3}]
[๐ 1, ๐ 2, ๐ 3]
=1
๐123{[๐ฃ1๐ฎ, ๐ฃ2๐ฎ, ๐ 3] + [๐ 1, ๐ฃ2๐ฎ, ๐ฃ3๐ฎ] + [๐ฃ1๐ฎ, ๐ 2, ๐ฃ3๐ฎ]} = 0. Vanishes โ collinearity.
The third invariant ๐ผ3(๐ฎ โ ๐ฏ) =1
๐123[๐ฃ1๐ฎ, ๐ฃ2๐ฎ, ๐ฃ3๐ฎ] which also vanishes on account of
collinearity
77. Define the cofactor of a tensor as, cof ๐ โก ๐c โก ๐โ๐ det ๐. Show that, for any
pair of independent vectors ๐ฎ and ๐ฏ the cofactor satisfies, ๐๐ฎ ร ๐๐ฏ = ๐c(๐ฎ ร ๐ฏ)
First note that if ๐ is invertible, the independence of the vectors ๐ and ๐ implies the
independence of vectors ๐๐ฎ and ๐๐ฏ. Consequently, we can define the non-vanishing
๐ง โก ๐๐ฎ ร ๐๐ฏ โ ๐.
It follows that ๐ง must be on the perpendicular line to both ๐๐ฎ and ๐๐ฏ. Therefore,
๐ง โ ๐๐ฎ = ๐ง โ ๐๐ฏ = ๐.
We can also take a transpose and write,
๐ฎ โ ๐๐๐ง = ๐ฏ โ ๐๐๐ง = ๐
Showing that the vector ๐๐๐ง is perpendicular to both ๐ฎ and ๐ฏ. It follows that โ ๐ผ โ๐ฝ
such that
๐๐๐ง = ๐ผ(๐ฎ ร ๐ฏ)
Therefore, ๐T(๐๐ฎ ร ๐๐ฏ) = ๐ผ(๐ฎ ร ๐ฏ).
Let ๐ฐ = ๐ฎ ร ๐ฏ so that ๐ฎ, ๐ฏ and ๐ฐ are linearly independent, then we can take a scalar
product of the above equation and obtain,
๐ฐ โ ๐T(๐๐ฎ ร ๐๐ฏ) = ๐ผ(๐ฎ ร ๐ฏ โ ๐ฐ)
The LHS is also ๐๐ฐ โ (๐๐ฎ ร ๐๐ฏ) = ๐๐ฎ ร ๐๐ฏ โ ๐๐ฐ. In the equation, ๐๐ฎ ร
๐๐ฏ โ ๐๐ฐ = ๐ผ(๐ฎ ร ๐ฏ โ ๐ฐ), it is clear that
๐ผ = det ๐
We have that, ๐๐ฎ ร ๐๐ฏ = ๐โT det ๐ (๐ฎ ร ๐ฏ). And therefore, we have that,
๐๐ฎ ร ๐๐ฏ = ๐โT det ๐ (๐ฎ ร ๐ฏ) = ๐c(๐ฎ ร ๐ฏ).
78. Find and expression for the components of the cofactor of tensor ๐ in terms of
the components of ๐.
We now express the cofactor in its general components.
๐c = (๐c)๐๐ผ๐ ๐ผ โ ๐ ๐ = (๐ ฮฑ โ ๐c๐ ๐)๐ ๐ผ โ ๐ ๐
=1
2๐๐๐๐[๐ ฮฑ โ ๐c(๐ ๐ ร ๐ ๐)]๐ ๐ผ โ ๐ ๐
=1
2๐๐๐๐[๐ ฮฑ โ (๐๐ ๐) ร (๐๐ ๐)]๐ ๐ผ โ ๐ ๐ .
The scalar in brackets,
๐ ฮฑ โ (๐๐ ๐) ร (๐๐ ๐) = ๐ ฮฑ โ ๐๐๐๐(๐ ๐ โ ๐๐ ๐)(๐ ๐ โ ๐๐ ๐)๐ ๐
= ๐ฟ๐๐ผ๐๐๐๐(๐ ๐ โ ๐๐ ๐)(๐ ๐ โ ๐๐ ๐)
= ๐ฟ๐๐ผ๐๐๐๐๐๐
๐๐๐
๐ = ๐๐ผ๐๐๐๐๐๐๐
๐
Inserting this above, we therefore have, in invariant component form,
๐c =1
2๐๐๐๐[๐ ฮฑ โ (๐๐ ๐) ร (๐๐ ๐)]๐ ๐ผ โ ๐ ๐
=1
2๐๐๐๐๐๐ผ๐๐๐๐
๐๐๐
๐๐ ๐ผ โ ๐ ๐
=1
2๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐ โ ๐ ๐
79. Given that cof ๐ =1
2๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐ โ ๐ ๐ if ๐ = ๐ฎ โ ๐ฉ + ๐ฏ โ ๐ช + ๐ฐ โ ๐ซ, show that
cof ๐ = ๐ฎ ร ๐ฏ โ ๐ฉ ร ๐ช + ๐ฏ ร ๐ฐ โ ๐ช ร ๐ซ + ๐ฐ ร ๐ฎ โ ๐ซ ร ๐ฉ, tr ๐ = ๐ฎ โ ๐ฉ + ๐ฏ โ ๐ช +
๐ฐ โ ๐ซ and ๐ผ2(๐) = ๐ฎ ร ๐ฏ โ ๐ฉ ร ๐ช + ๐ฏ ร ๐ฐ โ ๐ช ร ๐ซ + ๐ฐ ร ๐ฎ โ ๐ซ ร ๐ฉ
Clearly, ๐๐๐ = ๐ข๐๐
๐ + ๐ฃ๐๐๐ + ๐ค๐๐
๐. The cofactor
cof ๐ =1
2๐ฟ๐๐๐
๐๐๐(๐ข๐๐๐ + ๐ฃ๐๐
๐ + ๐ค๐๐๐)(๐ข๐๐๐ + ๐ฃ๐๐๐ + ๐ค๐๐๐)๐ ๐ โ ๐ ๐
=1
2๐ฟ๐๐๐
๐๐๐ ((๐ข๐๐๐๐ข๐๐๐ + ๐ฃ๐๐
๐๐ฃ๐๐๐ + ๐ค๐๐๐๐ค๐๐
๐)
+ 2(๐ข๐๐๐๐ฃ๐๐๐ + ๐ฃ๐๐
๐๐ค๐๐๐ + ๐ข๐๐๐๐ค๐๐
๐)) ๐ ๐ โ ๐ ๐
= ๐ฟ๐๐๐๐๐๐(๐ข๐๐
๐๐ฃ๐๐๐ + ๐ฃ๐๐๐๐ค๐๐
๐ + ๐ข๐๐๐๐ค๐๐๐)๐ ๐ โ ๐ ๐
= ๐ฎ ร ๐ฏ โ ๐ฉ ร ๐ช + ๐ฏ ร ๐ฐ โ ๐ช ร ๐ซ + ๐ฐ ร ๐ฎ โ ๐ซ ร ๐ฉ
as the terms in the first inner parentheses vanish on account of symmetry and anti-symmetry in
๐, ๐, as well as ๐, ๐. The rest of the results follow by a change in the tensor product to the dot
product once we take the traces of the tensor and its cofactor noting that
๐ผ2(๐) = tr(cof ๐) = ๐ฎ ร ๐ฏ โ ๐ฉ ร ๐ช + ๐ฏ ร ๐ฐ โ ๐ช ร ๐ซ + ๐ฐ ร ๐ฎ โ ๐ซ ร ๐ฉ
80. Find an expression for the inverse of the tensor ๐ in terms of the components of
๐.
The cofactor ๐c = ๐โT det ๐ . Transposing the equation, we have,
๐โ1 =1
det ๐(๐c)T
=1
2det ๐๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐ โ ๐ ๐
81. Let ๐ be invertible and assume we can select the vectors ๐ฎ and ๐ฏ such that ๐ฏ โ
๐โ1๐ฎ โ 1, Show that (๐ + ๐ฎ โ ๐ฏ)โ1 = ๐โ1 +๐โ1๐ฎโ๐ฏ๐โ1
1โ๐ฏโ ๐โ1๐ฎ.
Let (๐ + ๐ฎ โ ๐ฏ)๐ฑ = ๐ฒ. In our attempt to find ๐ฑ, it will be necessary to find the
inverse of ๐ + ๐ฎ โ ๐ฏ. We proceed indirectly and operate the tensor ๐โ1 on this
vector euation and obtain,
๐โ1(๐ + ๐ฎ โ ๐ฏ)๐ฑ = ๐โ1๐ฒ
= ๐ฑ + (๐โ1๐ฎ โ ๐ฏ)๐ฑ = ๐ฑ + ๐โ1๐ฎ(๐ฏ โ ๐ฑ)
= ๐ฑ + ๐โ1๐ฎ(๐ฏ โ ๐ฑ) = ๐โ1๐ฒ
Taking the dot product of both sides with ๐ฏ,
๐ฏ โ ๐ฑ + ๐ฏ โ ๐โ1๐ฎ(๐ฏ โ ๐ฑ) = ๐ฏ โ ๐โ1๐ฒ
so that the scalar ,
๐ฏ โ ๐ฑ =(๐ฏ โ ๐โ1๐ฒ)
(1 + ๐ฏ โ ๐โ1๐ฎ)
But we can write that,
๐ฑ = (๐ + ๐ฎ โ ๐ฏ)โ1๐ฒ
= ๐โ1๐ฒ โ ๐โ1๐ฎ(๐ฏ โ ๐ฑ)
Substituting for ๐ฏ โ ๐ฑ, we have,
(๐ + ๐ฎ โ ๐ฏ)โ1๐ฒ = ๐โ1๐ฒ โ๐โ1๐ฎ(๐ฏ โ ๐โ1๐ฒ)
1 + ๐ฏ โ ๐โ1๐ฎ
= ๐โ1๐ฒ โ(๐โ1(๐ฎ โ ๐ฏ)๐โ1)๐ฒ
1 + ๐ฏ โ ๐โ1๐ฎ
so that,
(๐ + ๐ฎ โ ๐ฏ)โ1 = ๐โ1 โ๐โ1(๐ฎ โ ๐ฏ)๐โ1
1 + ๐ฏ โ ๐โ1๐ฎ
Provided, as we have been given, that ๐ฏ โ ๐โ1๐ฎ โ 1.
82. For the invertible tensor ๐ and the tensors ๐ , ๐ and ๐, show that
(๐ + ๐๐ ๐)โ1 = ๐โ1 โ ๐โ1๐(๐ โ1 + ๐๐โ๐๐)โ1
๐๐โ1
First consider the matrix (๐ โ๐๐ ๐ โ1). Its inverse is obtained by solving the matrix
equation,
(๐ ๐๐ ๐
) (๐ โ๐๐ ๐ โ1) = (
๐ ๐๐ ๐
)
which yields,
๐ ๐ + ๐๐ = ๐
โ๐๐ + ๐๐ โ1 = ๐ โ ๐ = ๐๐๐
so that,
๐ ๐ + ๐๐๐ ๐ = ๐(๐ + ๐๐ ๐) = ๐
โ ๐ = (๐ + ๐๐ ๐)โ1
But ๐ = ๐โ1 โ ๐๐๐โ1 substituting in the second equation,
from which we can now write that (๐โ1 โ ๐๐๐โ1)๐ = ๐๐ โ1 so that
๐ = ๐โ1๐(๐ โ1 + ๐๐โ1๐)โ1
๐ = ๐โ1 โ ๐๐๐โ1 = ๐โ1 โ ๐โ1๐(๐ โ1 + ๐๐โ1๐)โ1๐๐โ1
Finally ๐ = (๐ + ๐๐ ๐)โ1 = ๐โ1 โ ๐โ1๐(๐ โ1 + ๐๐โ1๐)โ1๐๐โ1 as required
In the special case when ๐ is the unit tensor, we have,
(๐ + ๐๐)โ1 = ๐โ1 โ ๐โ1๐(๐ + ๐๐โ1๐)โ1๐๐โ1
83. Determinant is not a linear operation. For any two tensors ๐ and ๐, use the
component representation of the cofactor to show that the determinant of the sum
det(๐ + ๐) = det(๐) + tr(๐c๐T) + tr(๐c๐T) + det(๐)
det(๐ + ๐) = ๐๐๐๐(๐๐1 + ๐๐
1)(๐๐2 + ๐๐
2)(๐๐3 + ๐๐
3)
= ๐๐๐๐[๐๐1๐๐
2๐๐3 + (๐๐
1๐๐2๐๐
3 + ๐๐1๐๐
2๐๐3 + ๐๐
1๐๐2๐๐
3) + ๐๐1๐๐
2๐๐3
+ (๐๐1๐๐
2๐๐3 + ๐๐
1๐๐2๐๐
3 + ๐๐1๐๐
2๐๐3)]
Now, we can write,
๐T๐c = (๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ)T(1
2๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐ โ ๐ ๐)
=1
2๐ฟ๐๐๐
๐๐๐๐๐ผ๐ฝ๐๐๐๐๐
๐(๐ ๐ฝ โ ๐ ๐ผ)(๐ ๐ โ ๐ ๐)
=1
2๐ฟ๐๐๐
๐๐๐๐๐ผ๐ฝ๐๐๐๐๐
๐๐ฟ๐๐ผ(๐ ๐ฝ โ ๐ ๐) =
1
2๐ฟ๐๐๐
๐๐๐๐๐๐ฝ๐๐๐๐๐
๐(๐ ๐ฝ โ ๐ ๐)
So that tr(๐T๐c) =1
2๐ฟ๐๐๐
๐๐๐๐๐๐ฝ๐๐๐๐๐
๐ = ๐๐๐๐(๐๐1๐๐
2๐๐3 + ๐๐
1๐๐2๐๐
3 + ๐๐1๐๐
2๐๐3) after
expansion. Similarly, tr(๐T๐c) expands to ๐๐๐๐(๐๐1๐๐
2๐๐3 + ๐๐
1๐๐2๐๐
3 + ๐๐1๐๐
2๐๐3) from
which the result immediately follows.
84. Given that ๐c is the cofactor of the tensor ๐, show that ๐ผ1(๐c) = ๐ผ2(๐) that is,
that the trace of the cofactor is the second principal invariant of the original tensor:
tr(๐c) =1
2๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐ โ ๐ ๐ = ๐ผ1(๐c)
=1
2๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐๐ฟ๐๐ =
1
2๐ฟ๐๐๐
๐๐๐๐๐๐๐๐
๐
=1
2(๐ฟ๐
๐๐ฟ๐๐ โ ๐ฟ๐
๐๐ฟ๐๐)๐๐
๐๐๐
๐
=1
2(๐๐
๐๐๐
๐ โ ๐๐๐๐๐
๐) = ๐ผ2(๐)
85. For a scalar ๐ผ show that det ๐ผ๐ = ๐ผ3 det ๐
Given that det ๐ =[๐๐,๐๐,๐๐]
[๐,๐,๐], then
det ๐ผ๐ =[๐ผ๐๐, ๐ผ๐๐, ๐ผ๐๐]
[๐, ๐, ๐]= ๐ผ3
[๐๐, ๐๐, ๐๐]
[๐, ๐, ๐]= ๐ผ3 det ๐
86. Define the inner product of tensors ๐ and ๐ as ๐: ๐ = tr(๐๐๐) = tr(๐๐๐) show
that ๐ผ1(๐) = ๐: ๐
๐: ๐ = tr(๐๐๐) = tr(๐๐๐)
Let ๐ = ๐;
๐: ๐ = tr(๐๐๐) = tr(๐๐)
= tr(๐) = ๐ผ1(๐)
87. Given any scalar ๐ > 0, for the scalar-valued tensor function, ๐(๐) = tr(๐๐),
show that, ๐
๐๐ ๐(๐) = ๐(๐๐โ1)๐.
When ๐ = 1,
๐ท๐(๐บ, ๐๐บ) =๐
๐๐ผ๐(๐บ + ๐ผ๐๐บ)|
๐ผ=0
=๐
๐๐ผtr(๐บ + ๐ผ๐๐บ)|
๐ผ=0 = tr(๐ ๐๐บ)
= ๐: ๐๐บ
So that,
๐
๐๐บ tr(๐บ) = ๐.
When ๐ = 2, The Gateaux differential in this case,
๐ท๐(๐บ, ๐๐บ) =๐
๐๐ผ๐(๐บ + ๐ผ๐๐บ)|
๐ผ=0 =
๐
๐๐ผtr{(๐บ + ๐ผ๐๐บ)2}|
๐ผ=0
=๐
๐๐ผtr{(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)}|
๐ผ=0 = tr [
๐
๐๐ผ(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)]|
๐ผ=0
= tr[๐๐บ(๐บ + ๐ผ๐๐บ) + (๐บ + ๐ผ๐๐บ)๐๐บ]|๐ผ=0
= tr[๐๐บ ๐บ + ๐บ ๐๐บ] = 2(๐บ)T: ๐๐บ
When ๐ = 3,
๐ท๐(๐บ, ๐๐บ) =๐
๐๐ผ๐(๐บ + ๐ผ๐๐บ)|
๐ผ=0 =
๐
๐๐ผtr{(๐บ + ๐ผ๐๐บ)3}|
๐ผ=0
=๐
๐๐ผtr{(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)}|
๐ผ=0
= tr [๐
๐๐ผ(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)]|
๐ผ=0
= tr[๐๐บ(๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ) + (๐บ + ๐ผ๐๐บ)๐๐บ(๐บ + ๐ผ๐๐บ)
+ (๐บ + ๐ผ๐๐บ)(๐บ + ๐ผ๐๐บ)๐๐บ]|๐ผ=0
= tr[๐๐บ ๐บ ๐บ + ๐บ ๐๐บ ๐บ + ๐บ ๐บ ๐ ๐บ] = 3(๐บ2)T: ๐๐บ
It easily follows by induction that, ๐
๐๐ ๐(๐) = ๐(๐๐โ1)๐.
88. Find the derivative of the second principal invariant of the tensor ๐บ ๐
๐๐บ ๐ผ2(๐บ) =
1
2
๐
๐๐บ[tr2(๐บ) โ tr(๐บ2)]
=1
2[2tr(๐บ)๐ โ 2 ๐บT]
= tr(๐บ)๐ โ ๐บT
89. Find the derivative of the third principal invariant of the tensor ๐บ
By Cayley-Hamilton, in terms of traces only,
๐ผ3(๐บ) =1
6[tr3(๐บ) โ 3tr(๐บ)tr(๐บ2) + 2tr(๐บ3)]
๐
๐๐บ ๐ผ3(๐บ) =
1
6
๐
๐๐บ[tr3(๐บ) โ 3tr(๐บ)tr(๐บ2) + 2tr(๐บ3)]
=1
6[3tr2(๐บ)๐ โ 3tr(๐บ2)๐ โ 3tr(๐บ)2๐บ๐ + 2 ร 3(๐บ2)๐]
= ๐ผ2๐ โ ๐ผ1(๐บ)๐บ๐ + ๐บ2๐
90. By the representation theorem, every real isotropic tensor function is a function
of its principal invariants. Show that for every isotropic function ๐(๐) the derivative โ๐(๐)
โ๐ can be represented as a quadratic function of ๐T.
By the representation theorem,
๐(๐) = ๐(๐ผ1(๐), ๐ผ2(๐), ๐ผ3(๐))
consequently,
โ๐(๐)
โ๐=
โ๐(๐)
โ๐ผ1
โ๐ผ1โ๐
+โ๐(๐)
โ๐ผ2
โ๐ผ2โ๐
+โ๐(๐)
โ๐ผ3
โ๐ผ3โ๐
=โ๐(๐)
โ๐ผ1๐ +
โ๐(๐)
โ๐ผ2(tr(๐)๐ โ ๐T) +
โ๐(๐)
โ๐ผ3(๐ผ2๐ โ ๐ผ1(๐)๐๐ + ๐2๐)
= (โ๐(๐)
โ๐ผ1+
โ๐(๐)
โ๐ผ2๐ผ1(๐) +
โ๐(๐)
โ๐ผ3๐ผ2(๐)) ๐ โ (
โ๐(๐)
โ๐ผ2+
โ๐(๐)
โ๐ผ3๐ผ1(๐)) ๐T
+โ๐(๐)
โ๐ผ3๐2๐
= ๐ผ0๐ + ๐ผ1๐T + ฮฑ2๐
2๐
where ๐ผ0 =โ๐(๐)
โ๐ผ1+
โ๐(๐)
โ๐ผ2๐ผ1(๐) +
โ๐(๐)
โ๐ผ3๐ผ2(๐), ๐ผ1 = โ
โ๐(๐)
โ๐ผ2โ
โ๐(๐)
โ๐ผ3๐ผ1(๐) and ๐ผ2 =
โ๐(๐)
โ๐ผ3 is the desired
quadratic representation of the derivative.
91. Given arbitrary vectors ๐ and ๐ the tensor ๐ is said to be orthogonal if (๐๐) โ
(๐๐) = ๐ โ ๐ = ๐ โ ๐ show that the inverse of ๐ is its transpose. and that ๐ is the
cofactor of itself.
By definition of the transpose, we have that,
๐ช โ ๐๐ = ๐ โ ๐T๐ช = ๐ โ ๐๐๐๐ = ๐ โ ๐
Clearly, ๐๐๐ = 1 which makes the transpose the same as the inverse tensor.
A condition necessary and sufficient for a tensor ๐ to be orthogonal is that ๐ be
invertible and its inverse is equal to its transpose.
92. An orthogonal tensor ๐ is said to be โproper orthogonalโ if its determinant |๐| =
+1. Show that a proper orthogonal tensor is the cofactor of itself. Show also that its
first two invariants are equal.
cof๐ = det ๐ ๐โT = +1 (๐T)โ๐ = 1 (๐โ1)โ๐ = ๐
๐ผ2(๐) = ๐ผ1(๐c)
The second principal invariant for any vector is equal to the first principal invariant
of its co-factor. But we find here that ๐ = ๐c. It follows that the first two invariants
of a proper orthogonal tensor are equal. The third invariant, ๐ผ3(๐) = det(๐)=1. All
essential information on an orthogonal tensor is known once we know its trace!
93. For any invertible tensor ๐ show that det(๐c) = (det ๐)2
First note that the determinant of the product of a tensor C with a scalar ๐ผ is,
det ๐ผ๐ = ๐๐๐๐(๐ผ๐ถ๐1)(๐ผ๐ถ๐
2)(๐ผ๐ถ๐3) = ๐ผ3 det ๐
The inverse of tensor ๐,
๐โ1 = (det ๐)โ1(๐cT)
Let the scalar ๐ผ = det ๐. We can see clearly that,
๐c = ๐ผ๐โ๐
Taking the determinant of this equation, we have,
det(๐c) = ๐ผ3 det(๐โ๐) = ๐ผ3 det(๐โ1)
as the transpose operation has no effect on the value of a determinant. Noting that
the determinant of an inverse is the inverse of the determinant, we have,
det(๐c) = ๐ผ3 det(๐โ1) =๐ผ3
๐ผ= (det ๐)2
94. For any invertible tensor ๐ and a scalar ๐ผ show that show that the cofactor of the
product of ๐ผ and ๐ equals ๐ผ2 ร the cofactor of ๐ , that is, (๐ผ๐)c = ๐ผ2๐c
(๐ผ๐)c = (det(๐ผ๐))(๐ผ๐)โT
= (๐ผ3 det(๐))๐ผโ1๐บโT
= (๐ผ2 det(๐))๐โT
= ๐ผ2๐c
95. For any invertible tensor ๐ show that (๐โ1)c = (det ๐)โ1๐T
(๐โ1)c = det(๐โ1) (๐โ1)โ๐
= (det ๐)โ1๐T
96. For any invertible tensor ๐ show that ๐โc = (det ๐)โ1๐T, that is, the inverse of the
cofactor is the transpose divided by the determinant.
๐C = det(๐) ๐โ๐
Consequently,
๐โc = (det ๐)โ1(๐บโ๐)โ1 = (det ๐)โ1๐T
97. For an invertible tensor show that the cofactor of the cofactor is the product of
the original tensor and its determinant ๐cc = (det ๐)๐
๐c = det(๐) ๐โ๐
So that,
๐cc = (det ๐c)(๐c)โ๐
= (det ๐)2[(๐c)โ1]๐ = (det ๐)2[(det ๐)โ1๐T]๐
= (det ๐)2(det ๐)โ1๐
= (det ๐)๐
as required.
98. Show that for any invertible tensor ๐ and any vector ๐ฎ, [(๐๐ฎ) ร] = ๐c(๐ฎ ร)๐โ๐
where ๐c and ๐โ๐ are the cofactor and inverse of ๐ respectively.
By definition,
๐๐ = (det ๐)๐โT
We are to prove that,
[(๐๐ฎ) ร] = ๐๐(๐ฎ ร)๐โ๐ = (det ๐)๐โT(๐ฎ ร)๐โ๐
or that,
๐T[(๐๐ฎ) ร] = (๐ฎ ร)(det ๐)๐โ๐ = (๐ฎ ร)(๐๐)๐
On the RHS, the contravariant ๐๐ component of ๐ ร is
(๐ฎ ร)๐๐ = ๐๐๐ผ๐๐ข๐ผ
which is exactly the same as writing, (๐ฎ ร) = ๐๐๐ผ๐๐ข๐ผ๐ ๐ โ ๐ ๐ in the invariant form.
We now turn to the LHS;
[(๐๐ฎ) ร] = ๐๐๐ผ๐(๐๐ฎ)๐ผ๐ ๐ โ ๐ ๐ = ๐๐๐ผ๐๐๐ผ๐๐ข๐๐ ๐ โ ๐ ๐
Now, ๐ = ๐.๐ฝ๐. ๐ ๐ โ ๐ ๐ฝ so that its transpose, ๐T = ๐๐ฝ
๐ ๐ ฮฒ โ ๐ i = ๐๐๐ฝ๐ ๐ โ ๐ ๐ฝ so that
๐T[(๐๐ฎ) ร] = ๐๐๐ผ๐๐๐๐ฝ๐๐ผ
๐๐ข๐๐
๐ โ ๐ ๐ฝ โ ๐ ๐ โ ๐ ๐
= ๐๐๐ผ๐๐๐๐๐๐ผ๐๐ข๐๐
๐ โ ๐ ๐ = ๐๐๐ผ๐๐๐๐๐๐ผ
๐๐ข๐๐ ๐ โ ๐ ๐
= ๐๐ผ๐ฝ๐๐ข๐๐๐ผ๐ ๐๐ฝ
๐๐ ๐ โ ๐ ๐ = (๐ฎ ร)(๐c)T
99. Show that [(๐c๐ฎ) ร] = ๐(๐ฎ ร)๐T
The LHS in component invariant form can be written as:
[(๐c๐ฎ) ร] = ๐๐๐๐(๐c๐ฎ)๐๐ ๐ โ ๐ ๐
where (๐c)๐๐ฝ
=1
2๐๐๐๐๐
๐ฝ๐๐๐๐๐๐๐
๐ so that
(๐๐๐ฎ)๐ = (๐c)๐๐ฝ๐ข๐ฝ =
1
2๐๐๐๐๐
๐ฝ๐๐๐ข๐ฝ๐๐๐๐๐
๐
Consequently,
[(๐c๐ฎ) ร] =1
2๐๐๐๐๐๐๐๐๐
๐ฝ๐๐๐ข๐ฝ๐๐๐๐๐
๐๐ ๐ โ ๐ ๐
=1
2๐๐ฝ๐๐(๐ฟ๐
๐๐ฟ๐๐ โ ๐ฟ๐
๐๐ฟ๐๐ )๐ข๐ฝ๐๐
๐๐๐๐๐ ๐ โ ๐ ๐
=1
2๐๐ฝ๐๐๐ข๐ฝ(๐๐
๐๐๐๐ โ ๐๐
๐๐๐๐)๐ ๐ โ ๐ ๐
= ๐๐ฝ๐๐๐ข๐ฝ๐๐๐๐๐
๐ ๐ ๐ โ ๐ ๐
On the RHS, (๐ฎ ร)๐T = ๐๐ผ๐ฝ๐พ๐ข๐ฝ๐๐พ๐๐ ๐ โ ๐ ๐ . We can therefore write,
๐(๐ฎ ร)๐T = ๐๐ผ๐ฝ๐พ๐ข๐ฝ๐๐ผ๐ ๐๐พ
๐๐ ๐ โ ๐ ๐
Which on a closer look is exactly the same as the LHS so that, [(๐c๐ฎ) ร] = ๐(๐ฎ ร)๐T
as required.
100. For a tensor ๐, given that [(๐c๐ฎ) ร] = ๐(๐ฎ ร)๐T for any two vectors ๐ฎ and ๐ฏ,
show that ((cof ๐)๐ฎ ร)๐ฏ = ๐(๐ฎ ร ๐T๐ฏ)
The product of the given equation with the vector ๐ฏ immediately yields,
[(๐c๐ฎ) ร]๐ฏ = ๐(๐ฎ ร)๐T๐ฏ
โ ((cof ๐)๐ฎ ร)๐ฏ = ๐(๐ฎ ร ๐T๐ฏ)
101. Given that ๐ is a skew tensor with the corresponding axial vector ๐. Given
vectors ๐ฎ and ๐ฏ, show that ๐๐ฎ ร ๐๐ฏ = (๐ โ ๐)(๐ฎ ร ๐ฏ) and, hence conclude that
๐c = (๐ โ ๐). ๐๐ฎ ร ๐๐ฏ = (๐ ร ๐ฎ) ร (๐ ร ๐ฏ) = (๐ ร ๐ฎ) ร (๐ ร ๐ฏ)
= [(๐ ร ๐ฎ) โ ๐ฏ]๐ โ [(๐ ร ๐ฎ) โ ๐]๐ฏ = [๐ โ (๐ฎ ร ๐ฏ)]๐ = (๐ โ ๐)(๐ฎ ร ๐ฏ)
But by definition, the cofactor must satisfy,
๐๐ฎ ร ๐๐ฏ = ๐c(๐ฎ ร ๐ฏ)
which compared with the previous equation yields the desired result that
๐c = (๐ โ ๐).
102. Show, using indicial notation, that the cofactor of a tensor can be written as ๐c =
(๐2 โ ๐ผ1๐ + ๐ผ2๐)T even if ๐ is not invertible. ๐ผ1, ๐ผ2 are the first two invariants of ๐.
The above equation can be written more explicitly as,
๐๐ = (๐2 โ tr(๐)๐ + [1
2[tr2(๐) โ tr(๐2)]] ๐)
T
In the invariant component form, this is easily seen to be,
๐บc = (๐๐ ๐ ๐๐
๐โ ๐๐ผ
๐ผ๐๐๐ +
1
2(๐๐ผ
๐ผ๐๐ฝ๐ฝ
โ ๐๐ฝ๐ผ๐๐ผ
๐ฝ) ๐ฟ๐
๐) ๐ ๐ โ ๐ ๐
But we know that the cofactor can be obtained directly from the equation,
(๐c) =1
2๐๐๐ฝ๐พ๐๐๐๐๐๐ฝ
๐๐๐พ๐๐ ๐ โ ๐ ๐ =
1
2[ ๐ฟ๐
๐ ๐ฟ๐ ๐ ๐ฟ๐
๐
๐ฟ๐๐ฝ
๐ฟ๐ ๐ฝ
๐ฟ๐๐ฝ
๐ฟ๐๐พ
๐ฟ๐ ๐พ
๐ฟ๐๐พ] ๐๐ฝ
๐๐๐พ๐๐ ๐ โ ๐ ๐
=1
2(๐ฟ๐
๐ [๐ฟ๐
๐ฝ๐ฟ๐
๐ฝ
๐ฟ๐ ๐พ
๐ฟ๐๐พ] โ ๐ฟ๐
๐ [๐ฟ๐
๐ฝ๐ฟ๐
๐ฝ
๐ฟ๐๐พ
๐ฟ๐๐พ] + ๐ฟ๐
๐ [๐ฟ๐
๐ฝ๐ฟ๐
๐ฝ
๐ฟ๐๐พ
๐ฟ๐ ๐พ]) ๐๐ฝ
๐๐๐พ๐๐ ๐ โ ๐ ๐
=1
2[๐ฟ๐
๐ (๐ฟ๐ ๐ฝ๐ฟ๐
๐พโ ๐ฟ๐
๐ฝ๐ฟ๐
๐พ) โ ๐ฟ๐
๐ (๐ฟ๐๐ฝ๐ฟ๐
๐พโ ๐ฟ๐
๐ฝ๐ฟ๐
๐พ) + ๐ฟ๐
๐ (๐ฟ๐๐ฝ๐ฟ๐
๐พโ ๐ฟ๐
๐ฝ๐ฟ๐
๐พ)] ๐๐ฝ
๐๐๐พ๐๐ ๐
โ ๐ ๐
=1
2[๐ฟ๐
๐ (๐๐ผ๐ผ๐๐ฝ
๐ฝโ ๐๐
๐๐๐ ๐) โ 2๐๐
๐๐๐ผ๐ผ + 2๐๐
๐ ๐๐๐
] ๐ ๐ โ ๐ ๐
= [๐ผ2(๐)๐ โ ๐ผ1(๐)๐ + ๐2]T
103. Show, using direct notation, that the cofactor of a tensor can be written as ๐c =
(๐2 โ ๐ผ1๐ + ๐ผ2๐)T even if ๐ is not invertible. ๐ผ1, ๐ผ2 are the first two invariants of ๐.
For any three linearly independent vectors, the trace of a tensor ๐
tr ๐ โก ๐ผ1(๐) =[๐๐ 1, ๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
Replacing ๐ 1 by ๐๐ 1 in the above equation, we have,
tr ๐ [๐๐ 1, ๐ 2, ๐ 3] = [๐2๐ 1, ๐ 2, ๐ 3] + [๐๐ 1, ๐๐ 2, ๐ 3] + [๐๐ 1, ๐ 2, ๐๐ 3]
Or, upon rearrangement,
[๐๐ 1, ๐๐ 2, ๐ 3] + [๐๐ 1, ๐ 2, ๐๐ 3] = tr ๐ [๐๐ 1, ๐ 2, ๐ 3] โ [๐2๐ 1, ๐ 2, ๐ 3]
But, the second Invariant,
๐ผ2(๐) =[๐๐ 1, ๐๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐๐ 3] + [๐๐ 1, ๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
=tr ๐ [๐๐ 1, ๐ 2, ๐ 3] โ [๐2๐ 1, ๐ 2, ๐ 3] + [๐ 1, ๐๐ 2, ๐๐ 3]
[๐ 1, ๐ 2, ๐ 3]
=tr ๐ [๐๐ 1, ๐ 2, ๐ 3] โ [๐2๐ 1, ๐ 2, ๐ 3] + ๐ 1 โ ๐c(๐ 2 ร ๐ 3)
[๐ 1, ๐ 2, ๐ 3]
=[(tr ๐)๐๐ 1, ๐ 2, ๐ 3] โ [๐2๐ 1, ๐ 2, ๐ 3] + [๐cT๐ 1, ๐ 2, ๐ 3]
[๐ 1, ๐ 2, ๐ 3]
so that,
[(๐ผ2(๐)๐)๐ 1, ๐ 2, ๐ 3] = [(tr ๐)๐๐ 1, ๐ 2, ๐ 3] โ [๐2๐ 1, ๐ 2, ๐ 3] + [๐cT๐ 1, ๐ 2, ๐ 3]
From which we can write that
๐ผ2(๐)๐ = (tr ๐)๐ โ ๐2 + ๐cT
or,
๐c = (๐2 โ ๐ผ1(๐)๐ + ๐ผ2(๐)๐)T
104. Given a Euclidean Vector Space E, a tensor ๐ is said to be rotation if in addition to
satisfying (๐๐) โ (๐๐) = ๐ โ ๐ โ๐, ๐ โ E, its determinant (det ๐) = +1. For any pair
of vectors ๐ฎ, ๐ฏ show that ๐(๐ฎ ร ๐ฏ) = (๐๐ฎ) ร (๐๐ฏ) if ๐ is a rotation. That is, that the
cofactor of ๐ is ๐ itself
We can write that
๐(๐ฎ ร ๐ฏ) = (๐๐ฎ) ร (๐๐ฏ)
where
๐ = cof(๐) = det(๐)๐โ๐
Now that ๐ is a rotation, det(๐) = 1, and because it is orthogonal, its inverse is its
transpose:
๐โT = (๐โ1)T = (๐T)T = ๐
This implies that ๐ = ๐ and consequently,
๐(๐ฎ ร ๐ฏ) = (๐๐ฎ) ร (๐๐ฏ)
105. For a proper orthogonal tensor ๐, show that the eigenvalue equation always
yields an eigenvalue of +1. This means that there is always a solution for the
equation, ๐๐ฎ = ๐ฎ.
For any invertible tensor, note that the cofactor is defined as,
๐c = (det ๐)๐โT
For a proper orthogonal tensor ๐, det ๐ = 1, and its inverse is its transpose. It
therefore follows that,
๐๐ = (det๐)๐โT = ๐โT = ๐
It is easily shown that tr ๐c = ๐ผ2(๐). It follows from the above that, ๐ผ2(๐) = ๐ผ๐(๐).
For the general eigenvalue equation, ๐๐ฎ = ๐๐ฎ, the characteristic equation is,
๐et (๐ โ ๐๐) = ๐3 โ ๐2๐1 + ๐๐2 โ ๐3 = 0
where ๐1, ๐2(= ๐1) and ๐3(= 1) are the first second and third invariants of the
orthogonal tensor respectively. In this particular case, the characteristic equation
becomes,
๐3 โ ๐2๐1 + ๐๐1 โ 1 = 0
Which is obviously satisfied by ๐ = 1. Hence there is always a solution to ๐๐ฎ = ๐ฎ.
106. If for an arbitrary unit vector ๐, the tensor, ๐(๐) = cos (๐)๐ + (1 โ cos (๐))๐ โ
๐ + sin (๐)(๐ ร) where (๐ ร) is the vector cross of ๐. Show that ๐(๐)(๐ โ ๐ โ ๐) =
cos(๐)(๐ โ ๐ โ ๐) + sin(๐)(๐ ร) We first observe that,
๐(๐)(๐ โ ๐) = cos(๐)(๐ โ ๐) + (1 โ cos(๐))๐ โ ๐ + sin(๐)[๐ ร (๐ โ ๐)]
The last term vanishes immediately on account of the fact that ๐ โ ๐ is a symmetric
tensor. (The contraction of a symmetric and an antisymmetric tensor always
vanishes). Consequently, we have,
๐(๐)(๐ โ ๐) = cos(๐)(๐ โ ๐) + (1 โ cos(๐))๐ โ ๐ = ๐ โ ๐
which again means that ๐(๐) has no effect on ๐ โ ๐ so that,
๐(๐)(๐ โ ๐ โ ๐) = cos(๐)๐ + (1 โ cos(๐))๐ โ ๐ + sin(๐)(๐ ร) โ ๐ โ ๐
= cos(๐) (๐ โ ๐ โ ๐) + sin(๐)(๐ ร)
as required.
107. For an arbitrary unit vector ๐, show that the skew tensor, ๐ = (๐ ร) is such that
๐2 โก (๐ ร)(๐ ร) = (๐ โ ๐) โ ๐
(๐ ร)(๐ ร) = (๐๐๐๐๐๐๐ ๐ โ ๐ ๐)(๐๐ผ๐ฝ๐พ๐๐ฝ๐ ๐ผ โ ๐ ๐พ)
= ๐ฟ๐ผ๐ฝ๐พ๐๐๐
๐๐๐๐ฝ๐ ๐ โ ๐ ๐พ๐ฟ๐
๐ผ
= ๐ฟ๐๐ฝ๐พ๐๐๐
๐๐๐๐ฝ๐ ๐ โ ๐ ๐พ
= (๐ฟ๐ฝ๐ ๐ฟ๐พ
๐โ ๐ฟ๐ฝ
๐๐ฟ๐พ
๐) ๐๐๐๐ฝ๐ ๐ โ ๐ ๐พ
= ๐๐พ๐๐ฝ๐ ๐ฝ โ ๐ ๐พ โ ๐๐ฝ๐๐ฝ๐ ๐ โ ๐ ๐
= ๐๐พ๐๐ฝ๐ ๐ฝ โ ๐ ๐พ โ (๐ โ ๐)๐ ๐ โ ๐ ๐
= (๐ โ ๐) โ ๐
upon noting that the dot product of the unit vector with itself is unity.
108. If for an arbitrary unit vector ๐, the tensor, ๐(๐) = cos (๐)๐ + (1 โ cos (๐))๐ โ
๐ + sin(๐)(๐ ร) where (๐ ร) โก ๐ is the vector cross of ๐. Show that for, ๐(๐) =
๐ + ๐sin ๐ + ๐2(1 โ cos ๐) [Note that ๐ โ ๐ = ๐2 + ๐] Using the noted result,
๐(๐) = cos ๐ ๐ + (1 โ cos ๐)๐ โ ๐ + sin ๐ (๐ ร)
= cos ๐ ๐ + (1 โ cos ๐)(๐2 + ๐) + ๐sin ๐
= ๐ + ๐sin ๐ + ๐2(1 โ cos ๐)
109. Use the fact that the tensor ๐(๐) = ๐ + ๐sin ๐ + ๐2(1 โ cos ๐) where ๐ โก
(๐ ร) - the vector cross of the unit tensor, rotates every vector about the axis of ๐
by the angle ๐ to find the tensor that rotates {๐1, ๐2, ๐3} to {๐๐, โ๐1, ๐3}.
Clearly, the rotation axis here is the unit vector ๐3 and the angle of rotation is ๐
2.
Consequently, since ๐3 = {0,0,1},
๐ โก (๐3 ร) = (0 โ1 01 0 00 0 0
), and ๐2 = (โ1 0 00 โ1 00 0 0
)
๐(๐
2) = ๐ + ๐sin
๐
2+ ๐2 (1 โ cos
๐
2)
= (1 0 00 1 00 0 1
) + (0 โ1 01 0 00 0 0
) + (โ1 0 00 โ1 00 0 0
)
= (0 โ1 01 0 00 0 1
)
This same tensor can be found directly by recognizing that the tensor, ๐ = ๐1 โ
๐1 + ๐2 โ ๐2 + ๐3 โ ๐3 rotates {๐1, ๐2, ๐3} to {๐1, ๐2, ๐3} so that the tensor we seek
is,
๐ = ๐2 โ ๐1 โ ๐1 โ ๐2 + ๐3 โ ๐3 = (0 โ1 01 0 00 0 1
)
110. Given that ๐1 = {1,0,0}, ๐2 = {0,1,0}, ๐3 = {0,0,1}, ๐4 = {โ3
4,1
4,โ3
2} , ๐5 =
{3
4,โ3
4, โ
1
2} , ๐6 = {โ
1
2,โ3
2, 0}, Find the tensor that transforms from {๐๐, ๐1, โ๐3} to
{๐4, ๐5, ๐6}.
Tensor, ๐1 โ ๐1 + ๐2 โ ๐2 + ๐3 โ ๐3 rotates {๐1, ๐2, ๐3} to {๐1, ๐2, ๐3}. The tensor
we seek is,
๐ = ๐4 โ ๐2 + ๐5 โ ๐1 โ ๐6 โ ๐3
=
(
3
4
โ3
4
1
2
โ3
4
1
4โ
โ3
2
โ1
2
โ3
20 )
111. Find the rotation tensor around an axis parallel to the unit vector, ๐ =
{โ1
โ6,
1
โ6,
2
โ6} through an angle
๐
3.
The skew tensor (๐ ร) = ๐ =
(
0 โโ2
3
1
โ6
โ2
30
1
โ6
โ1
โ6โ
1
โ60)
.
(๐ ร)2 = ๐2 =
(
0 โโ
2
3
1
โ6
โ2
30
1
โ6
โ1
โ6โ
1
โ60
)
(
0 โโ
2
3
1
โ6
โ2
30
1
โ6
โ1
โ6โ
1
โ60
)
=
(
โ5
6โ
1
6โ
1
3
โ1
6โ
5
6
1
3
โ1
3
1
3โ
1
3)
๐(๐
6) = ๐ + ๐sin
๐
6+ ๐2 (1 โ cos
๐
6)
=
(
1 โ5
6(1 โ
โ3
2) โ
1
โ6+
1
6(โ1 +
โ3
2)
1
2โ6+
1
3(โ1 +
โ3
2)
1
โ6+
1
6(โ1 +
โ3
2) 1 โ
5
6(1 โ
โ3
2)
1
2โ6+
1
3(1 โ
โ3
2)
โ1
2โ6+
1
3(โ1 +
โ3
2) โ
1
2โ6+
1
3(1 โ
โ3
2) 1 +
1
3(โ1 +
โ3
2)
)
= (0.888354 โ0.430577 0.1594650.385919 0.888354 0.248782
โ0.248782 โ0.159465 0.955341)
The inverse of this tensor is its transpose and its determinant is unity. Clearly, it is
the rotation tensor we seek.
112. Given the unit vector, ๐ฐ = sin ๐ฝ cos ๐ผ ๐1 + sin ๐ฝ sin ๐ผ ๐2 + cos ๐ฝ ๐3. Find its
vector cross, ๐ โก (๐ฐ ร) and use the formula ๐(๐) = ๐ + ๐sin ๐ + ๐2(1 โ cos ๐)
to determine the rotation tensor around the bisector of the ๐1 โ ๐2 axis through an
angle ๐.
๐(๐ผ, ๐ฝ) = (๐ฐ ร) = (
0 โ cos ๐ฝ sin ๐ฝ sin ๐ผcos ๐ฝ 0 โ sin ๐ฝ cos ๐ผ
โ sin ๐ฝ sin ๐ผ sin ๐ฝ cos ๐ผ 0)
Along the bisector of the ๐1 โ ๐2 axis, ๐ผ =๐
4, ๐ฝ =
๐
2. Consequently, ๐ฐ =
1
โ2๐1 +
1
โ2๐2.
๐ (๐
4,๐
2) = (๐ฐ ร) =
(
0 01
โ2
0 0 โ1
โ2
โ1
โ2
1
โ20
)
, ๐2 (๐
4,๐
2) = (
โ1
2
1
20
1
2โ
1
20
0 0 โ1
)
And the rotation tensor for this axis is,
๐(๐) = ๐ + ๐sin ๐ + ๐2(1 โ cos ๐)
=
(
1
2(1 + cos ๐)
1
2(1 โ cos ๐)
sin ๐
โ21
2(1 โ cos ๐)
1
2(1 + cos ๐) โ
sin ๐
โ2
โsin ๐
โ2
sin ๐
โ2cos ๐
)
.
113. Given the unit vector, ๐ฐ = sin ๐ฝ cos ๐ผ ๐1 + sin ๐ฝ sin ๐ผ ๐2 + cos ๐ฝ ๐3. Find its
vector cross, ๐ โก (๐ฐ ร) and use the formula ๐(๐) = ๐ + ๐sin ๐ + ๐2(1 โ cos ๐)
to determine the general rotation through an angle ๐.
๐(๐ผ, ๐ฝ) = (๐ฐ ร) = (
0 โ cos ๐ฝ sin ๐ฝ sin ๐ผcos ๐ฝ 0 โ sin ๐ฝ cos ๐ผ
โ sin ๐ฝ sin ๐ผ sin ๐ฝ cos ๐ผ 0)
๐(๐ผ, ๐ฝ, ๐) = ๐ + ๐(๐ผ, ๐ฝ) sin ๐ + ๐2(๐ผ, ๐ฝ)(1 โ cos ๐) =
๐(๐ผ, ๐ฝ, ๐) Row 1:
{(1 โ cos (๐))(โsin2(๐ผ)sin2(๐ฝ) โ cos2(๐ฝ)) + 1,
sin(๐ผ) cos(๐ผ) sin2(๐ฝ)(1 โ cos(๐)) โ cos(๐ฝ) sin(๐) ,
sin (๐ผ)sin (๐ฝ)sin (๐) + cos (๐ผ)sin (๐ฝ)cos (๐ฝ)(1 โ cos (๐))}
๐(๐ผ, ๐ฝ, ๐) Row 2:
{sin (๐ผ)cos (๐ผ)sin2(๐ฝ)(1 โ cos (๐)) + cos (๐ฝ)sin (๐),
(1 โ cos(๐))(โcos2(๐ผ)sin2(๐ฝ) โ cos2(๐ฝ)) + 1,
sin (๐ผ)sin (๐ฝ)cos (๐ฝ)(1 โ cos (๐)) โ cos (๐ผ)sin (๐ฝ)sin (๐)}
๐(๐ผ, ๐ฝ, ๐) Row 3
{cos (๐ผ)sin (๐ฝ)cos (๐ฝ)(1 โ cos (๐)) โ sin (๐ผ)sin (๐ฝ)sin (๐),
sin(๐ผ) sin(๐ฝ) cos(๐ฝ) (1 โ cos(๐)) + cos(๐ผ) sin(๐ฝ) sin(๐) ,
(1 โ cos (๐))(โsin2(๐ผ)sin2(๐ฝ) โ cos2(๐ผ)sin2(๐ฝ)) + 1}
114. Given that the skew tensor (๐ ร) โก ๐, and that ๐(๐) โก ๐ + ๐sin ๐ +
๐2(1 โ cos ๐) is the rotation along the axis ๐ through the angle ๐, Find out if the
set {๐,๐,๐2} is linearly independent.
First note that ๐ is antisymmetric but ๐2 = (๐ โ ๐) โ ๐ is the linear combination
of two symmetric tensors, and therefore symmetric. Assume that {๐,๐,๐2} to be
linearly dependent. It means we can find ๐ผ, ๐ฝ and ๐พ not all equal to zero such that
๐ผ๐ + ๐ฝ๐ + ๐พ๐2 = 0
Since ๐ผ, ๐ฝ and ๐พ are not all equal to zero, we assume in particular that ๐ฝ โ 0.
Consequently, we can write,
๐ = โ๐ผ
๐ฝ๐ โ
๐พ
๐ฝ๐2
In which we have expressed the anti-symmetric tensor ๐ as a linear combination of
two symmetric tensors! A contradiction! We can conclude that the set {๐,๐,๐2} is
linearly independent.
115. Given that every rotation tensor ๐ can be expressed in terms of the skew tensor
๐(โก ๐ ร) as a function of the rotation angle ๐: ๐(๐) = ๐ + ๐sin ๐ +
๐2(1 โ cos ๐) and that {๐,๐,๐2} is linearly independent set of tensors, show that,
{๐, ๐, ๐T} is also a linearly independent set.
Assume that the tensor set, {๐, ๐, ๐T} is linearly dependent. It means we can find ๐ผ, ๐ฝ
and ๐พ not all equal to zero such that
๐ผ๐ + ๐ฝ๐ + ๐พ๐T = 0
Since ๐(๐) = ๐ + ๐sin ๐ + ๐2(1 โ cos ๐), we substitute and obtain,
๐ผ๐ + ๐ฝ๐ + ๐พ๐T =
= ๐ผ๐ + ๐ฝ(๐ + ๐ sin ๐ + ๐2(1 โ cos ๐)) + ๐พ(๐ โ ๐sin ๐ + ๐2(1 โ cos ๐))
= (๐ผ + ๐ฝ + ๐พ)๐ + ( ๐ฝ โ ๐พ) ๐ sin ๐ + ( ๐ฝ + ๐พ)๐2(1 โ cos ๐)
= ๐๐ + ๐๐ + ๐๐2 = 0
if we write (๐ผ + ๐ฝ + ๐พ) = ๐, ( ๐ฝ โ ๐พ) sin ๐ = ๐ and ( ๐ฝ + ๐พ)(1 โ cos ๐) = ๐ thereby
contradicting the well-known fact that {๐,๐,๐2} is a linearly independent set.
116. If for an arbitrary unit vector ๐, the tensor, ๐(๐) = cos(๐)๐ + (1 โ cos(๐))๐ โ
๐ + sin(๐)(๐ ร) where (๐ ร) is the vector cross of ๐. Given that for any vector ๐ฎ,
the vector ๐ฏ โก ๐(๐) ๐ฎ has the same magnitude as ๐ฎ, and that, for any scalar ๐ผ,
๐(๐)(๐ผ๐) = ๐ผ๐, What is the physical meaning of ๐(๐)?
๐(๐) is a rotation about the vector ๐ counterclockwise through an angle ๐. It
therefore does not alter the magnitude or direction of any vector in the direction of
๐; for any other vector, it has no effect on the magnitude but affects direction.
117. If for an arbitrary unit vector ๐, the tensor, ๐(๐) = cos(๐)๐ + (1 โ cos(๐))๐ โ
๐ + sin(๐)(๐ ร) where (๐ ร) is the vector cross of ๐. Show that for any vector ๐ฎ,
the vector ๐ฏ โก ๐(๐) ๐ฎ has the same magnitude as ๐ฎ. What is the physical meaning
of ๐(๐)?
Let the scalar ๐ฅ โก ๐ โ ๐ฎ be the projection of ๐ฎ onto the unit vector ๐. The square of
the magnitude of ๐ฏ is |๐ฏ|๐
= ๐ฏ โ ๐ฏ = ( cos๐(๐๐ฎ) + (1 โ cos๐)(๐ โ ๐)๐ฎ + sin๐(๐ ร ๐ฎ))
โ ( cos๐(๐๐ฎ) + (1 โ cos๐)(๐ โ ๐)๐ฎ + sin๐(๐ ร ๐ฎ)) โ
= (๐ฎ cos๐ + (1 โ cos๐)๐ฅ๐ + sin๐(๐ ร ๐ฎ))2
= (๐ฎ cos๐) โ (๐ฎ cos๐ + (1 โ cos๐)๐ฅ๐ + sin๐(๐ ร ๐ฎ))
+๐ฅ๐ โ (๐ฎ cos๐ + (1 โ cos๐)๐ฅ๐ + sin๐(๐ ร ๐ฎ))(1 โ cos๐)
+(๐ ร ๐ฎ) โ (๐ฎ cos๐ + (1 โ cos๐)๐ฅ๐ + sin๐(๐ ร ๐ฎ))sin๐
= ๐ฎ๐ cos2 ๐ + 2(cos๐ โ cos2 ๐)๐ฅ2 + 2(๐ ร ๐ฎ โ ๐ฎ)sin๐ cos๐ + (1 โ cos๐)2๐ฅ2
+ 2๐ฅ(๐ ร ๐ฎ โ ๐)(1 โ cos๐) sin๐ + sin2 ๐ (๐ ร ๐ฎ)2
= ๐ฎ๐ cos2 ๐ + 2(cos๐ โ cos2 ๐)๐ฅ2 + 2(๐ ร ๐ฎ โ ๐ฎ)sin๐ cos๐ + (1 โ cos๐)2๐ฅ2
+ 2๐ฅ(๐ ร ๐ฎ โ ๐)(1 โ cos๐) sin๐ + sin2 ๐ (๐ฎ2 โ ๐ฅ2)
= ๐ฎ๐(cos2 ๐ + sin2 ๐) + ๐ฅ2[2(cos๐ โ cos2 ๐) + (1 โ cos๐)2 โ sin2 ๐]
= ๐ฎ๐
As the term in square brackets vanish when expanded.
118. If for an arbitrary unit vector ๐, the tensor, ๐(๐) = cos(๐)๐ + (1 โ cos(๐))๐ โ
๐ + sin(๐)(๐ ร) where (๐ ร) is the vector cross of ๐. Show that for arbitrary 0 <
๐ผ, ๐ฝ โค 2๐, that ๐(๐ผ + ๐ฝ) = ๐(๐ผ)๐(๐ฝ).
It is convenient to write ๐(๐ผ) and ๐(๐ฝ) in terms of their ๐, ๐ components; we assume
that the unit vector ๐ = (๐ฅ1, ๐ฅ2, ๐ฅ3):
[๐(๐ผ)]๐๐ = cos๐ผ ๐ฟ๐๐ + (1 โ cos ๐ผ)๐ฅ๐๐ฅ๐ โ sin ๐ผ ๐๐๐๐๐ฅ๐
Consequently, we can write for the product ๐(๐ผ)๐(๐ฝ),
[๐(๐ผ)๐(๐ฝ)]๐๐ = [๐(๐ผ)]๐๐[๐(๐ฝ)]๐๐ =
= [cos๐ผ ๐ฟ๐๐ + (1 โ cos ๐ผ)๐ฅ๐๐ฅ๐ โ sin ๐ผ ๐๐๐๐๐ฅ๐][cos๐ฝ ๐ฟ๐๐ + (1 โ cos ๐ฝ)๐ฅ๐๐ฅ๐
โ sin ๐ฝ ๐๐๐๐๐ฅ๐]
= cos๐ผ cos ๐ฝ ๐ฟ๐๐๐ฟ๐๐ + cos๐ผ (1 โ cos ๐ฝ)๐ฟ๐๐๐ฅ๐๐ฅ๐ โ cos๐ผ sin ๐ฝ ๐ฟ๐๐๐๐๐๐๐ฅ๐
+ (1 โ cos ๐ผ) cos ๐ฝ ๐ฅ๐๐ฅ๐๐ฟ๐๐ + (1 โ cos ๐ผ)(1 โ cos ๐ฝ)๐ฅ๐๐ฅ๐2๐ฅ๐
โ (1 โ cos ๐ผ) sin ๐ฝ ๐ฅ๐๐ฅ๐๐ฅ๐๐๐๐๐ โ sin ๐ผ cos ๐ฝ ๐๐๐๐๐ฅ๐๐ฟ๐๐
โ sin ๐ผ (1 โ cos ๐ฝ)๐๐๐๐๐ฅ๐๐ฅ๐๐ฅ๐ + sin ๐ผ sin ๐ฝ ๐๐๐๐๐๐๐๐๐ฅ๐๐ฅ๐
= cos๐ผ cos ๐ฝ ๐ฟ๐๐ + cos ๐ผ (1 โ cos ๐ฝ)๐ฅ๐๐ฅ๐ โ cos๐ผ sin ๐ฝ ๐๐๐๐๐ฅ๐ + (1 โ cos ๐ผ) cos ๐ฝ ๐ฅ๐๐ฅ๐
+ (1 โ cos ๐ผ)(1 โ cos ๐ฝ)๐ฅ๐๐ฅ๐ โ (1 โ cos ๐ผ) sin ๐ฝ ๐ฅ๐๐ฅ๐๐ฅ๐๐๐๐๐
โ sin ๐ผ cos ๐ฝ ๐๐๐๐๐ฅ๐ โ sin ๐ผ (1 โ cos ๐ฝ)๐๐๐๐๐ฅ๐๐ฅ๐๐ฅ๐ + sin ๐ผ sin ๐ฝ ๐๐๐๐๐๐๐๐๐ฅ๐๐ฅ๐
= cos๐ผ cos ๐ฝ ๐ฟ๐๐ + cos ๐ผ (1 โ cos ๐ฝ)๐ฅ๐๐ฅ๐ โ cos๐ผ sin ๐ฝ ๐๐๐๐๐ฅ๐ + (1 โ cos ๐ผ) cos ๐ฝ ๐ฅ๐๐ฅ๐
+ (1 โ cos ๐ผ)(1 โ cos ๐ฝ)๐ฅ๐๐ฅ๐ โ (1 โ cos ๐ผ) sin ๐ฝ ๐ฅ๐๐ฅ๐๐ฅ๐๐๐๐๐
โ sin ๐ผ cos ๐ฝ ๐๐๐๐๐ฅ๐ โ sin ๐ผ (1 โ cos ๐ฝ)๐๐๐๐๐ฅ๐๐ฅ๐๐ฅ๐
+ sin ๐ผ sin ๐ฝ (๐ฟ๐๐๐ฟ๐๐ โ ๐ฟ๐๐๐ฟ๐๐)๐ฅ๐๐ฅ๐
= (cos ๐ผ cos ๐ฝ โ sin ๐ผ sin ๐ฝ)๐ฟ๐๐ + [1 โ (cos ๐ผ cos ๐ฝ โ sin ๐ผ sin ๐ฝ)]๐ฅ๐๐ฅ๐
โ [(cos ๐ผ sin ๐ฝ โ sin ๐ผ cos ๐ฝ)]๐๐๐๐๐ฅ๐
= [๐(๐ผ + ๐ฝ)]๐๐
With the boxed terms vanishing on account of antisymmetric contraction with
symmetry.
119. If for an arbitrary unit vector ๐, the tensor, ๐(๐) = cos(๐)๐ + (1 โ cos(๐))๐ โ
๐ + sin(๐)(๐ ร) where (๐ ร) is the vector cross of ๐. Show that . ๐(๐) is a periodic
tensor function with period 2๐. [Hint: ๐(๐ผ + ๐ฝ) = ๐(๐ผ)๐(๐ฝ)
Since ๐(๐ผ + ๐ฝ) = ๐(๐ผ)๐(๐ฝ) we can write that ๐(๐ผ + 2๐) = ๐(๐ผ)๐(2๐). But a
direct substitution shows that, ๐(0) = ๐(2ฯ) = ๐. We therefore have that,
๐(๐ผ + 2๐) = ๐(๐ผ)๐(2๐) = ๐(๐ผ)
which completes the proof. The above results show that ๐(๐ผ) is a rotation along the
unit vector ๐ through an angle ๐ผ.
120. Given that ๐ is an orthogonal tensor, show that the principal invariants of a
tensor ๐ satisfy ๐ผ๐(๐๐๐T) = ๐ผ๐(๐), ๐ = 1,2, or 3, that is, Rotations and orthogonal
transformations do not change the Invariants.
๐ผ1(๐๐๐T) = tr(๐๐๐T)
= tr(๐T๐๐) = tr(๐)
= ๐ผ1(๐)
๐ผ2(๐๐๐T) =1
2[tr2(๐๐๐T) โ tr(๐๐๐T๐๐๐T)]
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