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Homework 3 and solutions for February.
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1. 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
2. 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 ๐๐ฟ๐๐ = 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.
3. 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
4. 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
5. 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
= (โ๐) ร ๐ฏ + ๐ curl ๐ฏ
6. 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)๐
๐ ๐๐ฟ๐๐ = (๐ข๐๐ฃ๐)
๐๐ ๐
= ๐ข๐ ,๐ ๐ฃ๐๐ ๐ + ๐ข๐๐ฃ๐ ,๐ ๐ ๐
= (grad ๐ฎ)๐ฏ + (div ๐ฏ)๐ฎ
7. For a scalar field ๐ and a tensor field ๐ show that grad (๐๐) = ๐ grad 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)๐ ๐
= (๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ )๐ ๐๐ฟ๐๐
= ๐๐๐ ๐,๐ ๐ ๐ + ๐๐๐๐ ,๐ ๐ ๐
= ๐grad๐ + ๐ div ๐
8. 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 ๐๐๐๐๐ข๐)๐ ๐ โ ๐ ๐
= โ(๐ฏ ร)grad๐ฎ + (๐ฎ ร)grad๐ฏ
9. 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
10. 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
11. Show that curl (๐๐) = ( grad ๐) ร
Note that ๐๐ = (๐๐๐ผ๐ฝ)๐ ๐ผ โ ๐ ๐ฝ, and that curl ๐ป = 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
12. Show that curl (๐ฏ ร) = (div ๐ฏ)๐ โ grad ๐ฏ
(๐ฏ ร) = ๐๐ผ๐ฝ๐๐ฃ๐ฝ ๐ ๐ผ โ ๐ ๐
curl ๐ป = 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
13. 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
14. Given a scalar point function ๐ and a vector field ๐ฏ, show that curl (๐๐ฏ) =
๐ curl ๐ฏ + (โ๐) 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
= (โ๐) ร ๐ฏ + ๐ curl ๐ฏ
15. 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 = ๐.
The contraction of symmetric tensors with unsymmetric 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.
16. 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 unsymmetric 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.
17. Show that curl (grad ๐) = ๐
For a third order tensor ๐ต = ฮ๐ผ๐ฝ๐พ๐ ๐ผ โ ๐ ๐ฝ โ ๐ ๐พ
curl ๐ต = ๐๐๐๐ฮ๐ผ๐ฝ๐ ,๐ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ
Let ๐ต = grad ๐. Clearly, in this case, ฮ๐ผ๐ฝ๐พ = ๐๐ผ๐ฝ ,๐พ so that ฮ๐ผ๐ฝ๐ ,๐ = ๐๐ผ๐ฝ ,๐๐ . It
therefore follows that,
curl (grad ๐) = ๐๐๐๐๐๐ผ๐ฝ ,๐๐ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ = ๐.
The contraction of symmetric tensors with unsymmetric led to this conclusion.
Note that this presupposes that the order of differentiation in the tensor field ๐
is immaterial. This will be true only if the tensor field is continuous โ a
proposition we have assumed in the above.
18. 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
19. Given that ๐ and ๐are scalar fields, show that div (grad ๐ ร grad ๐) = 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.
20. 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 = ๐๐พ๐๐๐ผ๐ฝ๐พ๐๐๐๐๐ฃ๐ ,๐๐ฝ ๐ ๐ผ
Now, it can be shown (see below) that ๐๐พ๐๐๐ผ๐ฝ๐พ๐๐๐๐ = ๐๐ผ๐๐๐ฝ๐ โ ๐๐ผ๐๐๐ฝ๐ so that,
curl ๐ฐ = (๐๐ผ๐๐๐ฝ๐ โ ๐๐ผ๐๐๐ฝ๐)๐ฃ๐,๐๐ฝ ๐ ๐ผ
= ๐ฃ๐ฝ ,๐๐ฝ ๐ ๐ โ ๐ฃ๐ผ ,๐๐ ๐ ๐ผ = grad(div ๐ฏ) โ grad2๐ฏ
21. Show that ๐๐พ๐๐๐ผ๐ฝ๐พ๐๐๐๐ = ๐๐ผ๐๐๐ฝ๐ โ ๐๐ผ๐๐๐ฝ๐
Note that
๐๐พ๐๐๐ผ๐ฝ๐พ๐๐๐๐ = ๐๐พ๐ |
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
|
= |
๐๐พ๐๐๐๐ผ ๐๐พ๐๐๐๐ฝ ๐๐พ๐๐๐๐พ
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
| = |
๐ฟ๐พ๐ผ ๐ฟ๐พ
๐ฝ๐ฟ๐พ
๐พ
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
๐๐๐ผ ๐๐๐ฝ ๐๐๐พ
|
= ๐ฟ๐พ๐ผ |
๐๐๐ฝ ๐๐๐พ
๐๐๐ฝ ๐๐๐พ| โ ๐ฟ๐พ
๐ฝ|๐๐๐ผ ๐๐๐พ
๐๐๐ผ ๐๐๐พ| + ๐ฟ๐พ๐พ
|๐๐๐ผ ๐๐๐ฝ
๐๐๐ผ ๐๐๐ฝ|
= |๐๐๐ฝ ๐๐๐ผ
๐๐๐ฝ ๐๐๐ผ| โ |
๐๐๐ผ ๐๐๐ฝ
๐๐๐ผ ๐๐๐ฝ| + 3 |
๐๐๐ผ ๐๐๐ฝ
๐๐๐ผ ๐๐๐ฝ|
= |๐๐๐ผ ๐๐๐ฝ
๐๐๐ผ ๐๐๐ฝ| = ๐๐ผ๐๐๐ฝ๐ โ ๐๐ผ๐๐๐ฝ๐
22. 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.
23. Given a tensor field ๐, obtain the vector ๐ฐ โก ๐T๐ and show that its divergence
is ๐: (โ๐ฏ) + ๐ฏ โ div ๐
The divergence of ๐ is the scalar sum , (๐ป๐๐๐๐),๐. Expanding the product covariant
derivative we obtain,
div (๐T๐ฏ) = (๐๐๐๐ฃ๐),๐ = ๐๐๐ ,๐ ๐ฃ๐ + ๐๐๐๐ฃ๐ ,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
24. 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 โ (๐)
= ๐๐๐๐๐๐ผ๐ ,๐ ๐๐ฝ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ
= ๐๐๐๐๐๐ผ๐ ,๐ ๐๐ฝ๐ ๐๐ฟ๐ฝ๐ผ
= 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)
25. For any two vectors ๐ฎ and ๐ฏ, show that curl (๐ฎ โ ๐ฏ) = [(grad ๐ฎ)๐ฏ ร]๐ +
(curl ๐ฏ) โ ๐ฎ 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 (๐ โ ๐) = 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.
26. 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)๐ ๐ผ = (๐ข๐๐ฃ๐),๐ ๐ ๐.
27. Given a scalar point function ๐ and a second-order tensor field ๐, show that
curl (๐๐) = ๐ curl ๐ + ((โ๐) ร)๐T where [(โ๐) ร] is the skew tensor
๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐
curl (๐๐ป) 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 + ((โ๐) ร)๐T
28. 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 ๐ป) = ๐๐๐๐๐๐ผ๐ ,๐๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ = ๐๐๐๐๐๐ผ๐ ,๐๐ฝ ๐ ๐ ๐๐ผ๐ฝ
= ๐๐๐๐๐๐ฝ๐ ,๐๐ฝ ๐ ๐ = curl(div ๐T)
29. Show that if ๐ defined in the space spanned by orthogonal coordinates ๐ฅ๐, then
โ2(๐ฅ๐๐) = 2๐๐
๐๐ฅ๐ + ๐ฅ๐โ2๐ .
By definition, โ2(๐ฅ๐๐) = ๐๐๐(๐ฅ๐๐),๐๐
. Expanding, we have
๐๐๐(๐ฅ๐๐),๐๐
= ๐๐๐(๐ฅ๐,๐๐ + ๐ฅ๐๐,๐)
,๐= ๐๐๐(๐ฟ๐
๐๐ + ๐ฅ๐๐,๐),๐
= ๐๐๐(๐ฟ๐๐๐,๐ + ๐ฅ๐
,๐๐,๐ + ๐ฅ๐๐,๐๐)
= ๐๐๐(๐ฟ๐๐๐,๐ + ๐ฟ๐
๐ ๐,๐ + ๐ฅ๐๐,๐๐)
= ๐๐๐๐,๐ + ๐๐๐๐,๐ + ๐ฅ๐๐๐๐๐,๐๐
When the coordinates are orthogonal, this becomes,
2
(โ๐)2
๐ฮฆ
๐๐ฅ๐+ ๐ฅ๐โ2ฮฆ
where we have suspended the summation rule and โ๐ is the square root of the
appropriate metric tensor component.
30. In Cartesian coordinates, If the volume ๐ is enclosed by the surface ๐, the
position vector ๐ = ๐ฅ๐๐ ๐ and ๐ is the external unit normal to each surface element,
show that 1
6โซ โ(๐ โ ๐) โ ๐๐๐
๐ equals the volume contained in ๐.
๐ โ ๐ = ๐ฅ๐๐ฅ๐๐ ๐ โ ๐ ๐ = ๐ฅ๐๐ฅ๐๐๐๐
By the Divergence Theorem,
โซโ(๐ โ ๐) โ ๐๐๐๐
= โซ โ โ [โ(๐ โ ๐)]๐๐๐
= โซ ๐๐[๐๐(๐ฅ๐๐ฅ๐๐๐๐)] ๐ ๐ โ ๐ ๐ ๐๐๐
= โซ ๐๐[๐๐๐(๐ฅ๐ ,๐ ๐ฅ๐ + ๐ฅ๐๐ฅ๐ ,๐ )] ๐ ๐ โ ๐ ๐ ๐๐๐
= โซ ๐๐๐๐๐๐(๐ฟ๐๐ ๐ฅ๐ + ๐ฅ๐๐ฟ๐
๐),๐ ๐๐
๐
= โซ 2๐๐๐๐๐๐๐ฅ๐ ,๐ ๐๐๐
= โซ 2๐ฟ๐๐๐ฟ๐
๐ ๐๐๐
= 6 โซ ๐๐๐
31. For any Euclidean coordinate system, show that div ๐ฎ ร ๐ฏ = ๐ฏ curl ๐ฎ โ
๐ฎ curl ๐ฏ
Given the contravariant vector ๐ข๐ and ๐ฃ๐ with their associated vectors ๐ข๐ and ๐ฃ๐,
the contravariant component of the above cross product is ๐๐๐๐๐ข๐๐ฃ๐ .The
required divergence is simply the contraction of the covariant ๐ฅ๐ derivative of this
quantity:
(๐๐๐๐๐ข๐๐ฃ๐),๐
= ๐๐๐๐๐ข๐,๐๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐,๐
where we have treated the tensor ํ๐๐๐ as a constant under the covariant
derivative.
Cyclically rearranging the RHS we obtain,
(๐๐๐๐๐ข๐๐ฃ๐),๐
= ๐ฃ๐๐๐๐๐๐ข๐,๐ + ๐ข๐๐๐๐๐๐ฃ๐,๐ = ๐ฃ๐๐๐๐๐๐ข๐,๐ + ๐ข๐๐๐๐๐๐ฃ๐,๐
where we have used the anti-symmetric property of the tensor ๐๐๐๐. The last
expression shows clearly that
div ๐ฎ ร ๐ฏ = ๐ฏ curl ๐ฎ โ ๐ฎ curl ๐ฏ
as required.
32. For a general tensor field ๐ป show that, curl(curl ๐ป) = [โ2(tr ๐ป) โ
div(div ๐ป)]๐ฐ + grad(div ๐ป) + (grad(div ๐ป))T
โ grad(grad (tr๐ป)) โ โ2๐ปT
curl ๐ป = ๐๐ผ๐ ๐ก๐๐ฝ๐ก ,๐ ๐ ๐ผ โ ๐ ๐ฝ
= ๐ .๐ฝ๐ผ ๐ ๐ผ โ ๐ ๐ฝ
curl ๐บ = ๐๐๐๐๐ .๐๐ผ ,๐ ๐ ๐ โ ๐ ๐ผ
so that
curl ๐บ = curl(curl ๐ป) = ๐๐๐๐๐๐ผ๐ ๐ก๐๐๐ก ,๐ ๐ ๐ ๐ โ ๐ ๐ผ
= |
๐๐๐ผ ๐๐๐ ๐๐๐ก
๐๐๐ผ ๐๐๐ ๐๐๐ก
๐๐๐ผ ๐๐๐ ๐๐๐ก
| ๐๐๐ก ,๐ ๐ ๐ ๐ โ ๐ ๐ผ
= [๐๐๐ผ(๐๐๐ ๐๐๐ก โ ๐๐๐ก๐๐๐ ) + ๐๐๐ (๐๐๐ก๐๐๐ผ โ ๐๐๐ผ๐๐๐ก)
+๐๐๐ก(๐๐๐ผ๐๐๐ โ ๐๐๐ ๐๐๐ผ)] ๐๐๐ก ,๐ ๐ ๐ ๐ โ ๐ ๐ผ
= [๐๐๐ ๐ .๐ก๐ก ,๐ ๐โ ๐ ..
๐ ๐ ,๐ ๐ ](๐ ๐ผ โ ๐ ๐ผ) + [๐ ..๐ผ๐ ,๐ ๐โ ๐๐๐ผ๐ .๐ก
๐ก ,๐ ๐ ](๐ ๐ โ ๐ ๐ผ)
+ [๐๐๐ผ๐ .๐ก๐ .,๐ ๐โ ๐๐๐ ๐ .๐ก
๐ผ.,๐ ๐ ](๐ ๐ก โ ๐ ๐ผ)
= [โ2(tr ๐ป) โ div(div ๐ป)]๐ฐ + (grad(div ๐ป))T
โ grad(grad (tr๐))
+ (grad(div ๐)) โ โ2๐T
33. When ๐ is symmetric, show that tr(curl ๐) vanishes.
curl ๐ = ๐๐๐๐๐๐ฝ๐ ,๐ ๐ ๐ โ ๐ ๐ฝ
tr(curl ๐) = ๐๐๐๐๐๐ฝ๐ ,๐ ๐ ๐ โ ๐ ๐ฝ
= ๐๐๐๐๐๐ฝ๐ ,๐ ๐ฟ๐๐ฝ
= ๐๐๐๐๐๐๐ ,๐
which obviously vanishes on account of the symmetry and antisymmetry in ๐ and
๐. In this case,
curl(curl ๐)
= [โ2(tr ๐) โ div(div ๐)]๐ โ grad(grad (tr๐ป)) + 2(grad(div ๐ป))
โ โ2๐
as (grad(div ๐))T
= grad(div ๐) if the order of differentiation is immaterial and
๐ is symmetric.
34. For a scalar function ๐ท and a vector ๐๐ show that the divergence of the vector
๐๐๐ฝ is equal to, ๐ฏ โ ๐๐ท + ๐ท ๐๐๐ฃ ๐ฏ
(๐ฃ๐ฮฆ),๐
= ฮฆ๐ฃ๐,๐ + ๐ฃ๐ฮฆ,i
Hence the result.
35. Show that curl ๐ฎ ร ๐ฏ = (๐ฏ โ ๐๐ฎ) + (๐ฎ โ div ๐ฏ) โ (๐ฏ โ div ๐ฎ) โ (๐ฎ โ ๐ ๐ฏ)
Taking the associated (covariant) vector of the expression for the cross product in
the last example, it is straightforward to see that the LHS in indicial notation is,
๐๐๐๐(๐๐๐๐๐ข๐๐ฃ๐),๐
Expanding in the usual way, noting the relation between the alternating tensors
and the Kronecker deltas,
๐๐๐๐(ํ๐๐๐๐ข๐๐ฃ๐),๐
= ๐ฟ๐๐๐๐๐๐(๐ข๐
,๐๐ฃ๐ โ ๐ข๐๐ฃ๐,๐)
= ๐ฟ๐๐๐๐(๐ข๐
,๐๐ฃ๐ โ ๐ข๐๐ฃ๐,๐) = |
๐ฟ๐๐ ๐ฟ๐
๐
๐ฟ๐๐ ๐ฟ๐
๐| (๐ข๐
,๐๐ฃ๐ โ ๐ข๐๐ฃ๐,๐)
= (๐ฟ๐๐๐ฟ๐
๐ โ ๐ฟ๐๐ ๐ฟ๐
๐)(๐ข๐,๐๐ฃ๐ โ ๐ข๐๐ฃ๐
,๐)
= ๐ฟ๐๐๐ฟ๐
๐๐ข๐,๐๐ฃ๐ โ ๐ฟ๐
๐๐ฟ๐๐๐ข๐๐ฃ๐
,๐ + ๐ฟ๐๐ ๐ฟ๐
๐๐ข๐,๐๐ฃ๐ โ ๐ฟ๐
๐ ๐ฟ๐๐๐ข๐๐ฃ๐
,๐
= ๐ข๐,๐๐ฃ๐ โ ๐ข๐
,๐๐ฃ๐ + ๐ข๐๐ฃ๐,๐ โ ๐ข๐๐ฃ๐
,๐
Which is the result we seek in indicial notation.
36. . In Cartesian coordinates let ๐ฅ denote the magnitude of the position vector ๐ซ =
๐ฅ๐๐๐. Show that (a) ๐ฅ,๐ =๐๐
๐, (b) ๐ฅ,๐๐ =
๐
๐๐ฟ๐๐ โ
๐๐๐๐
(๐)๐, (c) ๐ฅ,๐๐ =๐
๐, (d) If ๐ผ =
๐
๐, then ๐ผ,๐๐ =
โ๐น๐๐
๐๐ +๐๐๐๐๐
๐๐ ๐ผ,๐๐ = ๐ and div (๐ซ
๐ฅ) =
2
๐ฅ.
(๐) ๐ฅ = โ๐ฅ๐๐ฅ๐
๐ฅ,๐ =๐โ๐ฅ๐๐ฅ๐
๐๐ฅ๐=
๐โ๐ฅ๐๐ฅ๐
๐(๐ฅ๐๐ฅ๐)ร
๐(๐ฅ๐๐ฅ๐)
๐๐ฅ๐=
1
2โ๐ฅ๐๐ฅ๐
[๐ฅ๐๐ฟ๐๐ + ๐ฅ๐๐ฟ๐๐] =๐ฅ๐
๐ฅ.
(๐) ๐ฅ,๐๐ =๐
๐๐ฅ๐(
๐โ๐ฅ๐๐ฅ๐
๐๐ฅ๐) =
๐
๐๐ฅ๐(
๐ฅ๐
๐ฅ) =
๐ฅ๐๐ฅ๐
๐๐ฅ๐โ ๐ฅ๐
๐๐ฅ๐๐ฅ๐
(๐ฅ)2=
๐ฅ๐ฟ๐๐ โ๐ฅ๐๐ฅ๐
๐ฅ(๐ฅ)2
=1
๐ฅ๐ฟ๐๐ โ
๐ฅ๐๐ฅ๐
(๐ฅ)3
(๐) ๐ฅ,๐๐ =1
๐ฅ๐ฟ๐๐ โ
๐ฅ๐๐ฅ๐
(๐ฅ)3=
3
๐ฅโ
(๐ฅ)2
(๐ฅ)3=
2
๐ฅ.
(๐) ๐ =1
๐ฅ so that
๐,๐ =๐
1๐ฅ
๐๐ฅ๐=
๐1๐ฅ
๐๐ฅร
๐๐ฅ
๐๐ฅ๐= โ
1
๐ฅ2
1
๐ฅ๐ฅ๐ = โ
๐ฅ๐
๐ฅ3
Consequently,
๐,๐๐ =๐
๐๐ฅ๐
(๐,๐ ) = โ๐
๐๐ฅ๐(
๐ฅ๐
๐ฅ3) =
๐ฅ3 (๐
๐๐ฅ๐(โ๐ฅ2)) + ๐ฅ๐
๐๐๐ฅ๐
(๐ฅ3)
๐ฅ6
=
๐ฅ3(โ๐ฟ๐๐) + ๐ฅ๐ (๐(๐ฅ3)
๐๐ฅ๐๐ฅ๐๐ฅ๐
)
๐ฅ6=
โ๐ฅ3๐ฟ๐๐ + ๐ฅ๐ (3๐ฅ2๐ฅ๐
๐ฅ )
๐ฅ6=
โ๐ฟ๐๐
๐ฅ3+
3๐ฅ๐๐ฅ๐
๐ฅ5
๐,๐๐ =โ๐ฟ๐๐
๐ฅ3+
3๐ฅ๐๐ฅ๐
๐ฅ5=
โ3
๐ฅ3+
3๐ฅ2
๐ฅ5= 0.
div (๐ซ
๐ฅ) = (
๐ฅ๐
๐ฅ) ,๐ =
1
๐ฅ๐ฅ๐ ,๐+ (
1
๐ฅ)
,๐=
3
๐ฅ+ ๐ฅ๐ (
๐
๐๐ฅ(
1
๐ฅ)
๐๐ฅ
๐๐ฅ๐)
=3
๐ฅ+ ๐ฅ๐ [โ (
1
๐ฅ2)
๐ฅ๐
๐ฅ] =
3
๐ฅโ
๐ฅ๐๐ฅ๐
๐ฅ3=
3
๐ฅโ
1
๐ฅ=
2
๐ฅ
37. For vectors ๐ฎ, ๐ฏ and ๐ฐ, show that (๐ฎ ร)(๐ฏ ร)(๐ฐ ร) = ๐ฎ โ(๐ฏ ร ๐ฐ) โ
(๐ฎ โ ๐ฏ)๐ฐ ร.
The tensor (๐ฎ ร) = โ๐๐๐๐๐ข๐๐ ๐ โ ๐ ๐ similarly, (๐ฏ ร) = โ๐๐ผ๐ฝ๐พ๐ฃ๐พ๐ ๐ผ โ ๐ ๐ฝ and
(๐ฐ ร) = โ๐๐๐๐๐ค๐๐ ๐ โ ๐ ๐. Clearly,
(๐ฎ ร)(๐ฏ ร)(๐ฐ ร) = โ๐๐๐๐๐๐ผ๐ฝ๐พ๐๐๐๐๐ข๐๐ฃ๐พ๐ค๐(๐ ๐ผ โ ๐ ๐ฝ)(๐ ๐ โ ๐ ๐)(๐ ๐ โ ๐ ๐)
= โ๐๐ผ๐ฝ๐พ๐๐๐๐๐๐๐๐๐ข๐๐ฃ๐พ๐ค๐(๐ ๐ผ โ ๐ ๐)๐ฟ๐ฝ๐ ๐ฟ๐
๐
= โ๐๐ผ๐๐พ๐๐๐๐๐๐๐๐๐ข๐๐ฃ๐พ๐ค๐(๐ ๐ผ โ ๐ ๐)
= โ๐๐๐ผ๐พ๐๐๐๐๐๐๐๐๐ข๐๐ฃ๐พ๐ค๐(๐ ๐ผ โ ๐ ๐)
= โ(๐ฟ๐๐ผ๐ฟ๐
๐พโ ๐ฟ๐
๐ผ๐ฟ๐๐พ
)๐๐๐๐๐ข๐๐ฃ๐พ๐ค๐(๐ ๐ผ โ ๐ ๐)
= โ๐๐๐๐๐ข๐ผ๐ฃ๐๐ค๐(๐ ๐ผ โ ๐ ๐) + ๐๐๐๐๐ข๐พ๐ฃ๐พ๐ค๐(๐ ๐ โ ๐ ๐)
= [๐ฎ โ (๐ฏ ร ๐ฐ) โ (๐ฎ โ ๐ฏ)๐ฐ ร]
38. Show that [๐ฎ, ๐ฏ, ๐ฐ] = tr[(๐ฎ ร)(๐ฏ ร)(๐ฐ ร)]
In the above we have shown that (๐ฎ ร)(๐ฏ ร)(๐ฐ ร) = [๐ฎ โ (๐ฏ ร ๐ฐ) โ
(๐ฎ โ ๐ฏ)๐ฐ ร]
Because the vector cross is traceless, the trace of [(๐ฎ โ ๐ฏ)๐ฐ ร] = 0. The trace of
the first term, ๐ฎ โ (๐ฏ ร ๐ฐ) is obviously the same as [๐ฎ, ๐ฏ, ๐ฐ] which completes
the proof.
39. Show that (๐ฎ ร)(๐ฏ ร) = (๐ฎ โ ๐ฏ)๐ โ ๐ฎ โ ๐ฏ and that tr[(๐ฎ ร)(๐ฏ ร)] =
2(๐ฎ โ ๐ฏ)
(๐ฎ ร)(๐ฏ ร) = โ๐๐๐๐๐๐ผ๐ฝ๐พ๐ข๐๐ฃ๐พ(๐ ๐ผ โ ๐ ๐ฝ)(๐ ๐ โ ๐ ๐)
= โ๐๐๐๐๐๐ผ๐ฝ๐พ๐ข๐๐ฃ๐พ(๐ ๐ผ โ ๐ ๐)๐ฟ๐ฝ๐ = โ๐๐ฝ๐๐๐๐ฝ๐พ๐ผ๐ข๐๐ฃ๐พ(๐ ๐ผ โ ๐ ๐)
= [๐ฟ๐๐พ
๐ฟ๐๐ผ โ ๐ฟ๐
๐พ๐ฟ๐
๐ผ]๐ข๐๐ฃ๐พ(๐ ๐ผ โ ๐ ๐)
= ๐ข๐๐ฃ๐(๐ ๐ผ โ ๐ ๐ผ) โ ๐ข๐๐ฃ๐(๐ ๐ โ ๐ ๐) = (๐ฎ โ ๐ฏ)๐ โ ๐ฎ โ ๐ฏ
Obviously, the trace of this tensor is 2(๐ฎ โ ๐ฏ)
40. The position vector in the above example ๐ = ๐ฅ๐๐๐. Show that (a) div ๐ = ๐, (b)
div (๐ โ ๐) = ๐๐, (c) div ๐ = 3, and (d) grad ๐ = ๐ and (e) curl (๐ โ ๐) = โ๐ 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
41. 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.
42. Show that ๐๐ฐ3(๐บ)
๐๐บ=
๐det(๐บ)
๐๐บ= ๐บ๐ the cofactor of ๐บ.
Clearly ๐บ๐ = det(๐บ) ๐บโT = ๐ฐ3(๐บ) ๐บโT. Details of this for the contravariant
components of a tensor is presented below. Let
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
๐ผ๐ฟ๐๐ฝ
๐๐๐ ๐๐๐ก + ๐๐๐๐ฟ๐๐ผ๐ฟ๐
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
43. 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
2๐ด๐3 ๐๐
๐)
= ๐๐๐๐ [(๐ด11๐๐
1 + ๐ด21๐๐
2 + ๐ด31๐๐
3 ) ๐๐2๐๐
3 + ๐๐1 ( ๐ด1
2๐๐1 + ๐ด2
2๐๐2
+ ๐ด32๐๐
3 ) ๐๐3 + ๐๐
1๐๐2 ( ๐ด1
3๐๐1 + ๐ด2
3๐๐2 + ๐ด3
3๐๐3)]
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.)
๐
๐๐ผdet(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.
44. Without breaking down into components, establish the fact that ๐det(๐)
๐๐= ๐๐
Start from Liouvilleโs Theorem, given a scalar parameter such that ๐ = ๐(๐ผ),
โ
โ๐ผ(det(๐)) = det(๐) tr [(
โ๐
โ๐ผ) ๐โ๐] = [det(๐) ๐โ๐]: (
โ๐
โ๐ผ)
By the simple rules of multiple derivative,
โ
โ๐ผ(det(๐)) = [
โ
โ๐(det(๐))]: (
โ๐
โ๐ผ)
It therefore follows that,
[โ
โ๐(det(๐)) โ [det(๐) ๐โ๐]]: (
โ๐
โ๐ผ) = 0
Hence
โ
โ๐(det(๐)) = [det(๐) ๐โ๐] = ๐๐
45. [Gurtin 3.4.2a] If T 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)๐๐ =
1
det(๐)det(๐) ๐โ๐
= ๐โ๐
46. [Gurtin 3.4.2a] 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
47. 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.
48. 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
49. 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
50. Let ๐บ be a symmetric and positive definite tensor and let ๐ผ1(๐บ), ๐ผ2(๐บ)and๐ผ3(๐บ)
be the three principal invariants of ๐บ show that (a) ๐๐ฐ1(๐บ)
๐๐บ= ๐ the identity tensor, (b)
๐๐ฐ2(๐บ)
๐๐บ= ๐ผ1(๐บ)๐ โ ๐บ and (c)
๐๐ผ3(๐บ)
๐๐บ= ๐ผ3(๐บ) ๐บโ1
๐๐ฐ1(๐บ)
๐๐บ can be written in the invariant component form as,
๐๐ผ1(๐บ)
๐๐บ=
๐๐ผ1(๐บ)
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
๐๐๐๐
๐ 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
[๐ฮฑฮฑ๐๐ฝ
๐ฝโ ๐๐ฝ
ฮฑ๐ฮฑ๐ฝ
] ๐ ๐ โ ๐ ๐
=1
2[๐ฟ๐ผ
๐ ๐ฟ๐๐ผ๐๐ฝ
๐ฝ+ ๐ฟ๐ฝ
๐ ๐ฟ๐๐ฝ
๐ฮฑฮฑ โ ๐ฟ๐ฝ
๐ ๐ฟ๐๐ผ๐ฮฑ
๐ฝโ ๐ฟ๐ผ
๐ ๐ฟ๐๐ฝ
๐๐ฝฮฑ] ๐ ๐ โ ๐ ๐
=1
2[๐ฟ๐
๐๐๐ฝ๐ฝ
+ ๐ฟ๐๐๐ฮฑ
ฮฑ โ ๐๐๐
โ ๐๐๐] 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
๐ผ๐ฟ๐๐ฝ
๐๐๐ ๐๐๐ก + ๐๐๐๐ฟ๐๐ผ๐ฟ๐
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
51. 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
52. 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
53. Show that Stokes Theorem leads to
โซ ๐ฏ โ ๐๐ฑ =๐ถ
โซ (grad ๐ฏ) : (๐ง ร) ๐๐๐
In the classical form Stokes Theorem states that,
โซ ๐ฏ โ ๐๐ฑ =๐ถ
โซ ๐ง โ curl ๐ฏ ๐๐๐
Now curl ๐ฏ = ๐๐๐๐๐ฃ๐,๐ ๐ ๐ so that
๐ง โ curl ๐ฏ = ๐๐๐๐๐๐๐ฃ๐,๐ = ๐ฃ๐,๐ ๐๐๐๐๐๐
= (๐ฃ๐ผ,๐ฝ ๐ ๐ผ โ ๐ ๐ฝ): (โ๐๐๐๐๐๐๐ ๐ โ ๐ ๐)
= (grad ๐ฏ): (๐ง ร)
Hence
โซ ๐ฏ โ ๐๐ฑ =๐ถ
โซ ๐ง โ curl ๐ฏ ๐๐๐
= โซ(grad ๐ฏ): (๐ง ร)๐๐๐
As required.
54. Establish the following Identities using Stokes Theorem:
โซ (๐ฎ โ ๐ฏ) ๐๐ฑ =๐ถ
โซ [(grad ๐ฎ) ๐ฏ ร ๐ง + (๐ง โ curl ๐ฏ) ๐ฎ] ๐๐๐
โซ ๐ฏ ร ๐๐ฑ =๐ถ
โซ [(div ๐ฏ) ๐ง โ (grad ๐ฏ)T
๐ง] ๐๐๐
To establish the first equation, observe that โซ ๐ฏ โ ๐๐ฑ =๐ถ โซ ๐ง โ curl ๐ฏ ๐๐๐ can be used
to show that for any tensor ๐
โซ ๐๐๐ฑ =๐ถ
โซ (curl ๐)T
๐ง ๐๐๐
Now let ๐ = ๐ฎ โ ๐ฏ = ๐ข๐๐ฃ๐๐ ๐ โ ๐ ๐. But curl ๐ = ๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ The
transpose of this,
(curl ๐)T = ๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ผ โ ๐ ๐
Clearly,
โซ(curl ๐)T๐ง ๐๐๐
= โซ๐๐๐๐๐๐ผ๐ ,๐ (๐ ๐ผ โ ๐ ๐)๐ง ๐๐๐
= โซ๐๐๐๐(๐ข๐ผ๐ฃ๐),๐ (๐ ๐ผ โ ๐ ๐)๐ง ๐๐๐
= โซ๐๐๐๐(๐ข๐ผ ,๐ ๐ฃ๐ + ๐ข๐ผ๐ฃ๐ ,๐ )(๐ ๐ผ โ ๐ ๐)๐ง ๐๐๐
= โซ๐๐๐๐(๐ข๐ผ ,๐ ๐ฃ๐ + ๐ข๐ผ๐ฃ๐ ,๐ )(๐ ๐ผ โ ๐ ๐)๐ง ๐๐๐
= โซ(๐ข๐ผ ,๐ ๐๐๐๐๐ฃ๐๐๐๐ ๐ผ + ๐๐๐๐๐๐๐ข๐ผ๐ฃ๐,๐ ๐ ๐ผ) ๐๐๐
= โซ[(grad ๐ฎ)๐ฏ ร ๐ง + (๐ง โ curl ๐ฏ)๐ฎ]๐๐๐
As required.
To establish the second result, we allow ๐ = ๐ฏ ร= ๐๐๐๐๐ฃ๐๐ ๐ โ ๐ ๐. As before, we
have that,
โซ(curl ๐)T๐ง ๐๐๐
= โซ๐๐๐๐๐๐ผ๐ ,๐ (๐ ๐ผ โ ๐ ๐)๐ง ๐๐๐
= โซ๐๐๐๐(๐๐ผ๐ฝ๐๐ฃ๐ฝ),๐ (๐ ๐ผ โ ๐ ๐)๐ง ๐๐๐
= โซ (๐ฟ๐ผ๐ ๐ฟ๐ฝ
๐โ ๐ฟ๐ฝ
๐ ๐ฟ๐ผ๐) ๐ฃ๐ฝ ,๐ (๐ ๐ผ โ ๐ ๐)๐ง ๐๐
๐
= โซ๐ฃ๐ ,๐ ๐๐๐ ๐ โ ๐ฃ๐ ,๐ ๐ ๐๐๐ ๐๐๐
โซ ๐ฏ ร ๐๐ฑ =๐ถ
โซ [(div ๐ฏ) ๐ง โ (grad ๐ฏ)T
๐ง] ๐๐๐
Because (grad ๐ฏ)T๐ง = ๐ฃ๐ ,๐ (๐ ๐ โ ๐ ๐)๐ง = ๐ฃ๐ ,๐ ๐ ๐๐๐ .