Mathematical Physics Vol 1

4.6 Examples

205

or shortly

| ∇ u p | = h − 1 p ,

p = 1 , 2 , 3 .

∇ u p | ∇ u p |

b) According to the definition E p =

. Using the result from this example

under a) we obtain

p = e

E p = h p ∇ u

p ,

what was to be proved.

Exercise 159 Prove that e 1 = h 2 h 3 ∇ u orthogonal coordinates.

2 × ∇ u 3 and similarly for e

1 , u 2 and u 3 are

2 and e 3 , where u

Solution From previous example we have ∇ u 1 = ,

e 1 h 1

e 2 h 2

e 3 h 3

∇ u 2 =

∇ u 3 =

,

,

and thus

e 2 × e 3 h 2 h 3

e 1 h 2 h 3

∇ u 2 × ∇ u 3 =

2 × ∇ u 3 .

e 1 = h 2 h 3 ∇ u

and

=

In the same way, for the remaining two vectors we obtain e 2 = h 3 h 1 ∇ u 3 × ∇ u 1 and e

1 × ∇ u 2 ,

3 = h 1 h 2 ∇ u

what was to be proved.

Exercise 160 Show that the following is true for orthogonal coordinates

∂ ∂ u 1

1 h 1 h 2 h 3

∇ · ( A 1 e 1 )= ∇ × ( A 1 e 1 )=

a )

( A 1 h 2 h 3 )

∂ ∂ u 3

∂ ∂ u 2

e 2 h 3 h 1

e 3 h 1 h 2

b )

( A 1 h 1 ) −

( A 1 h 1 ) ,

as well as that analogous relations are true for vectors A 2 e 2 and A 3 e 3 .

Solution