Mathematical Physics Vol 1

4.6 Examples

207

As A = ∑ i A i e i , using the results from the previous Example 160, we obtain ∇ × A = ∇ × ( A 1 e 1 + A 2 e 2 + A 3 e 3 )= = ∇ × ( A 1 e 1 )+ ∇ × ( A 2 e 2 )+ ∇ × ( A 3 e 3 )= = e 2 h 3 h 1 ∂ ∂ u 3 ( A 1 h 1 ) − e 3 h 1 h 2 ∂ ∂ u 2 ( A 1 h 1 )+

∂ ∂ u 1

∂ ∂ u 3

e 3 h 1 h 2

e 1 h 2 h 3

( A 2 h 2 ) −

( A 2 h 2 )+

+

∂ ∂ u 2

∂ ∂ u 1

e 1 h 2 h 3

e 2 h 3 h 1 e 2 h 3 h 1

( A 3 h 3 ) − ( A 2 h 2 ) +

( A 3 h 3 )=

+

e 1 h 2 h 3

( A 3 h 3 ) +

∂ ∂ u 2

∂ ∂ u 3

∂ ∂ u 3

∂ ∂ u 1

( A 3 h 3 ) −

( A 1 h 1 ) −

=

e 3 h 1 h 2

( A 1 h 1 ) =

∂ ∂ u 1

∂ ∂ u 2

( A 2 h 2 ) −

+

h 1 e 1 h 1 h 2 h 3

( A 2 h 2 ) +

h 2 e 2 h 3 h 1 h 2 ( A 1 h 1 ) ,

( A 3 h 3 ) +

∂ ∂ u 2

∂ ∂ u 3

∂ ∂ u 3

∂ ∂ u 1

( A 3 h 3 ) −

( A 1 h 1 ) −

=

h 3 e 3 h 1 h 2 h 3

∂ ∂ u 1

∂ ∂ u 2

( A 2 h 2 ) −

+

and thus rot A can be written in the following form

h 1 e 1 h 2 e 2 h 3 e 3 ∂ ∂ u 1 ∂ ∂ u 2 ∂ ∂ u 3 A 1 h 1 A 2 h 2 A 3 h 3

1 h 1 h 2 h 3

∇ × A =

.

Exercise 162 Calculate ∇ 2 ψ in orthogonal generalized coordinates, where ψ is a scalar function.

Solution According to Example 157 on p. 203 we have

∂ψ ∂ u 1

∂ψ ∂ u 2

∂ψ ∂ u 3

e 1 h 1

e 2 h 2

e 3 h 3

∇ ψ =

+

+

.

∂ψ ∂ u 1

∂ψ ∂ u 2

∂ψ ∂ u 3

1 h 1

1 h 2

1 h 3

Let A = ∇ ψ , then A 1 =

, A 2 =

, A 3 =

, and this example comes

down to Example 160a on p. 206.

∇ · A = ∇ · ∇ ψ = ∇ 2 ψ = 1 h 1 h 2 h 3 ∂ ∂ u 1

∂ψ ∂ u 1

∂ ∂ u 2

∂ψ ∂ u 2

h 2 h 3 h 1

h 3 h 1 h 2

+

+

∂ ∂ u 3

∂ψ ∂ u 3

h 1 h 2 h 3

+

.