Mathematical Physics Vol 1

Chapter 4. Field theory

206

a) Using the expressions for e i from Example 159, p. 205 we obtain ∇ · ( A 1 e 1 )= ∇ · ( A 1 h 2 h 3 ∇ u 2 × ∇ u 3 )= = ∇ ( A 1 h 2 h 3 ) · ∇ u 2 × ∇ u 3 + A 1 h 2 h 3 ∇ · ( ∇ u 2 × ∇ u 3 )= = ∇ ( A 1 h 2 h 3 ) · e 2 h 2 × e 3 h 3 + 0 = ∇ ( A 1 h 2 h 3 ) · e 1 h 2 h 3 =

=

( A 1 h 2 h 3 ) ·

∂ ∂ u 1

∂ ∂ u 2

∂ ∂ u 3

e 1 h 1

e 2 h 2

e 3 h 3

e 1 h 2 h 3

( A 1 h 2 h 3 )+

( A 1 h 2 h 3 )+

=

∂ ∂ u 1

1 h 1 h 2 h 3

( A 1 h 2 h 3 ) .

=

Similarly, we obtain

∂ ∂ u 2 ∂ ∂ u 3

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

∇ · ( A 2 e 2 )= ∇ · ( A 3 e 3 )=

( A 2 h 3 h 1 ) ,

( A 3 h 1 h 2 ) .

b) Using the expressions for e i from Example 159 we obtain ∇ × ( A 1 e 1 )= ∇ × ( A 1 h 1 ∇ u 1 )= = ∇ ( A 1 h 1 ) × ∇ u 1 + A 1 h 1 ∇ × ∇ u 1 = = ∇ ( A 1 h 1 ) × e 1 h 1 + 0 =

=

( A 1 h 1 ) ×

∂ ∂ u 1

∂ ∂ u 2

∂ ∂ u 3

e 1 h 1

e 2 h 2

e 3 h 3

e 1 h 1

( A 1 h 1 )+

( A 1 h 1 )+

=

∂ ∂ u 3

∂ ∂ u 2

e 2 h 3 h 1

e 3 h 1 h 2

( A 1 h 1 ) −

( A 1 h 1 ) .

=

Similarly, we also obtain

∂ ∂ u 1 ∂ ∂ u 2

∂ ∂ u 3 ∂ ∂ u 1

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

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

∇ × ( A 2 e 2 )= ∇ × ( A 3 e 3 )=

( A 2 h 2 ) − ( A 3 h 3 ) −

( A 2 h 2 ) ,

( A 3 h 3 ) .

Exercise 161 Express rot A (= ∇ × A ) in terms of orthogonal generalized coordinates, where A = ∑ i A i e i .

Solution