Mathematical Physics Vol 1

Chapter 4. Field theory

214

Exercise 169 Let ( u 1 , u 2 , u 3 ) be generalized coordinates. a) Show that the following stands

=

∂ r ∂ u 3

g 11 g 12 g 13 g 21 g 22 g 23 g 31 g 32 g 33

2

∂ r ∂ u 1 ·

∂ r ∂ u 2 ×

g =

,

p d u q in the expressions for d s 2 (see Example

where g pq are metric coefficients next to d u

156 on p. 203). b) Show that the volume element in generalized orthogonal coordinates is equal to √ g d u 1 d u 2 d u 3 .

Solution a) Given that

∂ r ∂ u p ·

∂ r ∂ u q

∂ x ∂ u p

∂ x ∂ u q

∂ y ∂ u p

∂ y ∂ u q

∂ z ∂ u p

∂ z ∂ u q

g pq = α p · α q =

p , q = 1 , 2 , 3 ,

=

+

+

using the theorem on multiplication of determinants a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 A 1 A 2 A 3 B 1 B 2 B 3 C 1 C 2 C 3

=

= a 1 A 1 + a 2 A 2 + a 3 A 3 a 1 B 1 + a 2 B 2 + a 3 B 3 a 1 C 1 + a 2 C 2 + a 3 C 3 b 1 A 1 + b 2 A 2 + b 3 A 3 b 1 B 1 + b 2 B 2 + b 3 B 3 b 1 C 1 + b 2 C 2 + b 3 C 3 c 1 A 1 + c 2 A 2 + c 3 A 3 c 1 B 1 + c 2 B 2 + c 3 B 3 c 1 C 1 + c 2 C 2 + c 3 C 3

we obtain

2

∂ x ∂ u 1 ∂ x ∂ u 2 ∂ x ∂ u 3

∂ y ∂ u 1 ∂ y ∂ u 2 ∂ y ∂ u 3

∂ z ∂ u 1 ∂ z ∂ u 2 ∂ z ∂ u 3

∂ r

∂ r ∂ u 3

2

∂ r ∂ u 2 ×

∂ u 1 ·

=

=

∂ y ∂ u 1 ∂ y ∂ u 2 ∂ y ∂ u 3

∂ z ∂ u 1 ∂ z ∂ u 2 ∂ z ∂ u 3

∂ x ∂ u 1 ∂ x ∂ u 2 ∂ x ∂ u 3

∂ y ∂ u 1 ∂ y ∂ u 2 ∂ y ∂ u 3

∂ z ∂ u 1 ∂ z ∂ u 2 ∂ z ∂ u 3

∂ x ∂ u 1 ∂ x ∂ u 2 ∂ x ∂ u 3

g 11 g 12 g 13 g 21 g 22 g 23 g 31 g 32 g 33

=

=

.

b) The volume element is given by the expression d V = ∂ r ∂ u 1 d u 1 · ∂ r ∂ u 2 d u 2 × ∂ r ∂ u 3

d u 3 =

∂ r ∂ u 1 ·

∂ r ∂ u 2 ×

∂ r ∂ u 3

d u 1 d u 2 d u 3

= √ g d u 1 d u 2 d u 3 . It should be noted that √ g is the absolute value of the Jacobian.