Mathematical Physics Vol 1

Chapter 4. Field theory

210

Exercise 166 Show that the element of the surface defined by the relation r = r ( u , v ) , can be expressed in the following form d S = p EG − F 2 d u d v .

Solution The surface element is given by d S = ∂ r ∂ u d u ×

d v =

∂ r ∂ v

∂ r ∂ u ×

∂ r ∂ v

d u d v =

= s

∂ r ∂ v ·

∂ r ∂ v

∂ r ∂ u ×

∂ r ∂ u ×

d u d v .

The value under the square root is ∂ r ∂ u · ∂ r ∂ u ∂ r ∂ v · ∂ r which represents the required result. ∂ v −

∂ r ∂ v

∂ r ∂ u

∂ r ∂ u ·

∂ r ∂ v ·

= EG − F ,

R Note. This idea can be demonstrated by observing the vector product, which yields a × b = ab sin α n ⇒ | a × b | 2 = a 2 b 2 sin 2 α = a 2 b 2 ( 1 − cos 2 α )= = a 2 b 2 − ( ab cos α ) 2 = a 2 b 2 − ( a · b ) 2 = =( a · a )( b · b ) − ( a · b ) 2 = a · a a · b b · a b · b ⇒ = | a × b | = q a 2 b 2 − ( a · b ) 2 .

4.6.11 Generalized systems

Exercise 167 Let A be a vector and ( u 1 , u 2 , u 3 ) and ( u 1 , u 2 , u 3 ) two orthogonal curvilinear coordinate systems. Find the relation between contravariant components of this vector in the two coordinate systems.

Solution Let us assume that the coordinate transformations between the Cartesian coordi nate system and the ( u 1 , u 2 , u 3 ) system, and the Cartesian coordinate system and the