Mathematical Physics Vol 1

Chapter 4. Field theory

198

By differentiating (4.221) we obtain d x = − a ch ξ cos η sin ϕ d ϕ − a ch ξ sin η cos ϕ d η + a sh ξ cos η cos ϕ d ξ , d y = a ch ξ cos η cos ϕ d ϕ − a ch ξ sin η sin ϕ d η + a sh ξ cos η sin ϕ d ξ , d z = a sh ξ cos η d η + a ch ξ sin η d ξ .

The square of the arc element is

( d s ) 2 = d x 2 + d y 2 + d z 2 = a 2 ( sh 2 ξ + sin 2 η )( d ξ ) 2

+ a 2 ( sh 2 ξ + sin 2 η )( d η ) 2 + a ch 2 ξ cos 2 η ( d ϕ ) 2 . ⇒

Lame’s coefficients are

h 1 = h ξ = a q sh h 2 = h η = a q sh h 3 = h ϕ = a ch ξ cos η .

2 ξ + sin 2 η ,

2 ξ + sin 2 η ,

The elementary volume is d V = a q sh 2 ξ + sin 2 η a q sh 2 ξ + sin 2 η ( a ch ξ cos η ) d ϕ d η d ξ = = a 3 sh 2 ξ + sin 2 η d ϕ d η d ξ .

Exercise 152 Express the elementary area in terms of generalized coordinates.

Solution We have shown that the differential of the position vector can be expressed in general ized coordinates as follows (see relation (4.138) on p. 112)

3 ∑ i = 1

1 e

2 e

3 e

i e

d r = h 1 d u

1 + h 2 d u

2 + h 3 d u

h i d u

3 =

i .

In particular, along the coordinate line u 1 the coordinates u 2 and u 3 are constant, and it follows that d r = h 1 d u 1 e 1 . The length of the arc element d s 1 , along the u 1 coordinate line, at point P , is d s 1 = h 1 d u 1 . Similarly, we obtain the expressions for d s 2 andd s 3 along the coordinate lines u 2 and u 3 , respectively. As the area can be expressed in terms of a vector product (see Example 10 on p.