Mathematical Physics Vol 1

4.6 Examples

209

Solution For parabolic coordinates ( ρ , φ , z ) u 1 = u , u 2 = v ,

u 3 = z ; e 3 = e z ;

e 1 = e u ,

e 2 = e v ,

h 1 = h u = p u

h 2 = h v = p u

2 + v 2 ,

2 + v 2 ,

h

3 = h z = 1 .

Then

1 u 2 + v 2 1 u 2 + v 2

∂ ∂ u ∂ 2 ψ ∂ u 2

∂ψ ∂ u

∂ ∂ v

∂ψ ∂ v

∂ ∂ z

∂ψ ∂ z

∇ 2 ψ =

( u 2 + v 2 )

+

+

=

∂ 2 ψ ∂ v 2

∂ 2 ψ ∂ z 2

=

+

+

.

and the Laplace equation is ∇ 2 ψ = 0, that is ∂ 2 ψ ∂ u 2 + ∂ 2 ψ ∂ v 2

∂ 2 ψ ∂ z 2

+( u 2 + v 2 )

= 0 .

4.6.10 Surfaces in terms of orthogonal generalized coordinates

Exercise 165 Show that the square of the arc element of the curve r = r ( u , v ) that lies in a plane, can be written in the following form d s 2 = E d u 2 + 2 F d u d v + G d v 2 .

Solution Given that

∂ r ∂ u

∂ r ∂ v

d r =

d u +

d v ,

it follows that

d s 2 = d r · d r = =

∂ r ∂ u

∂ r ∂ u

∂ r ∂ u

∂ r ∂ v

∂ r ∂ v

∂ r ∂ v

d u 2 + 2

d v 2 =

d u d v +

= E d u 2 + 2 F d u d v + G d v 2 ,

which yields

∂ r ∂ u

∂ r ∂ u

∂ r ∂ u

∂ r ∂ v

∂ r ∂ v

∂ r ∂ v

E =

F =

G =

.