Rotating fluids and the outer core - Earth Dynamics: Deformations and Oscillations of the Rotating Earth - page 393

If the container surface is given byξ = ξ 0 , by comparison of (6.51) for ξ = ξ 0 , with
(6.50), we find

− e 2 − e 2

2

1 1 σ

2
0

2

, k 2 a 2 ξ =

2 σ

= . (6.53)

e 2

2

1 − σ 1 − σ

Since e 2 > 0. Solving (6.51) and (6.52) for R 2 and z 2

in terms of the spheroidal co-ordinates (ξ,η), we find
R 2

2

0 < 1, and k 2

< 1, we have ξ

k 2 1

2 1 2 , z 2

k 2

2 2

= − ξ − η = ξ η . (6.54)

As before, the co-ordinate relations are

k 1

2 1 2 cosφ =

x = − ξ − η R cosφ, (6.55)

y = k 1

2 1 2 sinφ = R sinφ,

− ξ − η (6.56)

k τξη = τ z . z = (6.57)

The slope of the co-ordinate surface (6.51) may be found by di ff erentiation with

respect to R for fixed ξ, giving

dz
dR =−

2

ξ R

z .

(6.58)

2

1 − ξ

From (6.54),

1

2 1 2

R
z =

− ξ − η

, (6.59)

ξη

then,

1

dz

dR =− ξ

− η

2

2 .

(6.60)

η 1 − ξ

Similarly, di ff erentiating (6.52) with respect to R ,forfixedη,gives

1

dz
dR =−

2

η R

z =−

η
ξ

− ξ

2

2 .

(6.61)

2

1 − η 1 − η

Thus, at the points of intersection, the slopes of the co-ordinate surfaces are recip-
rocals, rather than negative reciprocals, and hence they are not orthogonal.
Elimination of η from the pair of equations (6.54) yields

k 2 R 2

− k 2 ξ

z 2
k 2 =

1

4

− z 2

2

ξ + + 0, (6.62)

while elimination of ξ yields

k 2 R 2

k 2 η

z 2
k 2 =

1

4

z 2

2

η + − − + 0. (6.63)