Rotating fluids and the outer core - Earth Dynamics: Deformations and Oscillations of the Rotating Earth - page 395

is normal to the surface. Di ff erentiating (6.16), we find

dR
dz =−

a 2
b 2

z
R ,

(6.72)

giving, for the normal vector,

k a 2
b 2

z
R =

1 z

R .

R R k

n = + + (6.73)

e 2 1 −

Then, from expression (6.69), the condition for the normal component of displace-
ment to vanish at the rigid boundary is

2

2 R ∂χ

σ −

e 2 z ∂χ

1

σ

∂ R +

m σχ +

∂ z =

0. (6.74)

1 −

From the derivatives (6.65) and (6.66), we find
R ∂χ

∂ R = R ∂χ ∂ R + R ∂χ

∂ξ ∂η

∂ R ∂ξ ∂η

1 1 1 1

2 2 2 2

− ξ − η − ξ − η

∂χ
∂ξ + η

∂χ
∂η ,

=− ξ (6.75)

ξ

2

− η

2

ξ

2

− η

2

and

z ∂χ
∂ξ

z ∂χ z ∂χ

∂ξ
∂ z +

∂χ
∂η

∂η
∂ z

∂ z = ∂ z =

2 2

2 1

− ξ

2 ∂χ

1 − η

2 ∂χ

2

=− ξη

∂ξ + ξ

η

∂η .

(6.76)

ξ

2

− η ξ

2

− η

Substitution of these expressions into the boundary condition (6.74) produces
ξ 1

1 ∂χ

∂ξ

2 2

e 2

2

− ξ η − σ

2

σ −

ξ

2

− η

2

1 − e 2

η 1

2 2 1 2

∂χ
∂η −

e 2 − η ξ − σ

2

− σ − m σχ = 0 (6.77)

ξ

2

− η

2

1 − e 2

as the condition for the vanishing of the normal displacement at the rigid container
boundary. From (6.53),

e 2

2 2

1 −

e 2 = σ

σ

0 .

(6.78)

1 −

ξ

The variables then separate to give the boundary condition as
ξ 1

2

η 1

2

− ξ ∂χ

∂ξ −

− η ∂χ

∂η

2 1

0

2 1

0

2 2

σ − ξ

2

/ξ σ − η

2

/ξ

⎝ ⎠ χ =

2
0

m ξ

1 1

−

0 −

0 (6.79)

2 1 2 1

0

σ

2 2 2 2

σ − ξ /ξ σ − η /ξ