Worked Example Geodesics on the Surface of a Sphere Example Geodesics on the Surface of a Sphere...
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Worked Example Geodesics on the Surface of a Sphere Recall that in orthogonal curvilinear coordinates (q 1 ,q 2 ,q 3 ), dr = h 1 dq 1 e 1 + h 2 dq 2 e 2 + h 3 dq 3 e 3 . In spherical polar coordinates, dr =dr e r + r dθ e θ + r sin θ dφ e φ . Without loss of generality, we may take the sphere to be of unit radius: the length of a path from A to B is then L = B A |dr| = B A dθ 2 + sin 2 θ dφ 2 [since dr = 0] = θ B θ A 1 + sin 2 θφ 2 dθ where the path is described by the function φ(θ). Using Euler’s equation, d dθ ∂ ∂φ 1 + sin 2 θφ 2 = ∂ ∂φ 1 + sin 2 θφ 2 =0 so that sin 2 θφ 1 + sin 2 θφ 2 is a constant, c say. Hence φ = c sin θ √ sin 2 θ - c 2 and the problem reduces to integrating this with respect to θ. Substitute u = cot θ so that du = - cosec 2 θ dθ. Then φ = -c du √ 1 - c 2 cosec 2 θ = -c du 1 - c 2 (1 + u 2 ) = -du √ a 2 - u 2 where a = √ 1 - c 2 c = cos -1 (u/a)+ φ 0 where φ 0 is a constant of integration. Hence the geodesic path is given by cot θ = a cos(φ - φ 0 ) and the arbitrary constants a and φ 0 must be found using the end-points. This is a great circle path. Mathematical Methods II Natural Sciences Tripos Part IB
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Transcript of Worked Example Geodesics on the Surface of a Sphere Example Geodesics on the Surface of a Sphere...
Worked ExampleGeodesics on the Surface of a Sphere
Recall that in orthogonal curvilinear coordinates (q1, q2, q3),
dr = h1 dq1 e1 + h2 dq2 e2 + h3 dq3 e3.
In spherical polar coordinates,
dr = dr er + r dθ eθ + r sin θ dφ eφ.
Without loss of generality, we may take the sphere to be ofunit radius: the length of a path from A to B is then
L =∫ B
A
|dr|
=∫ B
A
√dθ2 + sin2 θ dφ2 [since dr = 0]
=∫ θB
θA
√1 + sin2 θ φ′2 dθ
where the path is described by the function φ(θ). Using Euler’s equation,
ddθ
(∂
∂φ′
√1 + sin2 θ φ′2
)=
∂
∂φ
√1 + sin2 θ φ′2 = 0
so thatsin2 θ φ′√
1 + sin2 θ φ′2
is a constant, c say. Hence
φ′ =c
sin θ√
sin2 θ − c2
and the problem reduces to integrating this with respect to θ.
Substitute u = cot θ so that du = − cosec2 θ dθ. Then
φ =∫
−cdu√1− c2 cosec2 θ
=∫
−cdu√1− c2(1 + u2)
=∫
−du√a2 − u2
where a =√
1− c2
c
= cos−1(u/a) + φ0
where φ0 is a constant of integration. Hence the geodesic path is given by
cot θ = a cos(φ− φ0)
and the arbitrary constants a and φ0 must be found using the end-points. This is agreat circle path.
Mathematical Methods IINatural Sciences Tripos Part IB
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