Spherical GeodesicMay 2, 2011

By the orthogonality of spherical polar coordinates; ηµν = ∑l(∂µxl)2 = ηµ .ds2 = ∑

µηµdq2

µ

Take φ as the inclination angle, and θ as the azimuthal angle.x = r sinφ cosθy = r sinφ sinθz = r cosφSo we have,

ηr = sin2 φ cos2 θ + sin2 φ sin2 θ + cos2 φ = 1ηφ = r2 cos2 φ cos2 θ + r2 cos2 φ sin2 θ + r2 sin2 φ = r2ηθ = r2 sin2 θ sin2 φ + r2 sin2 φ cos2 θ = r2 sin2 φ

And therefor,ds = √dr2 + r2dφ2 + r2 sin2 φdθ2 = rdθ

√(dφdθ

)2 + sin2 φNote that dr vanishes, as it is invariant with respect to θ. And now, for anypath from p1 → p2 on the surface of a sphere;

J {λ} = p2∫p1ds = r

θ2∫θ1dθ

√(dφdθ

)2 + sin2 φwhich gives the arclength of a path φ(θ). The geodesic is said to be an ex-tremum of the action defined above. The function λ(dφdθ , φ(θ)) = √(dφdθ)2 + sin2 φwill be a critical path if and only if it satisfies the Euler-Lagrange equation;

∂λ∂φ −

ddθ

∂λ∂(dφdθ ) = 01

And now a slew of computation that I have checked 4 or 5 times now.∂λ∂φ = ∂φ

{√(dφdθ

)2 + sin2 φ} = sinφ cosφ√( dφdθ )2+sin2 φ= [( dφdθ )2+sin2 φ] sinφ cosφ[( dφdθ )2+sin2 φ]3/2 = ( dφdθ )2 sinφ cosφ+sin3 φ cosφ[( dφdθ )2+sin2 φ]3/2

∂λ∂( dφdθ ) = ∂

∂( dφdθ ){√(

dφdθ

)2 + sin2 φ} = dφ

dθ√( dφdθ )2+sin2 φddθ

∂λ∂( dφdθ ) = d

dθ

{dφdθ√( dφdθ )2+sin2 φ

} = d2φdθ2√( dφdθ )2+sin2 φ−

dφdθ

(dφdθ

d2φdθ2 +sinφ cosφ dφ

dθ

)[( dφdθ )2+sin2 φ]3/2 = d2φ

dθ2 [( dφdθ )2+sin2 φ][( dφdθ )2+sin2 φ]3/2 −dφdθ

(dφdθ

d2φdθ2 +sinφ cosφ dφ

dθ

)[( dφdθ )2+sin2 φ]3/2ddθ

∂λ∂( dφdθ ) = d2φ

dθ2 sin2 φ−sinφ cosφ( dφdθ )2[( dφdθ )2+sin2 φ]3/2The Euler-Lagrange equation becomes;(dφdθ

)2 sinφ cosφ + sin3 φ cosφ − d2φdθ2 sin2 φ + sinφ cosφ (dφdθ)2

[(dφdθ

)2 + sin2 φ]3/2 = 0

=⇒ 2(dφdθ)2 sinφ cosφ + sin3 φ cosφ − d2φ

dθ2 sin2 φ = 02(dφdθ

)2 cosφ − d2φdθ2 sinφ = − sin2 φ cosφ

d2φdθ2 − 2(dφdθ

)2 cotφ = sinφ cosφHow in the fuck do I solve this? It is clear that a constant solution of the form

φ = nπ (nεN) or φ = nπ2 (n odd) is valid, and they do correspond to circles onthe surface of the sphere, but they do not necessarilly need to share the origin ofthe surface.Maybe constraining paths only to the paramterization φ→ φ(θ) is not generalenough. Maybe both phi and theta need to be parameterized appropriately.

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