Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 37D / i / Solution

By spherical symmetry, place the geodesic in the equatorial plane

θ = π /2

. With

f (r) = 1 - \frac{2 M}{r},

the geodesic Lagrangian is

2 L = - f \dot{t}^{2} + f^{- 1} \overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2} = Q,

where

Q

is the constant squared norm of the tangent:

Q = - 1

for a unit-speed timelike geodesic and

Q = 0

for a null geodesic. The cyclic coordinates

t

and

ϕ

give the conserved quantities

E = f \dot{t}, L = r^{2} \dot{ϕ} .

Substitution into the normalization gives

Q = - \frac{E ^{2}}{f} + \frac{r ˙ ^{2}}{f} + \frac{L ^{2}}{r ^{2}} .

Multiplying by

f

and rearranging produces

\frac{1}{2} \overset{r}{˙}^{2} + V (r) = \frac{1}{2} E^{2}, V (r) = \frac{1}{2} (1 - \frac{2 M}{r}) (\frac{L ^{2}}{r ^{2}} - Q) .

This is the timelike geodesic effective potential form, with the same derivation applying to any affinely parametrized causal geodesic.

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