L = 1/2[(1 − 2m/r)c^{2}ṫ^{2} − (1 − 2m/r)^{−1}ṙ^{2} − r^{2}ḋϕ^{2}] [1 mark]

the Euler Lagrange equations for cyclic coordinates eg t is

\frac{d}{dτ} (\frac{∂L}{∂ṫ}) − \frac{∂L}{∂t} = 0 = \frac{d}{dτ} (\frac{∂L}{∂ṫ})

(1 − 2m/r)ṫ = k

where k is a constant. A similar analysis for ϕ gives r^{2}ϕ̇ = h where h is a constant. [1 mark]

substitute for both ṫ and ϕ̇ to get ṙ from the null metric so

(1 − 2m/r)c^{2}ṫ^{2} − (1 − 2m/r)^{−1}ṙ^{2} − r^{2}ϕ̇^{2} = 0  [1 mark]
(1 − 2m/r)c^{2} \frac{k^{2}}{(1 − 2m/r)^{2}} − (1 − 2m/r)^{−1}ṙ^{2} − r^{2} \frac{h^{2}}{r^{4}} = 0 [1 mark]
c^{2}k^{2} − ṙ^{2} − h^{2}(1 − 2m/r)/r^{2} = 0
ṙ^{2} = c^{2}k^{2} − h^{2}(1 − 2m/r)/r^{2} = c^{2}k^{2} − h^{2}V^{2}(r)  [1 mark]

maximum at

dV^{2}/dr = 0 = −2(1−2m/r)/r^{3}+1/r^{2}d(1−2m/r)/dr = −2(1−2m/r)/r^{3}+1/r^{2}2m/r^{2}

so at −2 + 4m/r + 2m/r = 0 i.e. r = 3m [1 mark]

The maxium value is V^{2}(3m) = (1 − 2/3)1/(9m^{2}) = 1/(27m^{2}) [1 mark]

null geodesics have

(\frac{du}{dϕ})^{2} + u^{2} = \frac{c^{2}k^{2}}{h^{2}} + 2mu^{3}

For m → 0 this goes to (du/dϕ)^{2} + u^{2} = c^{2}k^{2}/h^{2}. Substitute in u = u_{0} sin ϕ and this gives u_{0}^{2} cos^{2} ϕ + u_{0}^{2} sin^{2} ϕ = c^{2}k^{2}/h^{2} so u_{0} = ck/h or b = 1/u_{0} = h/ck as the impact parameter [1 mark]

ṙ^{2} > 0 means light can get everywhere (including down the black hole, where it is captured). So capture when c^{2}k^{2} − h^{2}V^{2}(r) > 0 ie V^{2}(r) < c^{2}k^{2}/h^{2} but b = h/ck so this is V^{2}(r) < 1/b^{2}. [1 mark]

But from above V^{2} ≤ 1/(27m^{2}), so if b ≤ \sqrt{27}m then the light is captured. [1 mark]