Greg Egan on Nostr: Given b, r₁ and r₂, we can compute how far to the left or right of the second ...
Given b, r₁ and r₂, we can compute how far to the left or right of the second astronaut the photon will end up when it reaches them, say D(r₁, r₂, b), by solving the geodesic equation for the photon (see Misner, Thorne and Wheeler section 25.6).
Given r₀ and r₂, we can compute the 4-velocity of the second astronaut at the moment they reach r₂, say u(r₀, r₂), using the general relativistic equivalent of conservation of energy.
Given b and r₂, we can compute the momentum vector of the photon [up to an overall factor] when it reaches r₂, say p(r₂, b), using the general relativistic equivalent of conservation of angular momentum.
We project p into the three-dimensional space orthogonal to u, and find the angle between that projection and the direction pointing radially down, say δ(r₀, r₂, b).
The apparent distance is then:
d(r₀, r₁, r₂, b) = D(r₁, r₂, b) / tan(δ(r₀, r₂, b))
These expressions depend on the precise value of b. But we are only interested in the limit as b grows very small, so that D is comparable to the length scale of an astronaut’s body. Both the numerator and denominator here are proportional to b for small b, so we can take the limit as b→0.
The final result turns out to be not too terrible:
d/M = (ρ₂–ρ₁)/(ρ₂–2) (ρ₂/ρ₁) √[(ρ₂/ρ₀) (2ρ₀ + ρ₂ (ρ₀-4) + 2 √[2 ρ₂ (ρ₀–ρ₂)(ρ₀–2)])]
where ρ₀, ρ₁, ρ₂ are just r₀, r₁, r₂ divided by M.
In general, we still need to find r₂ numerically to plug into this formula. But in the specific case of the apparent distance when the astronauts are crossing the event horizon, we know that both ρ₁ and ρ₂ will draw close to 2. Once they are both very close to 2, we can treat them as approaching 2 linearly, but at different rates, say:
ρ₁ = 2+ε
ρ₂ = 2+ α ε
Given r₀ and r₂, we can compute the 4-velocity of the second astronaut at the moment they reach r₂, say u(r₀, r₂), using the general relativistic equivalent of conservation of energy.
Given b and r₂, we can compute the momentum vector of the photon [up to an overall factor] when it reaches r₂, say p(r₂, b), using the general relativistic equivalent of conservation of angular momentum.
We project p into the three-dimensional space orthogonal to u, and find the angle between that projection and the direction pointing radially down, say δ(r₀, r₂, b).
The apparent distance is then:
d(r₀, r₁, r₂, b) = D(r₁, r₂, b) / tan(δ(r₀, r₂, b))
These expressions depend on the precise value of b. But we are only interested in the limit as b grows very small, so that D is comparable to the length scale of an astronaut’s body. Both the numerator and denominator here are proportional to b for small b, so we can take the limit as b→0.
The final result turns out to be not too terrible:
d/M = (ρ₂–ρ₁)/(ρ₂–2) (ρ₂/ρ₁) √[(ρ₂/ρ₀) (2ρ₀ + ρ₂ (ρ₀-4) + 2 √[2 ρ₂ (ρ₀–ρ₂)(ρ₀–2)])]
where ρ₀, ρ₁, ρ₂ are just r₀, r₁, r₂ divided by M.
In general, we still need to find r₂ numerically to plug into this formula. But in the specific case of the apparent distance when the astronauts are crossing the event horizon, we know that both ρ₁ and ρ₂ will draw close to 2. Once they are both very close to 2, we can treat them as approaching 2 linearly, but at different rates, say:
ρ₁ = 2+ε
ρ₂ = 2+ α ε