Schwarzschild coordinates

Question

Please help as I've been stuck on this for days!

 

An unlucky astronaut plunges radially into a Schwarzschild black hole, starting from very far away so that the astronaut's proper time τ is related to Schwarzschild coordinate time t by dτ/dt = (1-2M/r)

i) Compare the astronaut's experience of the journey with that seen by an external observer

ii) Show that the astronaut's world line satisfies dr/dτ = (2M/r)1/2 and hence compute the time experienced by the astronaut to reach the singularity after crossing the event horizon

Thank you to whoever can help!

Answer 1

Ok, so the problem is written in Planck units, which means that the Schwartzshield radius of a black hole is simply 2M.

i: Far away from the astronaut, the proper time is the coordinate time. However, for all r>2M, the time dilation of the astronaut dτ/dt is a number between 0 and 1, meaning less proper time will pass for the astronaut. This continues until r=2M, where the dτ/dt=0 implies that time will appear frozen for the astronaut according to faraway observers. However, the astronaut will perceive their own time normally.

ii: The derivation of this requires the geodesic equation d^2x^μ/dτ^2 +Γ^µ_αβ *(dx^α/dτ*dx^β/dτ) to solve for the world line of the astronaut. Looking at the r-coordinate, you wind up getting 0=r''(τ)+M(r-2M)/r^3*(dt/dτ)^2+M/(r(2M-r))(r'(τ))^2 when substituting the appropriate Christoffel Symbols, and ignoring the angular coordinates. Using the given dt/dτ term and inverting it, we get:

0=r''(τ)+M(r-2M)/r^3*(1-2M/r)^-2+M/(r(2M-r))(r'(τ))^2

Attempting to simplify, we get

0=r''(τ)+M/(r(r-2M))-M/(r(r-2M))(r'(τ))^2

r''(τ)=-M/(r(r-2M))(1-r'(τ)^2)

r''(τ)/(1-r'(τ)^2)=M/(r(2M-r))

From here, integrating will yield you your solution dr/dτ=r'(τ) = (2M/r)^1/2.

To find the time it takes to the singularity, we need to integrate the last equation one more time, with t bounds of 0 to T (the answer) and the radius from 2M to 0 (the corresponding locations).

int(dr (r/2M)^1/2)|_2M^0=int(dt)|_0^T

T=(2/3)(r^3/(2M))^1/2|_2M^0=-2/3(2M)

This means that the time to fall to the singularity is T=-4M/3. The time is negative because on the interior of a black hole, time is a "spacelike" dimension.