some homework problems for Farsight, part 3
This is the third set of homework problems.
- Part 1 went over the basic definitions of manifolds and charts, using the 2-sphere as an example. (ctamblyn made an important correction to one of my inequalities.)
- Part 2 defined Schwarzschild charts on the spacetime manifold M outside the event horizon of an isolated black hole.
Exercises 12 and 13 established that it takes at least two Schwarzschild charts to cover all of M. Exercises 11 and 16 suggested that the Schwarzschild metric's apparent singularity at the event horizon might be a mere coordinate singularity, with no more physical significance than the coordinate singularities at the poles of a spherical coordinate system.
In this part 3, we will use Lemaître coordinates to define charts on a larger manifold M' that includes all of M plus the event horizon and a region of spacetime inside the event horizon that extends all the way down to the essential singularity at the center of the black hole. That will prove there is no real singularity at the event horizon. Another exercise will establish that particles falling radially into the black hole pass through the event horizon with no special fanfare and reach the essential singularity in finite proper time.
[size=+1]
Schwarzschild coordinates and charts[/size]
With the (-,+,+,+) signature convention and natural units in which c=G=1, the Schwarzschild coordinates are (t, r, θ, φ), and the Schwarzschild metric is
[latex]
\[
ds^2 = - (1 - 2M/r)dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 d\Omega^2
\]
[/latex]
where
[latex]
\[
d \Omega^2 = d \theta^2 + \sin^2 \theta d \phi^2
\]
[/latex]
is the usual metric on a 2-sphere in spherical coordinates and
[latex]
\[
\begin{align*}
r &> 2M \\
0 &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]
[size=+1]
Lemaître coordinates and charts[/size]
The Lemaître coordinates are (τ, ρ, θ, φ), and the Lemaître metric is
[latex]
\[
ds^2 = - d \tau^2 + \frac{2M}{r} d \rho^2 + r^2 d\Omega^2
\]
[/latex]
where dΩ
2 is defined as above,
[latex]
\[
r = R(\tau, \rho) = \left[ \frac{3}{2} (\rho - \tau) \right]^{2/3} (2M)^{1/3}
\]
[/latex]
will turn out to coincide with the r coordinate of Schwarzschild coordinates, and
[latex]
\[
\begin{align*}
\tau &< \rho \\
0 &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]
Exercises 17 through 20 will prove that R(τ, ρ) can be identified with the r coordinate of a Schwarzschild chart. For now, mathematically scrupulous readers will eliminate r from the Lemaître metric above by writing R(τ, ρ) wherever r appears.
Define a Lemaître chart to be of the above form, and let M' be the manifold that's covered by the set of all Lemaître charts. It takes two Lemaître charts to cover all of M' (see exercises 12 and 13).
Exercise 17. Prove: If 0 < x < 1, then
[latex]
\[
(1/x)^{1/2} (1 - x)^{-1} - x^{1/2} (1 - x)^{-1} = (1/x)^{1/2}
\]
[/latex]
Hint: trivial algebra.
Exercise 18. Prove: If
[latex]
\begin{align*}
d \tau &= dt + \left[ \frac{2M}{R(\tau, \rho)} \right]^{1/2}
\left[ 1 - \frac{2M}{R(\tau, \rho)} \right]^{-1} dr \\
d \rho &= dt + \left[ \frac{R(\tau, \rho)}{2M} \right]^{1/2}
\left[ 1 - \frac{2M}{R(\tau, \rho)} \right]^{-1} dr
\end{align*}
[/latex]
then
[latex]
\[
\frac{d \rho}{dr} - \frac{d \tau}{dr} = \sqrt{ \frac{R(\tau, \rho)}{2M} }
\]
[/latex]
Hint: trivial algebra; use exercise 17.
Exercise 19. Prove: If dτ and dρ are defined as in exercise 18, then dR/dr = 1.
Hint: basic calculus; substitute and simplify, using exercise 18 once.
Exercise 20. Prove: Given any Schwarzschild chart on M, there exists a Lemaître chart on M' whose restriction to (4/3)M < (ρ - τ) is isometric to the Schwarzschild chart.
Hint: Guess how R(τ, ρ), dτ, and dρ might be translated into Schwarzschild coordinates. To confirm your guesses, substitute them for R(τ, ρ), dτ, and dρ within the Lemaître metric, simplify, and compare your simplified result against the Schwarzschild metric.
Exercise 21. Prove: M is (isometric to) a proper submanifold of M'.
Hint: Use exercise 20 to prove the Lemaître charts cover M, and show that Lemaître charts also cover the event horizon and its interior (excluding the essential singularity at τ=ρ).
Now that we know M' contains a submanifold that's isometric to M, we might as well identify M with that submanifold. (That gets rid of the pedantic "isometric to" stuff.)
Exercise 22. Prove: If ρ, θ, and φ are held constant, then
[latex]
\[
ds^2 = - d \tau^2
\]
[/latex]
Hint: trivial algebra/calculus.
If (dτ, dρ, dθ, dφ) are infinitesimal displacements from some point p = (τ, ρ, θ, φ), then ds
2 is the square of the pseudo-distance between p and q = (τ+dτ, ρ+dρ, θ+dθ, φ+dφ), and is calculated by applying the metric tensor at p to two copies of (dτ, dρ, dθ, dφ).
ds
2 is a pseudo-metric because (unlike Euclidean metrics) it can be negative. The physical significance of a negative value is that a massive particle's world line can go through both of the points p and q. According to exercise 22, the timelike separation between points p and q is exactly the same as the difference between the values of their timelike coordinate τ. That means τ measures the proper time of an observer that considers himself to be at rest with respect to the Lemaître chart (because his spatial coordinates aren't changing).
In particular, consider a point-like observer who considers himself to be at Lemaître coordinates (τ, ρ, θ, φ), is not moving in a way that would change his spatial coordinates ρ, θ, and φ, and is not experiencing any sensation of acceleration. Since his ρ coordinate is not changing, his proper time coordinate τ will eventually equal ρ, which means he will encounter the central τ=ρ singularity in finite proper time (proportional to ρ-τ). So long as (ρ-τ) > (4/3)M, he will be outside the event horizon but falling radially toward the center of the black hole. Nothing special happens to him when his proper time coordinate increases to τ = (ρ - (4/3)M) and he passes through the event horizon.
When his proper time coordinate increases to τ = ρ, however, he leaves the spacetime manifold M' and encounters the central singularity. I don't know what happens to him at that moment, or whether that even counts as a moment.
Exercise 23. Prove: If the Schwarzschild coordinates r, θ, and φ are held constant, then
[latex]
\[
ds^2 = (-1 + 2M/r) d t^2
\]
[/latex]
Hint: trivial.
The physical significance of exercise 23 is that the Schwarzschild time coordinate does
not correspond to the proper time of an observer who considers himself to be at rest with respect to the spatial coordinates of the Schwarzschild chart. What's more, an observer cannot remain at rest with respect to the spatial Schwarzschild coordinates without undergoing a sensation of acceleration away from the black hole. The magnitude of that acceleration is a monotonic function of 2M/r, so the acceleration is small when r is large. That's why the Schwarzschild time coordinate t approximates the proper time of observers who are far from the black hole. That approximation becomes worse as observers approach the event horizon, and goes off the charts at the event horizon, where Schwarzschild coordinates fail entirely due to a coordinate singularity.
To understand what happens to an observer falling into a black hole, we must switch to a chart that doesn't have the Schwarzschild coordinate singularity at the event horizon. Lemaître coordinates are suitable, as was shown by exercises 20 and 21 above. Many other coordinate systems could be used instead, such as Eddington-Finkelstein, Novikov, or Kruskal-Szekeres coordinates.
Kruska-Szekeres coordinates are especially interesting because they cover an even larger manifold M
*. It's my impression that M
* can be regarded as the manifold M' that's covered by Lemaître coordinates, glued to a time-reversed copy of M'.
But I'm not a physicist. I'd be interested to hear what the physicists and mathematicians have to say about M
*.
! 