## Black hole evaporation: remnants of the big bang

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.2.

We’ve seen that through radiation, a black hole can eventually evaporate in a time ${t_{0}}$ :

$\displaystyle \boxed{t_{0}=2.0903\times10^{67}\left(\frac{M}{M_{s}}\right)^{3}\mbox{ years}} \ \ \ \ \ (1)$

where ${M_{s}}$ is the solar mass. Clearly if we’re going to observe black hole evaporation, its mass ${M}$ must be considerably less than the sun’s. Suppose black holes were formed during the big bang at ${13.7\times10^{9}}$ years ago. If the black hole is just evaporating now, its mass would have been:

 $\displaystyle M$ $\displaystyle =$ $\displaystyle M_{s}\left(\frac{13.7\times10^{9}}{2.0903\times10^{67}}\right)^{1/3}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8.69\times10^{-20}M_{s}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(8.69\times10^{-20}\right)\left(1.989\times10^{30}\right)\mbox{ kg}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.728\times10^{11}\mbox{ kg} \ \ \ \ \ (5)$

To put this in perspective, this is equivalent to an asteroid, with a typical rocky density of ${\rho=5000\mbox{ kg m}^{-3}}$, with a radius of

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \left(\frac{3M}{4\pi\rho}\right)^{1/3}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 202\mbox{ m} \ \ \ \ \ (7)$

To see how much energy is released in the final second of the black hole’s life, we can start with the equation we had earlier from the Stefan-Boltzmann relation:

$\displaystyle M^{2}dM=-\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}dt$

If we integrate this from ${t=0}$ to a time ${t_{1}}$ one second before the present, at which time the black hole’s remaining mass is ${M_{1}}$, then

 $\displaystyle \int_{M_{0}}^{M_{1}}M^{2}dM$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\int_{0}^{t_{1}}dt$ $\displaystyle \frac{1}{3}\left(M_{1}^{3}-M_{0}^{3}\right)$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}t_{1}$ $\displaystyle M_{1}$ $\displaystyle =$ $\displaystyle \left[M_{0}^{3}-\frac{3\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}t_{1}\right]^{1/3}$

However, ${M_{0}^{3}}$ is just ${\frac{3\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}t_{p}}$, where ${t_{p}}$ is the present time, so that ${t_{p}-t_{1}=1\mbox{ s}}$. Therefore the mass remaining 1 second before complete evaporation is:

 $\displaystyle M_{1}$ $\displaystyle =$ $\displaystyle \left[\frac{3\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\left(t_{p}-t_{1}\right)\right]^{1/3}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 341.38\left(t_{p}-t_{1}\right)^{1/3}$

In GR units, a time of 1 second is ${1/\left(3\times10^{8}\right)\mbox{ m}}$ so we get for the mass

$\displaystyle M_{1}=0.51\mbox{ kg}$

This is equivalent to

$\displaystyle E=M_{1}c^{2}=4.59\times10^{16}\mbox{ J}$

which is about 100 times the energy released in an atomic bomb blast.

## Black hole radiation: energy of a particle from a solar mass black hole

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.1.

The crude calculation presented earlier gives the energy at infinity of a particle created near the event horizon of a black hole as

$\displaystyle E_{\infty}=\frac{\hbar}{4GM} \ \ \ \ \ (1)$

In GR units

 $\displaystyle \hbar$ $\displaystyle =$ $\displaystyle 3.5153\times10^{-43}\mbox{ kg m}\ \ \ \ \ (2)$ $\displaystyle E_{\infty}$ $\displaystyle =$ $\displaystyle \frac{8.788\times10^{-44}}{GM}\mbox{ kg} \ \ \ \ \ (3)$

For a solar mass black hole ${GM=1477\mbox{ m}}$, so

$\displaystyle E_{\infty}=5.95\times10^{-47}\mbox{ kg} \ \ \ \ \ (4)$

In other units, this is (using ${1\mbox{ eV}=1.602\times10^{-19}\mbox{ J}}$)

 $\displaystyle E_{\infty}$ $\displaystyle =$ $\displaystyle mc^{2}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.355\times10^{-30}\mbox{ J}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.34\times10^{-11}\mbox{ eV} \ \ \ \ \ (7)$

This is an almost unimaginably small energy. For comparison, the energy of a single photon of visible light (with a wavelength of 500 nm) is around 2.5 eV:

$\displaystyle E=h\nu=\frac{hc}{\lambda}=6.626\times10^{-34}\times\frac{3\times10^{8}}{5\times10^{-7}\times1.602\times10^{-19}}\mbox{ eV} \ \ \ \ \ (8)$

## Black hole evaporation: how long will a black hole live?

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Boxes 16.3, 16.4.

Since a black hole can radiate its energy away and the energy of a black hole is equivalent to its mass, it’s possible that a black hole will simply evaporate after a long enough time. We can get an estimate of this time as follows. First, we recall that, for particle-antiparticle pairs that are produced very close to the event horizon, the energy of the particle at infinity tends to a constant:

$\displaystyle E_{\infty}=\frac{\hbar}{4GM} \ \ \ \ \ (1)$

This is only a crude estimate based on oversimplifying the situation and a more accurate calculation, due to Stephen Hawking, results in

$\displaystyle E_{\infty}=\frac{\hbar}{8\pi GM} \ \ \ \ \ (2)$

In thermodynamics, a blackbody is a body that absorbs all frequencies of electromagnetic radiation (hence ‘black’) and radiates with a spectrum that depends only its temperature ${T}$. The wavelength of radiation peaks at a shorter value the higher the temperature (which is why an iron bar, for example, glows red when it gets hotter; at cooler temperatures it radiates in the infra-red). Hawking showed that the energy spectrum of a black hole is actually the same as that of a blackbody with temperature ${T}$ if we set ${E_{\infty}=k_{B}T}$, where ${k_{B}}$ is Boltzmann’s constant. We can therefore define the temperature of a black hole as

$\displaystyle \boxed{T=\frac{\hbar}{8\pi k_{B}GM}} \ \ \ \ \ (3)$

To use this in calculations, it’s best to convert the constants to relativistic form, where ${c=1}$. We start with their values in SI units. For Planck’s constant:

 $\displaystyle \hbar$ $\displaystyle =$ $\displaystyle 1.0546\times10^{-34}\mbox{ J s}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.0546\times10^{-34}\mbox{ kg m}^{2}\mbox{ s}^{-1}\times\left(3\times10^{8}\right)^{-1}\mbox{ s m}^{-1}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.5153\times10^{-43}\mbox{ kg m} \ \ \ \ \ (6)$

For Boltzmann’s constant:

 $\displaystyle k_{B}$ $\displaystyle =$ $\displaystyle 1.3807\times10^{-23}\mbox{ J K}^{-1}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.3807\times10^{-23}\mbox{ kg m}^{2}\mbox{ s}^{-2}\mbox{ K}^{-1}\times\left(3\times10^{8}\right)^{-2}\mbox{ s}^{2}\mbox{ m}^{-2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.534\times10^{-40}\mbox{ kg K}^{-1} \ \ \ \ \ (9)$

Thus the temperature is

$\displaystyle T=\frac{9.118\times10^{-5}}{GM}\mbox{ K} \ \ \ \ \ (10)$

In terms of the solar mass ${M_{s}}$, we can use ${GM_{s}=1477\mbox{ m}}$ so

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{9.118\times10^{-5}}{GM_{s}}\frac{M_{s}}{M}\mbox{ K}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.173\times10^{-8}\frac{M_{s}}{M}\mbox{ K} \ \ \ \ \ (12)$

A one solar mass black hole is thus almost at absolute zero.

The rate at which an object radiates energy is given by the Stefan-Boltzmann law, which states

$\displaystyle \frac{dE}{dt}=A\sigma T^{4} \ \ \ \ \ (13)$

where ${A}$ is the surface area of the object and ${\sigma=2.105\times10^{-33}\mbox{ kg m}^{-3}\mbox{ K}^{-4}}$ is the Stefan-Boltzmann constant, in relativistic units. In the case of a black hole, the energy is just the mass, so

$\displaystyle \frac{dE}{dt}=-\frac{dM}{dt} \ \ \ \ \ (14)$

and the area is that of a sphere with a radius ${r=2GM}$, so

$\displaystyle A=4\pi\left(2GM\right)^{2}=16\pi\left(GM\right)^{2} \ \ \ \ \ (15)$

Using 3, we then have

 $\displaystyle \frac{dM}{dt}$ $\displaystyle =$ $\displaystyle -16\pi\left(GM\right)^{2}\sigma\left(\frac{\hbar}{8\pi k_{B}GM}\right)^{4}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\frac{1}{M^{2}}\ \ \ \ \ (17)$ $\displaystyle M^{2}dM$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}dt\ \ \ \ \ (18)$ $\displaystyle \frac{1}{3}M^{3}$ $\displaystyle =$ $\displaystyle \frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\Delta t\ \ \ \ \ (19)$ $\displaystyle \Delta t$ $\displaystyle =$ $\displaystyle \frac{256\pi^{3}k_{B}^{4}}{3\sigma\hbar^{4}G}\left(GM\right)^{3}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{256\pi^{3}k_{B}^{4}}{3\sigma\hbar^{4}G}\left(\frac{GM}{GM_{s}}\right)^{3}\left(GM_{s}\right)^{3} \ \ \ \ \ (21)$

where ${\Delta t}$ is the time required for the black hole to evaporate completely. We can plug in the numbers, using ${G=7.426\times10^{-28}\mbox{ m kg}^{-1}}$ and ${1\mbox{ year}=9.461\times10^{15}\mbox{ m}}$ to get

 $\displaystyle \Delta t$ $\displaystyle =$ $\displaystyle 1.9777\times10^{83}\left(\frac{M}{M_{s}}\right)^{3}\mbox{ m}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.0903\times10^{67}\left(\frac{M}{M_{s}}\right)^{3}\mbox{ years} \ \ \ \ \ (23)$

In other words, for a one solar mass black hole, we’re not going to see it evaporate any time soon.

## Writing posts with Lyx / LaTeX

Recently, I’ve been tinkering with the Latex2WP program that I’ve been using to convert Latex files into the form accepted by WordPress.com that you’ve been reading. While this was a good program, there were several features that I needed so, after learning a bit of Python, I’ve added these features to the program. My version, with I call Lyx2Wordpress, is available for download if you feel like writing your own blog on WordPress.com that involves equations.

Please note that I offer this without any warranty so use at your own risk. Any bug reports are welcome in the comments.

Lyx2Wordpress is a program that takes the Latex (plain) output (in a .tex file) produced by the LyX editor for LaTeX and converts it into an HTML file which may be pasted into a WordPress.com blog editor. It is an enhancement of the original Latex2WP program written by Luca Trevisan. To use it, you need Python version 3.3 (it may work with earlier versions of Python, but I haven’t tested this).

At the command line within the folder into which you unpacked Lyx2Wordpress, assuming python is in your executable path, type

python Lyx2Wordpress ‘.tex file’ [foreground colour] [background colour]

Here the .tex file is that output by LyX (it should also work with ordinary Latex files), the ‘foreground colour’ is the
colour of the text in the displayed equations, and ‘background colour’ is the background in the displayed equations. Both
colours are specified in RGB format, and are optional. If omitted they default to a foreground of black and a background
of white. To produce the orange background on my physicspages.com blog, I used foreground = ’000000′ (which is black)
and background = ‘e5e4e8′ to get the pale orange background. Thus a command for this would look like:

python Lyx2Wordpress C:\PathToMyFile\MyFile.tex 000000 e5e4e8

The output is placed in a file with the same name but suffix .html.

Enhancements that I’ve added to the original Latex2WP program are:

1. All equations and eqnarrays are centred horizontally.
2. Equations in an eqnarray environment are displayed at full size.
3. All equations (both single line and eqnarray) may be numbered and used as references.
4. The ‘array’ environment may be used within an eqnarray, so you can put matrices into eqnarrays.
5. Lyx tags for theorem-like environments (which are different from standard AMS environment tags) are handled properly.
6. Options for different text/background colours (see above).

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Box 16.2.

In radiation from a black hole, we found an estimate for the energy of a radiated particle as measured in the frame of an observer who is momentarily at rest at the point at which the particle pair is created.

$\displaystyle E=\frac{\hbar}{2\sqrt{2GM\epsilon}} \ \ \ \ \ (1)$

To work out the energy at infinity in the Schwarzschild (S) frame, we can use the particle’s four-momentum in the form

$\displaystyle E=-\mathbf{o}_{t}\cdot\mathbf{p} \ \ \ \ \ (2)$

where ${\mathbf{o}_{t}}$ is the time basis vector and ${\mathbf{p}}$ is the four-momentum of the particle in the observer’s frame. In this frame, ${\mathbf{o}_{t}=\left[1,0,0,0\right]}$ and ${E}$ is the time component of ${\mathbf{p}}$ so 2 follows.

If we work out this equation in S coordinates, we have

$\displaystyle \mathbf{o}_{t}=\left[\left(1-\frac{2GM}{r}\right)^{-1/2},0,0,0\right] \ \ \ \ \ (3)$

so using the S metric, we have

 $\displaystyle E$ $\displaystyle =$ $\displaystyle -g_{ij}o_{t}^{i}p^{j}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\left(1-\frac{2GM}{r}\right)^{-1/2}p^{t}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{1-\frac{2GM}{r}}m\frac{dt}{d\tau} \ \ \ \ \ (6)$

Since the energy per unit mass of a particle is given by

$\displaystyle e=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau} \ \ \ \ \ (7)$

$\displaystyle E=\frac{me}{\sqrt{1-2GM/r}} \ \ \ \ \ (8)$

Since ${e}$ is the energy per unit mass at infinity, ${me=E_{\infty}}$ is the total energy of the particle at infinity, so

$\displaystyle E_{\infty}=\sqrt{1-\frac{2GM}{r}}E \ \ \ \ \ (9)$

For a particle that is created at ${r=2GM+\epsilon}$ with energy given by 1, the energy at infinity is

$\displaystyle E_{\infty}=\sqrt{1-\frac{2GM}{2GM+\epsilon}}\frac{\hbar}{2\sqrt{2GM\epsilon}}$

This can be expanded in a series around ${\epsilon=0}$ to get

$\displaystyle E_{\infty}=\frac{\hbar}{4GM}-\frac{\hbar}{\left(4GM\right)^{2}}\epsilon+\frac{3}{2}\frac{\hbar}{\left(4GM\right)^{3}}\epsilon^{2}+...$

Thus for very small distances from the event horizon, the energy the particle has at infinity tends to a constant.

## Black hole radiation: energy of emitted particles

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Box 16.1.

In Schwarzschild (S) space-time, the energy of a particleis given by

$\displaystyle e=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau} \ \ \ \ \ (1)$

Outside the event horizon ${r>2GM}$ and since the S coordinate ${t}$ represents time, it must constantly increase as the proper time ${\tau}$ increases, so ${dt/d\tau>0}$, with the result that ${e>0}$. Inside the event horizon ${r<2GM}$, but time and space swap roundso ${dt/d\tau}$ can be either positive or negative, since ${t}$ is now a space coordinate. The the energy can be positive or negative inside the event horizon. For a single particle moving along a geodesic, ${e}$ is a constant of the motion, so since ${1-\frac{2GM}{r}}$ becomes zero at the event horizon, ${dt/d\tau}$ becomes infinite which results from the fact that the S time becomes infinite at the horizon. After crossing the horizon, ${1-\frac{2GM}{r}<0}$ so ${dt/d\tau}$ must also be negative to keep ${e}$ constant. However, it is possible for the particle to interact with another particle inside the horizon, which could cause ${dt/d\tau}$ to change sign, resulting in a negative energy.

The idea behind black hole radiation is actually quite simple. Quantum field theory predicts that vacuum fluctuations can occur, in which a particle-antiparticle pair spontaneously appears, essentially out of nothing, provided that the energies of the two particles are equal and opposite; that is, one particle has positive energy ${E}$ while the other has negative energy ${-E}$. This phenomenon can occur only for a very short time interval of the order of ${\Delta t\sim\hbar/E}$ (at this stage, you can just accept all this as god-given, since we haven’t studied quantum field theory yet), after which the particle recombines with its antiparticle.

Now suppose this pair creation event occurs very close to the event horizon, and the negative energy particle crosses the horizon before it has a chance to recombine, and that the positive energy particle therefore escapes to infinity. The energy of the black hole is thus reduced by ${E}$ and the energy ${E}$ is radiated away to infinity. To see how this works in a simplified model, suppose the pair creation event occurs at some small distance ${\epsilon}$ above the event horizon. We need ${\epsilon}$ to be small enough that the negative energy particle can cross the horizon before it recombines, so we need to estimate how long it takes the particle to travel to the horizon. Since we’re outside the horizon, the S coordinate ${r}$ is still spatial, so we can use the following expressionto estimate the proper time required:

$\displaystyle \frac{dr}{d\tau}=-\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)\left(1+\frac{\ell^{2}}{r^{2}}\right)} \ \ \ \ \ (2)$

We’ve used the minus sign to indicate that ${r}$ is decreasing. If the particle falls radially, the angular momentum is ${\ell=0}$, so the proper time required is

$\displaystyle \Delta\tau=-\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)}} \ \ \ \ \ (3)$

To go further, we need to know ${e}$. It’s unlikely that the pair is created at rest, but suppose we’re in a locally flat reference frame that is released from rest with its origin at ${r=2GM+\epsilon}$ and that the pair creation event occurs at this location. An object in the observer’s frame has an energy per unit massof

$\displaystyle e=\sqrt{1-\frac{2GM}{2GM+\epsilon}} \ \ \ \ \ (4)$

so for such an object the integral becomes

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle -\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{1-\frac{2GM}{2GM+\epsilon}-\left(1-\frac{2GM}{r}\right)}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{\frac{2GM}{r}-\frac{2GM}{2GM+\epsilon}}} \ \ \ \ \ (6)$

This integral can be done using Maple although the result is a bit too complex to write out here, and then a series expansion of the result around ${\epsilon=0}$ gives the leading terms

$\displaystyle \Delta\tau=2\sqrt{2GM\epsilon}+\frac{5}{6}\sqrt{\frac{2}{GM}}\epsilon^{3/2}+\mathcal{O}\left(\epsilon^{5/2}\right) \ \ \ \ \ (7)$

If we want just the first term, we can transform the integral a bit using ${\rho\equiv r-2GM}$:

$\displaystyle \Delta\tau=\int_{0}^{\epsilon}\frac{d\rho}{\sqrt{\frac{2GM}{\rho+2GM}-\frac{2GM}{2GM+\epsilon}}} \ \ \ \ \ (8)$

If we expand each term in the square root in a series and keep terms up to first order in ${\rho}$ and ${\epsilon}$ we get

$\displaystyle \frac{2GM}{\rho+2GM}-\frac{2GM}{2GM+\epsilon}=\frac{1}{2GM}\left(\epsilon-\rho\right)+\mathcal{O}\left(\epsilon^{2}\right)+\mathcal{O}\left(\rho^{2}\right) \ \ \ \ \ (9)$

so to this order, we have

$\displaystyle \Delta\tau=\int_{0}^{\epsilon}\frac{d\rho}{\sqrt{\left(\epsilon-\rho\right)/2GM}} \ \ \ \ \ (10)$

We can use another substitution ${u\equiv\epsilon-\rho}$ to get

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \int_{0}^{\epsilon}\frac{du}{\sqrt{u/2GM}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\sqrt{2GM\epsilon} \ \ \ \ \ (12)$

This is the time required for an object released from rest at ${r=2GM+\epsilon}$ to reach the event horizon. Presumably if the particle is not at rest when it is created, but is heading towards the event horizon, it will require less time than this to reach it.

Plugging this into the field theory estimate above, we can get an estimate of the energy of the radiated particle:

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \frac{\hbar}{E}\ \ \ \ \ (13)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2\sqrt{2GM\epsilon}} \ \ \ \ \ (14)$

I have to admit I’m not particularly comfortable with this calculation, since ${\Delta\tau}$ is calculated for an object released at rest from ${r=2GM+\epsilon}$ whereas the pair of particles would probably not be produced at rest. However, as an order of magnitude estimate, I suppose it’s reasonable.

## Painlevé-Gullstrand coordinates: derivation using a local flat frame

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.1.

Earlier, we worked out the basis vectors in a locally flat frame for a freely falling observer near a black hole. These basis vectors are worked out by considering the four-velocity in two frames: the local, flat frame, and the Schwarzschild (S) frame. In particular, in the flat frame, ${\mathbf{u}=\mathbf{o}_{t}=\left[1,0,0,0\right]}$ so in the other frame, the four-velocity is the transformed time basis vector: ${\mathbf{u}^{\prime}=\mathbf{o}_{t}^{\prime}}$. Using this argument, we worked out ${\mathbf{o}_{t}^{\prime}}$ in the S frame for a freely falling observer and got

$\displaystyle \mathbf{o}_{t}^{\prime}=\left[\left(1-\frac{2GM}{r}\right)^{-1},-\sqrt{\frac{2GM}{r}},0,0\right] \ \ \ \ \ (1)$

In the flat frame, we can write the interval between two events as ${d\mathbf{s}=\left[d\tau,dx,dy,dz\right]}$. In the Painlevé-Gullstrand system, the time coordinate ${\mathring{t}}$ is just the proper time of a freely falling observer, so ${d\mathring{t}=d\tau}$. Still in the flat frame, we have therefore

$\displaystyle d\mathring{t}=-\eta_{ij}o_{t}^{i}ds^{j}=-\mathbf{o}_{t}\cdot d\mathbf{s} \ \ \ \ \ (2)$

since only the component ${o_{t}^{t}}$ is non-zero, and ${\eta_{tt}=-1}$ in flat space. Since this is a scalar product, it has the same value in any coordinate system, such as the S system where we have

 $\displaystyle d\mathring{t}$ $\displaystyle =$ $\displaystyle \mathbf{o}_{t}^{\prime}\cdot d\mathbf{s}^{\prime}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}o_{t}^{\prime i}ds^{\prime j} \ \ \ \ \ (4)$

In the S system, we have

$\displaystyle d\mathbf{s}^{\prime}=\left[dt,dr,d\theta,d\phi\right] \ \ \ \ \ (5)$

so

 $\displaystyle d\mathring{t}$ $\displaystyle =$ $\displaystyle -\left[-\left(1-\frac{2GM}{r}\right)\right]\left(1-\frac{2GM}{r}\right)^{-1}dt-\left(1-\frac{2GM}{r}\right)^{-1}\left(-\sqrt{\frac{2GM}{r}}\right)dr\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle dt+\left(1-\frac{2GM}{r}\right)^{-1}\sqrt{\frac{2GM}{r}}dr \ \ \ \ \ (7)$

This agrees with the earlier result for Painlevé-Gullstrand.

## Kruskal-Szekeres diagrams: another space ship disaster

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.8.

In this example, we have a spaceship originally at some fixed distance from a black hole, but the engines fail and the ship starts to fall vertically downward (tail first) towards the black hole.

In the KS diagram, the tail of the ship follows the black world line and the front of the ship follows the parallel magenta world line. The diagonal lines with slope ${+1}$ indicate photons emitted by the tail end (with the green line being also the event horizon). The violet photon is emitted before the tail crosses the horizon and is received by the front before it crosses the horizon. A photon emitted just as the tail crosses the horizon follows the green world line, and since this line also corresponds to a constant ${r=2GM}$, the photon remains at the event horizon and is received by the front when the front crosses the horizon. The brown photon leaves the tail after it crosses the horizon and is received by the front after it too crosses the horizon. Thus there isn’t any time during the period where the ship is crossing the horizon that the front receives no photons from the tail, but any photons received by the front are emitted by the tail when both the front and the tail are on the same side of the horizon. When the front is outside the horizon and the tail is inside, the front still sees the tail, but it is seeing the tail as it was before it crossed the horizon.

Closer to the singularity at ${r=0}$, however, there is a cutoff point beyond which any photons (such as the short red line) emitted by the tail are absorbed at ${r=0}$ before reaching the front, so the front end will not be able to see the tail hit ${r=0}$.

## Kruskal-Szekeres diagrams: saving a space shuttle

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problems 15.7 15.9.

In this example (Moore, problem P15.9), we have a spaceship maintaining a constant distance ${R}$ from a black hole, such that the Kruskal-Szekeres (KS) coordinates satisfy

$\displaystyle u^{2}-v^{2}=\left(\frac{R}{2GM}-1\right)e^{R/2GM}=1$

At Schwarzschild (S) time ${t=0}$ (corresponding to ${v=0}$), a shuttle leaves the spaceship and gets pulled towards the black hole along the line ${u=1}$ in the KS diagram. If the mother ship is capable of speeds up to light speed, what is the latest time that it can leave its orbit to intercept the shuttle before the shuttle crosses the event horizon?

In the diagram, the shuttle’s world line is shown in yellow and the ship’s world line is the grey hyperbola. If the ship suddenly breaks out of its orbit and travels at light speed towards the black hole, and intercepts the shuttle just before it crosses the event horizon, the ship will have to follow the turquoise diagonal. (Remember that photon world lines have slopes of ${\pm1}$ on a KS diagram.) We therefore need to find the intersection of the turquoise line and the grey hyperbola, and then find the S time ${t_{r}}$ at which the ship starts on its rescue mission. On a KS diagram, curves of constant ${t}$ are straight lines through the origin, so the time is the thin green line in the diagram.

The turquoise line has slope ${-1}$ and passes through the point ${\left(1,1\right)}$ so its equation is

 $\displaystyle v-1$ $\displaystyle =$ $\displaystyle -\left(u-1\right)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle -u+2$

The intersection with the hyperbola is

 $\displaystyle u^{2}-\left(-u+2\right)^{2}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle 4u-4$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle u$ $\displaystyle =$ $\displaystyle \frac{5}{4}$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{3}{4}$

The S time is

 $\displaystyle t_{r}$ $\displaystyle =$ $\displaystyle 2GM\ln\frac{u+v}{u-v}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2GM\ln\frac{8}{2}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4GM\ln2$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 2.77GM$

As a slight variant on this problem (Moore, problem P15.7), suppose you are working on the outside of the spaceship hovering at radius ${R}$ and you drop a valuable piece of equipment, which then falls towards the black hole. Even though an observer at infinity would say that the object gets stuck at the event horizon (since the S ${t}$ coordinate becomes infinite there), the KS diagram shows that you have a limited time (as measured by your own proper time) to go after the object if you are to catch it before it crosses the horizon. In this case, the dropped object has a world line similar to the yellow line on the diagram (not quite the same line, since the dropped object would follow a geodesic, which isn’t a straight line here), and to catch it you’d have to leave your orbit at light speed along a turquoise diagonal that intercepts the object just before it crosses the horizon.

## Kruskal-Szekeres metric: more fun with photons

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problems 15.5, 15.6.

Here’s another example of using a Kruskal-Szekeres (KS) diagram. We have the same observer in a space ship falling into a black hole as before, but this time he is firing photon torpedoes radially outward. If these torpedoes follow photon world lines is there any danger that they will hit the ship before it reaches ${r=0}$? It might seem this is possible, since everything (even light) moves inwards inside the event horizon, so both the torpedoes and the ship are heading for ${r=0}$. However, looking at the KS diagram, we see this won’t happen:

The ship is the heavy black curve. Photons fired radially outwards will have world lines with slope +1, so the three heavy red lines represent three torpedoes fired from the ship after it enters the event horizon. We can see that these torpedoes won’t hit the ship, although they will get absorbed by the singularity at ${r=0}$.

As a variant on this problem, suppose that the black curve represents the surface of a collapsing star instead of a falling space ship. After the star collapses through its event horizon, it can still emit photons radially outwards. If another ship fell into the event horizon as the star is collapsing, it might follow the thick purple world line.

Since the ship’s world line intersects the lines of the emitted photons, this observer would still see light from the surface of the star as it collapses, so the interior of a black hole isn’t dark, at least until the star has finished collapsing.