by Michael Brunnbauer, 2023-06-14

Check out Exploring
Black Holes, Second Edition by Edwin F. Taylor, John Archibald Wheeler
and Edmund Bertschinger for an accessible introduction to general relativity
using only calculus and algebra.

See also: special
relativity cheatsheet

Quantity | Old unit | New unit | Conversion |
---|---|---|---|

Mass | \( kg \) | \( m \) | \( M = m_{old} \cdot \frac{G}{c^2} \) |

A worldline is a timelike/lightlike sequence of events in spacetime. It exists independent of any reference frame (aka labelling of events or coordinate system). The theory should be constructed such that everyone agrees on the proper time along a worldline. But notions like \( \Delta t \) and \( \Delta x \) are now useless as they depend not on an inertial frame - but on the worldline. Every observer has a different "coordinate system". So an invariant in general relativity is a quantity that has the same value when calculated using different (almost) arbitrary global coordinate systems. One such coordinate system are Schwarzschild coordinates \( t, r, \) and \( \phi \). They apply to any non-spinning, uncharged massive center of attraction.

The choice of polar spatial coordinates \( r, \phi \) is motivated by the fact that undisturbed objects stay in a plane through the center of a non-spinning, uncharged massive object. \( t, r, \) and \( \phi \) do not correspond to any particular observer and should not be seen as measureable quantities - although \( r \) and \( \phi \) correspond to arc lengths measured on circles around the massive object. If the circumference of such a circle is measured as \( C \), then the circle is at \( r \) - coordinate \( \frac{C}{2 \cdot \pi} \). If a point on the circle is at arc length \( L \), then the point is at \( \phi \) - coordinate \( \frac{L}{r} \). The global coordinate \( t \) can be thought of as the clock of a bookkeeper at rest very far away from the center of attraction collecting local measurements relayed to him.

A metric for a specific global coordinate system will express the invariant differential of proper time \( \tau \) or ruler distance \( \sigma \) in terms of global coordinates and differentials of global coordinates. Integrating over global coordinates will yield \( \Delta \tau \) and \( \Delta \sigma \) for a specific observer.

- The \( \phi \) coordinate behaves just like it would in flat spacetime with polar coordinates, while \( t \) and \( r \) do not.
- For \( M = 0 \) or very large \( r \), the metric reduces to the metric of flat spacetime.
- The metric has a singularity at \( r = 0 \) and \( r = 2 \cdot M \). The surface described by the second condition is called the event horizon. It is a fictitious singularity caused by using Schwarzschild coordinates and does not occur with other coordinate systems.
- Although \( r \) and \( \phi \) can be connected to measurements, we have to be careful with assumptions about \( t \). The only thing we know about \( t \) is that it behaves like flat spacetime \( t \) for very large \( r \) or very small \( M \).

The first two equations do not work inside the event horizon as the quantities become imaginary. The second equation is not defined at the event horizon and the first is suspiciously zero. We will see later that no observer can remain at constant \( r \) at or inside the event horizon so these equations can only be used outside it. This is why the quantities have the subscript \( s \) for shell. They are measurements of observers on static shells built outside the event horizon. No such shells are possible at or inside the event horizon.

Rotate this around the axis named \( Z(r) \) and you get the famous embedding diagram illustrating stretching of space in a plane around a black hole.

You can verify with the metric that as \( 1 - \frac{2 \cdot M}{r} \) becomes negative inside the event horizon, a separation with \( d\phi = 0, dr = 0 \) and \( dt \neq 0 \) becomes spacelike and a separation with \( d\phi = 0, dt = 0 \) and \( dr \neq 0 \) becomes timelike. The arrow pointing to the future must indeed be horizontal. It is as if time and space have switched roles. Does the future arrow have to point towards the center? The math does not demand it - it could point away from the center. If the singularity spits matter out instead of swallowing it, we have a white hole, which is of course a controversial notion.

The global \( t \) coordinate can run backward, as you can see with the worldline coming from the top when \( t_1, r_1 \) is inside the black hole. Also, light seems to be infinitely slow near the event horizon on both sides - but it should be measured as moving with \( c \) anywhere. The global \( t \) coordinate has a problem. It gets ever more "diseased" near the event horizon - deviating more and more from what is actually measured. Like no map projection of the surface of earth can reflect actual distances correctly everywhere, no global coordinate system of curved spacetime can reflect measurements everywhere. The problem with \( t \) is a specific tradeoff of Schwarzschild coordinates.

However, we can see that time \( \tau \) passes and does not run backward for an observer inside a black hole and that all matter inside must inevitably fall towards the singularity at the center.

Let an object move from \( t_1, r_1, \phi_1 \) via \( t_2, r_2, \phi_2 \) to \( t_3, r_3, \phi_3 \). We will assume that each of the two spacetime segments \( A \) and \( B \) is small enough to be approximately flat such that we can replace differentials with differences and \( r \) with its average over the segment: $$ \overline{r_A} = \frac{r_1 + r_2}{2} $$ $$ \overline{r_B} = \frac{r_2 + r_3}{2} $$ $$ K_A = 1 - \frac{2 \cdot M}{\overline{r_A}} $$ $$ K_B = 1 - \frac{2 \cdot M}{\overline{r_B}} $$ $$ \tau_A \approx \sqrt{ K_A \cdot (t_2 - t_1)^2 - \frac{(r_2 - r_1)^2}{K_A} - \overline{r_A}^2 \cdot (\phi_2 - \phi_1)^2 } $$ $$ \tau_B \approx \sqrt{ K_B \cdot (t_3 - t_2)^2 - \frac{(r_3 - r_2)^2}{K_B} - \overline{r_B}^2 \cdot (\phi_3 - \phi_2)^2 } $$ $$ \frac{d\tau_A}{dt_2} \approx K_A \cdot \frac{(t_2 - t_1)}{\tau_A} $$ $$ \frac{d\tau_B}{dt_2} \approx - K_B \cdot \frac{(t_3 - t_2)}{\tau_B} $$ $$ \frac{d\tau_A}{dt_2} + \frac{d\tau_B}{dt_2} \approx K_A \cdot \frac{(t_2 - t_1)}{\tau_A} - K_B \cdot \frac{(t_3 - t_2)}{\tau_B} \approx 0 $$ $$ K_A \cdot \frac{(t_2 - t_1)}{\tau_A} \approx K_B \cdot \frac{(t_3 - t_2)}{\tau_B} $$ If we shrink both spacetime segments to differentials, the expression on both sides becomes: $$ \left( 1 - \frac{2 \cdot M}{r} \right) \cdot \frac{dt}{d\tau} $$ For \( M = 0 \) or very large \( r \), this reduces to \( \frac{dt}{d\tau} = \gamma \). Multiplied by the mass of the object, this is the relativistic energy in flat spacetime, so we conclude: $$ E \equiv \left( 1 - \frac{2 \cdot M}{r} \right) \cdot \frac{dt}{d\tau} \cdot m \quad \text{(9)} $$

- The equation is valid everywhere except \( r = 0 \) and \( r = 2 \cdot M \) - we can use it inside the event horizon.
- It describes every movement in a plane through the center of a non-spinning, uncharged massive object - not only radial movement.
- \( dt \) has to be negative inside the event horizon for positive energy. This corresponds to worldlines coming from the top inside the event horizon in the above chart (stuff falling radially inward). That an object with a worldline coming from the bottom inside the event horizon has negative energy should not bother us too much now. As already stated: map energy should not be confused with measured energy.

None of these functions are defined at the event horizon - but \( \frac{dr}{d\tau} \) makes the most convincing case that objects actually cross the event horizon.

\( \frac{E}{m}=1 \) corresponds with an object starting at rest infinitely far away from the massive center of attraction. We call such an object a raindrop. Objects with \( \frac{E}{m}>1 \) come in with greater velocity than the raindrop - we call them hailstones. Objects with \( \frac{E}{m}<1 \) start at rest from a finite distance and are called drips. Outside the event horizon, the chart also describes outgoing movement with \( dr > 0 \). It looks like all objects have the same speed when near the event horizon but that is not the case. Only in the limiting case \( r = 2 \cdot M \) is \( v_s = 1 \) and at that point, it is already immeasureable.

Why is \( \frac{dr}{d\tau} = v_s \) for \( \frac{E}{m}=1 \)? The math says so but can we explain it physically? The magnitude of \( v_s \) is equal to \( \frac{dy_r}{d\tau} \) - where \( dy_r \) is the change of distance as measured on the raindrop. In order to convert \( dy_r \) to \( dr \), we have to account for length contraction and space stretching. From equation 10 we can see that at \( \frac{E}{m}=1 \): $$ \gamma_s = \frac{1}{\sqrt{1-\frac{2 \cdot M}{r}}} $$ Length contraction (\( \gamma^{-1} \)) and space stretching cancel out. In the general case, the product of Length contraction and space stretching is \( \frac{m}{E} \), so: $$ \frac{dr}{d\tau} = v_s \cdot \frac{E}{m} \quad \text{(14)} $$ This can be easily verified with the equations 11 and 13 above. Should we worry that \( \frac{dr}{d\tau} \) can be greater than 1 outside the event horizon? No - we did not worry about rulers in motion being shorter than those at rest in special relativity either.

For an object starting at rest from \( r_0 \) we have \( \gamma_s = 1 \), so: $$ m \cdot 1 = \frac{1}{\sqrt{1-\frac{2 \cdot M}{r_0}}} \cdot E $$ $$ \frac{m^2}{E^2} = \frac{1}{1-\frac{2 \cdot M}{r_0}} $$ $$ |a_s| = \frac{M}{r^2} \cdot \frac{\sqrt{1 - \frac{2 \cdot M}{r}}}{1 - \frac{2 \cdot M}{r_0}} \quad \text{(16)} $$ If we only want the initial acceleration at rest at \( r \), we set \( r = r_0 \): $$ |a_s| = \frac{M}{r^2} \cdot \frac{1}{\sqrt{1 - \frac{2 \cdot M}{r}}} \quad \text{(17)} $$

The graph shows \( \frac{V_{eff}}{m} \) from \( 0 \le r \le 30 \cdot M \) in red and \( \frac{E}{m} \) in green. Any \( r \) where the red line is above the green line is forbidden. The dashed blue lines show radii with special significance: \( r = 2 \cdot M \), \( r = 3 \cdot M \) and \( r = 6 \cdot M \). Any intersection of \( \frac{E}{m} \) with \( \frac{V_{eff}}{m} \) is a turning point where \( \frac{dr}{d\tau} = 0 \). An intersection with a a minimum of \( \frac{V_{eff}}{m} \) marks a stable circular orbit and an intersection with a maximum marks an unstable circular orbit. Intersections at the two borders of a potential well are bounce points where \( \frac{dr}{d\tau} \) changes sign. These mark the elliptical orbits.

Here is another one where you can change tangential shell velocity and position of an orbiting satellite to see how the potential and its position inside it changes:

From 9.2 and 10: $$ \frac{L}{m} = r \cdot v_{ts} \cdot \gamma_s $$ $$ \frac{E}{m} = \gamma_s \cdot \sqrt{1 - \frac{2 \cdot M}{r}} $$

What's the shell speed for circular orbits? From 9.2: $$ v_{ts} = \frac{L}{m \cdot r} \cdot \sqrt{1 - v_{ts}^2} $$ $$ v_{ts}^2 = \frac{L^2}{m^2 \cdot r^2} \cdot (1 - v_{ts}^2) $$ $$ \frac{v_{ts}^2}{1 - v_{ts}^2} = \frac{L^2}{m^2 \cdot r^2} $$ $$ \frac{1 - v_{ts}^2}{v_{ts}^2} = \frac{m^2 \cdot r^2}{L^2} $$ $$ \frac{1}{v_{ts}^2} - 1 = \frac{m^2 \cdot r^2}{L^2} $$ $$ v_{ts}^2 = \frac{1}{1 + \frac{m^2 \cdot r^2}{L^2}} $$ From 17.3, we get: $$ \frac{m^2 \cdot r^2}{L^2} = \frac{r}{M} - 3 $$ Which we can substitute into the previous equation: $$ v_{ts}^2 = \frac{1}{\frac{r}{M} - 2} = \frac{M}{r - 2 \cdot M} $$ So for the innermost stable circular orbit at \( r = 6 \cdot M \), \( v_{ts} = 0.5 \) and for the innermost unstable circular orbit at \( r = 3 \cdot M \), \( v_{ts} = 1 \). Only light can orbit a black hole at \( r = 3 \cdot M \).

For a raindrop, we have from 12 and 13: $$ \frac{E}{m} = 1 $$ $$ \frac{dr}{dt} = - \left( 1-\frac{2 \cdot M}{r} \right) \cdot \sqrt{\frac{2 \cdot M}{r}} $$ $$ \frac{dt}{d\tau} = \frac{1}{1-\frac{2 \cdot M}{r}} $$ Global coordinates need to have an exact differential. This would be true if \( T = t + f(r) \) such that: $$ dT = d[t+f(r)] = dt + \frac{df(r)}{dr} \cdot dr $$ Such a form can be created from the equations above: $$ dT = d\tau = dt \cdot \left( 1-\frac{2 \cdot M}{r} \right) $$ $$ dT = dt \cdot \left( 1 - \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot \left( 1-\frac{2 \cdot M}{r} \right) \cdot \sqrt{\frac{2 \cdot M}{r}} \right) $$ $$ dT = dt \cdot \left( 1 + \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot \frac{dr}{dt} \right) $$ $$ dT = dt + \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot dr $$ $$ \frac{df(r)}{dr} = \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} $$

- If we replace \( t \) with \( t(r) \) for a raindrop, \( T = t(r) + f(r) \) should be differentiable everywhere except at r = 0.
- \( t(r) \) can be determined by solving the equation for \( \frac{dr}{dt} \) above for dt and integrating. But that integral does not converge if we integrate from infinity to \( r \). It takes the raindrop infinitely long to get to a finite \( r \). We can only calculate differences \( t(r_1) - t(r_2) \) between two radii with the integral.
- \( t(r_1) - t(r_2) \) has a positive term that blows up near \( r_1 = 2 \cdot M \) - so we need something in \( f(r_1) - f(r_2) \) countering that.
- It can be shown that \( f(r_1) - f(r_2) \) has a negative term that blows up near \( r_1 = 2 \cdot M \) and that at \( r_1 = 2 \cdot M \), this term is equal and opposite in sign to the corresponding troublemaker in \( t(r_1) - t(r_2) \).
- The equation for dT can also be derived by doing a Lorentz transformation for a shell observers \( dt_s \) into the raindrop frame.

- There are no singularities except at \( r = 0 \).
- The metric correctly reduces to flat spacetime For \( M = 0 \) or very large \( r \).
- The metric should predict the same measurement results as the Schwarzschild metric for identical situations.

Are raindrop frame coordinates valid at and inside the event horizon? Yes: Like the GP metric, they do not have a singularity except at \( r = 0 \) and they correctly reproduce the GP metric. The form of the GP metric as an algebraically equivalent sum/difference of squares - in this case local frame coordinates - is called a tetrad form (tetrad because a metric with all three spacial dimensions would have four terms).

What is the meaning of the equation for \( dt_s \)? The time elapsed in the raindrop frame \( dT \) is not sufficient to determine elapsed time in the shell frame - a Lorentz transformation needs information about time and space. We need to supply a matching \( dr \). We can of course plug in \( dr = 0 \) and get a correct equation where \( dT \) behaves exactly like \( dt \) (gravitational time dilation follows from that). But whenever \( dr \neq 0 \), we have to plug it in here to get the correct \( dt_s \).

Be careful about the size of the spacetime patch you treat as flat by using these as coordinates. They also fail at or inside the event horizon - again confirming that nothing can stay at rest there.

Inside the event horizon, even the outgoing solution moves inward and exactly at the event horizon, outgoing light gets stuck. Will somebody measure light as going "slower than light"? No. For the observer in the raindrop frame, use equation 22 to get: $$ \frac{dy_r}{dt_r} = \frac{dy_r}{dT} = \frac{dr}{dT} + \sqrt{\frac{2 \cdot M}{r}} $$ Using equation 27 yields: $$ \frac{dy_r}{dt_r} = \pm 1 $$ For the shell observer, use equation 24+25 to get: $$ \frac{dy_s}{dt_s} = \frac{dr}{ \left( 1-\frac{2 \cdot M}{r} \right) \cdot dT - \sqrt{\frac{2 \cdot M}{r}} \cdot dr } = \frac{dr}{dT} \cdot \frac{1}{1 - \frac{2 \cdot M}{r} - \sqrt{\frac{2 \cdot M}{r}} \cdot \frac{dr}{dT} } $$ Use equation 27 with \( \pm 1 \) replaced by \( Q \): $$ \frac{dy_s}{dt_s} = \frac{dr}{dT} \cdot \frac{1}{ 1 - Q \cdot \sqrt{\frac{2 \cdot M}{r}}} = \frac{Q - \sqrt{\frac{2 \cdot M}{r}}}{ 1 - Q \cdot \sqrt{\frac{2 \cdot M}{r}}} = \pm 1 $$