# General relativity cheatsheet

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.

### Conversion of units

Measuring mass in meters will greatly simplify the equations. This can be accomplished by multiplying it with the universal gravitational constant ($$\frac{m^3}{kg \cdot s^2}$$) and dividing by the speed of light ($$\frac{m^2}{s^2}$$). The resulting conversion factor $$\frac{G}{c^2}$$ is approximately $$7.42 \cdot 10^{-28} \frac{m}{kg}$$.

Quantity Old unit New unit Conversion
Mass $$kg$$ $$m$$ $$M = m_{old} \cdot \frac{G}{c^2}$$

### Schwarzschild coordinates

In curved spacetime, inertial frames with flat spacetime metric become approximations only valid for limited regions of spacetime. What you measure depends on where you are and where you are going. It depends on your worldline.

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.

### Schwarzschild metric

Schwarzschild coordinates $$t, r,$$ and $$\phi$$ yield the Schwarzschild metric when submitted to Einstein's field equations. It applies to all space outside a non-spinning, uncharged massive center of attraction. In the case of a black hole, the size of the center of attraction reduces to a point so that the metric applies everywhere outside the point singularity at the center. The objects moving outside the massive object are considered of negligible mass such that they do not curve spacetime. $$d\tau^2 = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 - \frac{dr^2}{1-\frac{2\cdot M}{r}} - r^2 \cdot d\phi^2 \quad \text{(1 - timelike)}$$ $$d\sigma^2 = - d\tau^2 = - \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 + \frac{dr^2}{1-\frac{2\cdot M}{r}} + r^2 \cdot d\phi^2 \quad \text{(2 - spacelike)}$$ Observations:
• 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 $$M$$.

### Local shell measurements

$$dr = 0$$ clearly describes and object with no radial motion and $$d\phi = 0$$ must be an object with no tangential motion. It is also safe to assume that $$dt = 0$$ describes simultaneity when at rest relative to the center of attraction. So we can calculate the result of local measurements by setting two of the differentials to zero in the metric: $$dt_s = \sqrt{1 - \frac{2 \cdot M}{r}} \cdot dt \quad \text{(3)}$$ $$dy_s = \frac{dr}{\sqrt{1-\frac{2\cdot M}{r}}} \quad \text{(4)}$$ $$dx_s = r \cdot d\phi \quad \text{(5)}$$ The spacetime patch described by $$dt_s, dy_s$$ and $$dx_s$$ is infinitely small so we can ignore curvature and use all laws of special relativity with the differentials (applying flat spacetime metric in fact recovers the Schwarzschild metric).

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.

### Space stretching

Radially measured separation between $$r_L$$ (lower shell) and $$r_H$$ (higher shell) is greater than $$r_H - r_L$$: $$dt = d\phi = 0$$ $$r_L > 2 \cdot M$$ $$r_H > r_L$$ $$d\sigma = \frac{dr}{\sqrt{1-\frac{2 \cdot M}{r}}}$$ Integrate from $$r_L$$ to $$r_H$$: $$\sigma = \int\limits_{r_L}^{r_H} \frac{dr}{\sqrt{1-\frac{2 \cdot M}{r}}} = \int\limits_{r_L}^{r_H} \frac{\sqrt{r} \cdot dr}{\sqrt{r-2 \cdot M}}$$ Substitute $$r = z^2$$ to enable Euler substitution : $$r = z^2$$ $$dr = 2 \cdot z \cdot dz$$ $$\sigma = \int\limits_{z_L}^{z_H} \frac{2 \cdot z^2 \cdot dz }{\sqrt{z^2 - 2 \cdot M}}$$ Do Euler's first substitution ($$t$$ is just a substitution variable - not the global $$t$$ coordinate): $$\sqrt{z^2 - 2 \cdot M} = t - z$$ $$z = \frac{2 \cdot M + t^2}{2 \cdot t} = \frac{M}{t} + \frac{t}{2}$$ $$dz = \left( \frac{1}{2} - \frac{M}{t^2} \right) \cdot dt$$ $$\int \frac{2 \cdot z^2 \cdot dz}{\sqrt{z^2 - 2 \cdot M}} = \int \frac{2 \cdot \left( \frac{M}{t} + \frac{t}{2} \right) ^2 \cdot \left( \frac{1}{2} - \frac{M}{t^2} \right) \cdot dt }{t - \left( \frac{M}{t} + \frac{t}{2} \right) } = \int \frac{2 \cdot \left( \frac{M}{t^2} + \frac{1}{2} \right) ^2 \cdot t \cdot \left( \frac{t}{2} -\frac{M}{t} \right) \cdot dt }{\frac{t}{2} - \frac{M}{t}}$$ $$= 2 \int \left( \frac{M}{t^2} + \frac{1}{2} \right) ^2 \cdot t \cdot dt = 2 \int \left( \frac{M^2}{t^4} + \frac{M}{t^2} + \frac{1}{4} \right) \cdot t \cdot dt = \int \left( \frac{2 \cdot M^2}{t^3} + \frac{2 \cdot M}{t} + \frac{t}{2} \right) \cdot dt$$ The antiderivative is: $$\frac{t^2}{4} - \frac{M^2}{t^2} + 2 \cdot M \cdot \ln |t|$$ We now show that $$\frac{t^2}{4} - \frac{M^2}{t^2} = \frac{t^4 - 4 \dot M^2}{4 \cdot t^2} = z \cdot \sqrt{z^2 - 2 \cdot M}$$ Proof: $$t^2 = (\sqrt{z^2 - 2 \cdot M} + z)^2 = z^2 - 2 \cdot M + 2 \cdot z \cdot \sqrt{z^2 - 2 \cdot M} + z^2 = 2 \cdot (z^2 - M + z \cdot \sqrt{z^2 - 2 \cdot M} )$$ $$t^4 = 4 \cdot ( z^4 - z^2 \cdot M + z^3 \cdot \sqrt{z^2 - 2 \cdot M} - z^2 \cdot M + M^2 - z \cdot M \cdot \sqrt{z^2 - 2 \cdot M} + z^3 \cdot \sqrt{z^2 - 2 \cdot M}$$ $$- z \cdot M \cdot \sqrt{z^2 - 2 \cdot M} + z^4 - z^2 \cdot 2 \cdot M )$$ $$t^4 - 4 \cdot M^2 = 8 \cdot ( z^4 - z^2 \cdot 2 \cdot M + z^3 \cdot \sqrt{z^2 - 2 \cdot M} - z \cdot M \cdot \sqrt{z^2 - 2 \cdot M} )$$ $$4 \cdot z \cdot \sqrt{z^2 - 2 \cdot M} \cdot t^2 = 8 \cdot ( z^3 \cdot \sqrt{z^2 - 2 \cdot M} - z \cdot M \cdot \sqrt{z^2 - 2 \cdot M} + z^4 - z^2 \cdot 2 \cdot M ) = t^4 - 4 \cdot M^2$$ So the antiderivative is: $$z \cdot \sqrt{z^2 - 2 \cdot M} + 2 \cdot M \cdot \ln | z + \sqrt{z^2 - 2 \cdot M} |$$ And we get: $$\sigma = \Big| z \cdot \sqrt{z^2 - 2 \cdot M} + 2 \cdot M \cdot \ln | z + \sqrt{z^2 - 2 \cdot M} | \Big|_{z_L}^{z_H}$$ $$\sigma = \Big| \sqrt{r} \cdot \sqrt{r - 2 \cdot M} + 2 \cdot M \cdot \ln | \sqrt{r} + \sqrt{r - 2 \cdot M} | \Big|_{r_L}^{r_H} \quad \text{(6)}$$ $$\sigma$$ is well defined for $$r_L = 2 \cdot M$$, suggesting that the event horizon can actually be reached despite space stretching growing without limit near it (the limit of the Riemann integral converges, see Improper integral).

### Visualizing space stretching

Imagine a function $$Z(r)$$ with $$r > 2 \cdot M$$, for which the arc length between two points $$r_L$$ and $$r_H$$ corresponds with measured $$\sigma$$. The derivative of this function would be $$\frac{dZ}{dr}$$. We can treat the differential arc length $$d\sigma$$ as a straight line and use the Pythagorean theorem: $$dZ^2 = d\sigma^2 - dr^2$$ $$\frac{dZ}{dr} = \frac{\sqrt{d\sigma^2 - dr^2}}{dr}$$ $$d\sigma^2 - dr^2 = \frac{dr^2}{1-\frac{2 \cdot M}{r}} - dr^2 = dr^2 \cdot \left( \frac{1}{1-\frac{2 \cdot M}{r}} - 1 \right) = dr^2 \cdot \frac{2 \cdot M}{r - 2 \cdot M}$$ $$\frac{dZ}{dr} = \sqrt{2 \cdot M } \cdot \frac{1}{\sqrt{r - 2 \cdot M}}$$ $$Z(r) = \sqrt{2 \cdot M } \cdot \int\limits_{2 \cdot M}^{r} \frac{1}{\sqrt{r - 2 \cdot M}} \cdot dr = \sqrt{2 \cdot M } \cdot \Big| 2 \cdot \sqrt{r - 2 \cdot M } \Big|_{2 \cdot M}^{r} = 2 \cdot \sqrt{2 \cdot M } \cdot \sqrt{r - 2 \cdot M }$$

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.

### The problem with the Schwarzschild t-coordinate

Let's examine the worldline of a photon in global $$r, t$$ coordinates. For a photon $$d\tau = 0$$. Let the movement be radial, so $$d\phi = 0$$: $$0 = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 - \frac{dr^2}{ \left( 1-\frac{2\cdot M}{r} \right) }$$ $$dt = \pm \frac{r}{r - 2 \cdot M} \cdot dr$$ $$t - t_1 = \pm \int\limits_{r_1}^{r} \frac{r}{r - 2 \cdot M} \cdot dr$$ Substitute $$r - 2 \cdot M$$ = z: $$dr = dz$$ $$\int\limits_{z_1}^{z} \frac{z + 2 \cdot M}{z} \cdot dz = \int\limits_{z_1}^{z} \left( 1 + \frac{2 \cdot M}{z} \right) \cdot dz = \Big| z + 2 \cdot M \cdot ln |z| \Big|_{z_1}^{z}$$ $$= r - r_1 + 2 \cdot M \cdot (\ln | r - 2 \cdot M | - \ln | r_1 - 2 \cdot M |)$$ $$t - t_1 = \pm \left( r - r_1 + 2 \cdot M \cdot \ln \left| \frac{r - 2 \cdot M}{r_1 - 2 \cdot M} \right| \right) \quad \text{(7)}$$ The following chart ($$M = 1$$) shows the possible worldlines of light in global coordinates according to our solution for a chosen $$t_1, r_1$$. Those form the light cones separating future and past, timelike and spacelike. The red arrow points to the direction of maximum advance of proper time (the future).

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.

### Gravitational time dilation

Despite all problems with with the $$t$$ coordinate, we can also establish that clocks run faster at higher $$r$$ coordinate. Shift the $$t_1$$ coordinate in the chart above. The worldlines are shifted up and down but do not change shape. This means that two photons emitted radially from a higher shell at $$r_H$$ with infinitely small global $$t$$ separation $$dt$$ will arrive at a lower shell at $$r_L > 2 \cdot M$$ with the same global $$t$$ separation $$dt$$. In both cases, $$dr = 0, d\phi = 0$$, so: $$d\tau_H^2 = \left( 1-\frac{2 \cdot M}{r_H} \right) \cdot dt^2$$ $$d\tau_L^2 = \left( 1-\frac{2 \cdot M}{r_L} \right) \cdot dt^2$$ $$\frac{d\tau_H}{d\tau_L} = \sqrt{\frac{1-\frac{2 \cdot M}{r_H}}{1-\frac{2 \cdot M}{r_L}}} \quad \text{(8)}$$ The energy of the emitted photon will also be lower than the energy of the received photon (any oszillation at higher $$r$$ will be faster as seen from lower $$r$$). This manifestation of gravitational time dilation is called gravitational blueshift (towards lower $$r$$) or gravitational redshift (towards higher $$r$$).

### Map energy

We can apply the the principle of maximal aging like we did in special relativity to derive the map energy. We call it so because it is a global quantity like our coordinates and corresponds with actually measured energy only in a limiting case.

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: $$E \equiv \left( 1 - \frac{2 \cdot M}{r} \right) \cdot \frac{dt}{d\tau} \cdot m \quad \text{(9)}$$ Observations:
• 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.

### Map angular momentum

By searching the maximum proper time for variable $$\phi_2$$, one can find another conserved quantity: $$\frac{d\tau_A}{d\phi_2} \approx \overline{r_A}^2 \cdot \frac{(\phi_2 - \phi_1)}{\tau_A}$$ $$\frac{d\tau_B}{d\phi_2} \approx - \overline{r_B}^2 \cdot \frac{(\phi_3 - \phi_2)}{\tau_A}$$ $$\frac{d\tau_A}{d\phi_2} + \frac{d\tau_B}{d\phi_2} \approx \overline{r_A}^2 \cdot \frac{(\phi_2 - \phi_1)}{\tau_A} - \overline{r_B}^2 \cdot \frac{(\phi_3 - \phi_2)}{\tau_A} \approx 0$$ $$\overline{r_A}^2 \cdot \frac{(\phi_2 - \phi_1)}{\tau_A} \approx \overline{r_B}^2 \cdot \frac{(\phi_3 - \phi_2)}{\tau_A}$$ $$L \equiv r^2 \cdot \frac{d\phi}{d\tau} \cdot m \quad \text{(9.1)}$$ This is equal to the definition of relativistic angular momentum in flat spacetime: Radius multiplied with the tangential component of relativistic momentum: $$L = r^2 \cdot \frac{d\phi}{d\tau} \cdot m = r \cdot r \cdot \frac{d\phi}{dt_s} \cdot \frac{dt_s}{d\tau} \cdot m = r \cdot v_{ts} \cdot \gamma_s \cdot m = r \cdot p_{ts} \quad \text{(9.2)}$$

### Velocity

The measured energy of the object outside the event horizon will be: $$E_s = m \cdot \gamma_s = m \cdot \frac{dt_s}{d\tau} = m \cdot \sqrt{1-\frac{2 \cdot M}{r}} \cdot \frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{2 \cdot M}{r}}} \cdot E \quad \text{(10)}$$ The measured velocity will be: $$m \cdot \gamma_s = \frac{1}{\sqrt{1-\frac{2 \cdot M}{r}}} \cdot E$$ $$\frac{1}{\sqrt{1-v_s^2}} = \frac{1}{\sqrt{1-\frac{2 \cdot M}{r}}} \cdot \frac{E}{m}$$ $$1-v_s^2 = \left( 1-\frac{2 \cdot M}{r} \right) \cdot \frac{m^2}{E^2}$$ $$v_s = \pm \sqrt{ 1 - \left( 1-\frac{2 \cdot M}{r} \right) \cdot \frac{m^2}{E^2} } \quad \text{(11)}$$ To get velocity of a radially moving object in global coordinates, we solve the definition for map energy for $$d\tau$$ and substitute into the Schwarzschild metric with $$d\phi = 0$$: $$d\tau^2 = \left( 1 - \frac{2 \cdot M}{r} \right) ^2 \cdot dt^2 \cdot \frac{m^2}{E^2} = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 - \frac{dr^2}{1-\frac{2\cdot M}{r}}$$ $$\left( 1 - \frac{2 \cdot M}{r} \right) ^2 \cdot dt^2 \cdot \frac{m^2}{E^2} - \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 = - \frac{dr^2}{1-\frac{2\cdot M}{r}}$$ $$\left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 \cdot \left( \left( 1 - \frac{2 \cdot M}{r} \right) \cdot \frac{m^2}{E^2} - 1 \right) = - \frac{dr^2}{1-\frac{2\cdot M}{r}}$$ $$- \left( 1 - \frac{2 \cdot M}{r} \right) ^2 \cdot (-v_s^2) = \frac{dr^2}{dt^2}$$ $$\frac{dr}{dt} = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot v_s \quad \text{(12)}$$ Finally, we calculate $$\frac{dr}{d\tau}$$ for a radially moving object by solving the definition for map energy for $$dt$$ and substituting into the Schwarzschild metric with $$d\phi = 0$$: $$dt^2 = \frac{d\tau^2}{ \left( 1-\frac{2 \cdot M}{r} \right) ^2} \cdot \frac{E^2}{m^2}$$ $$d\tau^2 = \frac{d\tau^2}{1-\frac{2 \cdot M}{r}} \cdot \frac{E^2}{m^2} - \frac{dr^2}{1-\frac{2\cdot M}{r}}$$ $$\frac{dr^2}{1-\frac{2\cdot M}{r}} = d\tau^2 \cdot \left( \frac{1}{1-\frac{2 \cdot M}{r}} \cdot \frac{E^2}{m^2} - 1 \right)$$ $$\frac{dr^2}{d\tau^2} = \frac{E^2}{m^2} - \left( 1 - \frac{2 \cdot M}{r} \right)$$ $$\frac{dr}{d\tau} = \pm \sqrt{ \frac{E^2}{m^2} - 1 + \frac{2 \cdot M}{r} } \quad \text{(13)}$$ The equations for $$\frac{dr}{dt}$$ and $$\frac{dr}{d\tau}$$ should be valid everywhere except at $$r = 0$$ and $$r = 2 \cdot M$$. But $$v_s$$ should not be seen as measureable quantity inside the event horizon.

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.

### Acceleration

The measured acceleration of the object outside the event horizon is the derivative of the velocity by shell time: $$|a_s| = |\frac{dv_s}{dt_s}| = \frac{1}{2 \cdot v_s} \cdot \left( -\frac{2\cdot M \cdot m^2}{E^2} \right) \cdot \left( -\frac{1}{r^2} \right) \cdot \frac{dr}{dt_s}$$ $$\frac{dr}{dt_s} = \frac{dr}{dt} \cdot \frac{dt}{dt_s}$$ $$\frac{dt}{dt_s} = \frac{1}{\sqrt{1-\frac{2 \cdot M}{r}}}$$ $$\frac{dr}{dt} = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot v_s$$ $$|a_s| = \frac{M}{r^2} \cdot \frac{m^2}{E^2} \cdot \sqrt{1 - \frac{2 \cdot M}{r}} \quad \text{(15)}$$ We are not considering solutions where objects are repelled, so $$a_s$$ will always be negative.

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 $$r = r_0$$: $$|a_s| = \frac{M}{r^2} \cdot \frac{1}{\sqrt{1 - \frac{2 \cdot M}{r}}} \quad \text{(17)}$$

### General orbits

From 1, we get: $$1 = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot \frac{dt^2}{d\tau^2} - \left( 1-\frac{2\cdot M}{r} \right) ^{-1} \cdot \frac{dr^2}{d\tau^2} - r^2 \cdot \frac{d\phi^2}{d\tau^2}$$ From 9 and 9.1, we get: $$\frac{dt}{d\tau} = \frac{E}{m} \cdot \left( 1-\frac{2\cdot M}{r} \right) ^{-1}$$ $$\frac{d\phi}{d\tau} = \frac{L}{r^2 \cdot m}$$ And substitute those into the previous equation: $$1 = \frac{E^2}{m^2} \cdot \left( 1-\frac{2\cdot M}{r} \right) ^{-1} - \left( 1-\frac{2\cdot M}{r} \right) ^{-1} \cdot \frac{dr^2}{d\tau^2} - \frac{L^2}{r^2 \cdot m^2}$$ $$\frac{dr^2}{d\tau^2} = \frac{E^2}{m^2} - \left( 1-\frac{2\cdot M}{r} \right) \cdot \frac{L^2}{r^2 \cdot m^2} - \left( 1-\frac{2\cdot M}{r} \right)$$ $$\frac{dr}{d\tau} = \pm \sqrt{ \frac{E^2}{m^2} - \left( 1-\frac{2\cdot M}{r} \right) \cdot \left( 1 + \frac{L^2}{r^2 \cdot m^2} \right) } \quad \text{(17.1)}$$ Analogous to the central-force problem in classical mechanics, we can say a lot about possible orbits by defining an effective potential $$V_{eff}$$: $$\frac{V_{eff}^2}{m^2} \equiv \left( 1-\frac{2\cdot M}{r} \right) \cdot \left( 1 + \frac{L^2}{r^2 \cdot m^2} \right) \quad \text{(17.2)}$$ $$\frac{V_{eff}}{m}$$ cannot grow bigger than $$\frac{E}{m}$$ - otherwise $$\frac{dr}{d\tau}$$ would be imaginary. A plot of $$\frac{V_{eff}}{m}$$ shows the possible orbits:

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

### Circular orbits

A circular orbit is possible where $$V_{eff}$$ has zero slope. Take the derivative of both sides of 17.2 by r: $$\frac{2}{m^2} \cdot V_{eff} \cdot \frac{dV_{eff}}{dr} = \frac{2 \cdot M}{r^2} \cdot \left( 1 + \frac{L^2}{r^2 \cdot m^2} \right) - \left( 1-\frac{2 \cdot M}{r} \right) \cdot \frac{2 \cdot L^2}{r^3 \cdot m^2}$$ $$V_{eff}$$ is zero only at $$r = 2 \cdot M$$ - but the right side is not zero at $$r = 2 \cdot M$$ - so it should be zero where $$\frac{dV_{eff}}{dr}$$ is zero. $$0 = \frac{2 \cdot M}{r^2} \cdot \left( 1 + \frac{L^2}{r^2 \cdot m^2} \right) - \left( 1-\frac{2 \cdot M}{r} \right) \cdot \frac{2 \cdot L^2}{r^3 \cdot m^2}$$ $$0 = \frac{2 \cdot M}{r^2} + \frac{2 \cdot M \cdot L^2}{r^4 \cdot m^2} - \frac{2 \cdot L^2}{r^3 \cdot m^2} + \frac{4 \cdot M \cdot L^2}{r^4 \cdot m^2}$$ $$0 = \frac{1}{r^2} + \frac{L^2}{r^4 \cdot m^2} - \frac{L^2}{M \cdot r^3 \cdot m^2} + \frac{2 \cdot L^2}{r^4 \cdot m^2}$$ $$0 = \frac{1}{r^2} - \frac{L^2}{M \cdot r^3 \cdot m^2} + 3 \cdot \frac{L^2}{r^4 \cdot m^2}$$ $$0 = r^2 - \frac{L^2}{M \cdot m^2} \cdot r + 3 \cdot \frac{L^2}{m^2}$$ Solving for $$\frac{L^2}{m^2}$$ yields: $$\frac{L^2}{m^2} = \frac{r^2}{\frac{r}{M}-3} = \frac{M \cdot r^2}{r - 3 \cdot M} \quad \text{(17.3)}$$ So there are no circular orbits for $$r < 3 \cdot M$$ - otherwise $$\frac{L}{m}$$ would be imaginary. Solving the quadradic equation for $$r$$ yields: $$r = \frac{1}{2} \left( \frac{L^2}{M \cdot m^2} \pm \sqrt{ \frac{L^4}{M^2 \cdot m^4} - \frac{12 \cdot L^2}{m^2} } \right) = \frac{1}{2} \left( \frac{L^2}{M \cdot m^2} \pm \frac{L^2}{M \cdot m^2} \cdot \sqrt{ 1 - \frac{12 \cdot M^2 \cdot m^2}{L^2} } \right)$$ $$r = \frac{L^2}{2 \cdot M \cdot m^2} \cdot \left( 1 \pm \sqrt{ 1 - \frac{12 \cdot M^2 \cdot m^2}{L^2} } \right)$$ The solution where $$\pm$$ is replaced with $$-$$ corresponds to the unstable circular orbit and the one with $$+$$ to the stable circular orbit. In order for the square root to be real, $$\frac{L}{m} \ge \sqrt{12} \cdot M$$ must hold. If we insert $$\frac{L}{m} = \sqrt{12} \cdot M$$, we get $$r = 6 \cdot M$$. This is the innermost stable circular orbit (ISCO), where the effective potential has a saddle point instead of a minimum or maximum. The ISCO can be called half-stable as a small decrease in r makes the object fall into the black hole.

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 $$r = 3 \cdot M$$.

### Gullstrand-Painlevé (GP) coordinates

Gullstrand-Painlevé coordinates - also called global rain coordinates - is a coordinate system where Schwarzschild coordinate $$t$$ is replaced with global coordinate $$T$$ representing proper time elapsed for a raindrop (an object in free fall starting at rest infinitely far away). The corresponding metric can be derived independently from Einstein's field equations and does not contain a singularity at $$r = 2 \cdot M$$. Here, we will derive it from the Schwarzschild metric, which might cause us to mistrust it at $$r = 2 \cdot M$$ - but we know it can be derived independently so we will finally get rid of the problems at the event horizon.

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: $$dT = d\tau = dt \cdot \left( 1-\frac{2 \cdot M}{r} \right)$$ $$dT = dt \cdot \left( 1 - \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot \left( 1-\frac{2 \cdot M}{r} \right) \cdot \sqrt{\frac{2 \cdot M}{r}} \right)$$ $$dT = dt \cdot \left( 1 + \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot \frac{dr}{dt} \right)$$ $$dT = dt + \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot dr$$ $$\frac{df(r)}{dr} = \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1}$$ Notes:
• 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.

### Gullstrand-Painlevé (GP) metric

We can solve for $$dt$$ and substitute the term for $$dt$$ into the Schwarzschild metric: $$dt = dT - \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot dr \quad \text{(18)}$$ $$\left( 1 - \frac{2 \cdot M}{r} \right) \cdot dt^2 =$$ $$\left( 1 - \frac{2 \cdot M}{r} \right) \cdot \left( dT^2 - 2 \cdot \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot dT \cdot dr + \frac{2 \cdot M}{r} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-2} \cdot dr^2 \right) =$$ $$\left( 1 - \frac{2 \cdot M}{r} \right) \cdot dT^2 - 2 \cdot \sqrt{\frac{2 \cdot M}{r}} \cdot dT \cdot dr + \frac{2 \cdot M}{r} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot dr^2$$ The rest of the Schwarzschild metric is: $$- \frac{dr^2}{1-\frac{2\cdot M}{r}} - r^2 \cdot d\phi^2$$ We can combine the two elements with $$dr^2$$: $$\left( \frac{\frac{2 \cdot M}{r}}{1-\frac{2 \cdot M}{r}} - \frac{1}{1-\frac{2\cdot M}{r}} \right) \cdot dr^2 = - dr^2$$ So the final metric is: $$d\tau^2 = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dT^2 - 2 \cdot \sqrt{\frac{2 \cdot M}{r}} \cdot dT \cdot dr - dr^2 - r^2 \cdot d\phi^2 \quad \text{(19)}$$ $$d\sigma^2 = - d\tau^2 = - \left( 1 - \frac{2 \cdot M}{r} \right) \cdot dT^2 + 2 \cdot \sqrt{\frac{2 \cdot M}{r}} \cdot dT \cdot dr + dr^2 + r^2 \cdot d\phi^2 \quad \text{(20)}$$ Observations:
• 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.
It's time to define $$r$$ at and inside the event horizon - remember that no static shells can be built there and that our definition of $$r$$ was the reduced circumference of a shell. Assume that two objects moving radially inward are separated by $$x = r \cdot d\phi$$. If $$d\phi$$ is known for some point outside the event horizon, then $$r \equiv \frac{x}{d\phi}$$

### Raindrop frame measurements

What are the measurements of an observer riding on the raindrop? $$dt_r = dT$$ $$dy_r = dr$$ $$dx_r = r \cdot d\phi$$ Time in the raindrop frame should be the same as $$dT$$, which can be verified by substitution of $$dr$$ with $$- \sqrt{\frac{2 \cdot M}{r}} \cdot dT$$ in the GP metric - which describes the movement of the raindrop. Setting $$dT = 0$$ now means simultaneity in the raindrop frame - where space stretching is compensated by length contraction. This is not what the shell frame observer measures! To get that, multiply with $$\gamma_s$$ of the raindrop to remove the effect of length contraction.

### Raindrop frame coordinates

There is a problem though. Applying flat spacetime metric does not recover the GP metric as it did for local shell measurements expressed in Schwarzschild coordinates. The second equation is not wrong - but it is not general enough to be treated as local coordinate. $$dr = 0$$ still means stationary on a shell. If $$dr = 0$$ but $$dT \neq 0$$, then that shell moves in the raindrop frame. $$dy_r$$ must advance by $$- v_s \cdot dT$$. $$dt_r = dT \quad \text{(21)}$$ $$dy_r = dr + \sqrt{\frac{2 \cdot M}{r}} \cdot dT \quad \text{(22)}$$ $$dx_r = r \cdot d\phi \quad \text{(23)}$$ When treated as coordinates, these equations will be valid only over a sufficiently small spacetime patch.

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).

### Local shell coordinates

What are the measurements of an observer at rest on a shell outside the event horizon expressed in GP coordinates? Substitution of the term for $$dt$$ in our corresponding Schwarzschild-equations yields the correct results ($$dy_s$$ and $$dx_s$$ stay unchanged because they to not contain $$dt$$): $$dt_s = \sqrt{1 - \frac{2 \cdot M}{r}} \cdot dT - \frac{\sqrt{\frac{2 \cdot M}{r}}}{\sqrt{1-\frac{2 \cdot M}{r}}} \cdot dr \quad \text{(24)}$$ $$dy_s = \frac{dr}{\sqrt{1-\frac{2\cdot M}{r}}} \quad \text{(25)}$$ $$dx_s = r \cdot d\phi \quad \text{(26)}$$ Another way to arrive at this is to apply a Lorentz transformation to the raindrop frame coordinates. Applying the flat spacetime metric again yields the GP metric - we have another tetrad form.

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.

### Worldlines of light in GP coordinates

An alternative form of the GP metric is: $$d\tau^2 = - \left( dr + \left( 1 + \sqrt{\frac{2 \cdot M}{r}} \right) \cdot dT \right) \cdot \left( dr - \left( 1 - \sqrt{\frac{2 \cdot M}{r}} \right) \cdot dT \right) - r^2 \cdot d\phi^2$$ So for radially moving light: $$d\tau = 0$$ $$d\phi = 0$$ $$\frac{dr}{dT} = - \sqrt{\frac{2 \cdot M}{r}} \pm 1 \quad \text{(27)}$$ We want to find $$T(r)$$: $$dT = \frac{dr}{-\sqrt{\frac{2 \cdot M}{r}} \pm 1} = \frac{\sqrt{r}}{-\sqrt{2 \cdot M} \pm \sqrt{r}} \cdot dr$$ Do a substitution to simplify the integration: $$Q = \pm 1$$ $$u = -\sqrt{2 \cdot M} + Q \cdot \sqrt{r}$$ $$\sqrt{r} = Q \cdot ( u + \sqrt{2 \cdot M} )$$ $$dr = 2 \cdot ( u + \sqrt{2 \cdot M} ) \cdot du$$ $$dT = Q \cdot 2 \cdot \frac{(u + \sqrt{2 \cdot M})^2}{u} \cdot du = Q \cdot ( 2 \cdot u + 4 \cdot \sqrt{2 \cdot M} + \frac{4 \cdot M}{u} ) \cdot du$$ $$\int\limits_{T_0}^{T} dT = T - T_0 = Q \cdot \Big| u^2 + 4 \cdot \sqrt{2 \cdot M} \cdot u + 4 \cdot M \cdot ln |u| \Big|_{u_0}^{u}$$ $$T - T_0 = Q \cdot \left( u^2 - u_0^2 + 4 \cdot \sqrt{2 \cdot M} \cdot (u - u_0) + 4 \cdot M \cdot ln |\frac{u}{u_0}| \right) =$$ $$Q \cdot \left( (Q \cdot \sqrt{r} - \sqrt{2 \cdot M})^2 - (Q \cdot \sqrt{r_0} - \sqrt{2 \cdot M})^2 + 4 \cdot \sqrt{2 \cdot M} \cdot (Q \cdot \sqrt{r} - \sqrt{2 \cdot M} - Q \cdot \sqrt{r_0} + \sqrt{2 \cdot M}) \right)$$ $$+ 4 \cdot Q \cdot M \cdot ln \left| \frac{Q \cdot \sqrt{r} - \sqrt{2 \cdot M}}{Q \cdot \sqrt{r_0} - \sqrt{2 \cdot M}} \right | =$$ $$Q \cdot \left( r - 2 \cdot Q \cdot \sqrt{2 \cdot M} \cdot \sqrt{r} + 2 \cdot M - ( r_0 - 2 \cdot Q \cdot \sqrt{2 \cdot M} \cdot \sqrt{r_0} + 2 \cdot M ) + 4 \cdot Q \cdot \sqrt{2 \cdot M} \cdot (\sqrt{r} - \sqrt{r_0}) \right)$$ $$+ 4 \cdot Q \cdot M \cdot ln \left| \frac{Q \cdot \sqrt{\frac{r}{2 \cdot M}} - 1}{Q \cdot \sqrt{\frac{r_0}{2 \cdot M}} - 1} \right| =$$ $$Q \cdot \left( r - r_0 - 2 \cdot Q \cdot \sqrt{2 \cdot M} \cdot (\sqrt{r} - \sqrt{r_0}) + 4 \cdot Q \cdot \sqrt{2 \cdot M} \cdot (\sqrt{r} - \sqrt{r_0}) \right)$$ $$+ 4 \cdot Q \cdot M \cdot ln \left| \frac{Q \cdot \sqrt{\frac{r}{2 \cdot M}} - 1}{Q \cdot \sqrt{\frac{r_0}{2 \cdot M}} - 1} \right| =$$ $$Q \cdot (r - r_0) + 2 \cdot \sqrt{2 \cdot M} \cdot (\sqrt{r} - \sqrt{r_0}) + 4 \cdot Q \cdot M \cdot ln \left| \frac{Q \cdot \sqrt{\frac{r}{2 \cdot M}} - 1}{Q \cdot \sqrt{\frac{r_0}{2 \cdot M}} - 1} \right| =$$ $$Q \cdot (r - r_0) + 4 \cdot M \cdot \left( \sqrt{\frac{r}{2 \cdot M}} - \sqrt{\frac{r_0}{2 \cdot M}} \right) + 4 \cdot Q \cdot M \cdot ln \left| \frac{1 - Q \cdot \sqrt{\frac{r}{2 \cdot M}}}{1 - Q \cdot \sqrt{\frac{r_0}{2 \cdot M}}} \right|$$ Case 1: $$Q = +1$$ (outgoing light): $$T - T_0 = r - r_0 + 4 \cdot M \cdot \left( \sqrt{\frac{r}{2 \cdot M}} - \sqrt{\frac{r_0}{2 \cdot M}} \right) + 4 \cdot M \cdot ln \left| \frac{1 - \sqrt{\frac{r}{2 \cdot M}}}{1 - \sqrt{\frac{r_0}{2 \cdot M}}} \right| \quad \text{(28)}$$ Case 2: $$Q = -1$$ (incoming light): $$T - T_0 = r_0 - r + 4 \cdot M \cdot \left( \sqrt{\frac{r}{2 \cdot M}} - \sqrt{\frac{r_0}{2 \cdot M}} \right) - 4 \cdot M \cdot ln \left| \frac{1 + \sqrt{\frac{r}{2 \cdot M}}}{1 + \sqrt{\frac{r_0}{2 \cdot M}}} \right| \quad \text{(29)}$$ In the following chart, the outgoing solution is green while the incoming solution is blue and the red arrow points to the future.

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 $$Q$$: $$\frac{dy_s}{dt_s} = \frac{dr}{dT} \cdot \frac{1}{ 1 - Q \cdot \sqrt{\frac{2 \cdot M}{r}}} = \frac{Q - \sqrt{\frac{2 \cdot M}{r}}}{ 1 - Q \cdot \sqrt{\frac{2 \cdot M}{r}}} = \pm 1$$

### Map energy

Substitution of the term for $$dt$$ into the definition of map energy yields the map energy expressed with GP coordinates: $$E = \left( \left( 1 - \frac{2 \cdot M}{r} \right) \cdot \frac{dT}{d\tau} - \sqrt{\frac{2 \cdot M}{r}} \cdot \frac{dr}{d\tau} \right) \cdot m \quad \text{(30)}$$

### Velocity and acceleration

Equations 10, 11, 13, 14, 15, 16 and 17 derived from the Schwarzschild metric do not contain $$dt$$ and carry over. We can derive $$\frac{dr}{dT}$$ for a radially moving object as usual by substitution of $$dt$$: $$\frac{dr}{dt} = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot v_s$$ $$dr = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot v_s \cdot \left( dT - \sqrt{\frac{2 \cdot M}{r}} \cdot \left( 1-\frac{2 \cdot M}{r} \right) ^{-1} \cdot dr \right)$$ $$dr + \sqrt{\frac{2 \cdot M}{r}} \cdot v_s \cdot dr = \left( 1 - \frac{2 \cdot M}{r} \right) \cdot v_s \cdot dT$$ $$\frac{dr}{dT} = \frac{ \left( 1 - \frac{2 \cdot M}{r} \right) \cdot v_s}{1 + \sqrt{\frac{2 \cdot M}{r}} \cdot v_s } \quad \text{(31)}$$ For a raindrop: $$v_s = - \sqrt{\frac{2 \cdot M}{r}}$$ $$\frac{dr}{dT} = - \sqrt{\frac{2 \cdot M}{r}} = \frac{dr}{d\tau}$$ Equation 30 can be easily solved for $$\frac{dT}{d\tau}$$, which is the energy / mass ratio of the object in the raindrop frame: $$\frac{dT}{d\tau} = \frac{\frac{E}{m} + \sqrt{\frac{2 \cdot M}{r}} \cdot \frac{dr}{d\tau}}{1 - \frac{2 \cdot M}{r}} = \frac{dt_r}{d\tau} = \gamma_r = \frac{E_r}{m} \quad \text{(32)}$$