Gullstrand-Painlevé coordinates

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GullStrand-Painlevé (GP) coordinates were proposed by Paul Painlevé^[1] and Allvar Gullstrand^[2] in 1921. Similar to Schwarzschild coordinates, GP coordinates can be used in the Schwarzschild metric to describe correctly the space-time physics outside the event horizon of a static black hole. However, Schwarzschild coordinates are only valid exterior to the event horizon. GP coordinates were the first coordinates that can extend the analysis of space-time physics, such as speed of particles and lights, across the event horizon. The simplicity of the GP form makes it very useful for pedagogical purposes.

1 Derivation
2 Speeds of raindrop
3 Speeds of light
4 A rain observer's view of the universe
5 See also
6 References
7 External links

[edit] Derivation

The derivation of GP coordinates requires defining the following coordinate systems and understanding of how data measured for events in one coordinate system is interpreted in another coordinate system.

Convention: The units for the variables are all geometrized. Time and mass have units in meters. The speed of light in flat spacetime has a value of 1.

[edit] Free-float coordinates

In special relativity, free-float frame is the classic inertial frame which obeys Einstein's postulates of special relativity. In general relativity, it rides along the space-time geodesic. The inside of an unpowered spaceship that plunges toward a black hole in its presence demonstrates a typical free-float frame. The space-time in a free-float frame is flat and can be expressed by the metric:

$d\tau ^2 = dt_{ff}^2-dr_{ff}^2- r_{ff}^2\ d\theta_{ff}^2-r_{ff}^2\sin^2\theta_{ff}\,d\phi_{ff}^2$ ,

where

$r_{ff}, \ t_{ff},\ \theta_{ff},\ \phi_{ff}$ are the free-float coordinates,

The flatness claim is only locally valid. When spacetime curvature increases, it may be necessary to decrease the dimensions of the frame to make sure the tidal acceleration associated with the curvature is not detectable.

[edit] Schwarzschild coordinates

A Schwarzschild observer is a far observer or a bookkeeper. He does not directly make measurements of events that occurs in different places. Instead, he is far away from the black hole and the events. Observers local to the events are enlisted to make measurements and send the results to him. The bookkeeper gathers and combines the reports from various places. The numbers in the reports are translated into data in Schwarzschild coordinates, which provide a systematic mean of evaluating and describing the events globally. Thus, the physicist can compare and interpret the data intelligently. He can find meaningful information from these data. The Schwarzschild form of Schwarzschild metric using Schwarzschild coordinates is given by

$d\tau^{2} = \left(1-\frac{2M}{r} \right) dt^2 -\frac{dr^2}{ \left(1-\frac{2M}{r} \right)}- r^2d\theta^2-r^2\sin^2\theta d\phi^2$ .

where

$t,\ r,\ \theta,\ \phi$ are the Schwarzschild coordinates,

M is the mass of black hole.

[edit] Shell coordinates

Imagine a series of spherical shells held stationary against the force of gravity by rockets outside the event horizon. The shell observer, standing on the spherical shell, can form his own shell frame. When constricted to small enough region, the space-time in the shell frame can be regarded as flat. Special relativity works in the shell frame. Consider two sequential clicks of a clock at rest on a spherical shell, $dr,\ d\theta ,\ d\phi$ of the Schwarzschild coordinates is 0. Therefore, $d t s$ , the time coordinate in the rain frame is the proper time. Substitute into the Schwarzschild metric,

$dt_s^2=d\tau^2 = \left(1-\frac{2M}{r} \right) dt^2$ .

Take the square root,

$dt_s=\sqrt{1-\frac{2M}{r}} dt$ .

Similarly, by considering two simultaneous events located in the same radial direction on two spherical shells separated by distance $d r$ , the equation for $d r s$ can be derived:

$dr_s=\frac{dr}{\sqrt{1-\frac{2M}{r}} }$ .

Locally, with the shell coordinates $\ t_s, r_s$ , the metric for the shell frame is thus expressed in hybrid form:

$d\tau^{2} = dt_s^2-dr_s^2 - r^2d\theta^2-r^2\sin^2\theta d\phi^2$ .

[edit] Rain coordinates

Define rain frame as a free-float frame plunging freely from rest at infinity toward a black hole.^[3] The time coordinate of the rain frame is obtained by Lorentz transformation from the shell coordinates outside the event horizon.

$dt_r = \gamma\left(dt_s - \beta dr_s\right )$ ,

$dr_r = \gamma\left(dr_s - \beta dt_s\right )$ ,

where

$\gamma = \frac{1}{\sqrt {1-\beta^2} }$ ,

$\beta = -\sqrt {\frac {2M}{r}}$ is the velocity of the rain frame relative to the shell frame.

Taking into consideration the space-time curvature between the shell coordinates and the schwarzschild coordinates due to the presence of black hole, the rain coordinates can be derived:

$dt_r = dt - \beta \gamma^2 dr \$ ,

$dr_r = \gamma^2 dr - \beta dt \$ .

Substitute into the Schwarzschild form. The rain form is given by

$d\tau^{2} = dt_r^2 - dr_r^2 - r^2d\theta^2-r^2\sin^2\theta d\phi^2$ .

[edit] Gullstrand-Painlevé coordinates

When only $d t r$ is plugged into the Schwarzschild form, the GP form is obtained

$d\tau^{2} = \left(1-\frac{2M}{r} \right) dt_r^2-2\sqrt{\frac{2M}{r}}dt_r dr-dr^2- r^2d\theta^2-r^2\sin^2\theta d\phi^2$ .

The r-coordinate of Schwarzschild coordinates is used here. By using a mixed coordinate system, GP form is able to interprete events globally.

GP form doesn't encounter singularity at the event horizon, $r = 2 M$ . The r-coordinate term does not blow up at the event horizon. The only singularity is at the center of the black hole, $r = 0$ . It's possible now to answer many more questions about static black hole.

[edit] Speeds of raindrop

Define a raindrop as an object which plunges radially toward a black hole from rest at infinity. A raindrop is at rest in a rain frame. Its velocity is also the relative velocity of the rain frame with respect to a shell observer and can be measured by the shell observer.

In Schwarzschild coordinates, the velocity of raindrop is given by

$\frac{dr}{dt}=-\left(1-\frac{2M}{r} \right) \sqrt{\frac{2M}{r}}$ .

The speed tends to 0 as r approaches the event horizon. The raindrop appears to have slowed as it gets nearer the event horizon and halted at the event horizon as measured by the bookkeeper. Indeed, an observer outside the event horizon would see the raindrop plunges slower and slower. It's images infinitely redshifted and never make it through the event horizon. However, the bookkeeper does not physically measure the speed directly. He translates data relayed by the shell observer into Schwarzschild values and compute the speed. The result is only an accounting entry.

In GP coordinates, the velocity is given by

$\frac{dr}{dt_r}=\beta = -\sqrt{\frac{2M}{r}}$ .

The speed of the raindrop is inversely proportional to the square root of radius. At places very far away from the black hole, the speed is extremely small. As the raindrop plunges toward the black hold, the speed increases. At the event horizon, the speed has the value 1, same as the speed of light. There is no discontinuity or singularity at the event horizon.
Inside the event horizon, $r < 2 M$ . The speed exceeds the speed of light. It increases as the raindrop gets ever more closer to to the singularity. Eventually, the speed becomes infinite at the singularity. The results may not be correctly predicted by the equation at and near the singularity. The physics may be quite different when quantum mechanics is incorporated.

Despite the problem with the singularity, it's still possible to compute the travel time for the raindrop from horizon to the center of black hole mathematically. Integrate the equation of motion: $\int_{0}^{T_r}\,dt=-\int_{2M}^{0} \left ( \sqrt{\frac{2M}{r}} \right )^{-1}\,dr$ . The result is $T_r=\frac{4}{3} M$ .

[edit] Speeds of light

Assume radial motion. For light, $d τ = 0$ . Therefore,

$0 = \left (dr+\left ( 1+\sqrt{\frac{2M}{r}}\ \right )dt_r \right)\left (dr-\left (1-\sqrt{\frac{2M}{r}}\ \right )dt_r\right )$ .

$\frac{dr}{dt_r}=\plusmn 1-\sqrt{\frac{2M}{r}}$ .

At places very far away from the black hole, $r \to \infty$ , $\frac{dr}{dt_r}=\plusmn 1$ . The speed of light is 1, same as in special relativity.

At the event horizon, $r = 2 M$ , the speed of light shining outward away from the center of black hole is $\frac{dr}{dt_r}=0$ . It can not escape from the event horizon. Instead, it gets stuck at the event horizon. Since light moves faster than all others, matter can only move inward at the event horizon. Everything inside the event horizon is hidden from the outside world.

Inside the event horizon, r < 2M, the rain observer measures that the light moves toward the center with speed greater than 2. This is plausible. Even in special relativity, the proper speed of a moving object is $\quad \frac{dr_{ff}}{d\tau}=\frac{v}{\sqrt{1-v^2}} \ge 1$ . There are 2 important points to consider:

1. No object should have speed greater than the speed of light as measured in the same reference frame. Thus, the principle of causality is preserved. Indeed, the speed of raindrop is less than that of light: $\frac{\left ( \dfrac{dr}{dt_r}\right )_{raindrop}} {\left ( \dfrac{dr}{dt_r} \right)_{light}}= \frac{\sqrt{\dfrac{2M}{r}}} {1+\sqrt{\dfrac{2M}{r}}} < 1.$
2. Universally, there can be only one value for the speed of light in the free-float frame and it should be 1. The rain frame is a free-float frame which works inside the event horizon. Indeed, the speed of light in rain frame is

$\frac{dr_r}{dt_r}=1$ .

The time of travel for light shining inward from event horizon to the center of black hole can be obtained by integrating the equation for the velocity of light, . The result is .
- The light travel time for a stellar black hole with a typical size of 3 solar masses is about 11 microseconds.
- Ignoring effects of rotation, for Sagittarius A*, the supermassive black hole residing at the center of the Milky Way, with mass of 3.7 million solar masses, the light travel time is about 14 seconds.
- The supermassive black hole at the center of Messier 87, a giant elliptical galaxy in the Virgo Cluster, is the biggest known black hole. It has a mass of approximately 3 billion solar masses. It would take about 3 hours for light to travel to the central singularity of such a supermassive black hole, and for raindrop, 5 hours.

[edit] A rain observer's view of the universe

How does the universe look like as seen by a rain observer plunging into the black hole? The view can be described by the following equations:

$\cos \boldsymbol{\Phi}_r =\frac{dr_r}{dt_r} = \frac{ \sqrt{\dfrac{2M}{r}}+ \cos \boldsymbol{\Phi}_s } {1+ \sqrt{\dfrac{2M}{r}}\ \cos \boldsymbol{\Phi}_s }$ ,

$\cos \boldsymbol{\Phi}_s = \frac{dr_s}{dt_s} =\pm \sqrt{1-\left(1-\frac{2M}{r}\right)\frac{\mathit{I}^2}{r^2}}$ ,

$\phi=\mathit{I}\int_{r_0}^{\infty}\frac{dr}{r^2 \ \cos \boldsymbol{\Phi}_s}$ ,

where

$\boldsymbol{\Phi}_r, \ \boldsymbol{\Phi}_s$ are the rain observer's and shell observer's viewing angles with respect to the radially outward direction.

$\ \phi$ is the angle between the distant star and the radially outward direction.

I

is the impact parameter. Each incoming light ray can be backtraced to a corresponding ray at infinity. The Impact parameter for the incoming light ray is the distance between the corresponding ray at infinity and a ray parallel to it that plunges directly into the black hole.

Because of spherical symmetry, the trajectory of light always lies in a plane passing through the center of sphere. It's possible to simplify the metric by assuming $\theta = \frac{\pi}{2}$ .

The impact parameter $I$ can be computed knowing the rain observer's r-coordinate $r 0$ and viewing angle $\ \boldsymbol{\Phi}_{r0}$ . Then, the actual angle $\ \phi$ of the distant star, is determined by numerically integrating $d r$ from $r 0$ to infinity. A chart of the sample results is shown at right.

At r/M = 500, the black hole is still very far away. It subtends a diametrical angle of ~ 1 degree in the sky. The stars are not distorted much by the presence of the black hole, except for the stars directly behind it. Due to gravitiational lensing, These obstructed stars are now deflected 5 degrees away from the back. In between these stars and the black hole is a circular band of secondary images of the stars. The duplicate images are instrumental in the identification of the black hole.

At r/M = 30, the black hole has become much bigger, spanning a diametrical angle of ~15 degrees in the sky. The band of secondary images has also grown to 10 degrees. It’s now possible to find faint tertiary images in the band, which are produced by the light rays that have looped around the black hole once already. The primary images are distributed more tightly in the rest of the sky. The pattern of distribution is similar to that previously exhibited.

At r/M = 2, the event horizon, the black hole now occupies a substantial portion of the sky. The rain observer would see an area up to 42 degrees from the radially inward direction that is pitch dark. The band of secondary and tertiary images, rather than increasing, has decreased in size to 5 degrees. The aberration effect is now quite dominant. The speed of plunging has reached the light speed. The distribution pattern of primary images is changing drastically. The primary images are shifting toward the boundary of the band. The edge near the band is now crowded with stars. Due to Doppler effect, the stars which were originally located behind the rain observer have their images appreciably red-shifted, while those in front are blue-shifted and appear very bright.

At r/M=0.001, the curve of distant star angle versus view angle appears to form a right angle at the 90 degrees view anagle. Almost all of the star images are congregated in a narrow ring 90 degrees from the radially inward direction. Between the ring and the radially inward direction is the enormous black hole. On the opposite side, only a few stars shine faintly.

As the rain observer approaches the singularity, $r \rightarrow 0$ , and $\cos \boldsymbol{\Phi}_r \rightarrow \sqrt{\frac{r}{2M}}$ . Most of the stars are squeezed to a narrow band at the 90° viewing angle. The observer sees a magnificent bright ring of stars bisecting the dark sky.

[edit] See also

[edit] References

Wikibooks has a book on the topic of

General relativity

^ Paul Painlevé, “La mécanique classique et la théorie de la relativité”, C. R. Acad. Sci. (Paris) 173, 677–680 (1921).
^ Allvar Gullstrand, “Allgemeine Lösung des statischen Einkörperproblems in der Einsteinschen Gravitationstheorie”, Arkiv. Mat. Astron. Fys. 16(8), 1–15 (1922).
^ Taylor, Edwin F.; John Archibald Wheeler (2000). Exploring Black Holes: Introduction to General Relativity. Addison Wesley Longman. ISBN 0-201-38423-X.