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I'm studying Eddington-Finkelstein coordinates for Schwarzschild metric. Adopting the coordinate set $(t,r,\theta,\phi)$, the line element assumes the form: $$ ds^2 = \left(1 - \frac{R_S}{r}\right)dt^2 - \left(1 - \frac{R_S}{r}\right)^{-1}dr^2 - r^2 [d\theta^2 + (\sin{\theta})^2d\phi^2], $$ where $R_S$ is the Schwarzschild radius and $t$ is the time given by a clock located at infinite distance from the source of the field and stationary with respect to it. Using this coordinate system we find that a massive radially infalling particle (coming from $r > R_S$) takes an infinite amount of coordinate time to reach the Schwarzschild radius. An observer located far away from the source "sees" the particle approaching $R_S$ more and more slowly.

Adopting Eddington-Finkelstein coordinates $(t',r,\theta,\phi)$, we write the line element in the form: $$ ds^2 = \left( 1- \frac{R_S}{r}\right)dt'^2 - \frac{2R_S}{r}dt'dr - \left( 1 +\frac{R_S}{r}\right)dr^2 - r^2 [d\theta^2 + (\sin{\theta})^2d\phi^2]. $$ When an observer is located far away from the source ($r \to \infty$) and is stationary with respect to it, $t'$ is the time given by the observer's clock, as well as $t$ is. If I calculate how $t'$ and $r$ are related for a radially infalling massive particle, I find out that the particle takes a finite time to cross the Schwarzschild radius from an initial position $r_0 > R_S$ and to reach the singularity $r=0$. So, after this calculation, I don't understand what is the situation which actually happens: my problem arises because for the "far away observer" $t'$ and $t$ have the same meaning, since by a suitable time translation they both represent the time given by his clock. Since the observer must see only one situation happening, does he see the particle cross the Schwarzschild radius or does he see the particle gradually "freeze" at the Schwarzschild radius?

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  • $\begingroup$ In Finkelstein coordinates the light rays emitted at the horizon also don't reach the outside observer, such things are independend of the coordinates. Schwarzschild Droste coordinates use local clocks which are stationary with respect to the black hole and relate to the coordinate time via the Lapse function, Raindrop coordinates ones which are free falling with the negative escape velocity and Finkelstein accelerated ones with the squared negative escape velocity, but who receives who's signal is invariant. $\endgroup$
    – Yukterez
    Sep 19, 2023 at 14:26
  • $\begingroup$ I mainly don't understand this fact: far away from the source,$ t$ and $ t'$ have the same meaning: they represent both the proper time given by the clock of an observer stationary with respect to the black hole and infinitely distant from it. So the observer can use a single clock to monitor both coordinate times, he only needs to choose a suitable origin of time in order to syncronize the two time flows. (Continues in the next comment) $\endgroup$
    – Al01
    Sep 19, 2023 at 17:49
  • $\begingroup$ But when the clock will mark a certain instant of time in the remote future, using E-F coordinates the observer will calculate that the particle at that time will have already crossed the Schwarzschild radius, whereas using Schwarzschild coordinates he concludes that the particle will never cross it... what I mean is that the use of Schwarzschild coordinates seems to suggest that the particle will never fall into the black-hole, whereas in E-F coordinates the particle falls into it. $\endgroup$
    – Al01
    Sep 19, 2023 at 17:49
  • $\begingroup$ t and t' are the same at r=∞, but not at smaller r. The local Schwarzschild clocks have v=0 (which isn't possible at r<2) and the Finkelstein ones accelerate with v=-2/r (which is also v=0 at r=∞, but not at r<∞). Here you see curves of constant Schwarzschild t on the Finkelstein t',r-spacetime diagram: click - as you see the t gets larger in comparison with t' the closer to the black hole you look. $\endgroup$
    – Yukterez
    Sep 20, 2023 at 0:47

1 Answer 1

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What your are encountering here is a core feature of general relativity: time is inherently, and fundamentally a local quantity. While each observer will have an unambiguously defined local proper time, there is generally no unique way to extend this notion of time to the rest of the universe. The global time coordinates $t$ and $t'$ both reduce to the proper time of a distant stationary observer, but differ away from that observer. Both can equally claim to be a generalization of the time of the observer, and so can an infinity of other generalizations.

Consequently, there is no (unambiguous) notion in general relativity of "the time at some event according to some (distant) observer".

The best one can hope to do, is describe what a particular observer actually observes. This generally requires considering how this information got to the observer. Typically this is done by transmitting light rays of some type, and therefore requires tracing how a light ray travels from the event to the observer. Generally, this is not as easy as simply comparing time coordinates at the different points, although some coordinates systems are better adapted to such an analysis as others.

Neither Schwarzschild nor advanced Eddington-Finkelstein coordinates (as used in the OP) are particularly well adapted to analyzing the signal received by a distant observer coming from an object crossing the (future) event horizon of a Schwarschild black hole. For this it is better to use retarded Eddington-Finkelstein coordinates. In these coordinates one straigthforwardly sees that it takes an infinite amount of time for a distant observer to receive the last signal sent by an object as it crosses the horizon, i.e. the distant observer will never (according to his own time) see the object cross the horizon.

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  • $\begingroup$ Your answer is correct, but I wouldn't sign your conclusion "Neither Schwarzschild nor advanced Eddington-Finkelstein coordinates (as used in the OP) are particularly well adapted to analyzing the signal received by a distant observer", since if you set ds²=0 and solve for dr/dt that is easy in ingoing/advanced Finkelstein coordinates where we get dr/dt=-1 for ingoing and dr/dt=(r-2)/(r+2) for outgoing photons, which is negative ingoing and outgoing at r<2, so it's easy to see they can't travel from r<2 to r>2. In regular Schwarzschild at r=2 we get dr/dt=0 $\endgroup$
    – Yukterez
    Sep 20, 2023 at 23:13
  • $\begingroup$ In the outgoing/retarded Finkelstein coordinates that you suggest on the other hand, you do get signals from r<2, but they are from the white hole, not the black hole, and they also don't exist in astrophysical black holes, so in my opinion in that coordinates it is less obvious that no signal can get out of a black hole since they require knowledge about white holes. $\endgroup$
    – Yukterez
    Sep 20, 2023 at 23:27

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