Eddington-Finkelstein coordinates for Schwarzschild metric

[andyl01]

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?

 

[Ibix]

What he sees is not related to what coordinates he uses. To work out what he sees you need to propagate null curves back from the infallibg object to an observer and calculate the time between reception of pulses emitted at regular intervals of the infaller's proper time. If you do that correctly you'll get the same answer in any coordinate system - the gaps between pulses will grow and grow.

The question is not what he sees, but how he interprets that. The way Schwarzschild coordinates fail as you approach the event horizon means that the infaller crosses infinitely many surfaces of constant $t$ before reaching the horizon. Eddington-Finkelstein coordinates are better behaved and the infaller crosses a finite number of surfaces of constant $t'$. In both cases he crosses the horizon in finite proper time - it's just that Schwarzschild coordinates become singular at the event horizon.

So your confusion is that you are interpreting the time coordinate (which tells you "what happens now" for some definition of "now") as telling you something about what you see. You just have two different definitions of "now", one better behaved than the other. Neither choice makes a difference to what you actually physically see.

 

[andyl01]

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. 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.

 

[PeterDonis]

far away from the source, and have the same meaning

Not in all respects, no. Surfaces of constant $t'$ are not the same as surfaces of constant $t$. So $t'$ and $t$ define different simultaneity conventions.

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.

No, this won't work. The observer can observe both time flows with one clock only on his own worldline. He can't extend both the same way off his worldline because, as above, the two time coordinates define different simultaneity conventions.

 

[PeroK]

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. 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.

When I was learning SR I abandoned the idea of an "observer" and "what an observer sees", in favour of how things are measured in a given frame of reference. Now you are studying GR, you seriously have to abandon the all-seeing, all-knowing observer in favour of coordinate systems. Coordinate systems are used to model the physical scenario and predict what local observers measure. Observers make local observations/measurement only: the energy of a particle; the wavelength of light etc.

It may take some time to adjust your thinking, but thinking in terms of global observers is a dead-end in GR.

 

[PeterDonis]

the use of Schwarzschild coordinates seems to suggest that the particle will never fall into the black-hole

No, it doesn't, because Schwarzschild coordinates are singular, i.e., not well defined, at the horizon. And if you look at the reason why, you find that it is because surfaces of constant $t$ all intersect each other at the horizon (more precisely at a single "bifurcation point" on the horizon). By contrast, surfaces of constant Eddington-Finkelstein time $t'$ do not intersect at all; they all cross the horizon at different points.

 

[Ibix]

Coordinates do not have any physical meaning. They are just a way to label events in spacetime with four numbers. You are not even obligated to have a time coordinate if you don't want one. Schwarzschild and Eddington-Finkelstein coordinates just have different procedures for how to label events that tend towards being the same at large $r$, that's all.

It is quite common for coordinates to have some relation to physical observables, like some multiple of the clock times for a hovering observer. But don't mistake that for anything beyond convenience.

Think of a piece of paper with a line drawn on it. Schwarzschild and Eddington-Finkelstein coordinates are different procedures for drawing grids on the piece of paper that let you "name" a point on the paper as being in some sense "level" with each other, but any old grid (orthogonal lines or not, straight lines or not) will work. These two procedures make lines similar at one edge of the paper and very different at the other. In fact, the Schwarzschild lines are discontinuous at one place (the event horizon) and that's why they can't assign a time coordinate to anything crossing the horizon.

You might like to plot a graph of the Schwarzschild coordinate lines on a graph with Eddington-Finkelstein coordinates on the vertical and horizontal axes and you'll see what I'm getting at. Set $\phi=0, \theta =\pi/2$.

 

[Nugatory]

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.

The key is that although t and t' have the same relationship to the proper time of a stationary clock at infinite distance, they do not have the same relationship to the proper time of a clock free-falling into the black hole.

The infalling clock ticks off a finite amount of proper time as it falls towards and through the horizon. Every point on the worldline of the infalling clock is mapped to an E-F time coordinate. In contrast, only the the points outside of the event horizon are mapped to a Schwarzchild time coordinate. The statement "infinite time for the infaller to reach the horizon" is better understood as "no matter how large the proper time measured by the outside clock, the corresponding Schwarzschild t coordinate will be that of an event on the infaller's worldine and outside the event horizon"

 

[pervect]

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. 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.

The short version is that Schwarzschild coordinates don't assign finite coordinate label to the event where the infalling observer crosses the event horizon. Which is different from saying it doesn't happen.

This is easy to grasp if one realizes that coordinates are just labels, and don't have any intrinsic physical significance.

Some people seem to resist this simple observation - I'm not sure why. If one person labels the location of a village on a map at being A6, and another person, using a different map, labels a village as being at G4, the location of the village doesn't change. The maps are different, not the location of the village. The map is not the territory. Changing a label on a map is a change in convention, not a change in anything physically real. As Misner observes in "Precis of general relativity", https://arxiv.org/abs/gr-qc/9508043, coordinates serve the purpose of being a "space-time map".

One can fairly observe that the Schwarzschild coordinates are ill-behaved at the event horizon. Mathematically, there is a "coordinate singularity". This is a shortcoming of the coordinates, not of the physics.

EF coordinates are a bit better behaved in some respects, but not all. Kruskal coordinates are probably the best behaved.

In this example (and others), it's sometimes helpful to get away from coordinates and ask questions that are framed in a manner that is independent of coordinates. In this example, one might ask if the proper time interval measured by a free-falling watch from release to reaching the event horizon is finite or infinite. The answer is that if the watch is released from a finite height, it's reading when it reaches the event horizon is finite.

 

[vanhees71]

When I was learning SR I abandoned the idea of an "observer" and "what an observer sees", in favour of how things are measured in a given frame of reference. Now you are studying GR, you seriously have to abandon the all-seeing, all-knowing observer in favour of coordinate systems. Coordinate systems are used to model the physical scenario and predict what local observers measure. Observers make local observations/measurement only: the energy of a particle; the wavelength of light etc.

It may take some time to adjust your thinking, but thinking in terms of global observers is a dead-end in GR.

One should keep in mind that observable quantities are independent of the choice of coordinates. It helps to remember that GR is a gauge theory with the full diffeomorphism group as the gauge symmetry. From this point of view it should be clear that gauge-dependent quantities cannot be easily interpreted as observables.