Time coordinate in general relativity [closed]

Question

In special relativity, an inertial observer, for instance, in 1+1 dimensions, a world line of an object could be paramatrized, and the time measurement of the inertial observer between two points of the space-time trajectory could be easily computed by comparing the time coordinates. In general relativity, for instance, in Schwarzschild spacetime, we can parametrized a photon path in coordinates r, t. Focusing on region outside of the horizon, I wonder, what is t? In the case of ingoing light rays, the worldline never crosses the horizon. Lectures say that is what an observer outside sees. So that means t is the time measure by a observer very far away from the horizon? Appearentely, r is not a problem, at least not outside.

Answer 1

Yes, both $r$ and $t$ in Schwarzschild coordinates have puzzling features, but perhaps $t$ more than $r$ as you say. The first part of the answer is simply to say that these are coordinates, which means they are labels, or number-valued markers, that we associate with the various events that fill up spacetime. The next step is to take a look at the metric. This tells as that for $r > r_s$ (where $r_s$ is the Schwarzschild radius), $t$ is timelike, because events separated only in $t$ have a timelike separation. The metric also allows to calculate how much proper time elapses between neighbouring events separated only in $t$. This proper time $d\tau = \sqrt{|g_{00}|} dt$ can be physically interpreted as the amount by which a certain clock will advance between the events at $(t,r,\theta,\phi)$ and $(t+dt,r,\theta,\phi)$. The clock is an ordinary clock, moving inertially, whose worldline happens to be the one that makes it momentarily at rest relative to the spatial coordinates at the event in question.

That completes my statement of what $t$ means. However, I notice that you often see it written that $t$ is the time elapsed as observed by an observer at infinity, or very far away, at large $r$. The idea of this further statement is that we assume there can be an observer who uses Schwarzschild coordinates to map spacetime, and we notice that for events at the observer, $t$ is the same as proper time. So if this observer chooses to say that all events at some given $t$ are 'simultaneous' then he might say that $t$ measures elapsed time according to his clock, even though his clock is very far away from the events at finite $r$. Now I put the word 'simultaneous' in inverted commas there because this is a somewhat arbitrary choice; it is just a convenience that such an observer can adopt if he wants, and it is not always convenient. But this is what people mean when they speak of 'the observer at infinity' and things like that.

I think my answer will only be fully understood by someone who already has made some progress with general relativity. I encourage you to keep going, and hopefully this answer will be somewhat helpful now, and more helpful as you learn more.

Answer 2

Interactions between photons and mass are complex. For this reason I prefer at first to answer your question with respect to an infalling mass particle.

First, you have to avoid some possible confusion: You must distinguish between your coordinate system and what you see. Both are different concepts: One simple example is a Minkowski space: If a Minkowski diagram represents your coordinates, you get a fourdimensional view of the whole spacetime. In contrast, what you s e e are elements which are located on your lightcone which is showing into the past.

The same is true for the spacetime around black holes. One instructive mean to show what happens around a black hole is the Kruskal metric. In the following Kruskal diagram it is important to notice that the simultaneity lines of an outside observer are the radial lines t=1, t=2 etc. As an example, particle A is infalling, and particle B is an arbitrary outside observer whose worldline remains outside the event horizon. Following the radial lines you will see that according to the Kruskal spacetime diagram of an outside observer, the spacetime position of B will never be simultaneous with the point where A is crossing the event horizon.

A different question is if B will s e e A crossing the event horizon. The answer is not provided by the radial Kruskal lines but by the arrows which are simulating communication at speed of light between A and B. And you will see that B never will see A crossing the event horizon.

In conclusion, both questions have the same answer, but in order to be precise they must be distinguished.

After this clarification, the answer is quite simple: the ideal case is an observer who is "far away" from the black hole and from the effects of its gravity. However, for any observer outside of the event horizon the same logic does apply: he will never see particle A crossing the event horizon, and according to his spacetime diagram he will never reach a position which is simultaneous with the moment of the crossing of the event horizon by particle A.

Regarding photons, it is sure that no observer will ever see a photon which disappears through an event horizon. And also I do not see any possibility for any outside observer to reach a point in spacetime which is simultaneous with the crossing of the event horizon by the photon.