Do particles in Eddington Finkelstein coordinate system take a finite amount of time to reach the horizon? Or do they take infinite time? How is the time coordinate used in the Eddington Finkelstein coordinate system be matched with the time measured by an observer?

What will an observer who is sitting on the particle would observe when he is using Eddington Finkelstein coordinates? How much time will he take in his reference frame?


1 Answer 1


A coordinate system is just a scheme for labelling events in spacetime, and it does not necessarily relate directly to what is experienced by an observer. So it doesn't make sense to ask about particles using the EF coordinate system or indeed any other coordinate system.

The Schwarzschild coordinates correspond to the experience of an observer an infinite distance from the black hole. We can see this because in the limit of $r \to\infty$ the Schwarzschild metric:

$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{1-\frac{r_s}{r}}+r^2(d\theta^2 + \sin^2\theta d\phi^2) $$

simplifies to the flat space Minkowski metric (in polar coordinates):

$$ ds^2 = -dt^2 + dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$

So for the observer at $r=\infty$ the time $dt$ is just the time shown by that observer's clock. In this sense the Schwarzschild coordinates are intuitively simple.

The coordinates that correspond to the experience of an observer freely falling into the black holes are the Fermi normal coordinates. Locally these look like flat spacetime so the falling observer considers the spacetime immediately around them to be flat.

But the Eddington-Finkelstein coordinates do not correspond to anything directly experienced by an observer. So for example the EF timelike coordinate $v$ is related to the Schwarzschild coordinates by:

$$ v = t \pm r + 2GM\log\left( \frac{r}{2GM} - 1\right) $$

This "time" $v$ is not a quantity that would be measured by any observer's clock.

The time experienced by the falling observer has a nice simple geometric interpretation. It is just the length of the observer's world line calculated using the metric. If you're interested I go into this in more detail in What is time dilation really?.

The length of the trajectory is called the proper time and it is an invariant, meaning that we can use any coordinates to calculate it and whatever coordinates we choose we will get the same result.

  • $\begingroup$ But we do calculate null geodesics in Eddington Finkelstein coordinate system, and we interpret them as the path of light. These travel at an angle of 45 with respect to r. How do we understand this if t does not represent physical time? $\endgroup$
    – Khushal
    Commented Sep 14, 2017 at 7:11
  • $\begingroup$ You can calculate a geodesic using any coordinates. The EF coordinates happen to be well suited to dealing with null geodesics. The geodesic will look different in different coordinate systems but the length of the geodesic (i.e. $\int ds$) measured between any two points will be the same in all coordinate systems. This length is the proper time i.e. the elapsed time as shown by a clock carried by the falling observer. For a null geodesic the elapsed time is always zero. $\endgroup$ Commented Sep 14, 2017 at 8:00
  • $\begingroup$ Can I say that when I move from Schwarzchild coordinate system to Kruskal coordinates, then my point of observation is changing from an observer at infinite distance to an observer who is infalling into the black hole. $\endgroup$
    – Khushal
    Commented Sep 15, 2017 at 13:41
  • $\begingroup$ @Aniket: no. The Kruskal-Szekeres coordinates do not correspond to anything an observer could experience. They are entirely abstract. The nearest to the the experience of a freely falling observer is probably the Gullstrand-Painlevé coordinates, though while they accurately reflect the time experienced by the falling observer they do not correspond to distance measurements made by the observer. $\endgroup$ Commented Sep 15, 2017 at 14:53
  • $\begingroup$ Sorry, this might be a stupid question. But, looking at a metric can I say anything about the observer, whether the observer is at infinity or infalling into the black hole? $\endgroup$
    – Khushal
    Commented Sep 17, 2017 at 4:37

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