Recall that freefall time (t_ff) has been defined as the proper time experienced by an observer free-falling towards a black hole from infinity, relative to the time (t) of a distant static observer. It is given by:

t_ff = t + 2 r^1/2 + ln|(r^1/2 - 1)/(r^1/2 + 1)|

where r is expressed in multiples of the event horizon radius of the black hole.

Likewise, Finkelstein time is given simply by:

t_F = t + ln|(r - 1)|

and recall that this ensures straight spacetime paths for infalling photons. Recall that this is the time measured by an observer riding on a the surface of mass that collapses to a black hole.

It is easiest to plot these two functions as the difference from Schwarzschild coordinate time, i.e.:

t_ff - t = 2 r^1/2 + ln|(r^1/2 - 1)/(r^1/2 + 1)|, and

t_F - t = ln|(r - 1)|

These functions are plotted in Fig.1 and are fairly self-explanatory, but what do they tell us?

Figure 1:

The curves imply that both functions diverge to -∞ at the Schwarzschild radius (r=1), because ln(0) is taken. This is so because the Schwarzschild chart's paths of light rays diverge to infinite time - refer to Fig. 1 of this Blog entry - so the difference is -∞.

Does this mean that there still is a coordinate singularity at the Schwarzschild radius in both freefall and Finkelstein time? No, because you subtract the right amount to make both times finite. Exactly at the event horizon, it smacks a little like sweeping the problem under the carpet - like the re-normalization used in physics.

Lastly, what is the use of this graph? It is just a graphical way to view the conversion between Schwarzschild time, freefall time and Finkelstein time, e.g. If it takes 10 years of Schwarzschild time for a photon to reach a distance of twice the event horizon radius from a black hole (r=2), what will the freefall and Finkelstein times be respectively?

It is clear from the graph that the Finkelstein time will also be 10 years, because the curve cuts the r-axis there, so the deviation is zero. In freefall time it would have taken the photon about one year more.

When we get closer to the event horizon, say r=1.01, it is easier to use the equations. Say the photon took 20 years to reach r=1.01, we get

t_ff = t + 2 r^1/2 + ln|(r^1/2 - 1)/(r^1/2 + 1)| = 20 - 3.886 = 16.014 years.

t_F = t + ln|(r - 1)| = 20 - 4.605 = 15.395 years.

In the next Blog entry we will look briefly at the Kerr black hole, i.e., black holes that are rotating.

Edit: As discussed with Guest below, here is a graph with a fourth time added - the observer's time when stationary at r event horizon radii from the hole.

Figure 2:

The stationary local observer's time t_loc = √(1-1/r) t, so that the equivalent of the other graphs is:

t_loc - t = (√[1-1/r] - 1) t

Since the r.h.s. is a function of t, there is a curve for every value of t. The curve was plotted for a value t = 10. It is clear that the local observer's time approaches zero at the event horizon (r -> 1). the graph also implies (correctly) that t_loc will approach t as r -> ∞.

Jorrie