# Will a distant observer really see an object that has fallen close to a black hole freeze in time?

I'm currently taking my first course in general relativity, and I was wondering:

We know from the schwarzschild metric that for a (far away) observer looking at an object falling towards a black hole, he would see the object gets closer and closer to the schwarzschild radius, but never cross it because of the singularity at $$r =r_s$$ in the $$g_{tt}$$ coefficient of the metric.

But my teacher told us that this singularity was a "coordinate illusion" and that if we switched to the "Eddington-Finkelstein" coordinate, this singularity would disappear.

Now my questions are:

Does the existence of this coordinate system contradicts the previous statement? And if so, why is the Schwarzschild metric so well known as following :

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

if it has this illusion of coordinates ? Is there a deeper reason that I can't figure out?

• Possible duplicates: physics.stackexchange.com/search?q=time+horizon+black+hole Commented Feb 1 at 20:47
• I believe that the falling observer would be seen to not only slow down as they approach the horizon but also red shift. So they would redshift away and fade as they asymptotically approach the event horizon Commented Feb 2 at 0:40
• Don't take this comment as expert opinion, but I've read (presumably based on simulation results) that if a human being witnessed an object fall into a BH, it would be visually indistinguishable from just "winking out of existence" at the horizon. But if we examined high speed camera footage and went frame by frame, we would see the object slow down, fade to red, then fade to black. Commented Feb 2 at 1:20
• Related: How can anything ever fall into a black hole as seen from an outside observer? and links therein. Commented Feb 3 at 14:14

One reason the Schwarzschild metric is so well known, is because it was the first exact solution found for a gravitational body in General Relativity and even Einstein did not think an exact solution was possible.
Even more remarkably, Schwarzschild found the solution while serving on the Russian war front and only a few months after Einstein publish his theory of General Relativity.

Now why does this event horizon or singularity disappear in a different set of coordinates? One way to visualise this is to consider the analogous Rindler Metric.

This is set in flat spacetime with no gravity involved. The Rindler observers are a set of observers (the hyperbolic curved pink world lines) accelerating in flat spacetime in such a way that they consider their spatial separation to be constant. An artificial event horizon is visible to them (The diagonal dashed black x=0 world line). The X and T coordinates are the normal coordinates of an inertial observer in flat spacetime, observing the accelerating observers accelerating relative to her.

If an observer falls off one of the accelerating rockets, she follows a straight trajectory (The blue worldline) in flat spacetime and passes through the apparent event horizon. Her colleagues on the accelerating spaceships never see her pass through the event horizon because light rays from the crossing event (The red wordline) never reach the accelerating rocket observers (eg the pink worldline of the observer at x=0.4).

Any inertial observer in the flat spacetime does not see any event horizon and nothing unusual about the location. However, just like falling into a black hole, once the observer crosses to the left of the apparent event horizon, he can never cross back to the right side of the apparent event horizon, because he would have to exceed the speed of light to do so. Also, just like a black hole, before he crosses the event horizon, the accelerating Rindler observers see light coming from the falling observer as progressively more red shifted due to their own acceleration.

To the accelerating rocket observers, the event horizon is real because they can never see anything behind it. To the rocket observers the falling observer seems to take forever to reach the event horizon and they never see him cross it. On the other hand, the falling inertial observer just passes straight it and the trip only takes a few minutes by her watch.

The chart below is the point of view of an accelerating observer that was initially at X=1. The vertical axis at X=0 is the apparent event horizon. This accelerating observer sees the free falling observer as following the curved blue world line that asymptotically approaches the horizon, never quite stopping and never reaching the horizon.
The equation for the curve is $$t = acosh(X/x)$$ where X is the initial height of the falling object, x is the height at time t and t is the proper time according to the clock of the accelerating observer.

In the next chart, the vertical axis is the velocity of the free falling observer as seen by an accelerating observer and the horizontal axis is now the proper time of the accelerating observer:

The accelerating observer sees the falling observer rapidly accelerate at first and then slow down, but again he never sees her come exactly to a stop, but this motion is eventually so slow that it appears to be "frozen". The motion is described by the equation $$v = X \times \frac{tanh(t)}{cosh(t)}$$ where X is the initial height of the free falling observer at t=0 and t is the proper time of the accelerating observer. As t approaches infinity the observed falling velocity asymptotically approaches zero.

The equations used in the charts can fairly easily be derived from:

• I am trying to understand what they would see instead (if they are still or in motion relative to the falling observer). Do they really see the falling observer just freeze and stay there forever? How can that happen when there is nothing there for light to fall on and be reflected? Commented Feb 2 at 8:46
• @Parrotmaster Why do you think about reflection here? It is NOT that you could shine light on the black hole to reflect on something seemingly "stuck" (in space and time) at the event horizon. It means that when something falls into a BH and constantly sends signals (via electromagnetic radiation), the signal will take longer and longer to reach you and redshift more and more. This signal could be the light a body reflects when it falls into the BH, but it is not that the reflection happening takes forever, just getting back to you is. Commented Feb 2 at 12:20
• @Koschi I was thinking of how a spectator could see the falling observer if they are not actually in that spot anymore (they fell into an area where light cannot escape), so light from any source could not hit them and be reflected back into the eyes of the viewers. So if I understand correctly at some point the light just doesn't reach us anymore, but wouldn't that equate more to the falling observer vanishing (either instantly or by fading away)? In my mind when someone says "freeze" I think of an astronaut floating at the edge of the event horizon, forever. Commented Feb 2 at 12:36
• Seeing something means you get a constant stream of light/EM waves from that something right? If this light takes longer and longer to reach you, how should, from YOUR perspective, this something just vanish... there is always some of the light reaching you, it just takes longer and longer. Also remember that it kind of "fades away" in the sense that the light will also get infinitely redshifted. Commented Feb 2 at 13:45
• @safesphere So falling into black holes is impossible? That's pretty counter to literally everything else I've ever read about black holes. Commented Feb 7 at 7:51

Tristan Diotte wrote: "if we switched to the "Eddington-Finkelstein" coordinate, this singularity would disapear"

That doesn't help the far away observer, since the time coordinates of the Eddington-Finkelstein or Gullstrand-Painlevé coordinates are not the proper time of the far away observer, but of local observers accelerating or freefalling in with v=-c rs/r or v=-c √(rs/r).

The far away observer who has velocity v=0 and whose proper time is the regular Schwarzschild time coordinate never sees the particle reach the horizon, only those oberservers who also fall into the black hole as well do while they cross the horizon themselves, since the light rays emitted outwards by an infalling object stay at the horizon and can therefore only reach an eye which itself is at the horizon.

• Commented Feb 2 at 0:04
• GP and Schwarzschild time coordinates both reduce to the proper time of a distant stationary observer. Commented Feb 2 at 0:16
• @TimRias - only for events infinitely far away from the black hole, events close to the black hole have different time coordinates; in the case of the horizon finite with GP and infinite with regular Schwarzschild coordinates. They have different hypersurfaces of simultaneity, only the latter uses that of the distant stationary observer. For the curves of constant Schwarzschild time in GP coordinates see here, they flatten at large r but at small r they bend. Commented Feb 2 at 1:41
• There is no unique hypersurface of simultaneity attached to a single observer. The Schwarzschild and GP case have equal (non)claim to that moniker. Commented Feb 2 at 7:23
• @TimRias There is no “hypersurfaces of simultaneity“ in GR at all. Equating spacelike slices to simultaneity in curved spaces is absurd. There is however the global hyperbolicity in the Schwarzschild spacetime obeying the causal order of events regardless of the chosen coordinates. And everything outside happens before anything at the horizon. So this answer is correct +1, but what exactly is the value of your comments for this question? Commented Feb 7 at 6:10

Choice of coordinates does not change what is measured in physics. A distant observer will see a falling object slow down as it approaches the event of horizon. They will not see it "freeze" because the object always has an inward velocity, even if it becomes arbitrarily small. They also cannot see anything "freeze" because the light emitted by the object is also redshifted by a factor that is asymptotic to infinity as $$r \rightarrow r_s$$.

There is a coordinate singularity in the Schwarzschild spacetime when using the $$r, \theta, \phi, t$$ coordinate set (Droste coordinates, though more commonly known as Schwarzschild coordinates), in the sense that you can see that the $$g_{11}$$ component of the metric tensor, which is $$(1 - r_s/r)^{-1}$$, blows up to infinity when $$r=r_s$$. This infinity (unlike the one at $$r=0$$) can be sidestepped using a transformed set of coordinates like the E-F or G-P set, but a change of coordinates does not change what is measured.

Schwarzschild coordinates (Droste coordinates) are a popular choice for initial study because of their apparent similarity to the more familiar spherical polar coordinates (though the meaning of $$r$$ is different) and also because the $$t$$ coordinate corresponds to the time measured by a distant observer at rest. In these coordinates, an object falling from infinity towards a black hole has a rate of change of $$r$$ coordinate of $$\frac{dr}{dt} = -c\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}\ .$$ The first bracketed term goes to zero as $$r \rightarrow r_s$$ and this term appears even if the object starts from closer to the black hole or even if it is flung towards the black hole.

Schwarzschild coordinates are generally a poor choice of coordinates for dealing with scenarios involving objects falling through the event horizon (although the calculations can be done in Schwarzschild coordinates) and the spacetime diagrams for such scenarios get quite messy and difficult to interpret. A switch of coordinates to something like Gullstrand-Painleve (or Painleve-Gullstrand) coordinates can be helpful. In this case, the $$t$$ coordinate is replaced with $$T$$, where $$dT = dt + \frac{1}{c}\left(1 - \frac{r_s}{r}\right)^{-1}\left(\frac{r_s}{r}\right)^{1/2} dr\ .$$ With this transformation then there is no singularity in the metric tensor coefficients when $$r=r_s$$ (at the expense of introducing a non-zero $$g_{01}$$, i.e. a term in in $$dTdr$$ in the spacetime interval), but a free-falling observer from infinity obeys $$\frac{dr}{dT} = -c\left(\frac{r_s}{r}\right)^{1/2}$$ and does not "stop" at the event horizon. However, this is when measured using a time $$T$$, which is not the time measured by a distant observer, it is the time measured on the clock of the falling object. The distance observer's time is related to $$T$$ by $$\frac{dt}{dT} = \left(1 - \frac{r_s}{r}\right)^{-1}\ ,$$ where $$r$$ is the coordinate of the falling object, and thus as the falling object approaches $$r_s$$ the ticks on the distant observers clock blow up to infinity and they can never observe the falling object reach $$r_s$$, exactly as per the situation using Schwarzschild coordinates.

• You are correct that what are now usually called Schwarzschild coordinates are not quite what Schwarzschild used @safesphere. Commented Feb 7 at 7:23