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

Question

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 :

\begin{equation} 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 \end{equation}

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

Answer 1

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 had published 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 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 (e.g. 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 $$t = \cosh^{-1}\left(\frac{X}{x}\right),$$ 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 $\infty$, the observed falling velocity asymptotically approaches zero.

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

The Relativistic Rocket Equations (Baez).
The Rindler Coordinates (Wikipedia)

Answer 2

Tristan Diotte wrote:

"if we switched to the "Eddington-Finkelstein" coordinate, this singularity would disappear."

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 $$v=-c \ \frac{r_s}{r} \quad \text{or} \quad v=-c \ \sqrt{\frac{r_s}{r}}.$$

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 observers 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.

Answer 3

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.