2
$\begingroup$

I have been told by many lecturers and many books that in the Schwarzschild metric

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

the singularity at $r=r_s$ purely comes from the bad choice of coordinate and that there is no physical singularity there.

I got really confused up to this point because I have also been told that in a black hole, the surface $r=r_s$ is called the event horizon and nothing can pass through it.

If this singularity purely comes from the bad choice of coordinate and not a physical one, how can we perceive the existence of this event horizon? They even tried to make analogy to the polar coordinate where the azimutal angle $\phi$ is ill defined at the poles . I can see that these ill-defined points purely come from the poor choice of coordinate since every point on the sphere are equal (due to spherical symmetric). The situation in the black hole case is clearly different from the sphere analogy.

So, the question is, if the singularity at $r=r_s$ is not a physical one (just simply a bad coordinate), how can we justify the exitence of the event horizon that never led anything to cross it?

$\endgroup$
1
  • 4
    $\begingroup$ It is not true that nothing can pass through the event horizon. See physics.stackexchange.com/questions/21319/…, but don't read the accepted answer (it's pretty controversial), read the other ones. $\endgroup$
    – Javier
    Jun 7 '20 at 20:21
8
$\begingroup$

Schwarzschild solution in the ordinary spherical coordinates that you present is the view of the spacetime by a faraway observer that is fixed at a certain distance from the black hole for all its proper time span. For this observer, anything that falls into the black hole can only ever reach $r=r_s$, and only at $t=\infty$. So what the singularity at $r=r_s$ means is that a faraway observer can't get any information from the region with $r<r_s$. In this sense, it's a very real singularity that has an exact physical meaning.

But this choice of an observer (the reference frame for which the metric is calculated) is not unique. For some purposes one may want to choose a different observer, e.g. free-fall coordinates, that describe metric from the point of view of a freely falling observer, are regular at $r_s$:

$$ds^2=-dt_{\text{ff}}+(dr-v\,dt_{\text{ff}})^2+r^2d\Omega^2.$$

This regularity reflects the fact that a freely falling observer doesn't notice the event horizon at $r=r_s$. This also is the reason why Schwartzschild coordinates are sometimes considered to be "bad": they don't let us see smooth structure of the spacetime in all the points where it is actually smooth.

There are also a bunch of other coordinates, each serving its purpose, and many of them are regular at the event horizon.


Actually, you don't even have to touch general relativity to see such a "seeming" singularity: in special relativity you can switch to an accelerating frame of reference and get an event horizon behind* the accelerating observer—a surface from beyond which nothing will ever reach the observer, even light. But should you reduce acceleration of the frame to zero, and the event horizon recedes into infinity, and you once again get spacetime where any object moving to the observer will eventually reach it.


*Actually it's not necessarily behind: the frame can be moving oppositely to the direction of acceleration, in which case the horizon will be in front of the observer.

$\endgroup$
1
$\begingroup$

A free falling (whether upwards or downwards!) observer sees no horizon. See here for a thorough analysis. The motion is even that same as the Newtonian case in terms of the faller's proper time. It is coordinate time that causes all the confusion and the horizons, but it is arbitrary, so not "real".

Gullstrand-Painleve coordinates exhibit no horizon.

After reading these you might feel less uncomfortable about horizons.

$\endgroup$
4
  • $\begingroup$ I understood that a free-falling observer would not observe any event horizon. However, an observer on a star faraway from a black hole does see this horizon, so it is really something real to me. $\endgroup$
    – Kimari
    Jun 8 '20 at 11:04
  • $\begingroup$ "Real to me" is from the realm of psychology, not physics. $\endgroup$
    – m4r35n357
    Jun 8 '20 at 12:54
  • $\begingroup$ Since it is something that is observable -> it is real, isn't right? $\endgroup$
    – Kimari
    Jun 8 '20 at 12:59
  • $\begingroup$ A mirage is "observable" in that sense. Isn't this a bit like "if a tree falls in a forest" territory; the observer proceeds to the singularity regardless of whether anyone sees her. Anyway I am now at the limits of my own knowledge so I have to defer to my sources at this point! $\endgroup$
    – m4r35n357
    Jun 8 '20 at 15:17
-4
$\begingroup$

It is purely a matter of opinion as to whether the singularity at the event horizon is a true singularity. If you define a radial coordinate $R = r - r_s$, then this will describe manifold with a singularity at the event horizon and with no interior.

Mathematically the event horizon is called a "removable singularity" because it is possible to define a manifold which is identical to that of the observable universe outside the event horizon and which also has an interior solution, and it is possible to define coordinates for such a manifold in which there is no singularity at the event horizon. That is mathematics, not physics. As there is, and can be, no physical evidence for us as exterior observers, the interior solution should not be regarded as a part of physical science. An infalling observer would not observe empty space at the event horizon, for the simple reason that he would already be dead before he got that far, and even if he were not the time between passing the event horizon and hitting the central singularity is so small as to be meaningless.

If one considers the physical process of the creation of a black hole, as described in the seminal paper by Oppenheimer & Snyder, then, from the perspective of an exterior observer, infalling matter creates a black hole only in the infinite future. The interior solution somehow implies that empty space will spontaneously appear in the interior of a collapsing star. Seems to me, therefore, that the "bad choice of coordinates" is the one which makes the interior solution appear meaningful. This is not the choice which Schwarzschild himself used.

$\endgroup$
1
  • 3
    $\begingroup$ In very massive black holes an infalling observer can not only be still alive at the event horizon, he can even live for a long (proper) time after crossing it. $\endgroup$
    – Ruslan
    Jun 7 '20 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.