# The meaning of the singularity in Schwarzschild metric

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?

• 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. Jun 7, 2020 at 20:21

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. 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.

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.

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.