# Outside observers of a black hole and spherical symmetry?

ScruffyNerf
TL;DR Summary
If an outside observer sees something fall into a black hole, how does that not break spherical symmetry?
I know that for the infalling observer the horizon is a fake singularity that can be removed via the Eddington-Finkelstein co-ordinates but wouldn't the classic Swartsheild co-ordinates still apply for the outside observer?

So, while for the infaller it takes a finite time, the outside observer will still forever see them on the outside of the black hole, or more appropriately, on a specific set of points outside the black hole. How does this observed mass forever on the side of the black hole not influence the gravity observed by the external observer and not break spherical symmetry? If the outside observer does observe the gravity at the singularity, how does that happen?

Mentor
If an outside observer sees something fall into a black hole, how does that not break spherical symmetry?

It does if the object that falls in has enough mass to affect the spacetime geometry. In thought experiments the usual assumption is that it doesn't; objects are considered "test objects" that have well-defined worldlines but negligible mass, so they don't change the spacetime geometry.

I know that for the infalling observer the horizon is a fake singularity that can be removed via the Eddington-Finkelstein co-ordinates but wouldn't the classic Swartsheild co-ordinates still apply for the outside observer?

You can choose whatever coordinates you want. You can choose to describe the outside observer using Eddington-Finkelstein coordinates. You can choose to describe the trajectory of the infalling observer inside the horizon using Schwarzschild coordinates on the interior region. Nor are those the only possibilities.

The key point is that what coordinates you choose have no effect on the physics; the physics is the same regardless.

So, while for the infaller it takes a finite time, the outside observer will still forever see them on the outside of the black hole

That's because of how the outgoing light signals from the infaller behave; the closer to the horizon the infaller is when a light signal is emitted outward, the longer it takes for that signal to get back out to the outside observer.

How does this observed mass forever on the side of the black hole not influence the gravity observed by the external observer

As above, it will influence the observed gravity if it's large enough. Roughly speaking, the outside observer will see the hole's gravity appear to increase as the infalling mass passes him on its way in.

vanhees71 and ScruffyNerf
ScruffyNerf
It does if the object that falls in has enough mass to affect the spacetime geometry. In thought experiments the usual assumption is that it doesn't; objects are considered "test objects" that have well-defined worldlines but negligible mass, so they don't change the spacetime geometry.

This was the answer to my question, thanks!

After the object falls into the black hole, what's stopping the no-hair theorem from applying? / what restores spherically symmetry so it can apply?

Mentor
After the object falls into the black hole, what's stopping the no-hair theorem from applying? / what restores spherically symmetry so it can apply?

Assuming the object does not add any angular momentum to the hole (i.e., it falls in radially), there will be gravitational waves emitted as the object falls in; the waves will carry away all the asymmetry and leave behind a spherical hole.

If the object adds angular momentum to the hole, the final hole will be spinning, which means it will be only axisymmetric, not spherically symmetric (i.e., a Kerr hole, not a Schwarzschild hole).

ScruffyNerf