Black Holes and Charged Particles

[PeterDonis]

For the outside observer, this never happens, right? It takes infinite time, as observed from outside, to cross the horizon?

No. The outside observer will never see the particle cross the horizon. But the outside observer cannot deduce from that that the horizon crossing never happens or that it takes infinite time.

 

[nikkkom]

No. The outside observer will never see the particle cross the horizon. But the outside observer cannot deduce from that that the horizon crossing never happens or that it takes infinite time.

Outside observer can bounce a radar pulse off the falling object.

There are practical problems both with trying to do that with the "object" being a single electron, with the long time to get the pulse come back, and with red/blueshifting ("radio" waves may be seen as gamma ray by the object, and even if they do bounce off it, they will be red-shifted on the way back), but theoretically existence of the falling object is not unobservable.

 

[PAllen]

Outside observer can bounce a radar pulse off the falling object.

There are practical problems both with trying to do that with the "object" being a single electron, with the long time to get the pulse come back, and with red/blueshifting ("radio" waves may be seen as gamma ray by the object, and even if they do bounce off it, they will be red-shifted on the way back), but theoretically existence of the falling object is not unobservable.

No they can’t. There is a well defined time after which any light pulse sent sent by outside observer towards infaller will reach infaller only inside the horizon or not at all

 

[dllahr]

So this is much simpler than @PeterDonis is making it. The charge in the Reissner–Nordström metric is located at the singularity:
https://en.wikipedia.org/wiki/Reissner–Nordström_metric#Charged_black_holes
In the section above, the electromagnetic potential is defined:

So yes, we know effectively where the charge is. We don't need to invoke a lot of obfuscating mumbo-jumbo.

 

[PeterDonis]

The charge in the Reissner–Nordström metric is located at the singularity

No, it isn't. The "singularity" isn't even a single point, and it's certainly not a spatial point at the "center" of the black hole. Spacetime geometry in this case does not work the way you are assuming it does.

 

[dllahr]

So I guess the equation is wrong?

 

[Ibix]

So I guess the equation is wrong?

I think what Peter is saying is that you are interpreting r as a distance from a centre. It doesn't really have that meaning in black hole spacetimes, since it changes from a space like to a time like coordinate as r varies.

 

[PeterDonis]

I think what Peter is saying is that you are interpreting r as a distance from a centre. It doesn't really have that meaning in black hole spacetimes

Yes, that's what I'm saying.

since it changes from a space like to a time like coordinate as r varies

That's not the real reason, since that property depends on your choice of coordinates. The real reason is that the locus $r = 0$ is not a "point at the center of the black hole". If you look at a Penrose diagram of Reissner-Nordstrom spacetime, you will see that there are actually two timelike lines that both correspond to $r = 0$, and they are both inside the inner horizon. In fact, in the maximally extended geometry, which we have to talk about if we talk about going inside the inner horizon at all, there are an infinite number of such pairs of timelike lines. And, for good measure, these timelike lines are not actually part of the spacetime at all; they are limit points that do not exist in the actual manifold.

 

[PeterDonis]

So I guess the equation is wrong?

The equation is fine; it just doesn't mean what you think it means.

And with that, this thread is closed.