What I am trying to ask is the thickness like a Dedekind cut? An infinitesimal distance to the left is outside, and an infinitesimal to the right is inside?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
The event horizon itself is not a physical object, rather, it is defined as the boundary between regions of spacetime.
The boundary of the region from which no escape is possible is called the event horizon.
https://en.wikipedia.org/wiki/Black_hole
Inside the event horizon, the escape velocity exceeds the speed of light, and outside the escape velocity is below the speed of light.
Since the event horizon is not a material surface but rather merely a mathematically defined demarcation boundary
What is the "Event Horizon" of a black hole
So the answer to your question is that the event horizon is not a physical object (that would have a classical thickness), but a boundary between regions of spacetime.
answered Jan 2, 2021 at 17:23
-
$\begingroup$ Black holes are a very interesting thing for Gedankenexperiments $\endgroup$ Commented Jan 2, 2021 at 17:46
$\begingroup$
$\endgroup$
2
I'm not nearly knowledgeable enough on this subject to give an objective answer, but maybe by giving you my thoughts on your question you can get closer to what you're looking for.
From what I understand (as an absolute novice in this subject), the event horizon simply denotes a location at which no amount of acceleration (within the theoretical bounds of the speed of light) would allow any amount of information to escape the "curvature" of space-time beyond that point.
I'm a high school student and I've just taken calculus, so I can't say much. I have no idea what a Dedekind Cut is, so I'm not going to try to address your question from that angle, although I hope someone else will. If you think of a black hole as a sphere, the highest density and therefore the highest requisite escape velocity (if you could go faster than C, the speed of light) would be at the center of the sphere. As a bit of information's distance from the center of the sphere (let's just ignore all the wackyness at the singularity as it's irrelevant to this discussion) increases, the requisite acceleration to escape the black hole's curvature will approach a limit, which is the speed of light. If you think about the event horizon conceptually, using a limit to define requisite accelerations at the event horizon would make sense. The event horizon really isn't a physical thing, it's not like some fluid barrier that has definite edges. Therefore, I think the way you're approaching this idea can be altered a little bit. The idea that "An infinitesimal distance to the left is outside, and an infinitesimal to the right is... [inside]", may be thought of in another way: the event horizon is simply an infinitesimal value, a point or plane in space, which can be denoted by zero distance, thickness, or difference in resultant velocities/accelerations.
Do not take my "answer", by any means, to be a biblical truth. This is just my stab at answering your question, and if I'm wrong please point it out.
-
$\begingroup$ A Dedekind cut is mathematics, not physics, per se. It's a simple way to construct a real number from the rationals. $\endgroup$– PM 2RingCommented Jan 2, 2021 at 17:51
-
$\begingroup$ The escape velocity of a BH is c at the EH (event horizon). But rather than defining the EH in terms of velocity, it's better to focus on the spacetime geometry. You can't escape because all worldlines at the EH take you closer to the centre of the BH. $\endgroup$– PM 2RingCommented Jan 2, 2021 at 17:59
$\begingroup$
$\endgroup$
3
This is an interesting question in my opinion that relates to our newtonian view os space and time. The thing is, nearby the event horizon of the simplest black hole (Schwarzschild), you will not see "the structure of space" as a 'spherical shell', so there would not be a sense of thickness around it.
Trying to look to the horizon in a classical view (classical GR), if you go further and further near it, the sense of darkness in space grows, that is, the solid angle that looks dark around your path, grows. As you keep going, this angle will be so small at one point (still before surpassing the horizon) that almost all the visible 'outside' Universe will be condensed (as viewed by your 'eyes'), in a tiny small solid angle, looking far far away.
If you keep going and pass the horizon, then the difference would be now, this light of the visible (outside) Universe goes off, and all you can see everyplace you look is the singularity. Off course classically, there is no much sense in trying to explain what would occur beyond the horizon, so this is pretty much a guess based on Einsteins solutions for BH's inside.
But at some point, you loose contact with the outside Universe.
-
1$\begingroup$ This is wrong, if you fall into the black hole you will always see the external universe, your view gets aberrated into a point when you hover above the horizon but not if you fall in, see Fig. 7 in sci-hub.st/10.1016/j.cpc.2014.04.013 or yukterez.net/org/relativistic.raytracer/… and madore.org/~david/math/kerr.html $\endgroup$– YukterezCommented Aug 21, 2020 at 23:31
-
$\begingroup$ I downvoted because I think it is wrong, because past light cones of events inside the horizon include large parts of the spacetime outside the horizon, therefore an observer inside can receive signals from outside. $\endgroup$ Commented Jan 1, 2021 at 23:13
-
$\begingroup$ You cannot see the singularity while you are inside a Schwarzschild black hole. $\endgroup$ Commented Feb 2, 2022 at 5:22