1
$\begingroup$

At the beginning I thought that the event horizon coincides with the surfaces, but then making a new name when you could just call it surface would seem a bit pointless.

Then where is the event horizon? Is it inside or outside the black hole?

Notice that I have a really basic knowledge about physics

$\endgroup$

3 Answers 3

1
$\begingroup$

An event horizon is a "sphere-shaped surface of influence" of inescapable gravity from/towards the black hole. It is the point where the object (victim?) cannot overcome the gravity of the black hole, and will be sucked into it. This is also the point where, theoretically, you would vanish to an outside observer, since the light crossing the event horizon also will not escape to bounce back to the observer.

There is not one particular location/altitude for an event horizon, it depends on the size and mass of the black hole. The more the mass, the more the gravity, thus a larger event horizon.

Wikipedia has a nice writeup about them, although this article is more about event horizons in general, not just for black holes.

Edit: I use the term "Sphere of Influence" loosely -- you would enter the sphere of influence (meaning your motion would be affected by the black hole's gravity) before you would cross the event horizon -- you could return from the SOI, but you couldn't return once past the event horizon.

$\endgroup$
2
  • $\begingroup$ Good answer, although "sphere of influence" is a tad misleading. The event horizon is a surface, not a sphere (well, it actually is a 2-sphere, so I suppose that's not accurate), and the influence of the black hole extends outward beyond it. $\endgroup$
    – HDE 226868
    Sep 9, 2015 at 0:21
  • $\begingroup$ Good point, I've been playing a lot of Kerbal Space Program lately.. one of the reasons I quoted it was that it's not exactly accurate, but a generalized description... I updated the answer for the differentiation. $\endgroup$
    – Tim S.
    Sep 9, 2015 at 0:22
1
$\begingroup$

The event horizon is not a place.

If it was a place you could stay there. A black hole has an inside and an outside and the event horizon is the boundary between them.

A black hole's event horizon is not the only kind of event horizon. If someone started accelerating to the right at constant acceleration and never stopped accelerating they would have an event horizon. There would be events (an event is defined as a combination of a place and a time, it is a when-where), events they eventually see and events that they would never ever see and the boundary between them is the event horizon. In this case the event horizon literally moves to the right at the speed of light and that is why when something crosses it it can't cross back.

For a black hole there are observers that stay outside and they share a common event horizon and it acts just like a surface that expands outwards at the speed of light because acceleration and gravity are very similar. What's weird is that it maintains the same size, and has the same surface area. It is like space is accelerating into the black hole and the surface rushes out at the speed if light and they perfectly balance.

But it isn't a place. That said, when you say surface I can't tell which surface you are taking about. The photon sphere (inside it you can't have orbits that don't crash in) the ergospshere (inside it you have to rotate with the star) and the event horizon (the point of no return).

$\endgroup$
2
  • $\begingroup$ Would you be willing to call these surfaces separatrices? After all, what they really do is to separate completely different regimes of dynamical solutions. I think we should use a mathematical terminology in this case, since these are not truly physical objects. $\endgroup$
    – CuriousOne
    Sep 9, 2015 at 3:55
  • $\begingroup$ The Rindler horizon of Minkowski space is not an event horizon. Event horizons are geometrically invariant features of their spacetimes and have an objective existence and location. However that location cannot be determined by observers in the spacetime since it depends on the future. $\endgroup$
    – AGML
    Nov 29, 2015 at 16:32
0
$\begingroup$

I understood what was my doubt, which was a more basic question:

A star has a clear surface, and within the surface it is all mass.

When a black hole is born, for example after a supernova, i couldn't understand if the event horizon contains all the mass like a normal star. Instead from my understanding, the radius of the surface which contains all the mass is much smaller than the radius of the event horizon.

Of course, we can't say for sure that the the radius of the mass object is smaller than the event horizon, since we can't see what happens within.

So in other words a black hole is by definition much bigger that I thought since I make it coincide with the surface of the mass object.

$\endgroup$

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.