# Black hole question

1. Jun 25, 2013

### MathJakob

I think blackholes are amazingly interesting and I hope we get some answers about what they are in the future.

I have a question I hope someone can help me with. The Schwarzschild radius is the radius at which the gravity becomes so strong there is no going back right? Is there an equation that describes how much the gravity increases the closer you get to the singularity?

The one I found was $$r_\mathrm{s} = \frac{2Gm}{c^2}$$ but I don't really understand what it's saying

If so can you write it here :)

and also is a blackhole an actual hole in space, kinda like this for example

or is the eye of the blackhole as flat as the singularity at the centre of it?

2. Jun 25, 2013

### Staff: Mentor

The Schwarzschild metric. It is not as simple as Newtonian gravity.

The Schwarzschild radius (rs) is just a parameter in the Schwarzschild metric, and it is related to the "size" of the black hole.

It is a region of extreme spacetime curvature, it is not something "in space" - it modifies space and time.

3. Jun 25, 2013

### MathJakob

So it curves it in the shape of a cone type of thing? Like the picture I linked? Eventually coming to a single point with no dimensions?

Also not sure you mentioned anything about the equation if any that shows how much the gravity increases by the further you get

4. Jun 25, 2013

### Staff: Mentor

No. It is impossible to model a 3-dimensional black hole in a flat spacetime of 3 dimensions. If the black hole is not rotating, it has a spherical symmetry.
It is possible to calculate the motion of objects based on that equation, but you need General Relativity for that.

5. Jun 25, 2013

### WannabeNewton

The shape of a black hole is quite a complicated subject. yenchin showed me this paper a while back and you might be interested in reading it: http://arxiv.org/pdf/1306.1019v1.pdf

6. Jun 28, 2013

### Naty1

you can get a decent description here:

http://en.wikipedia.org/wiki/Black_hole

so it appears dark from our perspective....See the illustration in the Link above....When you see illustrations of brilliantly lit black holes what you are looking at the radiation [light] from accelerating particles outside of the black hole....some of this is called the accretion disc...where gravity of the black hole sucks in surrounding gas which becomes heated....and then disappears behind the horizon....

7. Apr 30, 2014

### simplesimon11

The event horizon seen as a plane from a direction of specific direction of observation. How does this observational plane change if you view the event horizon from say 45° below the event horizon! By what I understand there is a jet of high energy being produced from a area of the black-hole and I was trying to find out if you viewed this black hole from another point of observation, say 45° difference, would the jet of energy be directed into another part of the universe? If such was to be viewed and the jet of energy that is now being seen going into another direction of the Universe, then from all point of viewing of this black hole would have many jets and the even horizon would be viewed as well different. This being said, then would not a model of a black hole show a spherical shape after all angles of view were place together to make this model and in such as well there would be jets of energy going into space in all direction equal!

Trying to obtain a view of how a black hole could exist in space/time and bend the fabric of this space time and be view able only from all planes to make a model of such a black hole.

I hope I have not again posted to a wrong place. If I have posted again to a wrong area, I truly am lost as to how to find the right area, as I have been reading now so many pages and have not found any with this question. Thanks again and excuse my fumbling of posts.

Last edited: Apr 30, 2014
8. Apr 30, 2014

### Mordred

Nice I definitely need to study this in greater detail, thanks

9. May 2, 2014

### phsopher

I think the closest thing to what you're asking would be the acceleration felt by an observer near a black hole. Admittedly I'm very rusty on non-cosmological applications of GR but I think the Newtonian relation

$$a = -\frac{GM}{r^2}$$

would become

$$a = -\frac{GM}{r^2}\left(1 - \frac{2GM}{r}\right)^{-1}$$

where G is Newton's constant, M is the mass of the black hole and r is the coordinate distance from the center of the black hole. Perhaps more knowledgeable people can confirm or deny this or otherwise elaborate.

10. May 2, 2014

### timmdeeg

First off the formula yields the Schwarzschild radius rs for a black hole with mass m. At this characteristic radius the escape speed equals the speed of light. The spherical shell defined by rs is called event horizon.
So, depending on the radius r where eventually a photon is emitted, it will escape (r > rs), "stay" at the event horizon (r = rs) or will be drawn to the singularity (r < rs).

11. May 2, 2014

### Staff: Mentor

The last one is sure, the other two depend on the direction of the emission. In the wrong direction, the photon will get drawn in as well. And the cone where it can escape gets narrower the closer to the event horizon you are.

12. May 2, 2014

### timmdeeg

I agree, but one should keep in mind that a point source is not a directional light source. So, if the point source is outside the event horizon, then the photon can escape and so forth.