1. Aug 17, 2009

### Joe436

Hi,

The way I understand it (and correct me if I'm wrong). Sometimes a star collapses on itself and forms a blackhole. Floating around in space getting bigger as it pulls in whatever gets too close to esacpe it's gravitational field. What I don't understand is why a blackhole is so strong even light can't escape it. If mass is what decides how strong the gravity is and the star's mass wasn't enough to pull in light before. Why is it strong enough after turning into a blackhole?

Thanks,
Joe

2. Aug 17, 2009

### sylas

Because it is smaller.

Gravitational attraction increases as you get close to the attracting mass. With a star, you bump into the edge of the star before you get close enough that light can't escape. If you could put all the mass of our Sun within a space about 2.7 km in diameter (I think) then you could get to within 2.7 km of 230 or so kilograms, and if you get that close, you can't ever back out again. Not even if you are light.

Cheers -- sylas

3. Aug 17, 2009

### Phrak

Really? How long does this take?

How long would it take, say, before there were Hawking radiation? The answer is never, isn't it?

4. Aug 17, 2009

### sylas

The answer is immediately. It's just that the Hawking radiation from a black hole is very very weak, until the black hole is quite tiny.

For a star to collapse into a black hole, it just needs to be large enough. As runs out of fuel, and cools, and dims, it will compress. If it is big enough, and doesn't blow apart in the meantime, it will compress to a black hole. Our Sun is not big enough to compress all the way to a black hole once it runs out of fuel.

Cheers -- sylas

5. Aug 17, 2009

### Phrak

Sure about that? In asymtotic time, the event horizon in in the infinite future, and the Hawking radition arrives an infinite amount of time after formation. Or do I have something confused?

6. Aug 17, 2009

### sylas

Yes, I am sure about that. It's easy to get mixed up by the different co-ordinate systems; but a black hole does emit Hawking radiation (according to current physics) and if you do the co-ordinate stuff right, you should be able to represent this in which co-ordinate system you like. Some co-ordinates diverge as you approach the horizon, and so make it a tad difficult to talk about particles that cross the horizon, as they most certainly do. With any luck, someone who knows this better will pop in before I make a mess of it in the details, but I am sure about what I've said so far.

Cheers -- sylas

7. Aug 17, 2009

### Phrak

Yes, well, as I've heard it takes an infinite amout of time for anything to reach an event horizon as measured by the clock by an observer above the event horizon, it seems that an equal amount of time would be required to communicate, via Hawing radiation, the existance of a horizon.

8. Aug 17, 2009

### Joe436

This kind of contradicts what I was taught in highschool physics (granted they really didn't go into black holes). That the size of the object doesn't matter, just the mass. You could have two objects that are the same size but a different mass. The greater the mass the greater the gravity. So why would compressing the size of our sun (or any star) to a smaller object make a difference if the mass is the same? Or am I not correctly remembering my highschool physics class.

Thanks,
Joe

9. Aug 17, 2009

### mikelepore

First you derive a formula for the escape speed of a projectile on any planet by doing this: initial potential energy plus kinetic energy equals the final potential plus kinetic energy, which is zero: (1/2) m v^2 - GMm/R = 0. Then you solve for the speed: v = square root [2GM/R]. Now you take this formula for escape speed and use it backwards, you set the escape speed to be the speed of light, and solve for the necessary ratio of mass to radius. So if you were to put that much mass or more into that much radius or less, you would have a place where the escape velocity equals or exceeds the speed of light.

Last edited: Aug 17, 2009
10. Aug 17, 2009

### jgens

Strictly speaking, this in only true if you are sufficiently far away from the object. Just consider Newton's Law of Universal Gravitation:

$$F = \frac{Gm_1m_2}{r^2}$$

11. Aug 18, 2009

### Staff: Mentor

The attractive force between two masses is given by this formula:

$$F = G \frac{{m_1}{m_2}}{R^2}$$

where G is a gravitational constant, the m's are the two masses, and R is the distance between them.

So given the same two masses, as you move them closer together (like make the sun smaller so you can move your test mass in closer), the attractive force goes up with the square of that separation.

EDIT -- here's a reference to that equation: http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

.

12. Aug 18, 2009

### sylas

You are remembering just fine, and you are correct. The size doesn't make any difference to the gravitational attraction.

You've misunderstood my point, which does not contradict your maths at all.

Suppose that somehow our Sun was compressed into a black hole. The hole would have a radius of a bit under 3 kilometers.

There would be no difference in the gravitational force that we would feel, because all that matters for gravitational force is the total mass and how far you are away from it. Earth's orbit would be unchanged.

The significance of the size is that you can now get up to 2 kilometers of the center of the Sun, and STILL have all that gravitational mass somewhere ahead of you. If you try to get with in 2 kilometers of the center of the Sun now, you would be inside the Sun. That's the difference. You can get up close to the mass, which makes the force enormous, without actually being inside the mass.

If you DID get up with 2 kilometers, of course, you've crossed the point of no return. You are inside the "hole", or inside the "event horizon"... and you can never get out; you are now inevitably dragged right up to the singularity at the dead center, where physics breaks down and we don't have a good description of what goes on. Long before then you'll have been torn into pieces by tidal forces; but we all have to die sometime.

Physics can describe what happens inside the event horizon; just not at the central singularity.

Cheers -- sylas

13. Aug 18, 2009

### stevebd1

In close proximity of an object of collapsed mass, when calculating gravity you would also take into account coordinate acceleration (or the metric tensor) which describes the structure of spacetime about an object of mass-

$$dr'=\frac{1}{\sqrt{1-\frac{2M}{r}}}\ dr$$

where $M=Gm/c^2$

so gravity in GR becomes-

$$\frac{Gm}{r^2}\frac{1}{\sqrt{1-\frac{2M}{r}}}$$

At the surface of Earth, dr' is almost undetectable (dr'=1+6.96e-10 but still significant enough to provide the escape velocity based on $v_{rel}=-\sqrt{(2M/r)}c$) but becomes significant around neutron stars and black holes (dr' being infinite at the event horizon of black holes). While long range gravity doesn't change, as the volume of an object reduces, the short range effects increase significantly in GR compared to the Newtonian equation for gravity which simply increases relative to r.

14. Aug 18, 2009

### Joe436

OK, I think I got it with what you said above and the two posts from berkeman and jgens. So basically, star (or our sun if it had enough fuel) collapses and it compresses into a smaller object. Which allows you to get closer to it's entire mass. The closer you get to it the stronger the gravitational force. So a black hole is an object with a lot of mass. But compressed to a small enough size allowing you to get close enough to it that the gravitational force is so strong nothing, not even light could escape it. Did I get it right?

Thanks,
Joe

15. Aug 18, 2009

### Joe436

I do remember that formula now. I think I'm beginning to understand this now.

Thanks,
Joe

16. Aug 18, 2009

### Joe436

So a black hole is an object with a "big" mass and compressed to a small enough size allowing you to get closer to it. And if you get close enough, you or anything else including light couldn't escape it's gravitational force?

Thanks,
Joe

17. Aug 18, 2009

### sylas

Yes, that's it in a nutshell.

Cheers -- sylas

18. Aug 18, 2009

### Joe436

Cool. Thanks for explaining this to me. Wasn't sure if this site was meant for people like me who are just curious about physics. Hope I didn't annoy you.

Thanks again,
Joe

19. Aug 18, 2009

### sylas

Not at all. It's a very good question, and my understanding is that this is the main purpose of the forum.