# Trying to understand black holes

1. Nov 13, 2014

### acesuv

i understand that a black hole occurs because the atoms are not strong enough to hold back gravity after a certain point. is the matter continually collapsing on itself or does it stop after the atoms have been crushed? like is it a runaway effect?

my next question is why does a large mass in a small area create more gravity? thinking about the rubber sheet analogy... if you have 2 balls of the same mass but different densities (therefore different volumes).... will the denser ball really create a larger indent in the rubber sheet? i dont understand why the required gravity only occurs when an amount of mass is compressed in a small enough space.

my last question... there are two different things that ive found to define black holes. 1. light doesnt escape and 2. matter has collapsed on itself. my question is...... do these occur at precicely the same moment or do they occur at different moments? are we able to have a gravity so powerful that light doesnt escape, yet matter isnt collapsed? are we able to have collapsed matter with light escaping from it?

thank u so much

2. Nov 14, 2014

### rtsswmdktbmhw

Large mass gives higher gravity, it doesn't necessarily have to be in a small area.

3. Nov 14, 2014

### A.T.

The denser ball will create a steeper slope, which represents stronger local gravity, if the sheet represents the gravitational potential well:
http://en.wikipedia.org/wiki/Gravity_well

4. Jan 2, 2015

### LachyP

To answer the first question, we don't exactly have a completed answer. When the matter collapses and electron degeneracy, neutron degeneracy and possibly even quark degeneracy still is insufficient strength to prevent the collapse, it just continues collapsing into a singularity. Now this is the process that we do not understand because it now has no physical parameters, but infinite density and infinite gravity (only at the singularity). This collapse may never stop, we just do not know.

To answer your second question, I will create a scenario. Let' say that I have two stars, each of one solar mass(1.9891 x 10^30 kg), and condense one of the two until it is about a quarter of its original size. This condensed star would have more intense gravity when at a close distance, but from a significant distance, such as the orbits of our planets, it would be completely unchanged. This would be the same for a one solar mass black hole if it sat at the centre of our solar system (replaced the Sun), none of the planetary orbits would change, but If anything came into a close enough proximity of it, they would experience intense gravity.

To answer your last question, during the collapse, the radius obviously decreases. Now once this radius becomes smaller than the Schwarzschild Radius (which can be calculated by "Rs = 2GM/c^2"), the collapsing star would appear to become 'black'. So, the matter is collapsing while this 'light barrier' (to put it simply) is reached. But even after this 'barrier' is reached, the collapse would still continue, but would just be invisible to us.

I hope this answers yours questions and helps you to understand this concept better :)

5. Jan 6, 2015

### stevebd1

Not only does gravity increase with the reduction of r in the Newtonian equation $(g=Gm/r^2)$ but in relativity, you also have the metric tensor (which '..captures all the geometric and causal structure of spacetime') which is represented in the second half of the equation below-

$$g=\frac{Gm}{r^2}\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}$$

where $G$ is the gravitational constant, $m$ is the mass of the object and $c$ is the speed of light. The more compact an object becomes, the more gravity begins to increase exponentially.

6. Jan 6, 2015

### DaveC426913

In case the other answers were insufficient I will attempt to answer this in my layperson's way, and I'm going to use super rounded numbers just for clarity.

Gravitational force (or curvature) is dependent only on the mass and the distance.

Imagine a star 10x solar mass. Imagine it has a radius of, say, 5 million miles.
If you hovered in a spaceship at a distance from the star's centre of 5.1 million miles, you would feel a pull of X.

Now take that star of 10x solar mass and crush it to the radius of the Earth: 4,000 miles.
If you hovered in a spaceship at a distance from this star's centre of 5.1 million miles, you would feel the exact same pull of X. (mass has not changed, distance to centre of mass has not changed. Note that, if you had no windows on your spacecraft, you could not tell, based on your orbit, the size of the body you were orbiting, or your altitude above its surface - you could only tell its mass and the distance to the mass's centre. At a distance, the pull from a 10 solar mass block of styrofoam is identical to the pull from a 10 solar mass black hole.)

But - you are now (5.1 million - 4,000 =) 5,996,000 miles above the surface of the star. You could drop to a lower altitude - a lot lower altitude - if you wanted to. You are able to get much MUCH closer to the centre of mass of the star.

To sum: gravity is not stronger around a black hole, it is solely due to the compactness of a black hole that you can get closer.

7. Jan 7, 2015

### stevebd1

Some neutron stars, if they are compact enough, are predicted to have a photon sphere which would reside at 3M (where M=Gm/c2) though this strictly speaking isn't capturing the light as it can still escape to infinity. Light follows the curvature of spacetime and for the light not to escape, the mass has to be within the boundary of 2M (which is often called the Schwarzschild radius and defines the event horizon for a static black hole), once mass falls within this boundary there is no stable radius and the mass is predicted to collapse to a singularity (though this may change once a theory of quantum gravity has been established).

According to the time component of the Schwarzschild interior metric (see below), the point of no return is when the radius of the object reaches 9/4M (or 2.25M), this is when the proper time ($\tau$) at the centre of the neutron star becomes zero, this is a product that defines the event horizon of a black hole. As the matter collapses past 9/4M, the event horizon moves outwards from the centre towards surface until the star surface and the EH meet at 2M (see this image, the pink line represents the event horizon, the blue lines the collapsing star). Anything within 2M will collapse towards the singularity.

$$d\tau=\left( \frac{3}{2}\sqrt{1-\frac{2M}{r_0}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{r_0^{3}}}\right)dt$$
where $r_0$ is the radius of the star, $M=Gm/c^2$ and $r$ is the radius where you want to calculate the time dilation

You might also find the following link of interest, particularly the The Schwarzschild Metric and Inside the Black Hole sections-

Spacetime Geometry Inside a Black Hole

8. Jan 10, 2015

### TEFLing

That seems like an important clarification

As for the interior metric, seems like the cores of collapsed objects are forevermore fossilized for want of worthier words