# Description of black hole

1. Jan 4, 2005

### DaveC426913

Still editing a student textbook (elementary).

"The densest material in the universe is a black hole. It is the densest kind of matter in the universe."

The description of a black hole is obviously wrong: a black hole is not really a "material", and it's not simply "matter". Can someone suggest rewording for a middle grade level?

Perhaps, "a black hole is the densest 'place' in the universe"?

2. Jan 4, 2005

### DB

Yes, a black hole is the densest place in the universe, but surely it consist of matter at extreme densities. For a middle grader a simple but informative explanation could be this. A black hole is the densest place in the universe, with matter condensed into a very tiny tiny space where densities are extreme and can be infinite, such density and such mass in such a small amount of space leads to a very very strong force of gravity. It gets its name "black hole" because once light (and everything else for that matter) enters its center (singularity) gravity forces it back, not letting light out to shine, therefore a black hole. But a black hole does not "suck" all matter in like a vaccum unless matter has passed it's event horizon, where there is point of no return. Black holes are cause by the "death" of giant stars, they power galaxies, keeping stars in orbit; stars keeping planets in orbit; planets keeping moons in orbit. They are the reason the universe is the way it is today.

3. Jan 4, 2005

### Integral

Staff Emeritus
To claim infinite density is a bit extreme. It may be better to admit that the matter is so dense that the laws of the universe as we understand them break down. So we are unable to even guess what happens in those regions.

It may be refreshing for a student to get a well worded "we don't know".. cause we don't.

4. Jan 4, 2005

### DB

True that is, though a black hole's singularity has potential to infinite density no?
From the glossary: "Singularity In classical general relativity, a location at which physical quantities such as density become infinite. A singularity lies at the center of a black hole." I'm assuming it's saying that a black hole has a finite density, but it's such a large number that it shouldn't be counted, just considered infinite?

5. Jan 4, 2005

### Mk

A new theory came out, I read it in the "Controllng Hurricanes" issue of
Scientific American, it talks about a new concept per se of "phantom matter," what a black hole's made of.

Calling the black hole matter? I guess so.

Maybe something like:

A concentration of mass with gravity so strong that to get away from it you must be moving faster than light.

:) It doesn't have to be perfect, and it almost can't be with out using huge words and vocabulary terms not designated for elementary to middle school students.

"The densest place?" There are many black holes, and some places are denser; there are many levels of black holes, from micro and electron black holes to SMBHs (Stellar Mass Black Holes), IMBHs (Intermediate Mass Black Holes), to the Super-massive.

-XD

Last edited: Jan 4, 2005
6. Jan 4, 2005

### DB

At the least faster than light. The problem with a black hole each has it's own characteristics, but we cant figure them out. We know that light cannot escape them, but each black hole has a different escape velocity. A seyfert galaxie agn for example problably has a much bigger escape velocity then our milky way center.

7. Jan 4, 2005

### Mk

... or a micro-black hole.

8. Jan 5, 2005

### SGT

If you consider the size of the black hole to be its event horizon, the more massive black holes are less dense than the less massive. A really massive black hole can be less dense than a tenuous gas.

9. Jan 5, 2005

### DaveC426913

One of the other questions I've had (from the editor) is: is it a black hole's density that gives it its huge gravity?

My reponse was: no.

A black hole has the same gravitational pull as any other mass in the universe. Gravity is gravity. If the Sun were replaced with a black hole of the same mass, the planets would orbit it just the same. A black hole usually has has strong gravity because they are typically very massive (they need to be in order to form at all).

This is a point that many laypeople often misunderstand.

If the Earth were replaced with a black hole of the same mass as the Earth, and we were standing on a ladder 4000 miles above it, we would not notice any difference. Of course, since that black hole is so small, we could climb down the ladder (which we can't do on the surface of the Earth), where we would begin to experience very strong gravitational forces.

10. Jan 5, 2005

### DB

Well I wouldn't say gravity is gravity. I would say it was accelerated force. And this force varies all over the universe. A Black hole's gravitational and magnetic fields are massive, but their singularity is very small, some say they can be near planck lenght. As a star's evolution is running to its end, it loses its outer layers and the hot core begins to cool and contract because it can no longer fuse nuclei to support the pressure of its gravity. A black hole is somewhat like this, though on massive star scales.

Gravity varies intierly on density and mass. When such a star is extremly dense, heavy and small, it forms its black hole. And as it does so it curves spacetime extremely: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/BlackHoles/transp8.2.gif

So if our distance from the black hole is great enough, yes we would be able to orbit the hole, though the change in our orbits would depend on the amount of time we took to make the switch. Maybe pluto and neptune were not effected, though mercury surely was, because gravity waves travel at light speed.

A black hole is the strongest force of gravity in the universe, with stars being caught in orbit.

11. Jan 5, 2005

### DaveC426913

"Well I wouldn't say gravity is gravity."

(FYI, I am not literally writing for children. The editor does that. What I do is provide a science reference.)

Yes, gravity is gravity. It is a function of the mass and its distance, nothing else.

The point I am trying to get across to the editor (who asked what makes the gravity of a black hole so strong) is that a "black hole's gravity" is not "strong" at all. A black hole's gravity is the same as gravity from any other body with the same mass.

There are lots of objects whose gravitational pull easily exceeds that of a black hole, like, oh, say the Andromeda Galaxy, with the mass of a half billion stars. But even single objects - some stars are more massive than some black holes, and will have a stronger gravitational pull (assuming a fixed distance from their centres).

The only thing that's different about a black hole is that it fits into a space that allows us to dramatically reduce that distance.

12. Jan 5, 2005

### DB

I understand this as: the force of gravity is the same anywhere in the universe. It's not at all. Gravity is an accelerated force and vise versa. This is the equivilance principle. Surely accelerated force can vary.

Not at all.
Gravity is not only a funtion of mass and distance (radial), it is most dependent on density. In the fabric of spacetime of a massive star (that will later create a black hole) has a strong gravititional field, though because of its large radial distance, it does not warp spacetime as much as it would if it were the size of a pee! (and had the same mass) Its gravitational force is more spread out due to its massive size. Take a look :http://nrumiano.free.fr/Images/bh_warp1_E.gif Notice how the neutron star deformes space time (warping of spacetime leads to stronger and stronger gravitational force) much more than the usual star. Why? Because it is denser! So dense that it's electrons and protons are packed so tightly they form neutrons. A black hole has almost the exact mass of the star it used to be. But it is billions of time denser, creating a much much strong accelerated force i.e gravity towards it.

And why are those star caught in orbit? They are not orbiting around nothing, the strong force of a black hole is keeping them in orbit. The same as our sun keeps our tiny planets in orbit. Yes all stars are more massive then a black hole's singularity. Much larger! But the gravitational force of a black hole is much stronger. Strong enough to keep the half a billions star in orbit!

13. Jan 5, 2005

### DB

SGT, a neutron star is the closet thing to a black hole. Because the density of a black hole is so large, it's very hard to measure and study. So I will use a neutron star.
The averge density of a neutron star is $$10^{15} g/cm^3$$! I have never heard of a gas that dense, certainly not on earth.
You would weigh 1000000000000 times more on the surface of a neutron star then on earth. Due the 10^11 times stronger accelerated force pushing down on your body.

Also, the size of a black hole is measured by the Schwarzchild radius. Conside a circle where the circumference is the event horizon and the radius is simply the Schwarzchild radius.

$$R_{schw}=2*G*M/c^2$$
G is the gravitational constant
M the mass
c the speed of light.

Because the G is so small and c^2 is so big, were talkin one small radius.

Edit:
No. Let me correct myself lol. The Radius of the singularity is very small, the Schwarzchild radius can be pretty big because the mass of the star is huge.

Last edited: Jan 5, 2005
14. Jan 5, 2005

### DaveC426913

Let's just establish what we agree on and what we don't.

If I am stationary at 10^6km from (the centre of) a planet of mass 10^23kg, I will experience a pull of X.
If I am stationary at 10^6km from (the centre of) a black hole of mass 10^23kg, I will experience a pull of X.
In a windowless spaceship, I will not be able to tell what object I am near.

If descend until I am now 6x10^3 km from the planet's centre (hovering 1 foot of the ground), I will experience a pull of Y.
If I stop next to a black hole, so that I am now hovering 6x10^3 km from its centre, I will experience a pull of Y.
I will still not be able to tell which object I am near.

However, if I now descend another 10^3 km, I can now tell that it couldn't be a normal mass I am hovering over, since my calcs would show my height above it to be below its surface. It must be something denser, though I can't tell what yet.

That graph is schematic only. It has no figures. And that's my point. If you graphed a neutron star and a regular star both of the same mass, the curves all the way down to the surface of the larger star would be identical.

15. Jan 6, 2005

### DB

Yes, that is making sense. As long as your space ship does not pass the event horizon (keeping a longer radial distance from the singularity than the Schwarzchild radius), yes he will not notice that it wasn't an ordinary star. That is why galaxis exist. Some people consider black holes to be vaccums that suck everything up. There not, and this we agree on.

I will take this into account. Though tell me why it is you think this?
I disagree with it. And I can back my points. A common anology of spacetime fabric is "space foam" (like those beds where you can leave a hand print) a rubberish fabric. If you were to place a bowling ball in the center of space foam is would sink, curving the fabric along it's circumference. Now if you took a ball with exactly the same mass but with 1/10th the circumference and radius it will sink to the same depth that the bowling ball did and curve the space foam around its circumference. But since the circumference is much smaller, it would lead to a greater curvature of space foam. It would look like a hole. I think you can picture it. So knowing this situation, general relativity tells us that: the greater the curvature of spacetime, the strong the gravitational force. So if we took a marble and placed it on the space foam for both situations (very important that we don't place it with a force or initial velocity because it would fall into "orbit" for a bit) Here's what would happen:

The bowling ball: the marble would roll towards the center of the bowling ball at a moderate accelerated speed.

The denser bowling ball of same mass: the marble would actually roll slower towards the "event horizon" of the hole, though once it passed it, it would fall towards the ball at much faster accelerated speed than our other situation.

Edit: I realize now surface area might be better to say than circumference.

Accelerated speed relates to gravity, gravity relates to curvature of spacetime, this is why I find gravity to be more than just gravity. So my points say that, yes; the denser the object the more curvature of space time is created.

Last edited: Jan 6, 2005
16. Jan 6, 2005

### DaveC426913

"...this is why I find gravity to be more than just gravity. So my points say that, yes; the denser the object the more curvature of space time is created..."

The difference in curvature is only when within the radius of the star! (where you can only get with the BH). Farther than the radius of the star, you cannot tell the diff.

If you are in a windowless spaceship And you find yourself 10^6km from an object pulling on you, and you measure the gravitiational force exerted upon you, the only thing it will tell you is the mass of the object ($$m = G_1*r_1 ^2$$).

If you then move yourself to within 10^4km of the object, and take a measurement again, you will still be able to tell the mass of the object ($$m = G_2*r_2 ^2$$).

But you still have no info about any other property of the body. You have no info because gravity is dependent on mass and distance. There is no variable for density in the formula.

BTW, it is important to note that one measures an object from its centre of mass, not from its surface. That's why size (and thus volume, and thus density) is irrelevant.

The graph:

The graph is inadequately drawn in a way that allows one to draw true parallels between the two components.

Here, I will redraw it to show the correct relation:
Black hole vs. massive star

Last edited: Jan 6, 2005
17. Jan 6, 2005

### DB

Once again, I agree with you there.

"Falling in
Consider a hapless astronaut falling feet first radially towards the center of a simple Schwarzschild-type (non-rotating) black hole. The closer he gets to the event horizon, the longer the photons he emits take to escape from the black hole's gravitational field. A distant observer will see the astronaut's descent slowing as he approaches the event horizon, which he never appears to reach.

However, in his own frame of reference, the astronaut will cross the event horizon and reach the singularity, all in a finite amount of time. Once he has crossed the event horizon he can no longer be observed from the outside universe. As he falls, he will notice his feet, then his knees, becoming increasingly red-shifted until they appear invisible. As he nears the singularity, the gradient of the gravitational field from head to foot will become considerable, and he will feel stretched, and finally torn. This process is known as spaghettification. This gradient becomes large enough, close to the singularity, to tear atoms apart. The point at which these tidal forces become fatal depends on the size of the black hole. For a very large black hole such as those found at the center of galaxies, this point will lie well inside the event horizon, so the astronaut may cross the event horizon painlessly. Conversely, for a small black hole, those tidal effects may become fatal long before the astronaut reaches the event horizon."(Wikipedia)

These are from Wikipedia:
"Black holes as they are most widely understood require general relativity's concept of a curved spacetime, since their most striking properties rely on a distortion of the geometry of the space surrounding them."

"At the center of the event horizon is a singularity, a place where general relativity predicts that spacetime becomes infinitely curved (i.e., where gravity becomes infinitely strong). "

Infinitely curved.

I have studied your graph, and it does seem to make sense, and I'm not an expert so I should say if it's right or wrong.
The last quote above states as I have said before that the more curvature of spacetime the stronger the gravitational field. So if a black hole and a massive star have the same gravitational field, as your graph seems to say, then how come light can escape the massive star?

Aswell I don't understand why spacetime fabric would not curve to the bottom of your massive star? Why does it seem like there is something beneath it that is curving spacetime?

18. Jan 6, 2005

### DB

Ah, I've found the jackpot!

http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag2.jpg [Broken]

Last edited by a moderator: May 1, 2017
19. Jan 6, 2005

### DB

This is another quote from http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag.html [Broken]

"It is hard for us to visualize a curved space or spacetime, especially if that space (spacetime) has three or more dimensions. However, physicists have invented a "trick" for visualizing a curved two-dimensional surface within an imaginary, flat, three-dimensional space, and that trick is called an "embedding diagram" (Misner et al 1973, 613ff; Thorn 1994, 129ff. â€“Mâ€“Ã³1997, 112ff.). You already know the distinction of intrinsic and extrinsic curvature. For instance, the surface of a sphere is a curved two-dimensional surface, and its curvature can be identified by measurements within that surface; thus we can say that the surface has an intrinsic curvature. However, we can of course visualize it within a three-dimensional space, by regarding it as the surface of a sphere in a flat, three-dimensional Euclidean space. The embedding diagram utilizes this relation. But do not imagine a higher dimensional space used for embedding as something "real"; it is only a fiction for enabling us to visualize the curvature.

Suppose a simple two-dimensional space, which has a concavity in the middle, as in the figure. Gravity in general relativity is expressed by the curvature of spacetime. This means that light rays, or free particles go along a geodesic, a straightest line allowed by the curvature. Thus if two light rays (red lines) come to the concavity, they will cross at the bottom (because of the curvature of this part), and this can be expressed by a two-dimensional figure. However, the image is somehow poor, and a better visualization is to embed this space into a three-dimensional Euclidean space, and express its curvature as an extrinsic curvature of a two-dimensional surface within this 3-space. The result is an embedding diagram.

The same trick can be applied to any two-dimensional surface in a curved 3-space or 4-space (spacetime). For instance, the equatorial plane crossing a star (spherical, without spin, and with a constant density) is a curved 2-space, and its embedding diagram is obtained as follows. For the sake of comparison, we consider stars with the same mass but with different sizes; you can see (intuitively) how the curvature of the equatorial plane (2-space) changes. On the left-hand side are pictures of physical (3-) space, and on the right-hand side are embedding diagrams of the equatorial plane (2-space). These embedding diagrams illustrate Schwarzshild's solution to the Einstein field equations. Such a diagram can be extended to a black hole too."

http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag2.jpg [Broken]

Last edited by a moderator: May 1, 2017
20. Jan 6, 2005

### DaveC426913

Yep. I see where we're diverging. We don;t disagree, we're just talking about towo different events.

As we get closer to the black hole than we can get to the star, gravity will increase. But the gravity of a black hole at any farther distance is exactly the same as the gravity from any other object of the same mass.

i.e. If the Earth suddenly imploded into a black hole the size of a pea, and I were left standing on a platform (i.e. didn't start falling), I would find myself ~6000km above its surface.

But - I would not feel any stronger pull. I would still weigh 175lbs - black hole or normal Earth. Same mass, same distance = same pull.

Yep. No argument here.

If we were to draw a light beam's path on the curves, it would be able to escape from the relatively shallow slope around the star, but it would not be able to climb the steeper slope near the black hole.

The star does not have the steep slope that the black hole does.

As you point out, next:

My mistake. You are correct. Gravity drops off in the centre of the star, thus the pit would have a rounded bottom. My graph is inaccurate within the radii of the objects.

(Thus, no point on the star's pit has a slope so steep that light can't climb up it.)

We seem to be reaching consensus.

Last edited: Jan 6, 2005