# Is there a density point which reduces anything to a black hole?

• delsaber8
In summary, you can create a black hole by squeezing an object down to a very dense state. The smaller the mass, the higher the density you need to achieve in order to collapse.
delsaber8
First off I couldn't decide whether to enter my question in the relativity section or this one, so if I'm in the wrong place my apologies. Anyway I have a question or perhaps a hole in my understanding. I was wondering if there is a density point at which anything becomes a black hole. Now I understand the conventional black hole where the mass of a neutron star becomes so great that the gravity will crush it into a singularity but I'm wondering if hypothetically we crushed something down to the point where it would be the size of the singularity would a black hole form? Of course it would be an everyday object nothing close to the mass of a star, so could it still form a black hole?

Sorry if what I'm explaining comes across a little vague or disjointed but I'm having a bit of trouble describing what I'm thinking of.

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Since the black hole "singularity" is by some definitions a point, you can't squeeze it down that far, BUT yes for any given amount of matter, if you squeeze it down small enough the local acceleration due to gravity exceeds the speed of light and it becomes a black hole. It is hypothesized that very small black holes evaporate quickly due to Hawking radiation, so to make one that lasts, you'd have to start off with a fair amount of matter.

Squeezing ANYTHING down that far is not currently within the capability of human technology. There was fear mongering in the popular press that when CERN came online it would create a tiny black hole that would eat the Earth.

phinds said:
Since the black hole "singularity" is by some definitions a point, you can't squeeze it down that far, BUT yes for any given amount of matter, if you squeeze it down small enough the local acceleration due to gravity exceeds the speed of light and it becomes a black hole. It is hypothesized that very small black holes evaporate quickly due to Hawking radiation, so to make one that lasts, you'd have to start off with a fair amount of matter.

Squeezing ANYTHING down that far is not currently within the capability of human technology. There was fear mongering in the popular press that when CERN came online it would create a tiny black hole that would eat the Earth.

Thank you, you answered my question, one last thing though, why exactly, does having matter compressed to this extremely dense state cause a curvature in space time? I was under the impression that gravity was a result of mass only?

delsaber8 said:
Thank you, you answered my question, one last thing though, why exactly, does having matter compressed to this extremely dense state cause a curvature in space time? I was under the impression that gravity was a result of mass only?

You put enough mass in a small space, yes, a warping of the fabric of space will occur close to where the mass is located.

If you could visit a neutron star and not be annihilated by the radiation and other lethal effects present, the intense gravity would cause some remarkable optical effects:

http://wordpress.mrreid.org/2012/08/12/you-can-see-more-than-half-of-a-neutron-star/

NASA has prepared some illustrations and videos showing what an observer near a neutron star would see due to gravitational distortion of space.

http://apod.nasa.gov/htmltest/rjn_bht.html

delsaber8 said:
Thank you, you answered my question, one last thing though, why exactly, does having matter compressed to this extremely dense state cause a curvature in space time? I was under the impression that gravity was a result of mass only?

ALL mass causes a "bend" or "warp", or whatever you want to call it (I think "curve" is the most common term), in space-time, it's just that it isn't noticeable until you get to HUGE amounts of mass such as a galaxy or a black hole. Google Einstein Rings for examples of galaxies causing such.

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delsaber8 said:
... I was wondering if there is a density... at which [any given mass] becomes a black hole. … [Suppose] it would be an everyday object nothing close to the mass of a star, so could it still form a black hole?
...

I think basically you are asking given some specified mass, how small do you have to crush it to so that a black hole forms? I suggested some edits to your question to bring out what I think your meaning is. Another way to say it would be

given some specified mass, how DENSE do you have to make it in order for a BH to form?

The fact is it depends on the mass! The smaller the mass the higher the density you have to achieve in order to trigger collapse. Very large masses, like billions of tons, can collapse more "easily" in the sense that they don't have to be squeezed to as high a density as little masses.

maybe you would like to know how small you'd need to squeeze a billion ton mass.

it's easier to work in KILOGRAMS. A (metric) ton is 1000 kg (about the same but a little more than a "2000 pound English ton" or socalled "short ton". Let's work in metric tons. A billion tons is 1012 kg. That is 109 (a billion) times 103 kg (a ton).

A trillion tons is 1015 kg. That's the mass of a goodsize asteroid, or a mountain, or a cube block of ice which is 10 kilometers on a side. It's nowhere near the mass of a STAR but it is certainly respectable.

You can calculate the radius of a little ball you would have to squeeze that down into so as to get collapse. Then if you like you can calculate the corresponding density (i.e. the mass divided by the volume of that little ball.)

It's easy to calculate, using the calculator built into Google. You just have to multiply that mass by 2G (twice the Newton grav constant) and divide by the speed of light squared. So go to Google and put this into the window:

"2G*10^(15)kg/c^2" and press return. Google knows G the multipurpose grav constant, so what that says to the built-in calculator is "take twice G and multiply by trillion tons, and divide by c^2 which is speed of light squared."

Why don't you type that into the window (without the quotes) and press return and see what you get?
You should get something like a trillionth of a meter. A trillionth is 10-12.
Something like that, maybe closer to 1.5 trillionths of a meter.

That then is the radius of the little ball you would have to squeeze the mountain or the asteroid down into in order for it to trigger full collapse and for a BH to form.

If you can do the calculation for one mass you can do it for other masses, like the mass of earth, or the mass of the sun. You just have to put different masses in.
Google knows the mass of the Earth so for example you could put in (without the quotes)
(2G*mass of earth)/c^2
You should get something roughly like a CENTIMETER

Of course compressing the whole Earth down into a centimeter radius ball would be an extremely high density. It is very hard to achieve collapse density with ordinary objects like the Earth, or smaller, like a used car or old television set. With star-size masses you don't have to reach such insanely high densities. you can check it out by calculating with the mass of the sun. Google knows "mass of sun" so you just put that in place of "mass of earth" and go.

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marcus said:
I think basically you are asking given some specified mass, how small do you have to crush it to so that a black hole forms? I suggested some edits to your question to bring out what I think your meaning is. Another way to say it would be

given some specified mass, how DENSE do you have to make it in order for a BH to form?

The fact is it depends on the mass! The smaller the mass the higher the density you have to achieve in order to trigger collapse. Very large masses, like billions of tons, can collapse more "easily" in the sense that they don't have to be squeezed to as high a density as little masses.

maybe you would like to know how small you'd need to squeeze a billion ton mass.

it's easier to work in KILOGRAMS. A (metric) ton is 1000 kg (about the same but a little more than a "2000 pound English ton" or socalled "short ton". Let's work in metric tons. A billion tons is 1012 kg. That is 109 (a billion) times 103 kg (a ton).

A trillion tons is 1015 kg. That's the mass of a goodsize asteroid, or a mountain, or a cube block of ice which is 10 kilometers on a side. It's nowhere near the mass of a STAR but it is certainly respectable.

You can calculate the radius of a little ball you would have to squeeze that down into so as to get collapse. Then if you like you can calculate the corresponding density (i.e. the mass divided by the volume of that little ball.)

It's easy to calculate, using the calculator built into Google. You just have to multiply that mass by 2G (twice the Newton grav constant) and divide by the speed of light squared. So go to Google and put this into the window:

"2G*10^(15)kg/c^2" and press return. Google knows G the multipurpose grav constant, so what that says to the built-in calculator is "take twice G and multiply by trillion tons, and divide by c^2 which is speed of light squared."

Why don't you type that into the window (without the quotes) and press return and see what you get?
You should get something like a trillionth of a meter. A trillionth is 10-12.
Something like that, maybe closer to 1.5 trillionths of a meter.

That then is the radius of the little ball you would have to squeeze the mountain or the asteroid down into in order for it to trigger full collapse and for a BH to form.

If you can do the calculation for one mass you can do it for other masses, like the mass of earth, or the mass of the sun. You just have to put different masses in.
Google knows the mass of the Earth so for example you could put in (without the quotes)
(2G*mass of earth)/c^2
You should get something roughly like a CENTIMETER

Of course compressing the whole Earth down into a centimeter radius ball would be an extremely high density. It is very hard to achieve collapse density with ordinary objects like the Earth, or smaller, like a used car or old television set. With star-size masses you don't have to reach such insanely high densities. you can check it out by calculating with the mass of the sun. Google knows "mass of sun" so you just put that in place of "mass of earth" and go.

Thank you, it is nice to have this equation to visualize the needed densities of different objects to become a black hole. I feel as though previously I may have misunderstood gravity. So just to clear things up, when an object of constant mass is squeezed down in size i.e increasing in density so to increases the curvature of space and time?

Just to complete this topic, take any density you want. Then, with enough volume at that density, you form an event horizon and thus a BH. A galaxy size ball of air would be black hole (and would then rapidly collapse to singularity per classical GR).

delsaber8 said:
Thank you, it is nice to have this equation to visualize the needed densities of different objects to become a black hole. I feel as though previously I may have misunderstood gravity. So just to clear things up, when an object of constant mass is squeezed down in size i.e increasing in density so to increases the curvature of space and time?

Basically that is what the 1915 GR equation says. Think of the LHS as curvature expressed in units of inverse area (units could be 1/meter2), and the RHS talking about energy density (units could be joules per cubic meter, energy per unit volume). So as one increases the other must increase because LHS = RHS.

But why is curvature expressed as 1/area ? think about different size spheres. A little sphere is strongly curved and has a small area. Big spheres are only slightly curved and have a large surface area. So its INVERSELY PROPORTIONAL. little area ⇔ big curvature
big area ⇔ tiny amount of curvature
so curvature ≈ 1/area

So what kind of PROPORTIONALITY CONSTANT is there between 1/area and energy/volume ?

That is something you might think about overnight or sometime when you have free time to just think.

A hint: the metric unit of energy joule = force unit x distance unit
One form of energy is WORK, the work of pushing with unit force for a unit of distance.

energy density = (force x distance)/distance3 = force/distance2
= force x 1/area

So if you multiply a curvature by a force you get an energy density. And Einstein discovered that energy density DIVIDED BY a certain force gives you the curvature that the force causes.

If you keep on asking questions, I or somebody else here will probably tell you what that force is that is the constant of proportionality between the LHS and RHS of the Einstein GR equation.

Also be warned that the Einstein GR equation might be just slightly wrong at extremely high density. People are working on that. When they "quantize" the classic 1915 GR equation some people have found that a "quantum correction term" appears which only has a significant effect at very very very high density. So something else besides the formation of a "singularity" might happen when a star collapses. These people talk about it in this recent research paper
http://arxiv.org/abs/1401.6562 just click on PDF, it's free.

But Einstein GR equation in original NOT quantized form is very nearly perfect and the possible error in the proportionality between density and curvature only comes in at extreme density. For almost all cases it is amazingly accurate.

Also I have oversimplified by narrowing the discussion down to just energy density and a simplified form of curvature. I wanted to get across the spirit of the equation without to much extra detail.

marcus said:
Basically that is what the 1915 GR equation says. Think of the LHS as curvature expressed in units of inverse area (units could be 1/meter2), and the RHS talking about energy density (units could be joules per cubic meter, energy per unit volume). So as one increases the other must increase because LHS = RHS.

But why is curvature expressed as 1/area ? think about different size spheres. A little sphere is strongly curved and has a small area. Big spheres are only slightly curved and have a large surface area. So its INVERSELY PROPORTIONAL. little area ⇔ big curvature
big area ⇔ tiny amount of curvature
so curvature ≈ 1/area

So what kind of PROPORTIONALITY CONSTANT is there between 1/area and energy/volume ?

That is something you might think about overnight or sometime when you have free time to just think.

A hint: the metric unit of energy joule = force unit x distance unit
One form of energy is WORK, the work of pushing with unit force for a unit of distance.

energy density = (force x distance)/distance3 = force/distance2
= force x 1/area

So if you multiply a curvature by a force you get an energy density. And Einstein discovered that energy density DIVIDED BY a certain force gives you the curvature that the force causes.

If you keep on asking questions, I or somebody else here will probably tell you what that force is that is the constant of proportionality between the LHS and RHS of the Einstein GR equation.

Also be warned that the Einstein GR equation might be just slightly wrong at extremely high density. People are working on that. When they "quantize" the classic 1915 GR equation some people have found that a "quantum correction term" appears which only has a significant effect at very very very high density. So something else besides the formation of a "singularity" might happen when a star collapses. These people talk about it in this recent research paper
http://arxiv.org/abs/1401.6562 just click on PDF, it's free.

But Einstein GR equation in original NOT quantized form is very nearly perfect and the possible error in the proportionality between density and curvature only comes in at extreme density. For almost all cases it is amazingly accurate.

Also I have oversimplified by narrowing the discussion down to just energy density and a simplified form of curvature. I wanted to get across the spirit of the equation without to much extra detail.

I think I'm kinda starting to understand this better, but of course things are still a little fuzzy. This constant that equates both sides of the formula isn't one of those tensors that I have come across on wikipedia is it? Perhaps I better continue thinking this through before I make any guesses at what this force is.

delsaber8 said:
...This constant that equates both sides of the formula isn't one of those tensors that I have come across on wikipedia is it? Perhaps I better continue thinking this through before I make any guesses at what this force is.

It's just an regular force quantity, the kind measured in units called "Newtons" in the metric system . why don't I just go ahead and say what it is in known terms and let you calculate it and see what size it is?

It's not anything complicated like a tensor. BTW do you use the metric system? One "Newton" of force is roughly equal to the weight, in normal Earth gravity, of a mass which is 1/10 kilogram.
If you have, say, a one kilogram block of cheese and you weigh it in your hands you feel a force of 9.8 Newtons. This is probably very familiar stuff to you, but I wanted to make sure.

This force has a really simple formula. It is made using just two natural constants, the speed of light c and the universal grav constant G. You can get google calculator to tell you what it comes to in metric terms, ie. in Newtons.

This force which s the proportionality between LHS and RHS of the Einstein GR equation is
c4/G

If you put "c^4/G" into the google window it should tell you some Newtons. This force is very big! Like you might imagine a "universe force" would be (whatever that might be.) It might be something like 1 followed by 44 zeros, written in terms of the metric unit that you can feel with your hands.

Remember that if you divide an energy density by a force, you get a simple curvature quantity: a reciprocal area. Dividng by c4/G is the same as multiplying by G/c4.

So in the Einstein GR equation you take an energy density on the RHS, and you multiply it by G/c4 and you get the curvature quantity on the LHS. It is the curvature that matter and energy put into geometry. It is how matter and geometry interact. (by that force, that you divide by or its reciprocal 1/F that you multiply by).

Also there is a factor of 8π in there that I didn't mention. You actually multiply by 8πG/c4. That's just conventional, it comes in because of the way curvature tensors on the LHS are traditionally defined.

Try using google calculator to calculate the Force, and see what you get. Put c^4/G into the window and press return. The fourth power of the speed of light divided by the universal gravity constant.

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marcus said:
It's just an regular force quantity, the kind measured in units called "Newtons" in the metric system . why don't I just go ahead and say what it is in known terms and let you calculate it and see what size it is?

It's not anything complicated like a tensor. BTW do you use the metric system? One "Newton" of force is roughly equal to the weight, in normal Earth gravity, of a mass which is 1/10 kilogram.
If you have, say, a one kilogram block of cheese and you weigh it in your hands you feel a force of 9.8 Newtons. This is probably very familiar stuff to you, but I wanted to make sure.

This force has a really simple formula. It is made using just two natural constants, the speed of light c and the universal grav constant G. You can get google calculator to tell you what it comes to in metric terms, ie. in Newtons.

This force which s the proportionality between LHS and RHS of the Einstein GR equation is
c4/G

If you put "c^4/G" into the google window it should tell you some Newtons. This force is very big! Like you might imagine a "universe force" would be (whatever that might be.) It might be something like 1 followed by 44 zeros, written in terms of the metric unit that you can feel with your hands.

Remember that if you divide an energy density by a force, you get a simple curvature quantity: a reciprocal area. Dividng by c4/G is the same as multiplying by G/c4.

So in the Einstein GR equation you take an energy density on the RHS, and you multiply it by G/c4 and you get the curvature quantity on the LHS. It is the curvature that matter and energy put into geometry. It is how matter and geometry interact. (by that force, that you divide by or its reciprocal 1/F that you multiply by).

Also there is a factor of 8π in there that I didn't mention. You actually multiply by 8πG/c4. That's just conventional, it comes in because of the way curvature tensors on the LHS are traditionally defined.

Try using google calculator to calculate the Force, and see what you get. Put c^4/G into the window and press return. The fourth power of the speed of light divided by the universal gravity constant.

So if I have this correct the equation would be:
Energy Density x 8πG/c^4 = 1/area

and then say if I wanted to calculate energy density I would just shuffle the equation around?

delsaber8 said:
So if I have this correct the equation would be:
Energy Density x 8πG/c^4 = 1/area

and then say if I wanted to calculate energy density I would just shuffle the equation around?

Yes, you got it!

c^4/G if you work it out with google-calculator is a FORCE, it comes to around 1044 Newtons and you can feel with your hand what force a Newton is (not all that much, like a quarter pound block of butter in the fridge.

so dividing by that force is like multiplying by G/c^4

I wrote it in that order because I think of energy density as CAUSING curvature, so given some energy density I want to know what curvature it causes in the geometry of nature.
So I multiply it by 8 pi G/c^4 and that tells what curvature it causes.

But you could swap it around, like you say and given a certain curvature you might ask "I wonder what concentration of matter and energy could have caused that?"

So you would multiply that given curvature by the Force, namely by c^4/(8 pi G).

My friend who has an unpredictable mind and uses simple-minded word-association to remember the important quantities in nature, like this force, when he saw that said:
"See! for I ate piG!"
Now I cannot forget this no matter how much I would like to.

marcus said:
Yes, you got it!

c^4/G if you work it out with google-calculator is a FORCE, it comes to around 1044 Newtons and you can feel with your hand what force a Newton is (not all that much, like a quarter pound block of butter in the fridge.

so dividing by that force is like multiplying by G/c^4

I wrote it in that order because I think of energy density as CAUSING curvature, so given some energy density I want to know what curvature it causes in the geometry of nature.
So I multiply it by 8 pi G/c^4 and that tells what curvature it causes.

But you could swap it around, like you say and given a certain curvature you might ask "I wonder what concentration of matter and energy could have caused that?"

So you would multiply that given curvature by the Force, namely by c^4/(8 pi G).

My friend who has an unpredictable mind and uses simple-minded word-association to remember the important quantities in nature, like this force, when he saw that said:
"See! for I ate piG!"
Now I cannot forget this no matter how much I would like to.

Actually that is a pretty nifty way of remembering the formula. I do appreciate your help, thank you!

## 1. What is a density point in relation to black holes?

A density point refers to the point at which the density of an object becomes so high that its gravitational pull becomes strong enough to overcome the forces that hold its particles together, resulting in the formation of a black hole.

## 2. Is there a specific density point that guarantees the formation of a black hole?

Yes, there is a critical density point called the Schwarzschild radius, which is the minimum density required for an object to become a black hole. This density point is directly related to the mass of the object and its radius.

## 3. Can any object reach the density point necessary to become a black hole?

Any object can theoretically reach the density point required to form a black hole, as long as its mass is large enough to create a strong gravitational pull. However, for most objects, this would require extreme amounts of mass and pressure.

## 4. How does the density point relate to the size of a black hole?

The density point is directly related to the size of a black hole. The larger the mass of an object, the smaller its radius needs to be to reach the critical density point and form a black hole. This is why black holes are often described as having infinite density at their center, known as a singularity.

## 5. Is there a way to prevent an object from reaching the density point and becoming a black hole?

In most cases, it is not possible to prevent an object from reaching the density point necessary to form a black hole. However, some theories suggest that extremely high energy collisions or the addition of large amounts of antimatter could potentially disrupt the formation of a black hole.

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