## Main Question or Discussion Point

Hello,

First off, I'm not a physics student by any means, but it does intrigue me. I've been wondering about the Schwarzschild radius, and how the density of an object dictates the escape speed. Why is it, that the more you compress an object, the more gravity it "owns?" If you were able to theoretically compress the earth down to 9.00mm, you'd have reached it's Schwarzschild radius, and essentially would have created a black hole... but why? I understand why that happens when stars collapse upon themselves due to mass, but I don't understand how taking a fixed amount of mass and compressing it, increases the amount of gravity to that object.

A "simple" explanation would be much appreciated. Again, I am not a physicist, and I only understand the basics.

Thanks

-Sean

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phinds
Gold Member
2019 Award
It doesn't increase the amount of gravity as felt by an object outside the Schwarzschild radius, but it warps space-time so much that objects inside that radius cannot escape. Think of it somewhat not as the AMOUNT of gravity but the DENSITY of gravity. Does that help?

mathman
The simplest explanation (not using General Relativity) is that the force of gravity is proportional to m/r2, where m is the mass of the earth (the attracting body) and r is its radius. Shrinking the radius down to 9 mm makes the force that much larger in comparison to that of the earth in its present size.

The gravitational field strength drops like 1/r^2 as you get further away, right? Imagine going closer to the mass, then the strength will keep increasing: get twice as close and the field is four times as strong. So what happens when you get too close?

In most cases, this trend is halted when you hit the surface of the object, because once you start digging to get even closer, some of the mass starts being above you and the effective mass is reduced. Once you pass the surface of the object (assuming homogeneous density), the 1/r^2 relationship no longer applies, and it becomes now a linear relationship with r. This is the key:

The gravitational field strength attains a maximum at this surface of the object.

But if you concentrate the same amount of mass into a smaller volume, you can keep going closer to the centre without hitting the surface, and the 1/r^2 increase in gravitational field strength continues. At the Schwarzschild radius (distance from the centre of mass), if you still haven't hit the surface, then the field strength at this point is so high you have the event horizon, and the result is a black hole.

It's essentially all down to this 1/r^2 relationship, which comes from the surface area of a sphere: the "flux" of gravity through a spherical surface centred on a mass is the same for all radii greater than the radius of the mass, and the surface area of a sphere is proportional to the square of the radius. So if you halve the radius of a sphere, you quarter its surface area, and to conserve gravitational flux, the gravitational field strength must be four times bigger. So that's where the 1/r^2 term comes from.

Ahh ok, so using Earth as an example, does this mean that gravity's force is stronger as I get closer to the core? As in, if an object weighs 100lbs on the earth's crust, it weighs more in the mantle, etc?

-Sean

Nugatory
Mentor
Ahh ok, so using Earth as an example, does this mean that gravity's force is stronger as I get closer to the core? As in, if an object weighs 100lbs on the earth's crust, it weighs more in the mantle, etc?

-Sean
No - that's in the second paragraph of mikeph's post #4 above.

"The gravitational field strength attains a maximum at this surface of the object. "

Assuming you can compress the object, you can go further, without touching the surface of the object, correct?

Hello,

First off, I'm not a physics student by any means, but it does intrigue me. I've been wondering about the Schwarzschild radius, and how the density of an object dictates the escape speed. Why is it, that the more you compress an object, the more gravity it "owns?" If you were able to theoretically compress the earth down to 9.00mm, you'd have reached it's Schwarzschild radius, and essentially would have created a black hole... but why? I understand why that happens when stars collapse upon themselves due to mass, but I don't understand how taking a fixed amount of mass and compressing it, increases the amount of gravity to that object.

A "simple" explanation would be much appreciated. Again, I am not a physicist, and I only understand the basics.

Thanks

-Sean
A very crude example. I think this is how it works. The weight of the entire Earth occupying only 9mm of space would cause space to warp so much that it would cause a blackhole. Just the same as say a marble compressed to the size of an atom would cause a blackhole (example)

Also to note that the gravity remains proportional, as in the gravity at the centre of the Earth relative to the surface would stay proportional even if the Earths radius was 9mm I'm pretty sure that anything with mass can cause a blackhole should you make it dense enough. Earth with a radius of 9mm would warp space the same as the Sun if it were the size of Earth. Both would create a blackhole (for example). The key to creating blackholes is if you take any object and increase is density, it will eventually become too dense and warp space so much that nothing can get out of the "dip"

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stevendaryl
Staff Emeritus
Hello,

First off, I'm not a physics student by any means, but it does intrigue me. I've been wondering about the Schwarzschild radius, and how the density of an object dictates the escape speed. Why is it, that the more you compress an object, the more gravity it "owns?" If you were able to theoretically compress the earth down to 9.00mm, you'd have reached it's Schwarzschild radius, and essentially would have created a black hole... but why? I understand why that happens when stars collapse upon themselves due to mass, but I don't understand how taking a fixed amount of mass and compressing it, increases the amount of gravity to that object.

A "simple" explanation would be much appreciated. Again, I am not a physicist, and I only understand the basics.

Thanks

-Sean
Actually, before Einstein developed his theory of General Relativity there was an argument that something weird would happen near a very massive star.

If you shoot a rocket straight up in the air, it will return to the Earth unless it's going fast enough to escape the Earth's gravity. To escape, the rocket must have a velocity greater than or equal to what's called the "escape velocity" v given by:

$v_{escape} = \sqrt{(2GM/r)}$

where M is the mass of the Earth, G is Newton's gravitational constant, and r is the radius of the Earth. The bigger M is, the bigger the escape velocity must be. But also, the smaller r is, the bigger the escape velocity must be. So based on this formula, there would seem to be a special radius $R$ such that even light can't escape:

If $R < 2GM/c^2$, where $c$ is the speed of light, then $v_{escape}> c$ and so not even light could escape.

This analysis (and I don't remember who first made it, but it was a long time ago) is based on an old theory of gravity, Newton's theory of gravity. Newton's theory is known to be wrong, only an approximation to the current best guess as to how gravity works, which is Einstein's General Relativity.

However, even though the argument uses Newton's theory, by one of those strange quirks, it turns out that if you use Einstein's theory instead, you get exactly the same answer: If the mass of a star or planet is squeezed down to a radius $R$ where $R < 2GM/c^2$, then it becomes a "black hole" where nothing, not even light, can escape.

This analysis (and I don't remember who first made it, but it was a long time ago) is based on an old theory of gravity, Newton's theory of gravity. Newton's theory is known to be wrong, only an approximation to the current best guess as to how gravity works, which is Einstein's General Relativity.
Michell-Laplace dark star. Note the crucial difference though: for dark star light can leave its surface much like a ball can leave earth's surface -- but it will fall back down due to the strong gravity. In GR, light cannot ever leave the horizon.

Michell-Laplace dark star. Note the crucial difference though: for dark star light can leave its surface much like a ball can leave earth's surface -- but it will fall back down due to the strong gravity.
This is an important difference. Light traveling radially does not have 'escape velocity' to reach infinity (i.e. never to return) given the dark star's mass, but can be 'observed' by someone outside the event horizon, and even provided additional momentum to attain escape velocity.
In GR, light cannot ever leave the horizon.
Interesting to note that gravity itself (a form of energy with a finite velocity of c) is able to escape a black hole (leave the event horizon) and be observed/measured from outside.

A.T.
Interesting to note that gravity itself (a form of energy with a finite velocity of c) is able to escape a black hole (leave the event horizon) and be observed/measured from outside.
Gravitational waves / disturbances have a finite velocity of c, and they cannot leave the event horizon.

Gravity itself having a velocity is a tricky concept, because gravity itself is not propagating, just its disturbances are. For example, an inertially moving mass pulls you towards its current position, not towards a retarded position due to finite signal speed.

Nugatory
Mentor
Interesting to note that gravity itself (a form of energy with a finite velocity of c) is able to escape a black hole (leave the event horizon) and be observed/measured from outside.
It doesn't have to escape because it wasn't inside to begin with. The body that collapsed to form the black hole had a gravitational field above its surface even before that surface fell through the Scwarzchild radius and the horizon appeared.

jambaugh
Gold Member
You can only go so far with "stretching space" analogies because space-time is not Euclidean but has a "hyperbolic" signature. Picturing fingers poking rubber sheets is just the wrong analog.

But think of it this way... even when an object is sitting still, it is moving through time so when we combine space and time considering 4-dim space-time it turns out that all objects are moving at the same "rate" of 1 second per second and their velocity is just the direction of their motion which ends up being what they see as their time direction.

You see an object starting to fall towards a mass and what is happening is that its time axis is getting turned toward the mass. You and I being farther away and unaffected by gravity see the objects time motion turn into spatial motion... it was sitting there ticking away and now it is falling toward the planet.

All futures for the object are a bit closer to the mass than if that mass wasn't there. That is true even if those futures involve motion away from the mass as for an object moving away from the mass. By this I mean if your moving away from the mass your future will be farther away but not so far as if the mass hadn't been there.

So at the Schwarzschild radius of a Black Hole what has happened is that the gravity has turned time so severely that at that point all futures lead explicitly closer to the mass. To get back out one would have to build the equivalent of a time machine/ FTL spacecraft (each equates to the other in relativity).

If you understand what a light cone is, the future light cones of all events near a mass get turned toward the mass until at the Sch.Rad. all future light-cones point inward.

The analog of a stretched sheet is not sufficient to understand gravitational forces one should rather imagine a twisting effect due to mass on the space vs time directions of space-time. This is harder (gross understatement) to visualize as it involves non-Euclidean geometry. You have to parse it logically and ultimately you have to "do the math" to get a deep understanding.

A.T.
You can only go so far with "stretching space" analogies because space-time is not Euclidean but has a "hyperbolic" signature. Picturing fingers poking rubber sheets is just the wrong analog.
Problem of the rubber sheet is lack of time dimension. As for the Euclidean/pseudo-Euclidean part, you can use space-propertime diagrams where coordinate time is the Euclidean path distance. But still you have to include the propertime as one coordinate of the sheet, to show gravity:

You see an object starting to fall towards a mass and what is happening is that its time axis is getting turned toward the mass. You and I being farther away and unaffected by gravity see the objects time motion turn into spatial motion... it was sitting there ticking away and now it is falling toward the planet.
You can also interpret it as distances between propertime coordiantes getting larger towards the mass (gravitational time dilation):

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Gravitational waves / disturbances have a finite velocity of c, and they cannot leave the event horizon.

Gravity itself having a velocity is a tricky concept, because gravity itself is not propagating, just its disturbances are. For example, an inertially moving mass pulls you towards its current position, not towards a retarded position due to finite signal speed.
I find this distinction hard to understand the way you word it. Since the parallelism with light is being considered (dark stars vs. black holes), if you were to say "disturbances of the EM field have a finite velocity but light itself is not propagating, just the disturbances are, it would sound odd, wouldn't it?
Also the "current versus retarded position" example obviates the difficulties that these terms have to be defined for spacelike separated points in the presence of the relativity of simultaneity, and if a gravitational field pulls you it is hard to maintain that this field is not propagating.
It doesn't have to escape because it wasn't inside to begin with. The body that collapsed to form the black hole had a gravitational field above its surface even before that surface fell through the Scwarzchild radius and the horizon appeared.
Again, I don't know how this is relevant to the analogy with light, how is this different from the EM fields and the light of the body that collapsed?

stevendaryl
Staff Emeritus
I find this distinction hard to understand the way you word it. Since the parallelism with light is being considered (dark stars vs. black holes), if you were to say "disturbances of the EM field have a finite velocity but light itself is not propagating, just the disturbances are, it would sound odd, wouldn't it?
The closer analogy with electromagnetism would be: The electric field is not propagating, only disturbances in it. Electromagnetic waves are, as the name indicates, waves in the electromagnetic field. But there are such things as static electric fields. For example, the electric field of a stationary charged particle. The electric field in that case isn't "propagating" from the particle to wherever it is measured. The electric field is everywhere in space, and it's static.

In the same way that a black hole can have a gravitational field outside the event horizon, even though it is impossible for gravity waves to propagate from inside the black hole to outside, a charged black hole can have an electric field, even though it is impossible for electromagnetic waves to propagate from inside to outside.

Fields are properties of space (or spacetime). It is only disturbances to fields that propagate, not the fields themselves.

A.T.
if you were to say "disturbances of the EM field have a finite velocity but light itself is not propagating, just the disturbances are
Light is a disturbance of the EM field, so light is propagating at a finite speed. But is a static E-field propagating?

Again, I don't know how this is relevant to the analogy with light,
You mean the retarded vs current position thing? If you have a charged object moving inertially, and you carry the opposite charge, then the Coulomb force will pull you towards its current position, while you still see it at the retarded position.

how is this different from the EM fields and the light of the body that collapsed?
If the body was charged, the E-field outside the horizon remains, just like the G-field. That doesn't require anything propagating from below the horizon. Neither EM-waves nor G-waves can leave the horizon.

.... but is a static E-field propagating?
Part of my questions lead to this question, yes. Is it safe to assume E-fields or gravitational fields don't propagate? Is that assumption just valid as a useful approximation, given the fact we know the universe is not static?
You mean the retarded vs current position thing?.......
No, that was meant for Nugatory's quoted comment.

stevendaryl
Staff Emeritus
Part of my questions lead to this question, yes. Is it safe to assume E-fields or gravitational fields don't propagate? Is that assumption just valid as a useful approximation, given the fact we know the universe is not static?
I don't think that changes anything: A charged black hole has an electric field. The field far from the black hole does not involve anything propagating from the black hole to the distant observer. The same is true for gravity.

A.T.
Is it safe to assume E-fields or gravitational fields don't propagate?
I guess it depends on what "propagate" means. My point in the previous posts was that you have to distinguish between the field and disturbances of the field, when it comes to propagation.

The closer analogy with electromagnetism would be: The electric field is not propagating, only disturbances in it. Electromagnetic waves are, as the name indicates, waves in the electromagnetic field. But there are such things as static electric fields. For example, the electric field of a stationary charged particle. The electric field in that case isn't "propagating" from the particle to wherever it is measured. The electric field is everywhere in space, and it's static.

In the same way that a black hole can have a gravitational field outside the event horizon, even though it is impossible for gravity waves to propagate from inside the black hole to outside, a charged black hole can have an electric field, even though it is impossible for electromagnetic waves to propagate from inside to outside.
All this is fine for sure, but see my questions in the post above.

Fields are properties of space (or spacetime). It is only disturbances to fields that propagate, not the fields themselves.
This is clear and understood as a general definition for undergrad physics, but I fear that it gives a somewhat simplistic account of the physics in GR and QFT.
Once you work with quantum fields the distinction wave-field is not so clear cut, and the same thing happens in GR, originating some confusion and debate even amongst quantum gravity experts regarding fields that include time(spacetime fields), background independence, the hole argument and all that.
Anyway, the term disturbances would need a clear definition here, if it is simply a way to refer to changes of the field in time, spacetime fields carry them within themselves.

WannabeNewton
Fields definitely propagate on space-time.

stevendaryl
Staff Emeritus
The term "propagation of such and such field" is a standard one. One says, for example, that the massive Klein Gordon field $\varphi$ propagates in a background Minkowski space-time $(\mathbb{R}^{4},\eta_{ab})$ according to $\partial^a \partial_a \varphi - m^2 \varphi = 0$ i.e. $\varphi$ evolves in $(\mathbb{R}^{4},\eta_{ab})$ according to the aforementioned evolution equations.