Schwarzschild Radius Explanation?

  • #1
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
 

Answers and Replies

  • #2
phinds
Science Advisor
Insights Author
Gold Member
2021 Award
17,712
9,640
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?
 
  • #3
mathman
Science Advisor
8,054
538
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.
 
  • #4
1,234
18
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.
 
  • #5
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
 
  • #6
Nugatory
Mentor
13,905
7,348
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.
 
  • #7
Oh I misread that.

"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?
 
  • #8
157
5
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

Rwhp9Ny.png


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"
 
Last edited:
  • #9
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
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:

[itex]v_{escape} = \sqrt{(2GM/r)}[/itex]

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 [itex]R[/itex] such that even light can't escape:

If [itex]R < 2GM/c^2[/itex], where [itex]c[/itex] is the speed of light, then [itex]v_{escape}> c[/itex] 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 [itex]R[/itex] where [itex]R < 2GM/c^2[/itex], then it becomes a "black hole" where nothing, not even light, can escape.
 
  • #10
545
3
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.
 
  • #11
38
1
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.
 
  • #12
A.T.
Science Advisor
11,529
2,898
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.
 
  • #13
Nugatory
Mentor
13,905
7,348
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.
 
  • #14
jambaugh
Science Advisor
Insights Author
Gold Member
2,328
309
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.
 
  • #15
A.T.
Science Advisor
11,529
2,898
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:
http://www.adamtoons.de/physics/gravitation.swf


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):

https://www.youtube.com/watch?v=DdC0QN6f3G4
 
Last edited:
  • #16
3,507
27
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?
 
  • #17
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
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.
 
  • #18
A.T.
Science Advisor
11,529
2,898
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.
 
  • #19
3,507
27
.... 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.
 
  • #20
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
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.
 
  • #21
A.T.
Science Advisor
11,529
2,898
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.
 
  • #22
3,507
27
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.
 
  • #23
WannabeNewton
Science Advisor
5,815
544
Fields definitely propagate on space-time.
 
  • #24
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
Fields definitely propagate on space-time.

I think maybe we're arguing about semantics. I would say that CHANGES to fields propagate. I don't know what it would mean for the fields themselves to propagate.

A toy model of a field might be a string stretched between two trees. It doesn't really make sense to say that the string propagates from one tree to the other. But if you tap the string near one tree, the tap will cause a vibration that will propagate to the other tree.
 
  • #25
WannabeNewton
Science Advisor
5,815
544
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.
 
  • #26
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
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.

"Evolution" and "propagation" are both about change. Static fields don't propagate.
 
  • #27
WannabeNewton
Science Advisor
5,815
544
Clearly but you were saying this for any field whatsoever as per the quote in post #22 and the statement in that quote is incorrect.
 
  • #28
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
Clearly but you were saying this for any field whatsoever as per the quote in post #22.

You're arguing about semantics, without really paying attention to the context. The original question was the old conundrum: If nothing can propagate from inside a black hole to outside, then how does gravity itself escape? (Same question for the electric field, in the case of a charged black hole.)

The notion of "propagate" in this conundrum is not "obeys a differential equation", but spreading out from one point to other points. The gravitational field DOESN'T propagate from inside the black hole to outside the black hole, in this sense of the word "propagate". Neither does the electric field.

The notion of "propagate" that is appropriate for talking about whether an effect inside a black hole can propagate outside is in terms of disturbances. If I make a tiny change to the distribution of mass inside a black hole (or distribution of charge), is that change visible from outside? The answer is "no".
 
  • #29
38,094
15,885
The gravitational field DOESN'T propagate from inside the black hole to outside the black hole, in this sense of the word "propagate". Neither does the electric field.

No, but there is a sense in which those fields "propagate" from the collapsing object that originally formed the hole. The law that governs the propagation is the vacuum Einstein Field Equation (coupled with the source-free Maxwell Equations if charge is present). For the idealized case of spherically symmetric collapse, the field in the entire vacuum region to the future of the collapsing object is completely determined by these laws of propagation from the source.
 
  • #30
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
No, but there is a sense in which those fields "propagate" from the collapsing object that originally formed the hole. The law that governs the propagation is the vacuum Einstein Field Equation (coupled with the source-free Maxwell Equations if charge is present). For the idealized case of spherically symmetric collapse, the field in the entire vacuum region to the future of the collapsing object is completely determined by these laws of propagation from the source.

Sure. But I think that the false intuition that is behind the question about how gravity (or the electric field) gets out of a black hole is something like this: We feel gravity because something (gravitons, or whatever) is emitted by the gravitational source and travels across space to hit us. We measure an electric field because something (photons) are emitted by charged particles and travel across space to hit us. So if we could only block those carrier particles, we could eliminate the gravitational effects of the mass, or we could eliminate the electric field due to a charge particle. But that's an incorrect intuition. Nothing has to travel from the source to us in order for us to detect gravity or an electric field.
 
  • #31
38,094
15,885
I think that the false intuition that is behind the question about how gravity (or the electric field) gets out of a black hole is something like this: We feel gravity because something (gravitons, or whatever) is emitted by the gravitational source and travels across space to hit us. We measure an electric field because something (photons) are emitted by charged particles and travel across space to hit us.

Well, on one view (the quantum field theory view of forces being mediated by virtual particles), this intuition is correct as far as it goes. We measure an electric field because of an exchange of virtual photons between the source of the field and our detector. We measure a gravitational field because of the exchange of virtual gravitons between the gravitating mass and our detector.

Where the intuition goes wrong, on this view, is thinking that the virtual particle exchange is limited to the speed of light. It isn't, which is why, on this view, it's perfectly possible for virtual gravitons to get out of a black hole and make us feel gravity from it.

The virtual particle picture certainly has limitations, one of which is that it's based on perturbation theory, and in the case of gravity, perturbation theory implies that the metric of spacetime can't be too different from a flat metric. I'm not sure how well that applies to spacetime near a black hole's horizon (much less inside it). But I think pop science discussions of the virtual particle picture are often the source (sometimes vague and indirect) of people's intuition that particles have to get out of the black hole to make us feel gravity from it, so I think it's important to recognize that there is a sense in which that picture is valid, if limited.

Nothing has to travel from the source to us in order for us to detect gravity or an electric field.

I don't agree, because this statement, as it stands, is even stronger than the one I quoted above; it rules out, not just the virtual particle view, but also the view I have advocated in a number of threads on PF, that the source of the black hole's gravity is the object that originally collapsed to form the hole. The metric in the vacuum region to the future of that collapsing object *is* caused by the object, and one way to describe how that causation occurs, as I said in my previous post, is that the spacetime metric propagates from the source to where you feel the gravity, governed by the vacuum EFE as the law of propagation.
 
  • #32
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,942
2,931
Well, on one view (the quantum field theory view of forces being mediated by virtual particles), this intuition is correct as far as it goes. We measure an electric field because of an exchange of virtual photons between the source of the field and our detector. We measure a gravitational field because of the exchange of virtual gravitons between the gravitating mass and our detector.

I think that's taking terms in Feynmann diagrams a little too literally. Virtual photons are calculational aids, they are terms in a Taylor series expansion. I don't think that they necessarily correspond to any real particles. Anyway, but that view, that forces are all due to the exchange of virtual photons or gravitons, is what is behind the intuition that it should be impossible for gravity to get out of a black hole.

Yeah, I know, you can say that virtual particles can escape in a way that real particles, but I think that's a little hoky. Virtual particles were developed for perturbation theory, which tries to understand interacting physics as small perturbations on free field theory. It's been a while since I've studied stuff like that, but I think that there are nonperturbative effects that can't be understood in terms of summing Feynmann diagrams. Bound states of particles I think is an example, and I would believe that black holes would be another example.

Where the intuition goes wrong, on this view, is thinking that the virtual particle exchange is limited to the speed of light. It isn't, which is why, on this view, it's perfectly possible for virtual gravitons to get out of a black hole and make us feel gravity from it.

I suppose that that's a way of viewing things, but I don't like it. As I said, I think that it's viewing terms of a Taylor series a little too literally.

The other reason I don't like that resolution in the case of black holes is that I feel like it misses out on the topological reasons that a charged star collapsing into a black hole can't possibly cause the electric field or gravitational field to disappear. Whatever is going on inside a collapsing star, we know that before collapse, the electric field has a certain value for the surface integral over a surface surrounding the star. According to Maxwell's equations, you can't change that value unless charge crosses the surface. So what goes on inside the surface is just irrelevant.

Now, that is a classical picture, which is certainly changed by quantum field theory, but the macroscopic predictions of quantum field theory have to agree with classical field theory.

The virtual particle picture certainly has limitations, one of which is that it's based on perturbation theory, and in the case of gravity, perturbation theory implies that the metric of spacetime can't be too different from a flat metric. I'm not sure how well that applies to spacetime near a black hole's horizon (much less inside it). But I think pop science discussions of the virtual particle picture are often the source (sometimes vague and indirect) of people's intuition that particles have to get out of the black hole to make us feel gravity from it, so I think it's important to recognize that there is a sense in which that picture is valid, if limited.


I don't agree, because this statement, as it stands, is even stronger than the one I quoted above; it rules out, not just the virtual particle view, but also the view I have advocated in a number of threads on PF, that the source of the black hole's gravity is the object that originally collapsed to form the hole. The metric in the vacuum region to the future of that collapsing object *is* caused by the object, and one way to describe how that causation occurs, as I said in my previous post, is that the spacetime metric propagates from the source to where you feel the gravity, governed by the vacuum EFE as the law of propagation.

Hmm. An eternal black hole that has always existed (rather than collapsing from a star) is certainly a solution to the Einstein Field Equations. So insisting that gravity has to have a "source" goes beyond GR. But it's probably more realistic. So maybe that point of view has something to be said for it.
 
  • #33
38,094
15,885
I think that's taking terms in Feynmann diagrams a little too literally.

I did say this viewpoint had limitations.

you can say that virtual particles can escape in a way that real particles, but I think that's a little hoky.

"Limited" might be a better word.

I think that there are nonperturbative effects that can't be understood in terms of summing Feynmann diagrams. Bound states of particles I think is an example, and I would believe that black holes would be another example.

Yes, those are among the limitations I was referring to.

The other reason I don't like that resolution in the case of black holes is that I feel like it misses out on the topological reasons that a charged star collapsing into a black hole can't possibly cause the electric field or gravitational field to disappear. Whatever is going on inside a collapsing star, we know that before collapse, the electric field has a certain value for the surface integral over a surface surrounding the star. According to Maxwell's equations, you can't change that value unless charge crosses the surface. So what goes on inside the surface is just irrelevant.

This is a good point, but note that it assumes that the spacetime is simply connected. A wormhole, for example, blurs the distinction between what is "inside" the surface and what is "outside"; charge could migrate through the wormhole from the region "inside" the surface to some point way "outside" it, without passing through the surface. (Also, on "charge without charge" views such as, IIRC, John Wheeler proposed, you could have a nonzero surface integral without having any charge at all; the field lines would pass through the wormhole and wrap back around without ever having to end on a source.) But that's certainly not a vindication of the virtual particle view either. :wink:

An eternal black hole that has always existed (rather than collapsing from a star) is certainly a solution to the Einstein Field Equations. So insisting that gravity has to have a "source" goes beyond GR.

I would say it goes beyond a "naive" application of the EFE, but is still within the purview of "GR"; GR is not just the EFE, it also includes discussion of which solutions of the EFE apply to which physical scenarios, and which solutions are not physically reasonable at all.
 
  • #34
38
1
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 understand that is so from GR theory. Also I think one explanation from Feynmann is that this is a matter of virtual particles whose speeds are not restricted to c (hope I got this correct!).

However, that still does mean that "some information" is passing from the event horizon to the outside - otherwise, without gravitational waves or anything leaving the event horizon, how does the 'disturbance' (presumably outside the horizon) know that it has to propagate? Isn't some sort of 'information' continuity from the event horizon to the outside necessary for this? How is this reconciled?
 
  • #35
A.T.
Science Advisor
11,529
2,898
However, that still does mean that "some information" is passing from the event horizon to the outside
No, no information can pass the horizon. The information that is outside was there, before the horizon formed.

how does the 'disturbance' (presumably outside the horizon) know that it has to propagate?
If the disturbance was created inside the horizon, it can never be outside of it.

The point is, there nothing special about gravity, compared to EM-fields here. In both cases:
- Waves propagate at a finite speed and cannot escape the horizon
- The static fields remain intact outside, even after the horizon forms.
 

Related Threads on Schwarzschild Radius Explanation?

  • Last Post
Replies
15
Views
726
  • Last Post
Replies
1
Views
735
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
904
T
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
7
Views
879
Replies
64
Views
6K
Replies
7
Views
7K
Replies
1
Views
919
Top