Schwarzschild Radius Explanation?

[PeterDonis]

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.

 

[stevendaryl]

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 Feynman 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 Feynman 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.

 

[PeterDonis]

I think that's taking terms in Feynman 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 Feynman 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.

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.

 

[adrian_m]

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 Feynman 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?

 

[A.T.]

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.

 

[TrickyDicky]

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".

The mass distribution is demanded to be spherically symmetric, the thing is that a change(increase) in mass inside the horizon can be theoretically detected outside the horizon, so the notion of propagation you highlight is also realized,

 

[yuiop]

Let us say we have a super-massive black hole that is being orbited by a number of satellites. Far away a star is on a course towards the SMBH but not on a direct collision course. On route it collapses to form a small black hole. After becoming a black hole, a relic of its gravity field is left outside the black hole event horizon which might be thought of as a distortion in spacetime. This distortion has an existence that is independent of the black hole and is not 'generated' by the mass of the black hole. On passing near the SMBH the mass of the black hole and the relic field are captured by the gravity of the SMBH and go into orbit around it. As a result the trajectories of the other bodies are disturbed by the either the relic field or the mass of the small black hole. In turn the trajectory of the small black hole is also disturbed by its interactions with the other bodies. Is it just coincidence that the complicated evolving trajectory of the black hole mass is identical to the trajectory of the relic gravitational field and that there is no causal connection between the two?

P.S. Here is an alternative thought experiment I just thought of. Imagine an object is in stable orbit around a black hole at an orbital radius of R. The black hole evaporates due to Hawking radiation. Eventually all the radiated energy and matter of the Hawking radiation is in a shell with a radius greater than R. It is well known that with a uniform shell the spacetime and gravitational curvature is flat. However, if the relic field of the black hole is independent of the mass of the black hole that created it, the relic field should still be there (as it has no need of the mass) and the body should continue orbiting at a radius of R from the position of the original black hole, even though there is no mass or energy within the orbital radius. Does this not cause a problem with energy balance of the universe?

 

[Bill_K]

Let us say we have a super-massive black hole that is being orbited by a number of satellites. Far away a star is on a course towards the SMBH but not on a direct collision course. On route it collapses to form a small black hole. After becoming a black hole, a relic of its gravity field is left outside the black hole event horizon which might be thought of as a distortion in spacetime. This distortion has an existence that is independent of the black hole and is not 'generated' by the mass of the black hole. On passing near the SMBH the mass of the black hole and the relic field are captured by the gravity of the SMBH and go into orbit around it. As a result the trajectories of the other bodies are disturbed by the either the relic field or the mass of the small black hole. In turn the trajectory of the small black hole is also disturbed by its interactions with the other bodies. Is it just coincidence that the complicated evolving trajectory of the black hole mass is identical to the trajectory of the relic gravitational field and that there is no causal connection between the two?

The mass of an isolated source is not independent of its external gravitational field, in fact it is defined in terms of the distant field. This is true regardless of whether the source is a star or a black hole. It is also true regardless of which of the several definitions of mass you refer to.

It is a complete mistake to think of the gravitational field of a mass as something that has to continuously "propagate out" from the source, presumably faster than light. The reason being that it must have been there already! You cannot change the mass of an isolated source without changing the distant field. The distant field defines the mass. To change it, either additional mass has to fall in from infinity (in the form of particles or waves), or some of the mass has to emerge and travel to infinity. In either case there is no superluminal propagation.

 

[stevendaryl]

The mass distribution is demanded to be spherically symmetric, the thing is that a change(increase) in mass inside the horizon can be theoretically detected outside the horizon, so the notion of propagation you highlight is also realized,

The first part of that is not true--a collapsing star does not need to be spherically symmetric in order for it to collapse into a black hole.

The second part I can't really make sense of. The mass of a black hole increases by something falling into the black hole. The external gravitational field changes long before the mass crosses the event horizon (infinitely long before, in the coordinate system of a distant observer), so it's certainly not an example of something propagating out of the event horizon.

 

[stevendaryl]

It is a complete mistake to think of mass as something that has to "propagate out" from the source, presumably faster than light, the reason being that it must have been there already! You cannot change the mass of an isolated source without changing the distant field. The distant field defines the mass. To change it, either additional mass has to fall in from infinity (in the form of particles or waves), or some of the mass has to emerge and travel to infinity. In either case there is no superluminal propagation.

That's what I've been getting at. People might be thinking in terms of: Suppose a massive object suddenly appears. How much time does it take for the gravity due to the object to propagate to a distant star?

But the premise makes no sense, in terms of GR. Mass can't suddenly appear at a spot in the universe, it can only come from other parts of the universe. (I'm not sure how the suggestion Peter made about the possibility of wormholes affects this.)

 

[Bill_K]

P.S. Here is an alternative thought experiment I just thought of. Imagine an object is in stable orbit around a black hole at an orbital radius of R. The black hole evaporates due to Hawking radiation. Eventually all the radiated energy and matter of the Hawking radiation is in a shell with a radius greater than R. It is well known that with a uniform shell the spacetime and gravitational curvature is flat. However, if the relic field of the black hole is independent of the mass of the black hole that created it, the relic field should still be there (as it has no need of the mass) and the body should continue orbiting at a radius of R from the position of the original black hole, even though there is no mass or energy within the orbital radius. Does this not cause a problem with energy balance of the universe?

GR uses different language than Newtonian gravity, and of course is more general. But despite that, many questions about GR can be answered quite simply by thinking of the Newtonian case, which GR must agree with in the weak field limit. Just because we have an interest in GR does not mean we should abandon our Newtonian intuition.

Take this example. At any time the expanding shell of Hawking radiation is at some large radius R₀. Outside R₀ the gravitational field is undisturbed and still corresponds to the same original mass M. Inside R₀ the field is zero and planets travel in straight lines.

 

[bobie]

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.
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.

Could anyone explain these points:
- can a neutron star emit light, since there are not electrons in orbits to "reflect" incoming light?
- if we do not consider escape velocity,
r_s (2Gm/c²) just says that PE around mass is roughly 10²⁰. If a photon had energy 10³⁴ hv, wouldn't it emerge from the BH with enough energy to be visible light?
- Planck unit of mass M_p is derived by L_p multiplying it by G/c², is it incorrect by a factor of 2 or is r_s wrong?

 

[jambaugh]

Could anyone explain these points:
- can a neutron star emit light, since there are not electrons in orbits to "reflect" incoming light?

Yes, a neutron star can and does emit light (but neutronium would be rather shiny and reflective). The neutron matter though neutral is polarizable (they way magnets are without us having a magnetic charge) and thus can create electromagnetic waves.

- if we do not consider escape velocity,
r_s (2Gm/c²) just says that PE around mass is roughly 10²⁰. If a photon had energy 10³⁴ hv, wouldn't it emerge from the BH with enough energy to be visible light?

Don't confuse potential energy with gravitational potential. In a given gravitational potential the potential energy of an object is its mass times the grav. potential. As to the photon business I don't follow your reasoning. You may if you like calculate the red shift for a photon climbing out of a gravity well, or the blue shift for photon falling in... but the event horizon does not itself "emit" photons. (Hawking radiation is a quantum phenomenon occurring over a spatial volume outside the horizon).

- Planck unit of mass M_p is derived by L_p multiplying it by G/c², is it incorrect by a factor of 2 or is r_s wrong?

The Planck scale is an order of magnitude, not a precise value... we may do unit conversion and so obtain a Plank mass, time, energy, momentum, or distance. But it isn't --in-and-of-itself-- a physical constant per se... as in a directly measurable or ratio of measurable quantities. The Planck scale defines the scale that when approached renders invalid assumptions of only classical effects... near the Planck scale we cannot ignore quantum phenomena. In such a statement a factor of 2 is not significant.

 

[bobie]

, not a precise value... we may do unit conversion and so obtain a Plank mass,.

I was referring to the way Planck derived his unit of mass and length , do you anything about it?

 

[bobie]

...neutronium would be rather shiny and reflective.

In ordinary matter light is reflected by electrons jumping up and down the orbits, how can matter without orbits reflect light?

we may do unit conversion and so obtain a Plank mass, time, energy, momentum, or distance. But it isn't --in-and-of-itself-- a physical constant per se... as in a directly measurable or ratio of measurable quantities.

We are not doing conversion.
It was Planck that established that L_p = G/c² * M_p

 

[stevebd1]

- can a neutron star emit light, since there are not electrons in orbits to "reflect" incoming light?

I recall reading somewhere that a lot of a neutron stars light is made up of the photons that were captured during the initial collapse and are slowly making their way to the surface and 'leaking' from the neutron star.

 

[PeterDonis]

In ordinary matter light is reflected by electrons jumping up and down the orbits

This isn't necessarily true; electrons transferring between energy levels in atoms is only one of a number of available degrees of freedom that can exchange energy with EM waves. Others include: free electrons that aren't bound to individual atoms (e.g., in metals); thermal vibrations of entire atoms/molecules (or whatever particles compose the object); collective vibrational modes involving many atoms/molecules (or whatever particles compose the object). The latter two, at least, are available in neutron stars, with neutrons playing the role of the atoms/molecules.

 

[nitsuj]

Ha neat distinction. Between fields and propagating, though how is it not all consicered propagating?

If I measure a bodys gravitational field and see its a perfect sphere, will its gravitaional field still be a perfect sphere if I measure while zooming by at near c? I thought it too would be contracted.