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WaxRubiks
2018-Jul-27, 12:04 AM
where the only the matter within radius r of a spherical ball counts towards gravitational pull, and the shell outside has no effect.

Ken G
2018-Jul-27, 03:27 AM
The problem is, the issue is not with the difference in the equations themselves, it is in the different treatments of the overall conditions, such as boundary conditions and spatial curvature and so on. For example, Newton thought that an infinite static universe with a constant mass density would have to have no gravity forces, because by symmetry the forces wouldn't know which way to point. But if you apply Newton's shell theorem, you can draw a sphere around any point and conclude there is a force toward that point, giving the strange result that the force depends on the imaginary sphere you draw.

You would need boundary conditions to distinguish "which sphere was the right one to draw," but that would be missing the point in a truly homogeneous universe.
Because there's really nothing wrong with not knowing what the "actual acceleration" is, if you simply notice the way Newton's laws also obey the equivalence principle. That means, you just say that if everything is in free-fall, and you can't see any boundaries, then nothing can determine which of the objects are actually accelerating, and which is the object at the "center" that is not accelerating. No observation can distinguish that, it was always pure assumption (unsupported assumption) by Newton that even objects in free fall had to have an absolute acceleration. Einstein noticed that even in Newton's laws, if all objects are in free fall, no experiment could ever determine anything you could call an absolute acceleration. So Einstein used that idea to build general relativity, but even Newton's laws produce that same effect, it just was never noticed.

grapes
2018-Jul-27, 04:12 AM
where the only the matter within radius r of a spherical ball counts towards gravitational pull, and the shell outside has no effect.
Besides Ken's comments, I want to point out that this is a newtonian conclusion only with spherically symmetric inner and outer shells. Without the spherical symmetry, things get a lot more complicated--approximations may hold, but you are asking about exact solutions right?

WaxRubiks
2018-Jul-27, 07:49 AM
I suppose I wanted to know whether it still roughly applied in relativistic situations, using GR.

Ken G
2018-Jul-28, 03:08 PM
When gravity is everywhere weak, both theories can produce similar experimental predictions for any given mass distribution. But because the theories are so different, the way you apply boundary conditions are different. This has always been a kind of "hidden" element of physics that is not always appreciated-- the equations themselves are never a complete theory, you always have to manually add something to the equations in the form of boundary conditions (like what is going on outside your sphere that is more than just spherical symmetry) and initial conditions (like the history of the universe in which you are finding your solution). Because those two theories take a very different stance on "what is happening" that produces the experimental predictions, they can differ in the boundary conditions they suggest, and hence differ in the predictions they make. If you handle those conditions carefully, even Newton's gravity can give you a Big Bang model that works more or less just the same as ours, shell theorem and all. Indeed, even the discovery of general relativity did not generate a Big Bang prediction, that still required Hubble's observations and would have also required that in Newton's day.

Hornblower
2018-Jul-28, 05:59 PM
So far I am seeing generalities, while I interpreted the OP question as asking whether a spherically symmetrical ball would have exactly the same gravitational signature, as a function of the distance from its center, as if its mass was concentrated at the center. In other words, if we magically compacted the Sun into a black hole, would there in principle be any change in the perihelion advance of Mercury?

George
2018-Jul-28, 06:19 PM
If you handle those conditions carefully, even Newton's gravity can give you a Big Bang model that works more or less just the same as ours, shell theorem and all.Yes, so given the acceptance of Newton's universal gravitation that all matter attracts, what justified their seemingly obdurate grip on the Static Theory? I can see angular momentum for some of it as a counter "force". I suppose it was a matter of timing since it took Hubble to confirm that there were "nebulae" like the Milky Way out there, which was about the time GR came along.


Indeed, even the discovery of general relativity did not generate a Big Bang prediction, that still required Hubble's observations and would have also required that in Newton's day. I assume you mean in acceptance since Lemaitre did introduce the theory strictly based upon GR. Ironically, Hubble avoided making any expansion claim, possibly due to deSitter's influence, IMO.

DaveC426913
2018-Jul-28, 10:33 PM
if we magically compacted the Sun into a black hole, would there in principle be any change in the perihelion advance of Mercury?
No, I'm pretty sure he's asking about Newton's shell theorem.

His OP mentions only mass from inside a given radius has an effect, and any mass outside that radius ("shell") has no effect.

Grey
2018-Jul-29, 01:38 AM
So, I think the answer is yes, in both directions. That is, changing the Sun into a black hole will not change the gravitational field far from the Sun, and also that spacetime behaves like Minkowski spacetime (i.e., no gravitational forces) inside a spherical shell. Birkhoff (https://en.m.wikipedia.org/wiki/Birkhoff%27s_theorem_(relativity)) demonstrated this in 1923.

Hornblower
2018-Jul-29, 04:17 AM
No, I'm pretty sure he's asking about Newton's shell theorem.

His OP mentions only mass from inside a given radius has an effect, and any mass outside that radius ("shell") has no effect.

I should have included a surrounding shell when asking my question. Yes, I understood his question as being specifically about the shell theorem. In Newton's theory it is exact. The generalities in some responses do not answer the question about whether or not it is exact in GR. Grey said yes and referred to Birkhoff, but the linked articles refer to the Newtonian limit, that is, very weak gravity. I am still uncertain as to whether it holds in strong gravity situations.

Ken G
2018-Jul-29, 01:52 PM
Yes, so given the acceptance of Newton's universal gravitation that all matter attracts, what justified their seemingly obdurate grip on the Static Theory? I can see angular momentum for some of it as a counter "force". I suppose it was a matter of timing since it took Hubble to confirm that there were "nebulae" like the Milky Way out there, which was about the time GR came along.Mathematicians applying Newton's law understood that angular momentum could be a counterforce-- for example, Laplace had a fairly complete model of solar system disk formation before 1800. But it requires extreme contraction, and allowing the universe to be dynamical enough to contract that much would defeat the purpose of trying to keep it static.


I assume you mean in acceptance since Lemaitre did introduce the theory strictly based upon GR. Ironically, Hubble avoided making any expansion claim, possibly due to deSitter's influence, IMO.
Yes, acceptance is what requires observations. Indeed, Edgar Allan Poe suggested a description that sounds remarkably like the Big Bang 70 years before relativity-- and was panned for it.

Ken G
2018-Jul-29, 01:59 PM
I am still uncertain as to whether it holds in strong gravity situations.What I was trying to point out in the homogeneous universe example was the importance of more than just the equations that one might use to prove the shell theorem, but also the boundary conditions. Newton understood the importance of boundary conditions on his own equations, but he probably did not have the mathematical sophistication to do it right. He thought that his shell theorem would be violated by a homogeneous infinite mass distribution, because he thought that there had to be absolute acceleration to have gravitational effects, and he could not see what direction such an absolute acceleration could point if the distribution was infinite and homogenous. So he thought some sort of boundary condition at infinity could violate the shell theorem (because a homogeneous distribution is spherically symmetric around any point so should obey that theorem around any point, leading to paradox) and allow for a static universe.

In GR, the nature of the Einstein field equations make it even more clear that boundary conditions will be needed in both space and time (the latter because of the finite speed of propagation of gravity, so you need to assert that the mass distribution has always been static, not just that it is static now). But a finite shell will not be a static structure, so the question is in some sense moot when applied to GR, unless you have an infinite homogeneous distribution of mass that has always been there, and even then you need a cosmological constant to keep it static. Of course, it would not be stable, so there really isn't any situation where GR formally obeys the shell theorem in an exact sense.

George
2018-Jul-29, 04:52 PM
Allowing rotation (hence frame dragging) would add a monkey wrench as well, I’d bet.

Grey
2018-Jul-30, 12:22 PM
I should have included a surrounding shell when asking my question. Yes, I understood his question as being specifically about the shell theorem. In Newton's theory it is exact. The generalities in some responses do not answer the question about whether or not it is exact in GR. Grey said yes and referred to Birkhoff, but the linked articles refer to the Newtonian limit, that is, very weak gravity. I am still uncertain as to whether it holds in strong gravity situations.The article just mentions that as an intuitive reason for thinking this is probably true, and that in several places the result agrees with the Newtonian one. But as I understand it, Birkhoff's theorem holds under general relativity regardless of the field strength. Birkhoff's result follows directly from the equations of general relativity.

Grey
2018-Jul-30, 12:30 PM
But a finite shell will not be a static structure, so the question is in some sense moot when applied to GR, unless you have an infinite homogeneous distribution of mass that has always been there, and even then you need a cosmological constant to keep it static. Of course, it would not be stable, so there really isn't any situation where GR formally obeys the shell theorem in an exact sense.Well, now, of course it's true that you won't find an idealized case where this will work exactly in the real universe, because you won't find a perfectly symmetrical spherical shell of matter anywhere, and if you did, it wouldn't stay there. But that's true for the Newtonian version, too. I think it's still a worthwhile question to ask whether general relativity works the same as Newtonian gravity in this one sense, and the answer turns out to be that it does.

Ken G
2018-Jul-31, 07:17 AM
It's more than the problem of perfect symmetry, that's just a matter of approximation. The problem is that with Newton, you can just say the mass is symmetrical now, and be done. But for GR, you need to specify the entire history of the mass distribution, or else you don't have a sufficient boundary condition on the spacetime to use with the field equations. Newton allows you to assume the spacetime, GR doesn't-- it's quite a fundamental difference , so even saying the space inside a spherical mass shell is Minkowski requires an assumption that is outside of GR itself. Newton needs no such outside assumptions, they are built into the theory.

Hornblower
2018-Jul-31, 11:45 AM
Let's cut to the chase. My question is, is the gravitational acceleration of a small object inside a spherically uniform shell exactly zero with respect to the shell in a GR model? If I accept the extended discussions here I would say, "Not necessarily. It depends on details that are a non-issue in the Newtonian model." Is that a valid response?

Grey
2018-Jul-31, 03:23 PM
Let's cut to the chase. My question is, is the gravitational acceleration of a small object inside a spherically uniform shell exactly zero with respect to the shell in a GR model? If I accept the extended discussions here I would say, "Not necessarily. It depends on details that are a non-issue in the Newtonian model." Is that a valid response?No, it's really not. I think Ken is confusing the issue unnecessarily. Yes, it's true that it's an idealized situation, but it's not something that's an assumption "outside of GR itself". It's just setting the mass distribution for the problem you want to analyze. We deal with semi-fictitious situations in physics all the time, including situations where we talk about something "static" when we know there's really no such thing. Saying that you need to know the whole history of the matter in the shell or you can't apply general relativity accurately would be just like suggesting that you can't use the Schwarzschild metric for a spherically symmetric object because you'd have to take into account what the matter making up the sphere was doing beforehand. The same issues would arise, and yet we can work with that as an ideal case without any problems. Of course in any real-universe situation, you won't have an idea case, but the discrepancies aren't somehow more fundamental in general relativity.

If you want to do a little more reading on Birkhoff's theorem, here (https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/)'s what looks like a pretty reasonable derivation (or you can go straight to Misner, Thorne, and Wheeler (https://archive.org/stream/GravitationMisnerThorneWheeler/Gravitation%20Misner%20Thorne%20Wheeler#page/n867), page 843. And here (https://arxiv.org/abs/0910.5194)'s a nice paper that talks about what some of the implications are. It really is accepted in the physics community as demonstrating that the shell theorem still holds in general relativity, and also that you can understand stellar collapse (as an idealized situation, again, of course) by looking at the Schwarzschild metric.

Ken G
2018-Aug-03, 01:13 AM
I actually don't think that's correct in GR, but it might just be because GR apparently requires dark energy in order to work. I think it might be more than just that, but I'm not sure. It seems to me that even without dark energy, if all you know is that you have a spherical mass distribution, you cannot use GR to say there is no gravitational effect inside the sphere. If that were true, then GR solutions of the Big Bang with a homogeneous mass distribution everywhere except inside an empty sphere would require the spacetime in the void be Minkowskian, and of course that's not true of a Big Bang solution. So we need to say more than just that the mass distribution is spherically symmetric, we need additional assumptions about it (such as that it is not only static, but that it lives in a static universe also, so has always been static).

It seems to me the situation is analogous to asking, what does GR say would happen if the Sun instantaneously disappeared? It's not just that this is technologically impossible, or requires some kind of idealized approximation, it is that one is asking to use GR to describe a situation that GR says doesn't happen. So we have to track our boundary conditions and make sure we are not introducing additional constraints beyond spherical symmetry as we derive Birkhoff's theorem.

Perhaps more technical insight can be found here: https://arxiv.org/abs/0910.5194, where it can certainly be seen that Birkhoff's theorem does not imply that the space inside a spherically symmetric mass distribution always has to be Minkowskian. In part this may be because GR appears to require dark energy; I don't know if that's the entirety of the reason that Birkhoff's theorem can lead to wrong conclusions about what GR asserts. A key phrase I would pick out from the abstract is "different topology than that of the maximal analytic extension of Schwarzschild", which to me quite clearly asserts that boundary conditions in the form of assumed topologies are of great importance to the conclusions.

Now one might counter that in this thread, we aren't interested in Big Bang solutions, we want some kind of idealization that applies to a star. But in that case we already know the answer, because we already know that GR gives all the same answers as Newton in that context, to a high degree of approximation. So to bring in GR, we must be talking about the technical differences between GR and Newton, otherwise they will always obviously agree to the contexts people are asking about. We cannot just be talking about practical idealizations where we already know the two agree, we must be talking about the structure of GR. We can say that the shell theorem continues to apply for black holes in GR, so it extends a bit past Newton, but we must also recognize it does not extend to Big Bang solutions, and they are GR also.

Grey
2018-Aug-03, 01:58 PM
I actually don't think that's correct in GR, but it might just be because GR apparently requires dark energy in order to work. I think it might be more than just that, but I'm not sure. It seems to me that even without dark energy, if all you know is that you have a spherical mass distribution, you cannot use GR to say there is no gravitational effect inside the sphere. If that were true, then GR solutions of the Big Bang with a homogeneous mass distribution everywhere except inside an empty sphere would require the spacetime in the void be Minkowskian, and of course that's not true of a Big Bang solution. So we need to say more than just that the mass distribution is spherically symmetric, we need additional assumptions about it (such as that it is not only static, but that it lives in a static universe also, so has always been static).I don't think that's relevant, though. There wouldn't be any point in the universe during the Big Bang where there would have been an empty sphere surrounded by a homogeneous mass distribution*. As far as we can tell, the universe would have been more or less uniformly filled with stuff** during the early stages of the Big Bang. If there would have been an empty sphere surrounded by a symmetric shell, general relativity appears to say that there would have been no net gravitational effects from that shell inside that void. I agree that it appears that dark energy or a very unusual topology can mean that there are gravitational effects inside such a void, but even in those cases, those gravitational effects are a result of the cosmological constant or topology, not really an effect of the matter surrounding the sphere. That is, if you have globally curved spacetime, that global curvature is still going to be present in an empty sphere surrounded by matter; the sphere won't somehow "shield" the interior from that global curvature and turn it into Minkowski spacetime. (By the way the paper you quoted is just the one that I had suggested as useful reading for understanding some of the consequences of Birkhoff's theorem.)

Technically, the same problem isn't limited to general relativity. You'd see the same issue with electromagnetism. A typical simple electrostatic problem might be, given a described charge distribution, determine the electric field it produces. But wait, moving and accelerating charges will produce a changing magnetic field which will in turn produce a changing electric field. I can't determine the field unless I know all the details of how the charge distribution was assembled, and even once I've done so, those charges will try to keep moving. So I could decide that I can't actually work out what the electric field is in any situation. Or I could remember that this is an idealized situation, and that I can describe the overall behavior simply, and then if I wish, I can try to address any discrepancies that arise because any real situation is not actually going to be that simple.

Hornblower asks "is the gravitational acceleration of a small object inside a spherically uniform shell exactly zero with respect to the shell in a GR model?" [my emphasis] And the answer there is "yes". Even if there is some gravitational effect from global curvature of spacetime or a cosmological constant, it's not because of the shell of material, it's because of other properties of spacetime in the universe that would be present even in the absence of that spherical shell. If I somehow manage to build a Dyson sphere around a black hole, the net gravitational effect between the sphere and the black hole will be zero (although the compression will be high; I'll need something stronger than scrith!).


* Of course, there isn't really now, either, but interstellar or intergalactic space can probably be considered close enough to empty for most purposes. One more acknowledgement that of course this is an idealized situation that will never be found exactly in the real universe.
** Probably a very high energy quark-gluon plasma kind of thing.

George
2018-Aug-03, 02:25 PM
Wouldn't frame dragging be one way to note differences between the two theories? Sometimes seeing the easier differences can lead to the more subtle ones when we simplify with having no rotation (relative to the idealized surrounding spacetime, I suppose). Or am I wrong? [I, of course, don't know squat about the Kerr metric.]

Ken G
2018-Aug-04, 05:18 AM
I don't think that's relevant, though. There wouldn't be any point in the universe during the Big Bang where there would have been an empty sphere surrounded by a homogeneous mass distribution*. All that matters is that there could have been, because the question is simply, is the shell theorem a theorem of GR? If it is, then it must always be true in every possible situation ruled by GR. If we agree that a Big Bang solution with a homogeneous universe except for a hollow sphere does not obey the shell theorem, then it's just plain not a general theorem of GR.

If there would have been an empty sphere surrounded by a symmetric shell, general relativity appears to say that there would have been no net gravitational effects from that shell inside that void. That's what I am saying is not true. If you pick the origin at the center of the void, and give all the mass a Hubble-type expansion, then that's spherical symmetry. But it will not be a solution to the EFE to have an expanding Minkowski sphere at the center, because Minkowski spheres cannot expand.


I agree that it appears that dark energy or a very unusual topology can mean that there are gravitational effects inside such a void, but even in those cases, those gravitational effects are a result of the cosmological constant or topology, not really an effect of the matter surrounding the sphere. But it makes no difference why GR violates the theorem, it only matters that it does. GR allows topologies that violate the theorem, so GR violates the theorem. This is what I was saying above-- to get the theorem, you must tack on additional assumptions that are not in GR. These additional assumptions are related to the history of the universe, in the form of its topological constraints and who knows what else. All that is true even if we don't add dark energy to GR, since we don't really know if GR should include dark energy at this point.


That is, if you have globally curved spacetime, that global curvature is still going to be present in an empty sphere surrounded by matter; the sphere won't somehow "shield" the interior from that global curvature and turn it into Minkowski spacetime. Exactly my point-- ergo, GR by itself does not produce a shell theorem like that.


Technically, the same problem isn't limited to general relativity. You'd see the same issue with electromagnetism. That's right, if you're not using Newton's action-at-a-distance in an electromagnetic situation, you have the same problem with boundary conditions and initial conditions that GR does-- which is the point I was making in my first post. Newton doubted his own shell theorem in the context of the whole universe, because it leads to paradoxes that he thought could be resolved via additional assumptions in regard to the boundary conditions (though at least you can assume the topology, as Newton's theory always uses flat infinite space with global absolute time). Newton thought that if you applied the shell theorem to a homogeneous infinite universe, it would get the wrong answer-- the theorem says that if you pick an origin anywhere, you'll then get accelerations toward the point you picked, and that made no sense to him because he thought the accelerations had to be absolute and independent of coordinates. So he must have thought the boundary condition at infinity violated the shell theorem and allowed the force to be everywhere zero (as required by the symmetry if the acceleration was to be absolute). Ironically, if you simply allow the acceleration to be coordinate dependent, as GR does, then even Newton's theory works pretty well for the Big Bang! You simply allow the acceleration to point toward and point you pick as your center, just as the shell theorem says, and you get the right answers for the dynamical behavior. Had Newton realized that, he could have anticipated Big Bang solutions himself! His shell theorem was better than he realized. But it's worse in GR, because GR allows for different topologies than Newton does.


A typical simple electrostatic problem might be, given a described charge distribution, determine the electric field it produces. But wait, moving and accelerating charges will produce a changing magnetic field which will in turn produce a changing electric field. I can't determine the field unless I know all the details of how the charge distribution was assembled, and even once I've done so, those charges will try to keep moving. So I could decide that I can't actually work out what the electric field is in any situation. Or I could remember that this is an idealized situation, and that I can describe the overall behavior simply, and then if I wish, I can try to address any discrepancies that arise because any real situation is not actually going to be that simple.The issue is not about using idealizations, it is about using the right idealization. Einstein used the wrong idealization when he sought a GR model that gave a static universe-- it was not a matter of some small adjustment, he missed the boat completely by using the same wrong idealization that Newton had used. The point is, GR is not restricted to any particular idealization, so it's important to recognize the full range of possibilities whenever considering its theorems. Only nature can tell us the right idealization to use, a mistake we've made many times in the history of physics and astronomy!


Hornblower asks "is the gravitational acceleration of a small object inside a spherically uniform shell exactly zero with respect to the shell in a GR model?" [my emphasis] And the answer there is "yes". Even if there is some gravitational effect from global curvature of spacetime or a cosmological constant, it's not because of the shell of material, it's because of other properties of spacetime in the universe that would be present even in the absence of that spherical shell.It is always impossible in physics to say "because"-- you only have the observations, either you have correct predictions or you don't-- the "because" is just a way to picture it to help get the prediction right, it's never part of what you predict and it's rarely uniquely true. The problem with Hornblower's question is that if he says the mass distribution is currently spherically symmetric, that's all you'd need to know for Newton's shell theorem, but it's not enough for GR. So if you instead tack on that the mass distribution has always been static as well as spherically symmetric, you run into problems with asserting a situation that GR already says is impossible, similar to the problems you get if you conclude that nothing ever can fall into a black hole because it would take an infinite time in the Schwarzschild coordinates reckoned from a large distance from the black hole. What's more, apply Hornblower's question to a Big Bang expansion with a spherical void, and even without dark energy GR will give a complicated answer to the motion of dust particles within that void-- violating the shell theorem.


If I somehow manage to build a Dyson sphere around a black hole, the net gravitational effect between the sphere and the black hole will be zero (although the compression will be high; I'll need something stronger than scrith!).Only if you choose the usual conditions in which the Schwarzschild metric will apply, but those assumptions are more than just GR with spherical symmetry. They implicitly assume something about the entire history of the universe-- a particular choice of boundary condition that is only approximately correct and actually is inconsistent with the Big Bang. Now don't get me wrong, choosing inconsistent idealizations is nothing new in physics and is not a wrong thing to do, it's used all the time. But it is wrong to forget you have done it, and think you have a more general theorem than you really do. This was essentially Newton's mistake in thinking his gravity could keep a homogeneous infinite universe static-- he thought there had to be a boundary condition that would allow the homogeneity to apply to the gravitational acceleration as well, but there isn't actually any such boundary condition, he forced an idealization onto a universe that didn't want it. Ironically, he would have been better off sticking to the shell theorem, because it works quite well in a homogeneous situation. But the shell theorem gets the wrong answer for the gravity in a spherical void in a homogeneous universe, and GR gets that right, expressly because the shell theorem is not a theorem of GR, even though it works for a spherical black hole if you apply asymptotically flat boundary conditions.

In short, when a theorem comes from the boundary conditions (topology, history, etc.) and not the structure of the theory by itself, then it is not a theorem of the theory, it is a theorem of the boundary conditions. That is how the shell theorem differs in GR versus Newton.

Ken G
2018-Aug-04, 05:22 AM
Wouldn't frame dragging be one way to note differences between the two theories? Sometimes seeing the easier differences can lead to the more subtle ones when we simplify with having no rotation (relative to the idealized surrounding spacetime, I suppose). Or am I wrong? [I, of course, don't know squat about the Kerr metric.]Frame dragging, and the Kerr metric, would violate the spherical symmetry. So yes, when there is not spherical symmetry, the two theories can diverge dramatically, but the question here is, how different can they be in strict spherical symmetry? My answer is, in Newton's theory, if you have a spherical void in an otherwise homogeneous mass distribution, you will get no gravitational affects inside that sphere under any circumstances. That's what is not true in GR, because it treats boundary conditions differently (it allows more complicated topologies because you cannot just assume the spacetime, and it also requires knowledge of the history of the mass distribution, not just it's current situation).