Is there a physical explanation for the relationship between light and space?

[PeterDonis]

Assume an empty universe (no stress-energy anywhere) and you have two rings with a distance between them and one is spinning relative to the other. Will either feel any forces of rotation?

Either one or both could be; there is no way to tell from the description you give. You would have to attach accelerometers and gyroscopes to each ring to see.

Note that "assume an empty universe" means "assume Minkowski spacetime", which means you are still assuming a spacetime geometry. So the general rule I gave--that the spacetime geometry determines which states of motion will show local acceleration and/or rotation using the acceleometers and gyroscopes, and which will not--is still true. It's just that your specification of the problem does not give enough information to know how each of the rings is situated in that spacetime geometry.

if you have two galaxies spinning relative to each other and they are a great distance apart and no other galaxies exist anywhere, can you say which one is spinning

Basically the same answer as above: the only difference is that you can't assume "an empty universe" in this case because the galaxies certainly have non-negligible stress-energy (whereas we could assume the rings above were idealized rings with negligible stress-energy). But there is still some spacetime geometry present, so the general rule I gave still applies. Again, the issue is that you haven't given enough information to know exactly how each galaxy relates to the spacetime geometry.

 

[PeterDonis]

I take it that such a state of affairs would not be possible in a flat spacetime?

Do you mean the observed acceleration of galaxies away from ours, but with all the galaxies being in free fall? Yes, that would not be possible in flat spacetime.

 

[PeterDonis]

The 2 rings have mass so there isn't null stress energy anywhere.

This is a quibble. You can assume that the rings have negligible stress-energy. We do this all the time when modeling ordinary objects (i.e., things that aren't galaxies, stars, or planets, or objects with similarly huge masses).

You cannot have 2 isolated objects spinning ( I think you mean rotating) relative to each other. Only one of them is rotating relative to the other taken to be stationary.

This is not correct. Where are you getting this from?

 

[puzzled fish]

This is a quibble. You can assume that the rings have negligible stress-energy. We do this all the time when modeling ordinary objects (i.e., things that aren't galaxies, stars, or planets, or objects with similarly huge masses).
This is not correct. Where are you getting this from?

Peter, I do not know what the OP means by two spinning disks in an empty universe. Are they the one inside the other? Are they separated and spinning as we view the distant galaxies spinning? Since he mentioned Mach, I assume he means something like two persons tied together with a piece of string, in an empty universe, and one is spinning around the other, taken to be at the center and stationary. Judged by their respective FORs, in both frames, one of them appears to be rotating whereas the other not. Yet only one of them experiences centifugal forces.
I maintain this is a completely untenable situation. Take the two persons to be stars.
1st. In the case of 2 stars rotating around their respective center of mass in an empty universe, where is the string binding them together? Only spacetime curvature and geodesics there, hence no acceleration.
2nd. You can't take the stars to have negligible mass. The two stars in an empty universe are all that there is, and they both create a well-known spacetime solution.

 

[PeterDonis]

Are they separated and spinning as we view the distant galaxies spinning?

That's what I assumed, but you're right that it would be nice to have more details from the OP.

However, that is irrelevant to point I was making when I responded to your post. You made a general claim that "you cannot have 2 isolated objects spinning". That claim is false, regardless of what the OP meant by his description.

I assume he means something like two persons tied together with a piece of string, in an empty universe, and one is spinning around the other, taken to be at the center and stationary. Judged by their respective FORs, in both frames, one of them appears to be rotating whereas the other not. Yet only one of them experiences centifugal forces.
I maintain this is a completely untenable situation.

I don't see why. This is just a version of the standard "rotating disk in flat spacetime" scenario in SR, which is perfectly valid. (And note that in this scenario, the observer that is "spinning around the other" does not have a global FOR in the usual sense.)

In any case, this scenario is not how I was interpreting the OP. And the claim you made that I responded to was much more general than the one just quoted. See above.

Take the two persons to be stars.

This is a completely different scenario, in which, as you note, spacetime is not flat. (The absence of the string in this scenario is actually a red herring. There doesn't have to be a string in the flat spacetime case either: the "spinning" observer could be using a rocket or some other self-contained means of propulsion.) There is no useful analogy between the two scenarios.

 

[PAllen]

I take it that such a state of affairs would not be possible in a flat spacetime? And thus we have evidence that the spacetime is curved, and the great mystery is finding the source of the curvature?

No, you can define an SR cosmology that has most of first order features of observed cosmology:

1) A flow of inertial bodies each of which sees isotriopic red shift of the others as a function of distance.
2) Superluminal recession rates as this is defined by cosmologists.

However, all of the quantitative relationships required by such flat spacetime model do not match observation. But if the total energy density of the universe were several orders of magnitude less than our universe, the SR model would be quite accurate as to red shifts and distance.

[edit: from seeing Peter's earlier response, some further technical clarification is needed. The flat spacetime model is maximally hyperbolic for the case zero cosmological constant. This, per cosmological terminology, is sometimes described as accelerated expansion. However, another sense of accelerated expansion is positive cosmological constant. The effects of this cannot be achieved in flat spacetime.]

 

[PeterDonis]

you can define an SR cosmology that has most of first order features of observed cosmology

But it does not have the feature of accelerating expansion (due to a positive cosmological constant), as you say. So we need clarification from Mister T on exactly what he meant by "this state of affairs" (I had assumed he meant accelerating expansion due to a positive cosmological constant).

 

[Buckethead]

With regard to the two rings, my scenario was not the rings orbiting each other (if there is such a thing in this scenario) but rather that their angular velocities were different and also they are (if they can be) just test objects so the stress-energy might be ignored.

(edit) It also just occurred to me that this scenario might be considered identical a scenario where one ring is orbiting the other with the orbiting ring in lock sync. The ring at the center of this orbit would be the one that is rotating relative to the line that connects the two objects. Or if you chose the ring that is rotating relative to the connecting line and have that orbit instead, then there would be no lock sync and that ring would just be spinning faster (or slower if spinning in the same direction as the orbit). In both cases however it seems neither gravity or a connecting string is necessary to maintain the orbit. In all 3 cases, if there are no forces on the objects then there must be no spacetime either.

 

[puzzled fish]

I don't see why. This is just a version of the standard "rotating disk in flat spacetime" scenario in SR, which is perfectly valid. (And note that in this scenario, the observer that is "spinning around the other" does not have a global FOR in the usual sense.)

No. Rotating disk affects space-time geometry and I do not think is good for such a simple example.
I still maintain that the example I've given above is valid, even for two very small massive objects rotating around their center of mass. Ok, you can fix one of them if you like and rotate the other like one mass on a disk rotates around another one at its center. Then I can find an angular velocity (no matter how small) for which there is no force measured for both. The one because is not rotating (if you like) and the other because it matches the gravitational force between them (no matter how small.) Of course this is attributed to ever so small spacetime curvature there. So, where's Mach's principle there? Still, I see no forces.

 

[PeterDonis]

Rotating disk affects space-time geometry

Not if we idealize it as having negligible stress-energy. That is the standard procedure in the SR model I referred to.

I still maintain that the example I've given above is valid

Can you give a reference for the specific solution of the Einstein Field Equation you are using for that example?

 

[Mister T]

But it does not have the feature of accelerating expansion (due to a positive cosmological constant), as you say. So we need clarification from Mister T on exactly what he meant by "this state of affairs" (I had assumed he meant accelerating expansion due to a positive cosmological constant).

I did mean accelerating expansion. Can we have an accelerating expansion, but have each galaxy experience a zero proper acceleration, in a flat spacetime.

 

[PeterDonis]

a scenario where one ring is orbiting the other with the orbiting ring in lock sync

Do you mean that the ring at the center has sufficient gravity to keep the other ring in orbit?

In both cases however it seems neither gravity or a connecting string is necessary to maintain the orbit.

I have no idea how this could be the case. Where are you getting this from?

In all 3 cases, if there are no forces on the objects then there must be no spacetime either.

I don't think your scenarios here are physically valid. Again, where are you getting this from?

 

[PAllen]

I did mean accelerating expansion. Can we have an accelerating expansion, but have each galaxy experience a zero proper acceleration, in a flat spacetime.

I think the most common meaning of accelerated expansion is that the function a(t) in the FLRW metric grows faster than a linear function. With that definition, it is not possible for this to occur in a flat spacetime cosmology (in such a cosmology, a(t) must be exactly linear). And by 'cosmology' I mean an isotropically expanding inertial congruence.

[edit: Let me make a few more points here. A flat spacetime cosmology requires linear a(t) and hyperbolic spatial slices of constant cosmological time. Our universe is pretty close to linear expansion (over periods not too long) with flat spatial slices of constant cosmological time. The observable difference is that our universe has much slower growth of red shift per distance defined by standard candles than would be the case for flat spacetime. This by itself establishes spacetime curvature, and this is a MUCH bigger effect than from the tiny cosmological constant. ]

 

[puzzled fish]

Not if we idealize it as having negligible stress-energy. That is the standard procedure in the SR model I referred to.
Can you give a reference for the specific solution of the Einstein Field Equation you are using for that example?

No, I am afraid I can't. I am not aware of any specific solutions to the two-body problem in GR, I am referring to, only numerical ones. The Schwarzschild solution when one of the bodies is substantially less massive than the other is a good approximation.

 

[PeterDonis]

I am not aware of any specific solutions to the two-body problem in GR, I am referring to, only numerical ones.

Ok. Then yes, you are correct that there are numerical solutions in which two bodies mutually orbit each other, and both are in free fall, feeling zero acceleration, and spacetime is curved in the region occupied by the orbits of the two bodies. (The question of vorticity is not so simple, but I don't think we need to go into it at this point.) However, those solutions are also asymptotically flat, which means there is a boundary condition at infinity that is required in order to derive the solution. That boundary condition at infinity, conceptually, represents the effect of all the other matter in the universe, on the idealized assumption that all of that matter is distributed in a spherically symmetric fashion about the isolated two-body system.

In other words, Mach's principle can be viewed as entering into this two-body solution as a boundary condition. Basically, the idea is that, if we have a region of spacetime outside of which everything is spherically symmetric, then the matter in that spherically symmetric outside region causes zero spacetime curvature in the inside region. (This is the GR version of the Newtonian shell theorem.) So we can put any isolated system we like in the inside region, and the spacetime curvature in that inside region will be solely due to that isolated system. But it also means that the spacetime geometry at the boundary of the inside region is determined by all the rest of the matter in the universe.

Of course this case is highly idealized--the matter in the actual universe is not exactly spherically symmetric about any isolated system, such as the solar system. But it turns out to be a very good approximation, which is why asymptotically flat solutions of the EFE are used so much--they are both mathematically tractable and physically reasonable for an isolated system.

 

[puzzled fish]

Ok. Then yes, you are correct that there are numerical solutions in which two bodies mutually orbit each other, and both are in free fall, feeling zero acceleration, and spacetime is curved in the region occupied by the orbits of the two bodies. (The question of vorticity is not so simple, but I don't think we need to go into it at this point.) However, those solutions are also asymptotically flat, which means there is a boundary condition at infinity that is required in order to derive the solution. That boundary condition at infinity, conceptually, represents the effect of all the other matter in the universe, on the idealized assumption that all of that matter is distributed in a spherically symmetric fashion about the isolated two-body system.

In other words, Mach's principle can be viewed as entering into this two-body solution as a boundary condition. Basically, the idea is that, if we have a region of spacetime outside of which everything is spherically symmetric, then the matter in that spherically symmetric outside region causes zero spacetime curvature in the inside region. (This is the GR version of the Newtonian shell theorem.) So we can put any isolated system we like in the inside region, and the spacetime curvature in that inside region will be solely due to that isolated system. But it also means that the spacetime geometry at the boundary of the inside region is determined by all the rest of the matter in the universe.

Of course this case is highly idealized--the matter in the actual universe is not exactly spherically symmetric about any isolated system, such as the solar system. But it turns out to be a very good approximation, which is why asymptotically flat solutions of the EFE are used so much--they are both mathematically tractable and physically reasonable for an isolated system.

Thank you. Great answer! Surely, Mach's Principle makes sense when viewed as boundary conditions of partial differential equations. I am not convinced though, whether these boundary conditions are dictated by matter. In the everywhere empty Schwarzschild solution which describes a black hole, (for example let us take the Kruskal metric) there isn't any mass anywhere to be found in the universe.
Anyway, I am not here to discuss Mach's Principles which I do not believe in, my idea was to explain to the OP that two bodies can fall towards each other without acceleration, in exactly the same way that two galaxies can move away from each other without acceleration, owing solely to the curvature of spacetime.

 

[PeterDonis]

In the everywhere empty Schwarzschild solution which describes a black hole, (for example let us take the Kruskal metric) there isn't any mass anywhere to be found in the universe.

Yes, but there is still an asymptotically flat boundary condition. In the idealized model you describe, which is vacuum everywhere, this boundary condition is simply declared by fiat; there is no other physical postulate from which you can deduce it. Of course it can't be due to the presence of matter, since there isn't any in this idealized model.

But any real isolated object, such as the Earth, of course is not alone in the universe: yet the Schwarzschild solution describes the spacetime geometry near the Earth to a very good approximation. So in the real universe, the asymptotically flat boundary condition turns out to be a very good approximation to the actual state of affairs in the spacetime around an isolated massive object. The shell theorem that I referred to is an explanation of how that can be the case even though the universe is filled with matter and is not, as a whole, asymptotically flat.

 

[Dale]

n all 3 cases, if there are no forces on the objects then there must be no spacetime either

Huh? No clue where you got this from, but a reminder about the forum rules seems appropriate.

 

[puzzled fish]

Yes, but there is still an asymptotically flat boundary condition. In the idealized model you describe, which is vacuum everywhere, this boundary condition is simply declared by fiat; there is no other physical postulate from which you can deduce it. Of course it can't be due to the presence of matter, since there isn't any in this idealized model.

But any real isolated object, such as the Earth, of course is not alone in the universe: yet the Schwarzschild solution describes the spacetime geometry near the Earth to a very good approximation. So in the real universe, the asymptotically flat boundary condition turns out to be a very good approximation to the actual state of affairs in the spacetime around an isolated massive object. The shell theorem that I referred to is an explanation of how that can be the case even though the universe is filled with matter and is not, as a whole, asymptotically flat.

Agreed. Thanks.

 

[nitsuj]

Well, that is what physicists (the usual non "conspiracy theory" ones) mean when they talk about space moving.

The laws of physics are written as differential equations, often $\partial/\partial t$ or $\partial/\partial x$. So if space moves then we would expect those laws of physics which depend on dx or dt (including Maxwells equations) to change as you change reference frame. In the more modern literature this is called Lorentz violation or CPT violation. It applies not just for the local laws governing the electromagnetic force, but also the strong and weak nuclear forces, and gravity.

So if you accept the invariance of c then you are basically 1/4 of the way to accepting that space doesn't move. All you have to do is check the strong and weak nuclear forces and gravity too.

I read some of those links to tests for moving space you kindly provided and that's what they did...test all the "interactions" we know of, and do a "Michelson Morley" experiment with them. I suppose with the ability to detect gravity waves, that one too can be added to the list? Oh that's tough one, "does space move?", "No, but spacetime waves!" lol

I think the strong nuclear was the most promising of them...for me that "limit" and mechanic of its "glue" makes it seem like it's "space"...what I mean I have no idea :D In coming joke...clearly that "elastic" mechanic of the strong nuclear forces is akin to the rubber seen in this vid.

 

[Dale]

that's what they did...test all the "interactions" we know of, and do a "Michelson Morley" experiment with them

Yes, that is the idea, except that it is not just limited to interferometry but all sorts of effects that should be sensitive to any variation in the corresponding law of physics.

 

[Buckethead]

The laws of physics are written as differential equations, often ∂/∂t\partial/\partial t or ∂/∂x\partial/\partial x. So if space moves then we would expect those laws of physics which depend on dx or dt (including Maxwells equations) to change as you change reference frame.

(Sorry for the delayed response) OK, I understand this. So you can't have a reference frame move through space because that would change the laws of physics as described by Maxwell's equations and because it would also imply a change in c. That makes sense. And I suppose in retrospect I expected this so I was thinking more outside of this reference frame. For example two galaxies in an otherwise empty universe were we can talk about the speed of light between those galaxies (but not at a local point in a galaxy) or whether or not two galaxies can accelerate toward each other without either feeling acceleration. Because here we are talking about not the relationship between a reference frame and space but rather between two reference frames and the effect space may have on that as I just described. Does this make sense?

 

[Buckethead]

Note that "assume an empty universe" means "assume Minkowski spacetime", which means you are still assuming a spacetime geometry. So the general rule I gave--that the spacetime geometry determines which states of motion will show local acceleration and/or rotation using the acceleometers and gyroscopes, and which will not--is still true.

OK, this is important because it implies a definite relationship between a geometry called spacetime and any objects within that space. So even a single ring in an otherwise empty universe can be in a condition of rotation (forces felt) or not. The elephant in the room is why there would be this relationship if spacetime is thought of as not absolute (with regard to for example whether spacetime itself is rotating or accelerating relative to the above ring) but rather only posseses properties of curvature and possibly expansion among other things. As Dale mentioned (if I understood him correctly), you can't think of the ring as moving through spacetime itself because that would imply c was not a constant and laws defined by Maxwell's equations would fail. What I'm getting at is spacetime is a geometry which seems to have a definite "absoluteness" if something inside that spacetime can be said to be spinning or not.

 

[PeterDonis]

with regard to for example whether spacetime itself is rotating or accelerating relative to the above ring

I think you are confusing yourself by focusing on words instead of physics. The ring in a condition of rotation in an otherwise empty universe is a physical system, with a definite mathematical description and a definite set of observations that it predicts. Whether you use the words "the ring is rotating" or "spacetime is rotating relative to the ring" to describe this physical system is a matter of words, not physics.

OTOH, if you are thinking of "the ring rotating in an otherwise empty universe" as a different physical system from "spacetime rotating relative to the ring", then you need to take a big step back and think about what the physical difference is between these two cases, what different mathematical descriptions they would have, and what different observations they would predict.

spacetime is a geometry which seems to have a definite "absoluteness" if something inside that spacetime can be said to be spinning or not

If this is just a way of saying what I said in my first paragraph above--that the ring in a condition of rotation in an otherwise empty universe is a physical system, with a definite mathematical description and a definite set of observations that it predicts--then it's fine. But I'm not sure if you are actually thinking of it that way.

 

[Buckethead]

Buckethead said: ↑
a scenario where one ring is orbiting the other with the orbiting ring in lock sync

Do you mean that the ring at the center has sufficient gravity to keep the other ring in orbit?

Sorry for the confusion. I was coming from the thought that if the geometry of spacetime is a mathematical relationship with matter, and if there is only for example one object in the universe (such as a ring), then its natural state might be to not be rotating. The reason I say this is because if it was rotating, then why wouldn't the whole of spacetime just rotate with it. I'm fighting the thought that spacetime is absolute. If it is, I just don't get it. Yet this is the feeling I'm getting through this thread. So if you have 2 rings (separated by some distance) and one is rotating, is it really rotating or is the other rotating? And which one and why? It seems it all comes down to which one is rotating relative to the absolute spacetime. If one is showing forces, then it is rotating, else it is not. And we don't need to use 2 rings, take just one single ring in an empty universe. Is is rotating or not? If it is, this secures spacetime as being absolute, doesn't it?

 

[PeterDonis]

if it was rotating, then why wouldn't the whole of spacetime just rotate with it

Again, I think you're confusing yourself by focusing on words instead of physics.

Consider two rings, both with negligible mass (so their effect on the spacetime geometry is negligible), and separated by a very large distance so their effect on each other is negligible. The rings are identical except that one has internal stresses of the sort that indicate rotation, and the other does not. The internal stresses are an observable physical difference, and their presence is what makes us say that the first ring is rotating and the second is not.

The only other question would be whether the first state is physically possible (assuming that the second one, with no stresses, is possible). It seems obvious that it should be, since we can create it in the lab. Granted, our universe is not empty, but it certainly seems like we could go out somewhere in deep space, far away from anything else, and still be able to get a ring into the state with internal stresses that indicate rotation. At any rate, GR predicts that we can.

 

[nitsuj]

... this secures spacetime as being absolute.

What does "spacetime is absolute" mean in terms of physical consequence? How does "absolute" spacetime fit into the "playing out" of physics.

 

[Buckethead]

If this is just a way of saying what I said in my first paragraph above--that the ring in a condition of rotation in an otherwise empty universe is a physical system, with a definite mathematical description and a definite set of observations that it predicts--then it's fine. But I'm not sure if you are actually thinking of it that way.

If there is a ring in an empty universe and it is spinning (forces felt) and you have a discription of that, and if then the ring stops spinning (no forces felt) and you have a description of that, . What part of these two descriptions would be different if not in relation to something else (like spacetime).

 

[nitsuj]

If there is a ring in an empty universe and it is spinning (forces felt) and you have a discription of that, and if then the ring stops spinning (no forces felt) and you have a description of that, . What part of these two descriptions would be different if not in relation to something else (like spacetime).

oh wow

 

[PAllen]

If there is a ring in an empty universe and it is spinning (forces felt) and you have a discription of that, and if then the ring stops spinning (no forces felt) and you have a description of that, . What part of these two descriptions would be different if not in relation to something else (like spacetime).

How would the ring stop spinning without forces felt??

 

[PeterDonis]

What part of these two descriptions would be different if not in relation to something else (like spacetime).

I don't know what you mean. It's perfectly obvious what's different about the two descriptions: in one case there are internal stresses, and in the other, there aren't. There's nothing in there about "in relation to something else"; the internal stresses can be measured even if you have no idea about the ring's relationship to anything else.

Once more: I think you are confusing yourself by focusing on words instead of physics.

 

[Buckethead]

Consider two rings, both with negligible mass (so their effect on the spacetime geometry is negligible), and separated by a very large distance so their effect on each other is negligible. The rings are identical except that one has internal stresses of the sort that indicate rotation, and the other does not. The internal stresses are an observable physical difference, and their presence is what makes us say that the first ring is rotating and the second is not.

I understand this but what I don't understand is why measuring these stresses is sufficient as being called a cause of these stresses, which it seems to me is what you are trying to say. There must be a reason for this measurement to show these stresses. What is the reason?

 

[PeterDonis]

How would the ring stop spinning without forces felt??

I think what he means is that after the ring has stopped spinning, no forces would be felt (no internal stresses).

 

[Buckethead]

How would the ring stop spinning without forces felt??

I meant two static states, one where the ring is spinning (and its associated description) and one where the ring is not spinning (and its associated description). I did not think that moving from one state to the other was relevant but maybe it is.

 

[PAllen]

I understand this but what I don't understand is why measuring these stresses is sufficient as being called a cause of these stresses, which it seems to me is what you are trying to say. There must be a reason for this measurement to show these stresses. What is the reason?

You could say the state with centrifugal stresses is in motion (rotating) relative to the possible state with no such stresses.