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

[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.

 

[PeterDonis]

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

No. What I'm saying is that the stresses are the physics. All this talk about "is it rotating relative to spacetime" is not physics, it's just words.

There must be a reason for this measurement to show these stresses. What is the reason?

According to GR, the reason is that in the first case (where the ring shows stresses), the worldlines of the particles in the ring have nonzero path curvature (which means they feel nonzero proper acceleration), whereas in the second case, the worldlines have zero path curvature (which means they feel zero proper acceleration). In other words, it's how the worldlines "sit" in the geometry of spacetime.

If you want to use the words "the ring is rotating relative to spacetime" to describe this physics (the stresses and GR's explanation of them), that's fine as long as you understand that those words are just words. They're not the physics; they're not the machinery you use to actually make predictions. The machinery you use to make predictions is the theory of GR; it's expressed in math, not words. So if you actually want to reason about physics, instead of just using words to label things, you have to learn the actual model.

 

[Buckethead]

No. What I'm saying is that the stresses are the physics. All this talk about "is it rotating relative to spacetime" is not physics, it's just words.
According to GR, the reason is that in the first case (where the ring shows stresses), the worldlines of the particles in the ring have nonzero path curvature (which means they feel nonzero proper acceleration), whereas in the second case, the worldlines have zero path curvature (which means they feel zero proper acceleration). In other words, it's how the worldlines "sit" in the geometry of spacetime.

If you want to use the words "the ring is rotating relative to spacetime" to describe this physics (the stresses and GR's explanation of them), that's fine as long as you understand that those words are just words. They're not the physics; they're not the machinery you use to actually make predictions. The machinery you use to make predictions is the theory of GR; it's expressed in math, not words. So if you actually want to reason about physics, instead of just using words to label things, you have to learn the actual model.

OK, that makes sense. So although I can say the ring is rotating relative to spacetime, I can't just say we have that simple relationship, the relationship is more complex and must be described using a mathematical relationship between the object and spacetime. All good! In the case of two rings, should the relationship between one ring and the other be first and foremost a relationship between one ring and spacetime and then that spacetime and the second ring? In other words I suppose it would be impossible to have a relationship just between 2 rings and not between 2 rings and spacetime as just two rings without spacetime gets us back to square one where we can't say if one ring is spinning and the other isn't anymore than we can say one inertial object is moving and another isn't. In other words, we need spacetime to build the relationship between one ring and the other and to say which one is spinning and which one isn't. Is that correct?

 

[Buzz Bloom]

Hi @Buckethead:

I strongly recommend you take a careful look at the following old thread.

https://www.physicsforums.com/threads/effort-to-get-us-all-on-the-same-page-balloon-analogy.261161​

The creator of this thread, Marcus, presented an excellent and simple non-mathematical explanation of the concepts of space, distances, light speeds, etc., using the expanding balloon analogy.

I hope you find it helpful.

Regards,
Buzz

 

[PAllen]

Look,if you have an isolated ring with stress, you can deduce a state without stress that the isolated ring rotates relative to. Thus, even though the ring is isolated, it physically defines its own nonrotating reference.

 

[Dale]

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.

Essentially, yes. Except that I would say that it could happen in principle, but we have measured it and found that it doesn't happen.

 

[Dale]

n other words I suppose it would be impossible to have a relationship just between 2 rings and not between 2 rings and spacetime as just two rings without spacetime

This doesn't make any sense. How could you even have two rings without spacetime? If there is no geometry then in what sense is there a ring, let alone two rings? Doesn't what you are describing as a "two rings" presuppose geometry?

 

[PeterDonis]

we need spacetime to build the relationship between one ring and the other and to say which one is spinning and which one isn't. Is that correct?

I would say we need spacetime in the GR model in order to explain the observations that tell us that one ring is rotating and the other isn't (as well as other observations that tell us, e.g., how far apart the rings are from each other, what size each ring is, etc.). But we don't need spacetime to tell us which one is rotating and which one isn't; we can just measure the stresses (or lack thereof) in the rings directly (and the same for other measurements). In other words, spacetime is a model that we use in order to tie together lots of different observations and give a compact explanation for all of them. But the model is not the observations. It's a model.

 

[nitsuj]

There would be a Doppler effect with the observation, I'll use my own choice of light. Even if the ring was luminescent, and uniformly, or at least consistently so, this would be seen.

 

[CarlM]

When viewed in easy terms what you are proposing is not too far from the "ether" theory, that light was transferred through an ether that permeated space. That was disproven in the Michelson-Morley experiment in the late 1800s. That disproof was one of the sticky points that lead to Einstein's thought about relativity.

The problem is that this physics, like quantum physics, is in physical terms but outside the boundaries of our day to day human experience. It is described in mathematics because everyday terms like door and chair and teddy bear really do not apply. Our everyday experience is pretty well described by Newton's physics for 90 percent of it and the other ten percent we gloss over till someone starts measuring and observing closely finding discrepancies. Then it gets weird and the Newtonian space of our daily experience cannot be reconciled. Common sense, another term for experience, does not apply in the depth of space or in a black hole or inside and atom because we cannot experience and survive these places.

 

[Buckethead]

Hi @Buckethead:

I strongly recommend you take a careful look at the following old thread.

https://www.physicsforums.com/threads/effort-to-get-us-all-on-the-same-page-balloon-analogy.261161​

The creator of this thread, Marcus, presented an excellent and simple non-mathematical explanation of the concepts of space, distances, light speeds, etc., using the expanding balloon analogy.

I hope you find it helpful.

Regards,
Buzz

Thanks Buzz. I actually read through the whole thread a couple years ago and it was very interesting. I think it might be time for a re-read.