I GR change in proximal tendancy

[name123]

Imagine a situation in which a space time curvature propagation reaches a zone (in space time) which it had no prior influence. Would the measured motion (for the objects in that rough zone) in spacetime (subsequently) be unaffected by the changes in spacetime curvature? If the answer was "yes" then was it achieved with no changes in force?

 

[PeterDonis]

Imagine a situation in which a space time curvature propagation reaches a zone (in space time) which it had no prior influence.

Can you give a concrete example?

Would the measured motion (for the objects in that rough zone) in spacetime (subsequently) be unaffected by the changes in spacetime curvature?

How are you measuring the motion? Motion is relative.

If the answer was "yes" then was it achieved with no changes in force?

What definition of "force" are you using? If all of the objects in question are in free fall, then there are no forces on them in the sense that GR uses that term, but that does not mean they are unaffected by spacetime curvature.

 

[name123]

Can you give a concrete example?

It is not so easy to provide a concrete example, as I was thinking of a theoretical circumstance. An orbiting planet for example though would do. As one could consider the propagation of the curvature linked to its mass, and a drifting satellite.

How are you measuring the motion? Motion is relative.

I was just thinking of measuring the distance between objects from an arbitrary frame of reference.

With general relativity would there exist a frame of reference where the planet and the satellite gravitating together could be explained in terms of the satellite being stationary and the planet gravitating towards it?

What definition of "force" are you using? If all of the objects in question are in free fall, then there are no forces on them in the sense that GR uses that term, but that does not mean they are unaffected by spacetime curvature.

So in GR what force curves spacetime (is mass a force?), or does the creating a curvature in spacetime not require any force?

 

[PeterDonis]

An orbiting planet for example though would do. As one could consider the propagation of the curvature linked to its mass, and a drifting satellite.

Ok, this will work. I think it's better to have a concrete example to illustrate the issues. To make it even more concrete, let's suppose we're looking at the Earth orbiting the Sun, and the Moon as the satellite.

With general relativity would there exist a frame of reference where the planet and the satellite gravitating together could be explained in terms of the satellite being stationary and the planet gravitating towards it?

Yes. In GR you can choose coordinates however you want, and you can choose them so any object you like is "at rest" in those coordinates. For example, you can choose coordinates such that the Moon is at rest and the Earth is "falling" towards it. But the Earth is also "falling" towards the Sun, so the Earth's motion in these Moon-centered coordinates will look rather complicated, as compared with Sun-centered coordinates (where the Earth's orbit is just an ellipse--or more precisely a slowly precessing almost ellipse) or Earth-centered coordinates (where the Earth is at rest).

As far as "propagation of curvature", that does happen in this case, but it's a very small effect, because the changes involved, while they might look very large to us in ordinary units, are really extremely tiny when we put them in the "natural" units of GR. The way you would measure it is with a gravitational wave detector, because that is what gravitational waves are: propagating changes in spacetime curvature. So LIGO, for example, recently detected the propagating changes in spacetime curvature due to two black holes merging (which was a much larger effect than the effect of, for example, the Earth moving in its orbit).

in GR what force curves spacetime

There is no force that curves spacetime. Stress-energy curves spacetime, but not by exerting a force on it; the concept does not make sense. Spacetime is not something that a force can be exerted on; your intuitions about "forces" and how they work simply don't apply.

is mass a force?

No.

does the creating a curvature in spacetime not require any force?

Correct. See above.

 

[name123]

Yes. In GR you can choose coordinates however you want, and you can choose them so any object you like is "at rest" in those coordinates. For example, you can choose coordinates such that the Moon is at rest and the Earth is "falling" towards it. But the Earth is also "falling" towards the Sun, so the Earth's motion in these Moon-centered coordinates will look rather complicated, as compared with Sun-centered coordinates (where the Earth's orbit is just an ellipse--or more precisely a slowly precessing almost ellipse) or Earth-centered coordinates (where the Earth is at rest).

But would GR be compatible with that in the sense that while I have no doubt you could choose whatever coordinate system you like, I would have thought that GR would predict spacetime curvature, and the curvature in turn would be thought to determine what happens, and it would not predict certain paths as being in freefall.

 

[PeterDonis]

would GR be compatible with that

Compatible with what? All of the equations of GR are written in covariant form, which means they are valid in any coordinate chart.

I would have thought that GR would predict spacetime curvature, and the curvature in turn would be thought to determine what happens, and it would not predict certain paths as being in freefall.

This is true, but I'm not sure why you think it's a potential problem.

 

[name123]

This is true, but I'm not sure why you think it's a potential problem.

Well consider a Sun centred coordinate system, I would have expected GR to have predicted the path that the Earth would be taking to be a freefall path, but if one were to consider an Earth centred coordinate system, I would not have expected GR to have predicted the path that the Sun would be taking to be a freefall path.

 

[pervect]

Imagine a situation in which a space time curvature propagation reaches a zone (in space time) which it had no prior influence. Would the measured motion (for the objects in that rough zone) in spacetime (subsequently) be unaffected by the changes in spacetime curvature? If the answer was "yes" then was it achieved with no changes in force?

I'm not sure what you mean by "propagation of curvature, unfortunately.

The closest thing I can think of that's physically interpretable as a "propagation of curvature" is a gravitational wave. I think one needs to take a few liberties to talk about gravitational waves "propagating", but it's done all the time. So what I think we need to understand is when we can talk about gravity / gravitational waves propagating, and why it makes sense to describe them in those terms.

So if we take this interpretation of "propagation of curvature" via the example of gravitational waves as "propagating", then we can say if we have a gravitational wave source, and we turn it on, that a distant object's motion won't be affected by the gravitational wave until the wave propagates through space and reaches said distant object. So, for example, the signal from the binary black hole inspiral observed by Ligo had to propagate through space to reach us before Ligo could measure it. To split some hairs, while Ligo was emitting gravitational radiation all the time, it only became significant and measurable for a few seconds near the end of the inspiral, so in effect it "turned on", and we could measure the effects on the motion resulting from the event.

Note that the whole idea of "propagation" implicitly assumes that there is some agreed-on notion of time for the propagation to occur in. This could be a source of confusion, as there are many possible ways one might choose to separate space-time into space and time. My understanding of the issue is that the usual way of splitting space-time into space+time is motivated by certain gauge choices in linearized gravity, this provides the fundamental structure to define an appropriate sense of "time" so that we can regard gravitational waves as "things that propagate through space". Mathematically, if we have something that satisfies the wave equation, we can regard that something as "propagating". I don't think I've ever seen a reference on this point, but if one accepts the notion that "propagating" means "satisfying a wave equation", I think it all makes sense and follows logically.

So, going by this viewpoint, we need to describe gravity by using a space-time structure that satisfies the wave equations, for the propagation description to make sense. This is possible with the right gauge choices. These gauge choices lead to a decription as a propagation of the spatial metric, as the spatial metric satisfies the wave equation. But - it's necessary to make the right gauge choices before the "propagation" analogy really works well. This is often assumed, and not stated as an assumption. Not being stated as assumption, people often set up the problem in a manner that violates the preconditions, and then get confused as to why the "propagation" idea doesn't work.

At an intermediate level analogy might be helpful, if one is familiar with Maxwell's equations, one can ask the question "what happens if a charge suddenly dissapears?" Does the charge disappearing "propagate" instantly? How is this compatible with special relativity - or for that matter, with Maxwell's equations?

The answer is basically that charges don't just dissappear, and in fact Gauss's law is a mathematical expression of the fact that they don't do that. So we can't imagine a charge disappearing without breaking Gass' law. This point of view leads into a discussion of the gauge choices in electromagnetism (the Coulomb gauge and the Lorentz gauge), and how we view EM waves as propagating in the Lorentz gauge - they satisfy the wave equation in that gauge, so the conceptual image of them as "propagating" is useful. If we adopt the Coulomb gauge, the idea that static electric fields propagate from charges isn't useful - they don't, assuming they do leads to confusion and wrong answers.

The gravitational analogy is that if we ask "what happens when a mass suddenly disssapears", the answer is the same. Masses can't suddenly dissapear. Things are a bit tricker in gravity, but there are continuity conditions built into Einstein's field equations that prevent masses from suddenly dissapearing, much as Gauss's law prevents charges from suddenly dissapearing, though the details are more complicated.

I think the analogy is very helpful, but it requires a fair knowledge of E&M to appreciate. And things are more complicated in gravity, we don't have anything that's quite like Gauss' law to make things so simple. But I think the basic observation that the mathematical description of "propagation" is "satisfying a wave equation" brings the issue more into focus. And the point is the same - static gravitational fields don't "propagate", any more than static electric Coulomb fields do. There isn't any wave equation in the coulomb gauge, so the "propagation" idea leads us astray.

 

[PeterDonis]

consider a Sun centred coordinate system, I would have expected GR to have predicted the path that the Earth would be taking to be a freefall path, but if one were to consider an Earth centred coordinate system, I would not have expected GR to have predicted the path that the Sun would be taking to be a freefall path.

Why not? I assume you agree that in a Sun-centered coordinate system, the Sun's path is predicted to be a free-fall path. Given that, all you need to show that the Sun's path will also be a free-fall path in an Earth-centered coordinate system is:

All of the equations of GR are written in covariant form, which means they are valid in any coordinate chart.

That means that if GR predicts that the Sun's path (or the Earth's path, or any other path) is a free fall path in one coordinate system, it will predict that that same path is a free-fall path in any coordinate system.

In fact, we can turn this around: you agree that in a Sun-centered coordinate system, the Earth's path is predicted to be a free-fall path. Do you also agree that the Earth's path is predicted to be a free-fall path in an Earth centered coordinate system? If not, why not?

(Bear in mind that in Earth-centered coordinates, if you are including the Sun in your calculations, you certainly cannot just model gravity as a Newtonian force centered on the Earth--not even as an approximation with small corrections. You can't even do that in Newtonian physics.)

 

[name123]

I'm not sure what you mean by "propagation of curvature, unfortunately.

The closest thing I can think of that's physically interpretable as a "propagation of curvature" is a gravitational wave. I think one needs to take a few liberties to talk about gravitational waves "propagating", but it's done all the time. So what I think we need to understand is when we can talk about gravity / gravitational waves propagating, and why it makes sense to describe them in those terms.

So if we take this interpretation of "propagation of curvature" via the example of gravitational waves as "propagating", then we can say if we have a gravitational wave source, and we turn it on, that a distant object's motion won't be affected by the gravitational wave until the wave propagates through space and reaches said distant object. So, for example, the signal from the binary black hole inspiral observed by Ligo had to propagate through space to reach us before Ligo could measure it. To split some hairs, while Ligo was emitting gravitational radiation all the time, it only became significant and measurable for a few seconds near the end of the inspiral, so in effect it "turned on", and we could measure the effects on the motion resulting from the event.

Note that the whole idea of "propagation" implicitly assumes that there is some agreed-on notion of time for the propagation to occur in. This could be a source of confusion, as there are many possible ways one might choose to separate space-time into space and time. My understanding of the issue is that the usual way of splitting space-time into space+time is motivated by certain gauge choices in linearized gravity, this provides the fundamental structure to define an appropriate sense of "time" so that we can regard gravitational waves as "things that propagate through space". Mathematically, if we have something that satisfies the wave equation, we can regard that something as "propagating". I don't think I've ever seen a reference on this point, but if one accepts the notion that "propagating" means "satisfying a wave equation", I think it all makes sense and follows logically.

So, going by this viewpoint, we need to describe gravity by using a space-time structure that satisfies the wave equations, for the propagation description to make sense. This is possible with the right gauge choices. These gauge choices lead to a decription as a propagation of the spatial metric, as the spatial metric satisfies the wave equation. But - it's necessary to make the right gauge choices before the "propagation" analogy really works well. This is often assumed, and not stated as an assumption. Not being stated as assumption, people often set up the problem in a manner that violates the preconditions, and then get confused as to why the "propagation" idea doesn't work.

At an intermediate level analogy might be helpful, if one is familiar with Maxwell's equations, one can ask the question "what happens if a charge suddenly dissapears?" Does the charge disappearing "propagate" instantly? How is this compatible with special relativity - or for that matter, with Maxwell's equations?

The answer is basically that charges don't just dissappear, and in fact Gauss's law is a mathematical expression of the fact that they don't do that. So we can't imagine a charge disappearing without breaking Gass' law. This point of view leads into a discussion of the gauge choices in electromagnetism (the Coulomb gauge and the Lorentz gauge), and how we view EM waves as propagating in the Lorentz gauge - they satisfy the wave equation in that gauge, so the conceptual image of them as "propagating" is useful. If we adopt the Coulomb gauge, the idea that static electric fields propagate from charges isn't useful - they don't, assuming they do leads to confusion and wrong answers.

The gravitational analogy is that if we ask "what happens when a mass suddenly disssapears", the answer is the same. Masses can't suddenly dissapear. Things are a bit tricker in gravity, but there are continuity conditions built into Einstein's field equations that prevent masses from suddenly dissapearing, much as Gauss's law prevents charges from suddenly dissapearing, though the details are more complicated.

I think the analogy is very helpful, but it requires a fair knowledge of E&M to appreciate. And things are more complicated in gravity, we don't have anything that's quite like Gauss' law to make things so simple. But I think the basic observation that the mathematical description of "propagation" is "satisfying a wave equation" brings the issue more into focus. And the point is the same - static gravitational fields don't "propagate", any more than static electric Coulomb fields do. There isn't any wave equation in the coulomb gauge, so the "propagation" idea leads us astray.

Thanks for the response.

 

[name123]

That means that if GR predicts that the Sun's path (or the Earth's path, or any other path) is a free fall path in one coordinate system, it will predict that that same path is a free-fall path in any coordinate system.

In fact, we can turn this around: you agree that in a Sun-centered coordinate system, the Earth's path is predicted to be a free-fall path. Do you also agree that the Earth's path is predicted to be a free-fall path in an Earth centered coordinate system? If not, why not?

(Bear in mind that in Earth-centered coordinates, if you are including the Sun in your calculations, you certainly cannot just model gravity as a Newtonian force centered on the Earth--not even as an approximation with small corrections. You can't even do that in Newtonian physics.)

It might just be that I have seen basic analogies such as weights on an elastic sheet. But with such analogies, while one can see why the Earth would move around the Sun (it being in freefall, following the slope down), however it wouldn't be so clear to me why the Sun would move around the Earth given the curvature (it would be "downhill" from the Earth, not "uphill" from it). On an Earth centered coordinate system, I would expect the Earth to be regarded as being at rest (in that frame of reference) and so not in freefall (unless the correct use of freefall just means no force being applied to it, whereas I had meant in motion (in the frame of reference) with no force being applied). So on the Earth centred coordinate system, I was assuming that the Sun would be regarded as being in motion with no force applied, but that the motion could not be explained in terms of the spacetime curvature. If I have used the wrong terminology regarding freefall, then perhaps I could rephrase and instead discuss whether motion not due to any force,in a certain frame of reference, can be described as explainable given the spacetime curvature predicted by GR. What would it be about the predicted spacetime curvature that would explain the Earth remaining at rest while the Sun orbited it?

 

[PeterDonis]

it wouldn't be so clear to me why the Sun would move around the Earth given the curvature (it would be "downhill" from the Earth, not "uphill" from it)

You are ignoring the spacetime curvature caused by the Sun. You can't do that; the Sun's mass is much larger than the Earth's mass. The spatial geometry in Earth-centered coordinates is not the Flamm paraboloid ("dented rubber sheet") you are imagining, not even approximately. The spacetime geometry is the same, but that doesn't mean it appears the same when you change the way spacetime is split into space and time and shift the spatial origin, which is what you are doing when you switch from Sun-centered to Earth-centered coordinates.

unless the correct use of freefall just means no force being applied to it, whereas I had meant in motion (in the frame of reference) with no force being applied

Free fall means zero proper acceleration; that is an invariant, independent of coordinates. It has nothing to do with whether the object is at rest or in motion in a particular set of coordinates.

What would it be about the predicted spacetime curvature that would explain the Earth remaining at rest while the Sun orbited it?

"Being at rest" is not an invariant, so the laws of physics don't have to "explain" it. It's just an artifact of your choice of coordinates. The laws of physics can't "explain" your choice of coordinates.

The relative motion of the Sun and the Earth, and the spacetime (not space!) curvature in the solar system, is the same in Earth-centered coordinates as in Sun-centered coordinates. That's the only thing the laws of physics have to explain.

 

[name123]

You are ignoring the spacetime curvature caused by the Sun. You can't do that; the Sun's mass is much larger than the Earth's mass. The spatial geometry in Earth-centered coordinates is not the Flamm paraboloid ("dented rubber sheet") you are imagining, not even approximately. The spacetime geometry is the same, but that doesn't mean it appears the same when you change the way spacetime is split into space and time and shift the spatial origin, which is what you are doing when you switch from Sun-centered to Earth-centered coordinates.

I didn't think I was ignoring the spacetime curvature caused by the Sun. My point was that on the rubber sheet example, the Sun having more mass would make a bigger dent. Which is why I was saying the Sun would be "downhill" from the Earth (it had a bigger mass and therefore made a bigger dent). So I am not sure why you thought I ignored that the spacetime curvature caused by the Sun's mass (as opposed to maybe misunderstood what influence that would have on the spacetime curvature).

I have assumed that with the Earth frame of reference and the Sun frame of reference, event synchronicity would be in pretty close agreement. So I am an not clear what big spatial difference you are suggesting there would be when spacetime is split into space and time and the spatial origin is moved in this example (as I have mentioned I assume the event synchronicity would be pretty close in both situations given the slow speeds). Is it that the dented rubber sheet analogy does not work with multiple objects on the rubber sheet, or are you perhaps stating that while it would be analogous using Sun-centred coordinates, it would not be using Earth-centred coordinates, or something else? It would perhaps be useful if you could explain it using a rubber sheet type analogy even if it is just explaining how it will appear different from what many would have thought if the rubber sheet analogy had of been useful.

 

[PeterDonis]

I didn't think I was ignoring the spacetime curvature caused by the Sun.

Well, you're clearly doing something wrong since you are getting the wrong answer. You are trying to reason intuitively from a framework that doesn't work well, so I'm not sure how much progress we'll be able to make unless you're willing to just give up that framework altogether.

on the rubber sheet example

The rubber sheet analogy only works if you are using coordinates centered on the Sun--or more generally on the overall center of mass of the system (which is close enough to being the Sun in this case). In Earth-centered coordinates you can only use the rubber sheet analogy close to the Earth; if you extend things far enough that the Sun is included, then trying to center the rubber sheet on the Earth simply does not work. So the only advice I can give is to drop that analogy.

I have assumed that with the Earth frame of reference and the Sun frame of reference, event synchronicity would be in pretty close agreement.

It's not just "event synchronicity". See my comments on the rubber sheet above.

It would perhaps be useful if you could explain it using a rubber sheet type analogy

I can't; that analogy does not work for what you are trying to do.

 

[name123]

The rubber sheet analogy only works if you are using coordinates centered on the Sun--or more generally on the overall center of mass of the system (which is close enough to being the Sun in this case). In Earth-centered coordinates you can only use the rubber sheet analogy close to the Earth; if you extend things far enough that the Sun is included, then trying to center the rubber sheet on the Earth simply does not work. So the only advice I can give is to drop that analogy.

I had mentioned that it would perhaps be useful if you could explain it using a rubber sheet type analogy even if it is just explaining how it will appear different from what many would have thought if the rubber sheet analogy had of been useful. And you replied that you couldn't because that analogy does not work for what you are trying to do. I did not quite understand why you could not explain the difference in how it does work and how people would have expected it to work if the rubber sheet analogy had of been useful.

Can spacetime curvature be thought of as having having a space part and a time part such that if you were to think of the coordinates as cartesian space + time? For example if the change in the x coordinate curvature in one coordinate system was equal to the change in t coordinate curvature in another coordinate system and vice verca, could in such a circumstance the spacetime curvature be regarded as the same but in one coordinate system there be more curvature in time-like spacetime and in the other more curvature in space-like spacetime?

Could you perhaps explain what the GR predictions are regards to how spacetime curvature affects free-fall paths?

Otherwise perhaps choose to explain the difference in whatever way you think would be most useful for a person looking to understand (such that they could answer those questions).

 

[PeterDonis]

I did not quite understand why you could not explain the difference in how it does work and how people would have expected it to work if the rubber sheet analogy had of been useful.

Because the rubber sheet analogy does not work for the case you are trying to apply it to. It doesn't work because it gives the wrong answer: you showed that by trying to use it and getting the wrong answer. There is no way to use it for that case to get the right answer. That's why I can't explain how to get the right answer using it for that case. I can't explain how to do something that is impossible.

Could you perhaps explain what the GR predictions are regards to how spacetime curvature affects free-fall paths?

That's way too much for a PF thread; you need to consult a GR textbook. The simplest short answer I can give is that spacetime curvature is tidal gravity, so if you can imagine the effects of tidal gravity, you are imagining the effects of spacetime curvature.

 

[name123]

That's way too much for a PF thread; you need to consult a GR textbook. The simplest short answer I can give is that spacetime curvature is tidal gravity, so if you can imagine the effects of tidal gravity, you are imagining the effects of spacetime curvature.

So if for example there were three marbles A, B & C with each with 2 small lasers attached. All in the same rest frame, separated from each other by a light year, and forming an isosceles triangle (where AB and AC are the same length, and the centre of BC is a kilometre from A), and that the lasers of each pointed to the marbles adjacent to them in the triangle (so like pointing to the adjacent vertices). And imagine, for the sake of discussion, that physicists had managed to transport fundamental particles to points in space, and that they very quickly transported a load of particles to mid-way between BC (a kilometre from A) to form a cylinder, D, which was 100m in diameter with its axis orientated along BC, and that D had a mass similar to Earth's.

Now if I understand you correctly, if you use a D centred coordinate system, the rubber sheet analogy works fine, the marble A simply free falls towards D, and B and C are unaffected (for at least a year). But that the rubber sheet analogy is inappropriate using an A centred coordinate system, because D has a larger mass. But what is the GR explanation using an A centred coordinate system, are B and C also pulled towards A along with D, if so, how is that explained, if not, what is the explanation for the light from each still pretty much pointing at the axis of D (even when A and D touch)?

 

[PeterDonis]

So if for example

Your whole scenario has nothing to do with tidal gravity, so I'm not sure why you proposed it in response to my comment about tidal gravity.

imagine, for the sake of discussion, that physicists had managed to transport fundamental particles to points in space

This is against the laws of physics so it's useless to imagine it. You're asking what the laws of physics predict for a scenario that violates the laws of physics. That's a meaningless question.

It's actually highly nontrivial to set up an experiment to test how fast changes in spacetime curvature propagate, precisely because you can't create or destroy mass or energy, so you can't create or destroy the source of spacetime curvature. The best tests we have currently are gravitational wave detectors like LIGO. There have been a number of PF threads on LIGO and there are good references on it online.

if you use a D centred coordinate system, the rubber sheet analogy works fine

Not if the spacetime curvature is changing, which is what, as far as I can tell, you are trying to model. The rubber sheet analogy only works if the spacetime geometry is static--more precisely, it is a good approximation only to the extent that changes in spacetime curvature are small enough or slow enough to ignore when computing the trajectories of the objects you are interested in. So for scenarios where seeing the effects of spacetime curvature is the point, the rubber sheet analogy is useless.

Once again, you need to consult a GR textbook. You are asking for more than PF can give you with the background you currently appear to have. Thread closed.