Modeling the effects of GW and the "Earth Frame"

[PeterDonis]

I'm not aware of any notion of "non-rigid" congruence in Pseudo-Riemannian geometry.

Then you evidently haven't read much about congruences. A non-rigid congruence is any congruence that has nonzero expansion or shear. See here for a brief description of the terminology:

https://en.wikipedia.org/wiki/Congruence_(general_relativity)

 

[GeorgeDishman]

I don't see how, but perhaps we're not viewing the problem the same way at all.

No, the problem is of my making. I'm trying to be circumspect in how I explain the problem.

What we want to do is measure fractional distance changes on the order of 10^-21 or so of, say, 3km or so.

No, that part is not a concern for me, the usual calculations work fine. What I am looking at is the motion realtive to the Earth frame. Since the material response of that is too complex to work through, I'm looking instead at the motion of one LIGO-in-space as you describe it relative to two others, each half a wavelength away. It is the common motion of the C-D pair relative to the others rather than that of D relative to C that I'm trying to quantify but only at a very crude level.
I've been revising the videos I posted before to try to visualise the transverse traceless approach in terms of non-moving test masses and a variable speed of light which, if I've followed your comments, is equivalent to the more conventional combination of moving masses and isotropic speed (i.e. both result in the same variation of phase at the photodiode).
I'll link the new videos but the descriptions are very crude at the moment:

 

[pervect]

See for instance: "The Rich Structure of Minkowskii Space" http://arxiv.org/abs/0802.4345

Let u be a normalised timelike vector field. The motion described by its flow is rigid iff u is of vanishing shear and expansion ...

"Born Rigid flow and the ADS-CFT correspondence" http://arxiv.org/abs/1010.3847

Here Born-rigid has the same meaning as dissipationless that we have been using. That dissipationless flows can be called “rigid” is seen by the equivalence of shear-free and expansion-free conditions and $\mathcal{L}_{\mu\nu} h = 0$ i.e., the orthogonal distance along the fluid flows are preserved (the derivation is not difficult and can be found in, e.g., [20]). This definition, originally formulated for flat spacetime, remains valid in curved spacetime.

"A Modern view of the classical Herglotz-Noether theorem" https://arxiv.org/abs/1004.1935

The fluid is shown to be in dissipationless (shear and expansion free) motion, which (see Section 3.3 in the present paper) coincides with the condition for Born rigidity.

and from the same paper, one of the less techinical definitions of Born rigidity

Definition Born rigidity)
.
A body is called rigid if the distance between every neighbouring pair of particles, measured orthogonal to the worldlines of either of them, remains constant along the worldline.

 

[pervect]

No, the problem is of my making. I'm trying to be circumspect in how I explain the problem.

No, that part is not a concern for me, the usual calculations work fine. What I am looking at is the motion realtive to the Earth frame. Since the material response of that is too complex to work through, I'm looking instead at the motion of one LIGO-in-space as you describe it relative to two others, each half a wavelength away. It is the common motion of the C-D pair relative to the others rather than that of D relative to C that I'm trying to quantify but only at a very crude level.
I've been revising the videos I posted before to try to visualise the transverse traceless approach in terms of non-moving test masses and a variable speed of light which, if I've followed your comments, is equivalent to the more conventional combination of moving masses and isotropic speed (i.e. both result in the same variation of phase at the photodiode).
I'll link the new videos but the descriptions are very crude at the moment:

What sort of description are you looking for? Any visual representation you can model in an Euclidean three-dimensional space is not going to capture every aspect of a curved, dynamical, four-dimensional geometry that is the mathematical description of a gravitational wave.

BTW, we can (and have) provided the mathematical description of the GW - which is its metric.

We started talking about representations that would be rigid, and determined that they don't exist - or rather that they exist in a limited region of space-time, but you run into problems if the region is too big. The answer doesn't seem to satisfy you.

It's rather like the issue of asking "how do we draw a map of the Earth on a flat sheet of paper". The basic answer is that you can't do it accuarately. Some of the issues are roughly parallel - the Earth is a higher dimensional curved manifold than the piece of paper is, and it's instructive to think about how we can have a perfectly good map of Chicago that is to scale, and a perfectly good map of Paris that is to scale, but we can't have map that shows both and the intervening territory that is also to scale.

Now, if we had some better idea of how your representation were to be applied, we could try to find one that was the least misleading in the particular case of interest. This is done in map-making, one can find different representatons of the globe with different sorts of problems, and get some idea of geography even though each individual representation is imperfect. As an aside, the mathematics are somewhat similar - the techniques of "projection" are used to make a 2d map from a 3d model, and the same projection techniques are used to get a 3d space out of a 4d space-time.

 

[PeterDonis]

the Earth is a higher dimensional curved manifold than the piece of paper is

Not if you're only considering the surface of the Earth; that is two-dimensional, just like the piece of paper. The only difference is the curvature.

If you're trying to figure out the structure of the Earth as a whole, interior as well as surface, then yes, you need to drop a dimension to represent things on a piece of paper. But such a representation--for example, a "cross section" of the Earth showing the various interior portions, crust, mantle, core, etc.--isn't usually called a "map".

 

[GeorgeDishman]

If you're trying to figure out the structure of the Earth as a whole, interior as well as surface, then yes, you need to drop a dimension to represent things on a piece of paper. But such a representation--for example, a "cross section" of the Earth showing the various interior portions, crust, mantle, core, etc.--isn't usually called a "map".

Excellent analogy Peter, I'll try to answer pervect's question in those terms.

What sort of description are you looking for? Any visual representation you can model in an Euclidean three-dimensional space is not going to capture every aspect of a curved, dynamical, four-dimensional geometry that is the mathematical description of a gravitational wave.

What I was trying to do was to represent the effect of GW on a surface which is a sphere round the source in a manner similar to depicting the motion of surface ocean currents on the Earth by drawing the motion of freely-moving floats on a globe while ignoring any bobbing up and down, so there is no attempt to map a sphere onto a plane. I could do this physically by drawing some sort of vector displacement field on a beach ball but a 3D animation should do it just as well. What I had read about gravitational waves before I started this was that they were transverse with no component in the direction of propagation (which I took to mean the radial direction when thinking in spherical coordinates centred on the source provided you are far enough away). Dropping any radial motion seemed therefore not to be a problem (but pervect's discovery of a z component complicates that).

That was all looking fine until I thought about a second identical sphere half a GW wavelength closer to the source. To put that in terms of the analogy, thinking about a similar map of currents at a fixed depth (again over the whole surface of the globe) and what I found is that comparing surface current with the current at depth gives a stupid answer. In GR terms though, I'm sure the problem isn't real, it's an artefact arising from my bad choice of coordinates so now I'm trying a different representation based on your suggestion of the transverse traceless approach. I just want to be sure I'm not "squeezing the balloon" in the sense of resolving one problem but creating another in some other aspect. To do that, I only need reasonable confidence that the common motion (not the measured differential effect) of the C-D pair of masses in my diagram is less than the few mm of compliance over which the suspension provides isolation.

My reluctance to go into details on the artefact is because I don't want crackpots taken my post out of context. Contact me privately if you want that aspect.

 

[RockyMarciano]

Then you evidently haven't read much about congruences. A non-rigid congruence is any congruence that has nonzero expansion or shear. See here for a brief description of the terminology:
https://en.wikipedia.org/wiki/Congruence_(general_relativity)

See for instance: "The Rich Structure of Minkowskii Space" http://arxiv.org/abs/0802.4345
"Born Rigid flow and the ADS-CFT correspondence" http://arxiv.org/abs/1010.3847
"A Modern view of the classical Herglotz-Noether theorem" https://arxiv.org/abs/1004.1935
and from the same paper, one of the less techinical definitions of Born rigidity

Sure, for timelike congruences that's understood. But the context of my post was Pervect's discussion of proper distances, and if I'm not missing something obvious these are computed from spacelike paths, so that's why I referred to spacelike congruences and since the spacelike geometry is obviously four-dimensional euclidean geometry it is in that sense that I said it was obviously rigid.

Now of course timelike congruences in GR are a different story, but I'm not seeing how they relate to the discussion about distances. I can see how it relates to the so called "expanding space" point of view and its "non-rigid" distances, but this point of view cannot be used to determine distances. So I don't know how the claim of Pervect that one can use the non-rigid congruence to measure distances can be sustained. It would need to use spacelike paths which are rigid, and there woud be no unique path between two spacetime points due to GR's curvature, and the only way to pick one would be coordinate dependent.

 

[PeterDonis]

the context of my post was Pervect's discussion of proper distances, and if I'm not missing something obvious these are computed from spacelike paths

They're computed from spacelike geodesics between neighboring worldlines in the same congruence, that are orthogonal to the worldlines.

that's why I referred to spacelike congruences

Which are irrelevant because there is no spacelike congruence involved. The spacelike geodesics given above do not form a well-defined congruence in the general case, because they will cross at some finite distance from the neighboring worldlines. And even if they did, the properties of that congruence are not the properties we are interested in; we are interested in the properties of the congruence of worldlines, i.e., the timelike (or null) congruence.

timelike congruences in GR are a different story

Um, what? Timelike congruences in GR are exactly what we are discussing, and the only thing we are discussing.

this point of view cannot be used to determine distances

Sure it can. See above. There is no requirement that the timelike congruence be rigid for the above method to work.

It would need to use spacelike paths which are rigid

This makes no sense. The "rigid" property (which is not even required, see above) does not apply to the spacelike paths. It applies to the timelike congruence.

there woud be no unique path between two spacetime points due to GR's curvature

This is true, but it has nothing to do with the rigidity or lack thereof of any congruence.

the only way to pick one would be coordinate dependent

Yes, that's correct. The method I described above is equivalent to constructing a particular coordinate chart--in the general case, it's basically Fermi normal coordinates centered on a particular chosen worldline in the congruence.

 

[RockyMarciano]

They're computed from spacelike geodesics between neighboring worldlines in the same congruence, that are orthogonal to the worldlines.

But can one compute proper distances from spacelike paths if they are not unique in curved spacetimes and therefore there are no unique spacelike geodesics between neighboring timelike worldlines, at least in the absence of a certain family of coordinates with time orthogonality?

And If the orthogonality of the spacelike geodesics with respect to the worldlines in the timelike congruence depends on the coordinate choice I'm not seeing how it can be used to compute invariant quantities.

On the other hand timelike congruences are not completely free from the problem of crossing at some finite distance(singularity theorems).

This is true, but it has nothing to do with the rigidity or lack thereof of any congruence.

But it has to do with the issue about proper distances.

Yes, that's correct. The method I described above is equivalent to constructing a particular coordinate chart--in the general case, it's basically Fermi normal coordinates centered on a particular chosen worldline in the congruence.

So how is any invariant obtained from having to pick a certain set of coordinates?

 

[PeterDonis]

can one compute proper distances from spacelike paths if they are not unique in curved spacetimes

Certainly; you just pick which spacelike paths you are going to use. In a practical sense, there is usually some physical way of picking out the ones you want to use--in this case, for example, we are picking the spacelike curves that are orthogonal to the worldlines in the congruence.

Also, bear in mind that the non-uniqueness here only shows up in particular cases; it is not the case that there are multiple spacelike geodesics between every single pair of events in a curved spacetime. So if we are only interested in a small enough region of spacetime, we can often find unique spacelike geodesics within that region.

If the orthogonality of the spacelike geodesics with respect to the worldlines in the timelike congruence depends on the coordinate choice

It doesn't. Whether or not a given pair of curves are orthogonal at a given event is an invariant, independent of your choice of coordinates.

 

[RockyMarciano]

Certainly; you just pick which spacelike paths you are going to use. In a practical sense, there is usually some physical way of picking out the ones you want to use--in this case, for example, we are picking the spacelike curves that are orthogonal to the worldlines in the congruence.

Also, bear in mind that the non-uniqueness here only shows up in particular cases; it is not the case that there are multiple spacelike geodesics between every single pair of events in a curved spacetime. So if we are only interested in a small enough region of spacetime, we can often find unique spacelike geodesics within that region.
It doesn't. Whether or not a given pair of curves are orthogonal at a given event is an invariant, independent of your choice of coordinates.

At a given event. Wich takes us back to the discussion in the previous thread, where your question was: how is this property of worldlines and spacelike geodesics at a point extended to the globality that covers the congruence?

 

[PeterDonis]

how is this property of worldlines and spacelike geodesics at a point extended to the globality that covers the congruence?

And the answer is, that might not be possible, depending on the congruence. Pervect's posts have been exploring how closely this goal can be met for the case of a gravitational wave detector like LIGO.

 

[pervect]

My concern about GeorgeD's map is that I expect he wants to draw them to scale. The issue is - it's mathematically impossible to do so exactly, though one can do it over a small region.

I'm not sure how to describe the magnitude of the error precisely without math, except to say that the effect becomes signficant compared to the already noted stretching motion of the rings when one approaches a wavelength.

If one is willing to drop the notion of drawing the map to scale, there's an especially simple map that doesn't need a fancy diagram. This is a latice of points that don't move at all. On the side, one puts a note saying "this map is not to scale - in fact, the scale changes with time". Perhaps one illustrates the change of scale in the map by stretching a rubber sheet, as several of the lecturers already did.

 

[GeorgeDishman]

My concern about GeorgeD's map is that I expect he wants to draw them to scale.

Not at all. It's a generic representation and the two cases I'm considering are first HM Cancri which is 1600 light years away (10¹⁶ km) and then GW150914 at an estimated range of 410 Mpc (1.3 billion light years or 10²² km). That is represented by the radius of the animation sphere and of the animated annulus in the orbital plane version. I'll probably show 2 or 3 cycles of the GW which have a wavelength of 161 light seconds (4.8*10⁶ km) for HM Cancri or about 5 light milliseconds for GW150914 (1500 km) but that forms the annulus and is shown covering perhaps quarter of the radial distance so it is clearly on a very different scale. The text will make that explicit eventually.

The surface lateral displacement in my early version of the sphere was large enough to make the displacement visible but that again represents a scaled up version of the actual displacement of the test masses from their location in the absence of the GW in the version where there is movement (in the coordinate system I was using which I think this thread has confirmed is inappropriate). Making a displacement of 75 microns visible on a sphere of 3200 light years diameter again cannot be done to scale.

In the newer version based on your advice on the "transverse traceless" approach, nothing moves over the surface of the sphere, the test masses (indicated by the nodes on fine black grid) are static while the anisotropy in the speed of light responsible for the LIGO detector output is indicated by the colouring.

The time is also scaled of course, the animations will take around 30 seconds for 3 cycles regardless of the true period which should be easy to follow on the screen.

The final two remaining questions I need to resolve are:

• (a) should I show a sine wave displacement of the test masses in the orbital plane or not? That's a yes or no question, if so the size will be shown as just be enough to be visible.
• (b) If there is some motion transverse to the direction of propagation, what is formula for the magnitude of that displacement in these coordinates? That will only be included in the text for the two examples and can be quite approximate, within a few orders of magnitude would be more than adequate for my purposes.

Those can I think be related to the "three LIGOs in space" simplification.

If one is willing to drop the notion of drawing the map to scale, there's an especially simple map that doesn't need a fancy diagram. This is a lattice of points that don't move at all.

That is what I hope I can apply, it solves my problem, but while scaling motion can change its amplitude, it doesn't change "moving" into "static", hence question (a) is yes/no.

[edit p.s.]I feel I have to apologise again for the somewhat circuituous way I'm posing this question, from sad experience, I'm only too aware of how an incautious comment can be quoted out of context by those of malicious intent and the world is sadly overstocked with crackpots.

 

[RockyMarciano]

And the answer is, that might not be possible, depending on the congruence. Pervect's posts have been exploring how closely this goal can be met for the case of a gravitational wave detector like LIGO.

AFAIK, any timelike congruence guarantees the possibility of a global orthogonal decomposition using projection tensor , so it confuses me the claim that it might not be possible or that is something to be explored.

 

[pervect]

Sure, for timelike congruences that's understood. But the context of my post was Pervect's discussion of proper distances, and if I'm not missing something obvious these are computed from spacelike paths, so that's why I referred to spacelike congruences and since the spacelike geometry is obviously four-dimensional euclidean geometry it is in that sense that I said it was obviously rigid.

Now of course timelike congruences in GR are a different story, but I'm not seeing how they relate to the discussion about distances. I can see how it relates to the so called "expanding space" point of view and its "non-rigid" distances, but this point of view cannot be used to determine distances.

It is true that distances are ultimately calculated along space-like paths. The question is, as always, which path does one measure the length of to determine the distance?

The time-like congruence can be regarded as one way of specifying the path- when one specifies the congruence, and one specifies that distances (which are measured along some space-like path ) should be measured orthogonal to the congruence. It can also be regarded as a way of specifying "an observer", or as a means of specifying the necessary simultaneity convention to split space-time into space+time.

Applications for this technique include cosmology, where the cosmological notion of "proper distance" can be regarded as the notion of distance associated with the special time-like congruence called the "hubble flow", the congruence of observers from which the universe appears isotropic.

Measuring distance along non-rigid congruences is not only possible, it's quite common, as the cosmological example illustrates.

 

[PeterDonis]

AFAIK, any timelike congruence guarantees the possibility of a global orthogonal decomposition using projection tensor

No, this is not correct. Any timelike congruence with nonzero vorticity is an obvious counterexample, since there is a theorem (the Frobenius theorem, IIRC) that states that no such congruence can be hypersurface orthogonal. For a specific concrete example, consider the Langevin congruence in flat spacetime--i.e., the congruence of observers who are at rest on a rotating disk. There is no way to foliate Minkowski spacetime by a family of hypersurfaces that are orthogonal to all of the worldlines in this congruence.

 

[RockyMarciano]

No, this is not correct. Any timelike congruence with nonzero vorticity is an obvious counterexample, since there is a theorem (the Frobenius theorem, IIRC) that states that no such congruence can be hypersurface orthogonal. For a specific concrete example, consider the Langevin congruence in flat spacetime--i.e., the congruence of observers who are at rest on a rotating disk. There is no way to foliate Minkowski spacetime by a family of hypersurfaces that are orthogonal to all of the worldlines in this congruence.

We've covered this before, your example is again the stationary but not static case and the timelike vector is a Killing vector. The decomposition I was talking about is the one described in https://en.wikipedia.org/wiki/Congruence_(general_relativity): "Note that this orthogonality relation holds only when X is a timelike unit vector of a Lorenzian Manifold. It does not hold in more general setting."

 

[RockyMarciano]

It is true that distances are ultimately calculated along space-like paths. The question is, as always, which path does one measure the length of to determine the distance?

The time-like congruence can be regarded as one way of specifying the path- when one specifies the congruence, and one specifies that distances (which are measured along some space-like path ) should be measured orthogonal to the congruence. It can also be regarded as a way of specifying "an observer", or as a means of specifying the necessary simultaneity convention to split space-time into space+time.

Applications for this technique include cosmology, where the cosmological notion of "proper distance" can be regarded as the notion of distance associated with the special time-like congruence called the "hubble flow", the congruence of observers from which the universe appears isotropic.

Measuring distance along non-rigid congruences is not only possible, it's quite common, as the cosmological example illustrates.

This is correct. I just had in mid a different notion of rigidity. In any case the concept of proper distance obtained with the timelike congruence is not without issues.

 

[PeterDonis]

your example is again the stationary but not static case and the timelike vector is a Killing vector.

No, it isn't, it's a timelike unit vector field that happens to have the same integral curves as a Killing vector field.

the concept of proper distance obtained with the timelike congruence is not without issues.

The "issues" you are raising are not issues at all, and if you keep on persisting with your mistaken claims about timelike congruences you will receive a warning.

 

[RockyMarciano]

No, it isn't, it's a timelike unit vector field that happens to have the same integral curves as a Killing vector field.

From https://en.wikipedia.org/wiki/Born_coordinates: "the Langevin congruence is stationary"
If this that shows where I got my claims about timelike congruences is also wrong I am glad to receive a warning but maybe someone should correct the wikipedia page too.

 

[PeterDonis]

From https://en.wikipedia.org/wiki/Born_coordinates: "the Langevin congruence is stationary"
If this that shows where I got my claims about timelike congruences is also wrong I am glad to receive a warning but maybe someone should correct the wikipedia page too.

You are very confused. The statement that the Langevin congruence is stationary (but not static) is true. What is not true is your apparent belief that that means the timelike vector field whose integral curves make up that congruence is not composed of unit timelike vectors. Not to mention your apparent belief that if we did use timelike unit vectors, that would guarantee that the congruence was hypersurface orthogonal. Or your apparent belief that Killing vector fields are somehow involved in the stationary case but not the static case.

Please, before posting further on this topic, take some time to learn what the terms involved actually mean and what their implications are and are not.

 

[RockyMarciano]

You are very confused. The statement that the Langevin congruence is stationary (but not static) is true. What is not true is your apparent belief that that means the timelike vector field whose integral curves make up that congruence is not composed of unit timelike vectors. Not to mention your apparent belief that if we did use timelike unit vectors, that would guarantee that the congruence was hypersurface orthogonal. Or your apparent belief that Killing vector fields are somehow involved in the stationary case but not the static case.

Please, before posting further on this topic, take some time to learn what the terms involved actually mean and what their implications are and are not.

Nothing of what you say after the first sentence was said or implied by me. I'm leaving the site for good.

 

[PeterDonis]

Nothing of what you say after the first sentence was said or implied by me.

Here are some quotes from you:

any timelike congruence guarantees the possibility of a global orthogonal decomposition using projection tensor

Note first that the "orthogonality relation" referred to in what you quoted from Wikipedia (in the next quote below) is local, not global. So having a congruence generated by a timelike unit vector field does not guarantee global hypersurface orthogonality.

your example is again the stationary but not static case and the timelike vector is a Killing vector. The decomposition I was talking about is the one described in https://en.wikipedia.org/wiki/Congruence_(general_relativity): "Note that this orthogonality relation holds only when X is a timelike unit vector of a Lorenzian Manifold. It does not hold in more general setting."

If you did not mean here that the stationary but not static case did not use a timelike unit vector field, and that the use of a Killing vector field was involved in the stationary case but not the static case, then this quote makes no sense. If that really wasn't what you meant, then you need to take some time to learn the proper terminology, because it is what the words you used meant, at least to anyone familiar with the field.

 

[GeorgeDishman]

The final two remaining questions I need to resolve are:

• (a) should I show a sine wave displacement of the test masses in the orbital plane or not? That's a yes or no question, if so the size will be shown as just be enough to be visible.
• (b) If there is some motion transverse to the direction of propagation, ...

Ok, I figured this out by considering another thought experiment. The result is that the answer to (a) must be "no", all the test masses must be motionless. Question (b) therefore does not apply.

Thanks ever so much to all for the patient assistance, I am now at the point where I can return to developing the animations and thinking how I can write this up into something that can be understood.

 

[PeterDonis]

the answer to (a) must be "no", all the test masses must be motionless.

I don't think this is correct. I haven't been following all the details of your exchange with pervect, but a couple of items should be noted:

(1) The "transverse-traceless" approach only applies in a small patch of spacetime in which the gravitational wave can be idealized as a purely transverse plane wave. It certainly can't be applied in a global coordinate chart that includes an entire sphere at some distance from the source.

(2) If you are trying to visualize the whole gravitational wave in a global coordinate chart that includes an entire sphere at some distance from the source, you cannot simply assume that the entire wave front is just a sphere (or annulus) to which the transverse plane wave in a small patch, described in #1 above, is a local approximation. In other words, you cannot assume that the wave amplitude at a given radius from the source, at a given time in a chart in which the source is at rest, is the same at all angular coordinates around a sphere at that radius. (I'm not positive that you can even assume this everywhere on a circle of a given radius in the source's orbital plane.)

 

[GeorgeDishman]

I don't think this is correct. I haven't been following all the details of your exchange with pervect, but a couple of items should be noted:

(1) The "transverse-traceless" approach only applies in a small patch of spacetime in which the gravitational wave can be idealized as a purely transverse plane wave. It certainly can't be applied in a global coordinate chart that includes an entire sphere at some distance from the source.

I think the local conditions can be iterated around the equator and perhaps over the whole surface again creating the global view by overlapping small regions but it is certainly a point where I have an ongoing concern. However, any problems are significantly smaller than in the alternative.

(2) If you are trying to visualize the whole gravitational wave in a global coordinate chart that includes an entire sphere at some distance from the source, you cannot simply assume that the entire wave front is just a sphere (or annulus) to which the transverse plane wave in a small patch, described in #1 above, is a local approximation. In other words, you cannot assume that the wave amplitude at a given radius from the source, at a given time in a chart in which the source is at rest, is the same at all angular coordinates around a sphere at that radius. (I'm not positive that you can even assume this everywhere on a circle of a given radius in the source's orbital plane.)

I totally agree, as you go round the circle, the phase of the GW changes and remember the GW has half the period of the binary orbit. I've attached the diagram I included some time ago which shows how two cycles fit round the "equator".

Based on what has been said, what I am suggesting is that we should not think of test particles "moving together then apart" but instead staying static with an equivalent variation of the speed of light causing the variation in the interferometer output.

The thought experiment that helped me on this is to imagine a variant of the RingWorld concept.

Usually that is shown with living space on the inner surface, which presupposes rotation to create artificial gravity through "centrifugal force". Instead, think of a non-rotating RingWorld with suspended test masses hanging towards the star by suspensions that leave them free to move in the direction along the ring, then make the star a hard binary. How do the masses move relative to the material of RingWorld if that is nearly rigid (as close as GR allows)?

Sorry, I just drew that image by hand so it's a bit rough but I'm sure you'll get the idea.

To avoid the complexity of the material behaviour, think of three sets of test masses half a GW wavelength apart in radius and ignore the RingWorld material. Do the middle set of masses move relative to the top and bottom sets (which must act in unison). The Ringworld material is shown at the top this time in cross section with three test masses, one from each set.

A little thought should show that either the middle ring moves in opposition to the outer two with equal magnitude or none move at all.

Do you have a metric that can describe this setup?

 

[PeterDonis]

I think the local conditions can be iterated around the equator and perhaps over the whole surface again creating the global view by overlapping small regions but it is certainly a point where I have an ongoing concern.

I have a concern about this too, as I've expressed before. Unfortunately I'm not familiar enough with the field to know what, if any, math has been developed to deal with this.

 

[GeorgeDishman]

Let me put it another way. Suppose that, relative to one of the mass suspension points (mass 0), the mass beneath it has some motion x₀=f(t) in the direction of the ring. Because the effects come from nothing more than the rotation of the binary, another mass n farther round the ring by θ must have the same motion but delayed relative to the first mass, x_n=f(t(1-2θ)) The separation of two adjacent masses is what LIGO measures and we can think of a LIGO between between every adjacent mass. I think this argument is the same as the definition of distance in the Hubble Law, as explained in Ned Wright's tutorial:

http://www.astro.ucla.edu/~wright/cosmo_02.htm#MD

D_now = D(us to Z) = D(us to A) + D(A to B) + ... D(X to Y) + D(Y to Z)​

If θ is small then the beam length L is small in comparison to the circumference and the strain dL/L in the limit must be the derivative of the displacement. Now I'm not sure that the usual symmetry holds in GR but naively I have been assuming that if the measured strain is the derivative of the displacement, then I can calculate the displacement by integrating the strain.

Incidentally, if I use Wright's diagram locally to illustrate the "expanding space" version of the light in the LIGO beam tube, the fact that "the lightcones must tip over" as he puts it means the speed of the light isn't quite c for the whole length relative to an end, but that's a second order effect so not significant, just a curiosity.

 

[pervect]

Do you have a metric that can describe this setup?

Currently, no :(

I've been doing some reading, but no answer yet - I may start a technical thread on the issue.

 

[GeorgeDishman]

Currently, no :(

I've been doing some reading, but no answer yet - I may start a technical thread on the issue.

That would certainly be interesting but probably beyond my present mathematical level. I do think I've got enough of an understanding to put the rambling conversation together as a more coherent write-up now, if only at a qualitative level, but I'll have to find the time between some DIY tasks to make an attempt at that.

I am very grateful for the effort you've put in already in helping me get past my mental logjam, thank you.

 

[pervect]

I'm curious myself about some of the issues that were raised. For quite a while I've thought of a metric (or line element) as defining the space-time geometry. In this case, though, I realized I don't have one and so far I haven't found one. So it's an interesting question in and of itself.

I have a feeling we'd differ about how well the metric / line element can be represented in 3d - but I think I've already explained my concerns about that as well as I can.