I Modeling the effects of GW and the "Earth Frame"

[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.