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

[GeorgeDishman]

Thanks for replying, the result surprises me but as I said, I'm not at all familiar with the workings of GR so this is fascinating.

At a peak chirp frequency of 300 hz, the Ligo burst can have wavelengths as small as a million meters, or 1000 km. So it's problematical to define an rigid frame the size of the Earth, though it's not problematical to define such a frame the size of Ligo.

Thank you, that is the problem I had seen albeit in my invalid approach.

.. it turns out (I believe the reason can be traced to the relativity of simultaneity) that when we do this, as we must to have a rigid flow, $\nabla_z u^a$ will not be zero.

Basically, we can choose our congruence to make $\nabla_x$ (and the equivalent in the y direction) zero at or near z=0, but due to the time and space variation of our metric, the choice that makes it zero at z=0 doesn't make it zero at different values of z. If we restrict the range of concern to a region of "small change in z", where the GW amplitude doesn't change much, we have a reasonable approximation to a rigid flow. If we don't restrict our range, we don't have a reasonable approximation to a rigid flow. So the shape of our almost-rigid region is more like a pancake than a sphere - it's limited to the dimension of the GW in the direction of propagation of the GW, but is large in the other two directions.

Yes, understood.

This is way too much work, ..

Also understood.

.. and is totally irrelevant to the Ligo result anyway, because the whole purpose of the Ligo apparatus is to isolate the test masses from the environment of the Earth.

That they are isolated is true but the range over which the isolation is effective is limited to a small fraction of a wavelength of the interferometer laser, so if the motion of the masses is too large then it will exceed the ability of the system to remain locked. That displacement limit is what I am looking at.

However, I take your point, the materials aspect makes this much more onerous so perhaps a simplified approach based on the space version might be more amenable to analysis (below).

Conceptually these issues wouldn't matter much to GW detection of what I've been calling "Ligo in space", because there's no problem with the size of the frame for that problem, most especially if you orient the detector optimally. Then it's easy to talk about how the test masses move relative to the beam tube, because we can approximate it in the familiar context of a rigid frame.

How about three copies of LIGO-in-space separated in the z direction by λ/2, would it be possible to say how that would behave? I've attached a sketch showing three tubes as black line pairs with red test masses (not to scale, λ≈2000km, tube length=4km).

Each tube individually, with its pair of masses, should be within a "pancake" region. In that view, I think the associated tubes would stretch and shrink, equivalent to the masses moving relative to the tube ends in the more usual representation. Since the tubes containing masses A-B and E-F are a wavelength apart, I am hoping we can define a congruence in which all four are always at rest and the distances A-E and B-F are also constant.
The question then is what happens to masses C and D and their associated tube? I think there should be a third "pancake region" covering them so, within that, the relative motion of the masses and tube should be as for the other two regions (just 180 degrees later), but is your calculation able to say how those two masses move relative to the congruence in which the other four are always at rest?

P.s. I'm also assuming we are simplifying by calculating for a constant wave frequency whereas the LIGO detection swept from a wavelength greater than the diameter of the Earth to 3 times smaller in less than 20ms, and that we are just working with plane waves.

 

[pervect]

How about three copies of LIGO-in-space separated in the z direction by λ/2, would it be possible to say how that would behave? I've attached a sketch showing three tubes as black line pairs with red test masses (not to scale, λ≈2000km, tube length=4km).
View attachment 103376
Each tube individually, with its pair of masses, should be within a "pancake" region. In that view, I think the associated tubes would stretch and shrink, equivalent to the masses moving relative to the tube ends in the more usual representation. Since the tubes containing masses A-B and E-F are a wavelength apart, I am hoping we can define a congruence in which all four are always at rest and the distances A-E and B-F are also constant.
The question then is what happens to masses C and D and their associated tube? I think there should be a third "pancake region" covering them so, within that, the relative motion of the masses and tube should be as for the other two regions (just 180 degrees later), but is your calculation able to say how those two masses move relative to the congruence in which the other four are always at rest?

P.s. I'm also assuming we are simplifying by calculating for a constant wave frequency whereas the LIGO detection swept from a wavelength greater than the diameter of the Earth to 3 times smaller in less than 20ms, and that we are just working with plane waves.

My gut reaction is that you could create a different congruence for each tube in the stack, but they would be slightly different, having different simultaneity conventions. So you wouldn't have a coherent picture of the whole stack sharing a single, common, convention of simultaneity.

 

[GeorgeDishman]

My gut reaction is that you could create a different congruence for each tube in the stack, but they would be slightly different, having different simultaneity conventions. So you wouldn't have a coherent picture of the whole stack sharing a single, common, convention of simultaneity.

If the assumption that the outer four masses which share the same wave phase can all be at rest at all times in this congruence, the question of simultaneity can be of lesser concern. Each of the z, y and z components of the C and D masses would, to first order, be a sine wave so all we need to characterise the motion is the three amplitudes and perhaps phases of those components. In terms of Earth-frame related limits, those amplitudes are probably a good proxy for the maximum local relative displacement. Is it possible to find out those amplitudes from your calculation? I'm sure the differential mode numbers (D relative to C etc.) are as obtained from the classic calculation but it's C relative to A and E that is of interest, especially in the X direction of course.

 

[pervect]

If the assumption that the outer four masses which share the same wave phase can all be at rest at all times in this congruence, the question of simultaneity can be of lesser concern.

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

What we want to do is measure fractional distance changes on the order of 10^-21 or so of, say, 3km or so. That's about 3*10^-18 meters, which, if we use relativistic methods to measure the distance based on light propagation, demands synchronization accuracies on the order of 10^-26 seconds over that 3km region.

So we wind up having to worry about non-intuitive seemingly very small effects due to the relativity of simultaneity if we wish to use relativistic based methods to measure the distance.

Perhaps you have some other idea of how to measure distance at the back of your mind? I can't really say, but I can say that if you want to compare whatever method, notion, or definition of distance you might have in mind to relativistic based method based on light travel time, you'll need synchronizations of the stated accuracy to do it in a meaningful manner.

And a lot of the mathematics has been in support of determining the theoretical existence of rigid congruences with the necessary syncrhonization accuracies.

One way of viewing the whole congruence issue is as of a means to specify a synchronization convention. "Specifying an synchronziation convention" sounds rather abstract, but physically specifying one specifies the other.

On a somewhat related note, it's perfectly reasonable to use non-rigid congruences to measure distances, but you'll get different results than you do if you use rigid ones. The "expanding space" point of view in fact does use non-rigid congruences to define distances, and it turns out to not make much difference. But in order to determine that it doesn't make any difference, it's helpful to compare the distance measured by a rigid congruence to the distance measured by a non-rigid one. But in order to do that, first one needs a rigid congruence to exist in the first place, otherwise you don't have anything to compare against. Fortunately, in this problem, the needed congruence does exist over a useful region.

 

[RockyMarciano]

On a somewhat related note, it's perfectly reasonable to use non-rigid congruences to measure distances, but you'll get different results than you do if you use rigid ones. The "expanding space" point of view in fact does use non-rigid congruences to define distances, and it turns out to not make much difference. But in order to determine that it doesn't make any difference, it's helpful to compare the distance measured by a rigid congruence to the distance measured by a non-rigid one. But in order to do that, first one needs a rigid congruence to exist in the first place, otherwise you don't have anything to compare against. Fortunately, in this problem, the needed congruence does exist over a useful region.

I think you are getting too far away from what is mathematically meaningful here. At least I'm not aware of any notion of "non-rigid" congruence in Pseudo-Riemannian geometry. Congruences in general are just integral curves defined by vector fields, perfectly rigid and nothing changes in the GR case of spacelike congruences.

 

[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.