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

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1. Jul 9, 2016

### GeorgeDishman

First let me be clear, I am not questioning GR or the detection of gravitational waves by LIGO, I am trying to improve my own understanding of GW to the point where I can offer a graphic illustration (web video) showing what they do as they pass us to help others understand them. I started this a couple of years ago and thought I had a handle on it but have hit a problem with the LIGO parameters that tells me I have a misunderstanding somewhere. I don't know how to use GR mathematically (I have only a qualitative grasp) so I can't resolve the problem myself, and it probably comes from some remnant of Newtonian thinking that I've not recognised yet.

OK, the starting point was a pair of Wikipedia animations of the effects of GW:
https://en.wikipedia.org/wiki/File:GravitationalWave_PlusPolarization.gif
https://en.wikipedia.org/wiki/File:GravitationalWave_CrossPolarization.gif

The two particles at the left and right end can be thought of roughly as the two test mass mirrors in one LIGO arm while the top and bottom particles would be the mirrors in the second arm.

Those animations are just in a plane, to see how that is related to a wave, you need a 3D version like this:
http://www.einstein-online.info/images/spotlights/gw_wavesI/cyl_plus
Combining the polarisations might give this:
http://www.einstein-online.info/images/spotlights/gw_wavesI/gw_elliptic

http://www.einstein-online.info/spotlights/gw_waves

The Wikipedia images can be thought of as a slice from the end of the einstein-online 3D tubes.

Those the effect of GW on a "ring of test particles" and I wanted to extend that to show the effect on a complete geodesic sphere enclosing a binary system system. The effect over any small region would be given by the equations in this part of the article (it's about the Earth-Sun system but the equations should be general):
https://en.wikipedia.org/wiki/Gravitational_wave#Wave_amplitudes_from_the_Earth.E2.80.93Sun_system

I intend to build up my understanding in steps so the simplest beginning was to consider a number of copies of the Wikipedia ring of particles with the horizontal ends sharing a particles (all are considered free-falling). It seemed to me that if one ring expanded, that next to it would have to expand too, alternating rings like this couldn't make sense:

However, if all the rings are the same, then there is not only distortion but also displacement as shown by the coloured lines relating three of the particles. The red particle is used as a fiducial datum but see later:

That view seemed to be borne out when, some months later, I saw this explanation by Rainer Weiss in the LIGO press conference. The link should take you to 23:06 and you only need to watch 45 seconds or so:

Now if we extend Rainer's mesh into a ribbon that runs completely around the GW source at uniform radius, there is an obvious problem, how can the whole circumference expand or contract? The answer is that it doesn't. The frequency of the waves is double the orbital frequency of the binary system because, in the simplest case of equal masses, it doesn't matter which star is which. That means if we see expansion then at the same time an observer on the opposite side from us but at the same radius would also see expansion while observers half way round the circle would see contraction like this:

The numbers in the top left corner are for HM Cancri, the fastest nearby binary system. Now the expanded regions are balanced by contracted regions so the proper length round the circumference is constant.To find the displacement at any location relative to our fiducial point, we need to integrate the strain around the circumference. The strain is h sin(2θ) so the integral should be -h/2 cos(2θ) but that implies no part of the circumference is static.

Putting that together as a video (the start of my ultimate goal) gives me this image of Rainer's mesh extended all the way round (not to scale, think of it as light years round, 1m high and displacements of microns):

Again, it's only 30 seconds long so please watch it, the text will make a lot more sense if you do.

So why don't we measure the displacement? My view was that the two LIGO test masses can be thought of as lying at two adjacent points on the circumference and the beam tube is essentially the chord joining them. Since the beam tube and even the whole Earth are also displaced, there is no relative motion to be detected.

It should be clear that one point will be affected slightly before the other as the band of compression (say) moves round the circumference, hence what is measured can be thought of as the distance variation caused by the slight phase difference between the masses displacements.

In case anyone thinks I'm suggesting superluminal motion of the compression band round the circumference, I'll add this similar video showing how that is an illusion and what is really happening is a transverse wave propagating out from the binary system at the speed of light:

OK, I've typed enough, can anyone tell me where have I gone wrong or is this OK so far?

Last edited: Jul 9, 2016
2. Jul 9, 2016

### Staff: Mentor

You are missing a crucial point about the Wikipedia animations and the description of gravitational waves that they come from: that description is a linearized approximation, valid only in a small region of spacetime far from the source in which the small piece of the whole GW that is passing can be idealized as a perfectly transverse plane wave.

That means the following:

(1) You can't extend the "rings" indefinitely in any direction, because as you do, the direction in which the plane waves are propagating will change (think small pieces of an expanding spherical wavefront at slighly different angles from the source). That means the linearized approximation breaks down and the visualization you are using is no longer applicable.

(2) You can't visualize a big "ring" all the way around the source this way, because such a ring is obviously not a small region of spacetime far from the source in which a small piece of the whole GW can be idealiszed as a perfectly transverse plane wave. If you want to describe the global configuration of GWs on some sphere that completely encloses the source, you have to do it some other way; the linearized approximation is not valid and the visualization you are using is no longer applicable.

3. Jul 9, 2016

### GeorgeDishman

I suspect I'm missing more than one.
That is true, the plane wave is an approximation to a variation of the distances on the surface of a sphere, but I understand the GR equations are differentials which need to be integrated, and if I take the limit as the ring size tends to zero then the chord length (which is the plane approximation) tends to the same as the arc length which would be the accurate version. Think of it as a piecewise linear approximation to a circle then taking the limit as the chord length goes to zero.
The direction of propagation is perpendicular to the plane of each circle so each distance should still be valid as an approximation which becomes perfect when we take infinitesimals. However I can see that there would be a radial discrepancy using the plane wave approximation, but I think that is the nature of the approximation itself.
OK so that's where I need the help from those members of the forum who understand GR. I've integrated the widely published equations at constant radius round the orbital plane and what that gives me seems logical but the numbers don't work out in the context of an "Earth frame" so what are the equations should I use instead? As I've said, I'm not someone who thinks GR is wrong, my aim is to produce a set of graphics like the ones you've seem that shows how a complete geodesic sphere of test masses would be displaced by the GW from a binary system. Finding out how to generate the graphics is the immediate task so if I can get the equations from a text book or web reference in polar coordinates, that will let me move on. Just point me in the right direction please.

4. Jul 9, 2016

### Staff: Mentor

Not just that. It is a linearized approximation to the nonlinear Einstein Field Equation. And the linearized approximation breaks down when you try to use it to cover too large an area of spacetime or a region in which spacetime curvature is too large.

I'm not sure what you mean by this. If you are referring to the "line element", the usual expression for the metric, yes that is written in the form of differentials because it describes a differential line element--the spacetime "length" of an infinitesimal curve between two events that are very close together. To get a "length" over a curve between two events at finite separation, you integrate the line element along the curve. But that only covers a very small part of what GR is used for.

If you are referring to the fact that the Einstein Field Equation is a differential equation, yes, that is true. But the primary thing that is done with the EFE in GR is to solve it, not to "integrate" it.

I already understand how to get a linear approximation to a circle's chord. I'm pointing out that this method only applies in a limited set of scenarios, and the scenarios you are trying to consider (such as a "ring" going all the way around a source of GWs) are outside that limited set.

As a bunch of individual approximations, each only applying to one particular small piece of spacetime, yes. But you are trying to put them all together into a single picture that covers more than a small piece of spacetime, but still assuming that all the laws of the linearized approximation will hold. That doesn't work.

There would, yes, but that is not the only discrepancy. See above.

I don't know, because I don't know of any global solution describing a complete spacetime containing a source emitting GWs and the GWs themselves, over a large region, that can be expressed as closed-form equations. The only solutions of this kind that I'm aware of are numerical ones. You might look into the papers on binary pulsars to see if they give any descriptions of the kinds of numerical methods they're using; those are probably the most developed numerical GW solutions in existence.

5. Jul 9, 2016

### pervect

Staff Emeritus
I think this is where your problems start. If I'm understanding correctly (and I may not be), you are assuming that the gravitational waves are emitted in a spherically symmetric fashion from the central source.

But the problem lacks spherical symmetry. The problem has a cylindrical symmetry. The inspiral occurs in some spatial plane, the orbital plane of two large masses m1 and m2 which are orbiting each other. If we draw a sphere around this, the plane of the inspiral will define an "equator". The response of the gravitational wave detectors on the equator will differ from the response of the detectors on the poles.
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You can see this lack of spherical symmetry in the Wiki equations in the above article for $h_+$ and $h_x$, which represent the amplitude of the two components of the gravitational waves (GW's). The total amplitude of the GW would be $\sqrt{h^2_+ \,+\, h^2_x}$. Setting $\theta = 90^{\circ}$ corresponds to what I'm calling the equator, $0^{\circ}$ and $180^{\circ}$ correspond to what I'm calling the poles. If you inspect the above equations, you'll see that the amplitude of the components (and the total amplitude of the GW's) depends on the lattitude.

Not a good idea - trying to have multiple centers in this manner is going to lead to confusion.

There is displacement, yes. I would go so far as to say that that's what Ligo is measuring - it compares the displacements between the two arms.

How I would describe what I think you're doing is that you're trying to describe the motion of the test particles relative to a rigid frame of reference.

The good news is that (assuming that you're doing what I think you're doing) there is an exact mathematical description of this viewpoint, it would be called a Born-rigid congruence. This viewpoint inherently has a "center" though. Roughly speaking, in this viewpoint we have one central object, and we define the motion of all nearby objects by saying that they are "at rest" relative to the central object, i.e. that their distance to the central object doesn't change. Under some conditions (including no rotation, though I won't get into the details), the objects at rest relative to the central object are also at rest relative to each other, and we have a rigid frame. Usually there are some fundamental constraints on the size of such a rigid frame, we don't expect it to cover all of space-time, just some local area that's "large enough" for some particular purpose we have in mind.

I believe you're trying to get rid of this center of your rigid frame. A possible motivation for what you're trying to do might be this. We've established that the GW's and the metric do depend on "lattitude", but does not depend on the longitude. So we might want a viewpoint that also doesn't depend on longitude. But the problem is - the "rigid frame" viewpoint, though highly intiutive, doesn't give us a viewpoint with these properties. The rigid frame viewpoint has a center.

So basically you're attempting to take a perfectly valid local viewpoint, that works great in a local area, and which has a center, and leverage it into a global viewpoint that doesn't have a center. The problem is that the local viewpoint just doesn't leverage in this manner.

The existence of a center is a property of the viewpoint and one that can't be gotten rid of by drawing "multiple circles", and attempting to create multiple centers. The whole structure collapses when you try to do this :(. To follow the "rigid frame" idea, you need to stick with one center, that defines your viewpoint, and realize that the viewpoint itself has limits. To expand those limits requires a different starting viewpoint.

The good news is that while the viewpoint has limits, it should be perfectly adequate to encompass the entire Earth.

I won't talk about alternatives to rigid frames - it's already confusing enough I think. But if I'm totally off track on what you're trying to do, I suppose that we'd have to discuss just what it is that you're trying to do to get anywhere.

Last edited: Jul 9, 2016
6. Jul 10, 2016

### GeorgeDishman

No, I am assuming that the simplest form is axisymmetric and that would apply only for a specific configuration of the spins of the binary components. It might be that they are not rotating or perhaps tidally locked or possibly any rotation as long as the angular momentum vectors of the components are in the same direction as that of the orbit.

In general, I appreciate that there is no symmetry at all. For example, thinking about the merger phase, for a binary orbiting in the x-y plane, GW can be preferentially emitted in the -z direction giving the resultant product a kick in the +z direction, and the earlier inspiral must show the same effect though to a much lesser extent.

What I am modelling is not a fully general solution but only the simplest subset.
Exactly. I'm starting on the equator but the aim is to extend that to show how the two polarisations combine to alter the distortion as the 'latitude' varies.
Yes, that is what I felt was confirmed by Rainer Weiss's presentation, although obviously that is very much aimed at the public.
That's precisely what I'm trying to do, at least for the animations.
Not at all, sorry if I gave that impression. The viewpoint for the movies is a Born rigid sphere whose centre is the barycentre of the binary system. The example I started with about a year ago was HM Cancri which is 1600 light years away (but I'm ignoring the other stars in the galaxy of course) so it would be a non-rotating sphere of 1600 light years radius.
Yes, and eventually I hope to model that variation with latitude, but I hit a problem just on the equatorial picture when I started thinking about how the "Earth frame" that was mentioned in the previous thread would relate to the picture relative to the Born rigid picture so I'm trying to resolve that before proceeding.
You've got 99%, it just needed the clarification that the centre of the Born-rigid sphere is the barycentre of the binary system, and I appreciate the solution is axisymmetric. It is perhaps also worth noting that I'm only using the rigid sphere as a viewpoint, essentially a set of fiducial points that have no input to the physics. Applying this to the real world will then note that the Earth is also an object that is free to be displaced and to distort in response to the GW, that's where my problems arise.

I think my concern at this stage would be to take Peter's point and question whether it is valid to sum (integrate) a large number of width changes to get the total displacement relative to the rigid reference sphere. I'm just starting to learn MATLAB to do the modelling so I have used a simple square mesh in my video, but each cell of that can be likened to Wikipedia ring of test particles, and for size perhaps think of two adjacent nodes on the mesh being the test masses in LIGO.

My understanding was that the Wikipedia images show a ring of particles in a plane relative to a fiducial point in the centre. My series of rings treats each ring as being in a slightly different flat plane, only the end particles of adjacent rings are common to both rings. Integrating the end particles separations round the equator suggests to me that we can think of each whole ring as also translating round the 'equator', but of course that isn't detectable locally. For example an infinitely sensitive accelerometer in space would always show zero even though it was being displaced 'east' and 'west' in a sine wave the particles are all following geodesics (other than ignoring the low decrease of radius).

P.s. I should say I am also considering the situation of a steadily orbiting system. HM Cancri for example is several hundred thousand years from merging (IIRC) and has an orbital period of 321.5 seconds so I am ignoring evolution of the orbit and GW.

Last edited: Jul 10, 2016
7. Jul 10, 2016

### Staff: Mentor

I don't think that's what pervect meant when he brought up Born rigidity (he will correct me if I'm wrong, of course). I think he was talking about a Born rigid congruence describing a local reference frame in which the linearized approximation holds.That certainly isn't true for a large, 1600 light year sphere centered on some binary system. (Even for the localized case there are issues with trying to make use of Born rigidity, since the test masses in a localized GW detector do not form a Born rigid congruence--if they did they could not detect GWs.)

Moreover, the binary system itself is not Born rigid and can't be described by a Born rigid congruence.

8. Jul 10, 2016

### pervect

Staff Emeritus
I basically agree with Peter's comments.

We can find an approximately Born-rigid congruences in a sphere around, say, the Earth, which is a large distant away from the gravitational wave source, where we make the approximations (linearizations) that Peter is talking about. Whether or not we can find such a scheme that works around the strongly-gravitating source is problematical.

I've had some discussions (and perhaps even arguments) here on PF recently over similar points, and found what looks like a good reference on this point. http://arxiv.org/pdf/gr-qc/0511041.pdf. But it may be rather advanced.

The GW source, does not have constant curvature.

Based on this paper, I now believe that we cannot (as others have suggested) find a perfectly Born-rigid congruence for a GW with the usual (Fermat) definition of the spatial metric (i.e. the usual definition of distance). The theorem mentioned by Cartan would seem to prevent it. The authors of the above paper suggest one way around this difficulty by modifying our notion of distance so that we can find an exactly "rigid" metric and congruence (using their new defintion which is very similar to the old one.)

But more importantly, however, by taking advantage of the fact that the we describe GW's by a linearization process as a pertubation of some flat spatial metric $\eta_{\mu\nu}$, we can find an "approximately rigid" congruence by ignoring the perturbative part of the gravity wave to create an approximately rigid frame. In another thread I've done some calculations for the "almost rigid" congruence. No expansion would represent ideal rigidity (we can ignore shear for now, as it turned out to be zero). The expansion measured by Ligo for the freely-floating test masses (which are not rigid, we are measuring the distance changes!) is on the order of $10^-21$ or so - barely detectable with our best equipment. The expansion of the "almost rigid" congruence would be second order, i.e. 10^-42 or so. So our approximation is more than good enough to draw a diagram, even if a mathematician might tell you, quite honestly, that there was no mathematical solution that was perfectly rigid.

Going from this approximate notion far from the source to saying that we can do the same in the strong-field region around the GW source is a bit of a stretch. I'm rather skeptical, I'd need to see the numbers. I rather doubt anyone has tried, though. One might get an idea of how accurate such a scheme would be by looking at the metric close-in (where you want to draw your diagram) and looking at how large the pertubations from flatness are. The one thing that might save you is that the approximations are so good distant from the source that they still might be "good enough" up closer. However, I don't have the data to do so, and I'm not sure if it's available or not.

So basically my suggestion is to consider a different problem - not around the source, but around the receiver. I'm pretty sure that's the intended use of the various illustrations you cite - not a sphere around the source, but rather a sphere around the receiver.

You can view the source as a curved 4-dimensional geometry using the "block universe" approach. But visualizing this directly doesn't seem promising.

...

9. Jul 10, 2016

### pervect

Staff Emeritus
Some more technical details of a new calculation: the last analysis I did was only 1 space + 1 time, at this point I felt that it would be productive to check the full 3space 1time case. I used the following metric for a linearized plus polarized plane GW:

$$g = -dt^2 + (1+2 \,k \,f(t-z))dx^2 + (1-2\,k\,f(t-z))dy^2 + dz^2$$

This is a plane wave metric far from the emission source - it's approximate because the metric itself is only accurate to linear order (terms of order k), and it's approximate because we've assumed the GW is a plane wave, whereas the actual wavefronts would have a spherical symmetry.

k represents the strength of the wave, a small dimesionless constant of approximately 10^-21 or so for Ligo. f and it's derivatives are considered to be not large, are x,y, and z.

Then I looked at the expansion and shear for the geodesic congruence (i.e. a congruence of freely floating test masses). For the given metric/line element, geodesic curves are just curves where x,y, and z have constant coordinate values. Formally, we can write the congruence as $\partial_t$. This calculation is for informational purposes - we know the test masses are moving relative to each other, that's what Ligo is measuring. We calculate it to have some idea of the significance of the numbers we calculate. The expansion has terms of order k^2, the shear has terms of order k. So the presence of a shear term of order k means that the GW's stretch our congruence by a factor of k, as in the illustrational videoes with the mesh being pulled apart, and the GW's change the area by a smaller amount, a factor of k^2.

Next I looked at the "approximately rigid" congruence. This attempts to compensate for the motion of the test masses in what in retrospect is an obvious manner - basically one assumes that the changing separation in the x direction is due to some velocity v in the x direction, a velocity which is proportional to the distance from some refererence point which is taken as "not moving". One adds a similar velocity in the y direction. Formally we can write this congruence as:

$$\partial_t -k\,\frac{\partial f(t-z)}{\partial t} \,x\,\partial_x + \frac{\partial f(t-z)}{\partial t} \, y \, \partial_y$$

After normalizing the above vector to unit magnitude, we find (with the aid of a computer to do the scut work) that the expansion of the almost-rigid congruence is of order k^3 and the shear is of order k^2. So both expansion and shear were multiplied by a factor of k, making them "small". Informally, my argumemnt for why this works is that the GW metric is "almost Minkowskian", and the "almost Minkowskian" property gives rise to an "almost rigid" congruence.

[add]There's some issues. It appears there are some linear-order terms in the xz and yz components of the shear tensor. So it's not quite accurate to second order :(

Last edited: Jul 12, 2016
10. Jul 11, 2016

### GeorgeDishman

This is just a quick note, I've downloaded the paper you cited and I need to mull over your other post too, but I think what I am trying to do might be similar to turning the above into polar coordinates and then solving in 2 space + time for constant radius (i.e. z -> R = constant).

This next bit is honestly beyond my level but I think roughly, for a binary a long time from merger (compared to the orbital period), what I am doing is treating the waves far from the binary as a perturbation on the Schwarzschild metric where the mass is something like the ADM mass but excluding the energy outside radius R. I think this is similar to the "chirp mass" of the system.

Also, I think you're on the right track when you say "Based on this paper, I now believe that we cannot (as others have suggested) find a perfectly Born-rigid congruence for a GW with the usual (Fermat) definition of the spatial metric (i.e. the usual definition of distance).". That is probably the key to my problem

I may also be able to reformulate the question into a small volume containing the Earth but unfortunately it will still have to relate back to the bigger picture.

All the external sources I've cited are for a small region far from the source and illustrate the local strain for the linear, plane wave solution. The two Vimeo animations are my own and represent the picture with the binary in the centre. If you look at just one cell in the cylindrical mesh, that relates to all the other commonly available animations.

Last edited: Jul 11, 2016
11. Jul 11, 2016

### GeorgeDishman

Is your 'k' the same as LIGO's 'h' term?
That puzzles me a bit, I would have thought "stretch" related to "expansion" rather than shear so I guess I'm unfamiliar with the terms here. Could you add a little guidance on that please.
That definitely confuses me, I thought any volume would be 'stretched' in say x but squished in y so that the area remained constant. That seems to fit the animations and it made sense if there is no z variation because the waves are akin to a tidal effect and I thought tidal variations preserved volume. If our small test ring of particles having radius r is distorted by a strain of k, it should be an ellipse with semi-major axis a=kr and semi-minor axis b=r/k. The area is πab so remains πr2.

The more I learn, the less it seems I know (which I take as a good thing BTW, exposing one's lack of knowledge is the first step to learning).

Last edited: Jul 11, 2016
12. Jul 11, 2016

### GeorgeDishman

This is extreme guesswork as is no doubt obvious, but:
• the waves can have different rates, should f(t-z) be f(ωt-2π(z/λ)) where ω is the angular frequency and λ is the wavelength?
• can that metric be adjusted to take account of the phase around the equator of the circle with the source at the centre by using f(ωt-2φ-2π(z/λ)) where φ is the azimuth in spherical coordinates (hence dx=R dφ and dy=R dθ with z≡R) ?
P.S.
I agree and I'm not trying to do that, I only want to model a spherical shell with a thickness of just a few AU and at a uniform radius no less than a thousand light years from the source. Any sort of strong field or "near field" effects (analogous to EM) are well beyond my scope.

Last edited: Jul 11, 2016
13. Jul 11, 2016

### Staff: Mentor

If what you mean by polar coordinates is polar coordinates centered on the source, this will not work. The metric pervect gave is only valid in a small piece of spacetime far from the source in which the GW can be idealized as a purely transverse plane wave. That will obviously not be true in polar coordinates centered on the source.

14. Jul 11, 2016

### GeorgeDishman

I agree completely, I have only been talking about a Born rigid congruence in the sense of a set of fiducial points for reference formed as a sphere around the binary at uniform radius but nowhere have I suggested that the plane wave equations would be able to describe the bigger picture.
Of course not, nothing physical in this scenario is Born rigid.
So do I.

The only purpose of introducing the concept of the Born rigid congruence is to provide a notional reference against which true geodesics can be measured. Ultimately, the only measurables are between physical objects (test masses, the Earth or eLISA satellites), each of which can hopefully be described via deviation from this common reference congruence. If we know how A and B each move relative to R, it should be possible to find out how A moves relative to B.
Sorry, I should have said spherical coordinates.

You are quite right again, but I have nowhere suggested using that metric which is defined for plane waves over a sphere. What I think is needed is a different metric defined in spherical coordinates but which is locally asymptotic to the one pervect quoted. I've already suggested a possible modification but I doubt it is correct, I am just hoping it is something that can prompt some constructive suggestions as to how to resolve the problem.

15. Jul 11, 2016

### Staff: Mentor

But this "notional reference" doesn't exist, so you can't measure anything against it.

16. Jul 11, 2016

### GeorgeDishman

Caught me, yes that was a bit of bad typing. I said it correctly a few lines later:
Nothing gets measured relative to R, A only gets measured relative to B, but R can be a useful intermediate in calculating the predicted motions. This goes back to the previous thread where we noted that a "fiducial point" does not have a physical significance:

17. Jul 11, 2016

### Staff: Mentor

Ah, I see, you are interpreting "R" as a coordinate (or something that gets defined relative to a coordinate chart), not an actual object.

18. Jul 11, 2016

### pervect

Staff Emeritus
If we look at some of the material and diagrams posted by the OP, for instance: https://en.wikipedia.org/wiki/File:GravitationalWave_PlusPolarization.gif we can ask the question: what motivates these diagrams as a 2-d cross section of a GW?

If our notional rigid reference frame doesn't exactly exist, what motivates them?

I've suggested variously that they could be motivated by looking at the appropriate tangent space, or they could be motivated by a frame that is rigid in the unperturbed flat space-time (but is only aproximately rigid in the actual GW space-time).

19. Jul 11, 2016

### Staff: Mentor

The way the metric looks in the transverse-traceless (TT) coordinate chart, which is the one in which you wrote down the linearized metric. In that chart, each little dot in the diagrams has constant spatial coordinates (x, y, z), so the only source of variation in the physical distance between them is the variation in the correction terms in $g_{xx}$ and $g_{yy}$. The variation in physical distance is what the diagrams portray.

This is more or less what the TT coordinate chart does--constant spatial coordinates in this chart basically amounts to "would be rigid if spacetime were unperturbed" (at least to linear order).

20. Jul 11, 2016

### pervect

Staff Emeritus
But in those coordinates, the test masses follow geodesics, which means that x, y, and z are all constant. So if we take a slice z=0, and we plot the coordinates x and y for a set of test masses in this coordinate system, we would represent the motion of the test masses by points on the diagram with constant coordinates - which makes for a boring diagram where the points don't move.

So the first thought that comes to my mind is that we are plotting the motion of the slice (z=0), but we are not plotting it in a coordinate basis, but in an orthonormal basis with unit vectors $\hat{x}$ and $\hat{y}$ given in terms of the TT coordinates as $\hat{x} = \sqrt{1+2kf(t-z)} \, \partial_x \approx (1+kf(t-z)) \, \partial_x$ and $\hat{y} = \sqrt{1-2kf(t-z)} \, \partial y \approx (1-kf(t-z)) \, \partial_y$. But this interpretation means we are plotting the points not on the manifold, but in the tangent space, where $\hat{x}$ and $\hat{y}$ are defined.

A second thought is that there should be some actual coordinate basis, different from the ones I wrote down, where the motion looks like the diagram, and that these coordinates should be physically interesting.