I Gravitational wave time variation

[andrew s 1905]

I have tried to discover if the local time as well as the local space is varied by the passage of a gravitational wave. I have seen animations and discussion of the effects of gravitational wave on space and test particles but can't find a reference to the changes in the time component of space-time.

Thanks in advance for any illumination on this topic.

 

[PeterDonis]

Changes in "space" are changes in "time"--you just have to look at them in a different frame.

To put it another way, there is no such thing, in any invariant sense, as a change that is just a change in "space" (or just a change in "time"). All changes are changes in spacetime. For GWs, we focus on talking about these changes as changes in "space" because that's how we model our measuring devices--we model them as measuring tiny changes in the lengths of the arms of an interferometer. But that model implicitly adopts a particular frame; it doesn't mean that the changes are just "changes in space" in any absolute sense. The focus on "space" in description is a convenience, to help us with visualizing and modeling what is going on; that's all.

 

[Lino]

Can I post my read of the OP's question in a slightly different way in order to ask for clarification: If the GW is causing a measurable distortion of space (and the frequencies being detected), and the total distortion is in space-time, how does one determine the "space" only element in order to be able to affirm the changes and what these changes mean in terms of mass and distance? To put it another way, if the wave is passing through the detector and distorts distance in one space dimension, why does the distortion in time not enhance or diminish the result, and in the worst case scenario for the experiment, one cancel the other out?

Thanks in advance,

Noel.

 

[PeterDonis]

how does one determine the "space" only element

By picking a particular set of coordinates. There is no such thing as "the space only element" in any absolute sense. How much of the wave's effect is "changes in space" and how much is "changes in time" depends on your choice of coordinates.

if the wave is passing through the detector and distorts distance in one space dimension

It can't. More precisely, there are no coordinates you can pick in which the distortion of distance caused by the wave will only be in one space dimension. The distortion will always be in at least two space dimensions.

why does the distortion in time not enhance or diminish the result

Because, once again, "distortion in time" and "distortion in space" are not absolutes; they're just artifacts of how you pick your coordinates. What the waves cause is "distortion in spacetime", and there's no way to make that distortion go away by choosing coordinates. It's there. All you can do by choosing coordinates is pick what "angle" in spacetime you are viewing the distortion from, so to speak. From one "angle", it looks like a distortion in two space dimensions, transverse to the direction of wave propagation; from another "angle", it could look like a distortion in two or three space dimensions, combined with a distortion in the time dimension. But there's no way to choose coordinates in which there's no distortion at all.

 

[Lino]

... There is no such thing as "the space only element" in any absolute sense. ... ... "distortion in time" and "distortion in space" are not absolutes; they're just artifacts of how you pick your coordinates. What the waves cause is "distortion in spacetime", and there's no way to make that distortion go away by choosing coordinates. ...

Thanks for the reply Peter.
Most of this makes sense, but if I may confirm ... LIGO has picked a set of coordinates by building the experiment (3d and time), and so they are measuring the wave from a specific angle. Is that correct?

Also, am I correct in thinking that none of the coordinates / angles interfere with each other?Thanks again,

Lino.

 

[PeterDonis]

LIGO has picked a set of coordinates by building the experiment (3d and time)

More precisely, the experimental apparatus of LIGO defines coordinates in its local region of spacetime, and the apparatus is only set up to detect distortions that are "spatial" in those coordinates, and along only two spatial dimensions.

they are measuring the wave from a specific angle.

At whatever angle the wave is coming in, relative to the orientation of the apparatus. That angle does not depend on any choice of coordinates.

am I correct in thinking that none of the coordinates / angles interfere with each other?

I'm not sure what you mean by this.

 

[pervect]

It may be a bit picky, but I wouldn't say the Ligo apparatus itself picks out a particular coordinate system. The human analyzing it do that.

Usually synchronous coordinates are used to describe gravitational waves. Since by definition they have $g_{00}$ constant, they could be described as having no variation in the metric components that describe time. Only the metric components which describe space, i.e. $g_{ij}, i,j>0$, vary.

For gravitational waves, synchronous coordinates have some nice features, like the ability to describe the whole space-time of the wave, and they're the coordinates most of the popularizations attempt to describe in imprecise informal language.

This is overall a very good description of things, but it leads to some confusion when trying to figure out problems like 'what does a passing gravitational wave do to a a material object such as a human body'. The lay reader tends to assume that rulers and clocks directly give the coordinate values, at least to a reasonable degree of approximation. This is unfortunately not the case. While the synchronous coordinates provide a good mathematical description of what's going on, even idealized rulers will appear to shrink and stretch in these coordinates. One might even say that that's the whole point of these coordinates, and the point of the explanations of the gravitational waves - but it causes a lot of confusion when one tries to understand what happens to physical bodies as a result of these waves.

It's possible to set up coordinates where idealized rulers do not shrink and stretch, i.e. Fermi normal coordinates. (See for instance
http://arxiv.org/abs/gr-qc/0010096 for a highly technical example of the mathematical details). Unfortunately these sorts of coordinates can only describe a small region of space-time. But they do so in a way that's a lot more intuitive, as idealized rulers do not stretch in these coordinates. Since we are used to rulers not stretching, that makes the physical consequence of passing gravity waves much easier to understand. In these coordinates, one has tidal forces, tidal forces that that would stretch bodies by differing amounts depending on their rigidity. Thus, marshmallow's would stretch a lot, a platinum bars - not so much, though they'd stretch some. Ligo, by the way, has free-floating test masses, so the stretch is maximized - there is nothing to oppose it.

So in summary, one can describe gravitational waves being "ripples in space" and this is a correct view (and the view that the popularizations try to convery), but the view isn't compatible with our normal view of physics in which rulers do not stretch. It is possible to use math to present a different view of gravitational waves in which rulers do not stretch - this view is convenient for describing how the waves interact with physical objects, but is inconvenient for other purposes (such as describing the entire space-time geometry of the wave).

 

[PeterDonis]

I wouldn't say the Ligo apparatus itself picks out a particular coordinate system

It defines one in the sense that the center of mass of the apparatus has a particular worldline, and the directions of the arms pick out particular spatial directions in the hypersurface that is orthogonal to the CoM worldline at a particular event. So basically the apparatus can be used to set up Fermi normal coordinates centered on the CoM worldline. Whether or not a human chooses to use these coordinates is, of course, a different question. There is also the issue that these coordinates might not be the same, or even a good approximation, to the synchronous coordinates in which GWs are most simply described, at a particular event.

 

[Ibix]

There is also the issue that these coordinates might not be the same, or even a good approximation, to the synchronous coordinates in which GWs are most simply described, at a particular event.

What are the synchronous coordinates, strictly speaking? Presumably something like "stationary with respect to the black holes". But what does stationary mean over cosmological distances? That I've parallel transported the velocity w.r.t the CMB of the black holes' center of mass along the worldline of the gravity wave? Or just the same velocity in co-moving coordinates? Or something else?

 

[PeterDonis]

What are the synchronous coordinates, strictly speaking? Presumably something like "stationary with respect to the black holes".

No; they're coordinates that are transverse to the planes of constant phase of the waves. A given system of coordinates will only be synchronous in this sense in a local region of spacetime; it's not a global concept. (In fact the concept doesn't make sense at all except in regions far enough away from the source that the waves can locally be considered to be plane waves.)

 

[Ibix]

No; they're coordinates that are transverse to the planes of constant phase of the waves. A given system of coordinates will only be synchronous in this sense in a local region of spacetime; it's not a global concept. (In fact the concept doesn't make sense at all except in regions far enough away from the source that the waves can locally be considered to be plane waves.)

Doesn't this leave me some freedom of choice? The surface of constant phase of gravitational waves is a light cone. There is a space-like plane picked out (locally, as you say) by the plane of the wave (say x-y). But don't all observers who have no motion in the x-y plane have freedom to choose their z and t directions?

For context, I was reading your these coordinates might not be the same, or even a good approximation, to the synchronous coordinates to mean that I could build another LIGO, accelerate it to relativistic speed perpendicular to the gravitational waves and would find that they were not purely spatial. This seems to follow from the notion that I can pick coordinates such that (at some event) the perturbation $h_{\mu\nu}$ is purely spatial, but an observer in relative motion at that same event would see $\Lambda_{\mu'}{}^{\mu}\Lambda_{\nu'}{}^{\nu}h_{\mu\nu}$, which would not be purely spatial.

Since these paragraphs seem to me to be contradictory, I deduce that I'm misunderstanding something.

 

[PeterDonis]

don't all observers who have no motion in the x-y plane have freedom to choose their z and t directions?

Yes; just as with EM waves, a pure boost in the z-t plane will only change the observed frequency of the waves; it won't change the fact that they're purely spatial.

an observer in relative motion at that same event would see $Lambda_{\mu'}{}^{\mu}\Lambda_{\nu'}{}^{\nu}h_{\mu\nu}$, which would not be purely spatial.

It would be if the transformation matrix $\Lambda$ was a pure boost in the z-t plane and $h_{\mu \nu}$ was purely in the x-y plane. All that would change is the observed frequency and wave number of the waves (because of contraction/dilation of the distance between the planes of constant phase in the new frame as compared to the old one). They would still be purely spatial in the x-y plane. See below.

I was reading your these coordinates might not be the same, or even a good approximation, to the synchronous coordinates to mean that I could build another LIGO, accelerate it to relativistic speed perpendicular to the gravitational waves and would find that they were not purely spatial.

No. As above, two LIGOs in relative motion but with the proper orientation to the incoming waves would both measure them to be purely spatial; they would just measure different frequencies and wave numbers. The local inertial rest frames of both of these LIGOs would be considered "synchronous coordinates".

However, we also need to clear up a point regarding what LIGO actually measures. LIGO only measures the spatial part of any incoming GW. If LIGO happens to be oriented in spacetime in such a way that its arms are not purely transverse to the incoming waves, then what will happen is that LIGO will only measure part of the total wave--the part that is purely spatial in LIGO's local frame. Any portion of the wave that ends up being temporal instead of spatial in the LIGO local frame will not be measured by LIGO. (It could be measured by other detectors, but not by LIGO.) So if we have two LIGOs that are oriented differently relative to an incoming GW, they could measure different amplitudes, as well as different frequencies and wave numbers if they also happened to be in relative motion. The different amplitudes would show that at least one of the LIGOs was not oriented transverse to the incoming waves, meaning that its local inertial coordinates were not synchronous. But LIGO would not give any direct measurement of the "time variation" due to its local coordinates not being synchronous.

 

[Ibix]

Thanks, Peter. I'm doing a small amount of self-kicking because some of that is obvious and I got caught up in the tensors and forgot about the physics (immediately before citing Feynman's sticky beads in another thread, which was apparently an argument prompted by exactly that issue).

So, the plane of constant phase picks out (locally) a plane which I called x-y. The direction of motion of the wave picks out a velocity in that plane that you can call "at rest". The polarisation of the waves pick out a sensible (non-unique) choice of x and y directions, parallel to the maximum stretch-and-squish.

If I have a LIGO interferometer "at rest" with its arms parallel to x and y then it is maximally sensitive. If I rotate it away from this the sensitivity falls off because the arms are no longer aligned with the maximum stretch-and-squich. If I start it moving in the x-y plane (i.e. rotate it in the x-t and/or y-t plane) then the sensitivity falls off because some of the stretch-and-squish is in the interferometer's time-like direction. There are zeroes of sensitivity under the purely spatial rotations, and in the limit as the x-y velocity tends to c. The former is why we're planning on building 4+ interferometers, which will all have different orientations which will, presumably, cover each other's blindspots.

Velocity in the z direction doesn't affect sensitivity except that it may Doppler shift the signal outside the instrument's frequency range.

Is that about right?

 

[PeterDonis]

Feynman's sticky beads in another thread, which was apparently an argument prompted by exactly that issue

Yes. His frustrated comment was something like "this is what comes of looking for conserved tensors, etc., instead of asking, can the waves do work?"

Is that about right?

Looks right to me, yes.

 

[RockyMarciano]

Changes in "space" are changes in "time"--you just have to look at them in a different frame.

To put it another way, there is no such thing, in any invariant sense, as a change that is just a change in "space" (or just a change in "time"). All changes are changes in spacetime. For GWs, we focus on talking about these changes as changes in "space" because that's how we model our measuring devices--we model them as measuring tiny changes in the lengths of the arms of an interferometer. But that model implicitly adopts a particular frame; it doesn't mean that the changes are just "changes in space" in any absolute sense. The focus on "space" in description is a convenience, to help us with visualizing and modeling what is going on; that's all.

There is no such thing as "the space only element" in any absolute sense.

However, we also need to clear up a point regarding what LIGO actually measures. LIGO only measures the spatial part of any incoming GW.

I think this explanation might be confusing. If there is no invariant sense in which a change is only a change in space, no such thing as a "space only element", how is it possible to measure(and measurements are usually considered as invariants) only a spatial part if there is no such thing in any invariant sense? It looks as if you were saying clocks and rulers only inform of "artifacts of how you pick your coordinates" and therefore wouldn't measure invariants, since they measure time only and space only elements. I guess this is not what you are saying but I can't figure out what you mean.

 

[PeterDonis]

I think this explanation might be confusing.

Yes, sorry about that. It's because ordinary language is confusing when used for this purpose. Math is much more precise and unambiguous.

The confusion here is that, while the split of spacetime into "space" and "time" has no invariant meaning (it depends on your choice of coordinates), the property of vectors being "spacelike" or "timelike" (or null) does have invariant meaning. Also, the property of a particular spacelike vector at a given event being orthogonal to a particular timelike vector at a given event is invariant. The word "spatial" can be used to refer to the latter properties.

So a more accurate, but more cumbersome, way of saying what I was trying to say about LIGO is this: pick an event on the worldline of the laser source/interferometer at the base of LIGO. The 4-velocity of the source/interferometer at that event defines a timelike vector. There will be a particular pair of spacelike vectors which are orthogonal to that timelike vector and which point along the directions of LIGO's two arms; this pair of vectors defines a spacelike plane. Those three vectors together then pick out a unique fourth (spacelike) vector which is orthogonal to all three. These four vectors together form what is called a "tetrad".

Now consider an incoming GW which passes through LIGO. At our chosen event, the wave vector of the GW (assuming it can be approximated as a plane wave) is a null vector, and there is a unique spacelike plane that is orthogonal to it. If that spacelike plane happens to be exactly parallel to the spacelike plane defined by LIGO's arms at that event, as described above, then the GW amplitude detected by LIGO will be the maximum possible. But if, as is more likely, the spacelike plane orthogonal to the wave vector is at some angle to the plane defined by LIGO's arms, then the GW amplitude detected by LIGO will be smaller, by an amount depending on the angle between the two planes. This is the sense in which LIGO only detects the portion of the amplitude that is "purely spatial" in its own rest frame.

Similarly, the projection of the null wave vector of the GW onto the timelike 4-velocity of LIGO at the chosen event determines the frequency of the GW that LIGO detects; and the projection of the null wave vector onto the fourth vector of the tetrad described above determines the wavelength of the GW that LIGO detects.

 

[Lino]

... Now consider an incoming GW which passes through LIGO. At our chosen event, the wave vector of the GW (assuming it can be approximated as a plane wave) is a null vector, and there is a unique spacelike plane that is orthogonal to it. If that spacelike plane happens to be exactly parallel to the spacelike plane defined by LIGO's arms at that event, as described above, then the GW amplitude detected by LIGO will be the maximum possible. But if, as is more likely, the spacelike plane orthogonal to the wave vector is at some angle to the plane defined by LIGO's arms, then the GW amplitude detected by LIGO will be smaller, by an amount depending on the angle between the two planes. This is the sense in which LIGO only detects the portion of the amplitude that is "purely spatial" in its own rest frame.

Similarly, the projection of the null wave vector of the GW onto the timelike 4-velocity of LIGO at the chosen event determines the frequency of the GW that LIGO detects; and the projection of the null wave vector onto the fourth vector of the tetrad described above determines the wavelength of the GW that LIGO detects.

Thank you Peter (and everyone),

This does make sense to me, not so much that I could explain it again but enough to inform my direction of reading :)

But it does kind of lead me back to the original question ... (I'll try again but please forgive my clumsy language) if the spacelike plane orthogonal to the wave vector is at some angle to the timelike 4-velocity at the chosen event, will the frequency the LIGO detects be impacted (ie the frequency reduced) ... or does the fact that you have described the wave vectors and timelike 4-velocity as "null", mean that there is no impact on the spacelike plane defined by LIGO's arms at that event?

(And if the later "or" question is correct, how come the timelike 4-velocity can be separated from spacetime like this? It reads counter to what you have been saying previously. Please note that I appreciate that this question is likely to be the result of my lack of understanding of null vectors and timelike 4-velocities, so the answer to this one could be as simple as "it is possible - read this reference".)

Thanks in advance,

Noel.

 

[Ibix]

The frequency is c divided by the distance between successive peaks. This isn't affected by the spatial orientation of the detector. However, two detectors moving at different speeds in the direction the wave propagates will see different frequencies. You can read this as the Doppler effect or as the detectors having different spacetime orientations. This isn't related to the null character of the gravitational waves' wave vector - the same is true of a microphone detecting a sound wave. If you turn your microphone you may be better able to detect a sound, but its frequency only changes if you or the source are in motion.

The sound wave analogy works quite well for your last paragraph, too. The waves aren't separate from spacetime. What they do is let me define a direction without reference to anything else. I can talk about "the direction the wave is propagating" without mentioning north, south, sideways, or anything else. The wave "picks out" a direction and, implicitly, a perpendicular plane. A microphone also has a direction in which it is most sensitive, which likewise picks out a direction without reference to anything but the microphone. The reaction of the microphone to the sound wave depends on the relative orientation of the two "picked out" vectors. There's nothing special about the two vectors in any general sense, but they are important in the specific case of figuring out how the detector will react to the signal.

Life is more complex in 4d spacetime, and there are flaws in the analogy because of this and because a sound wave is a wave in a medium. For example LIGO's sensitivity depends on its velocity perpendicular to the direction of the wave. But many of the basic concepts carry across.

 

[PeterDonis]

if the spacelike plane orthogonal to the wave vector is at some angle to the timelike 4-velocity at the chosen event, will the frequency the LIGO detects be impacted (ie the frequency reduced)

Yes, because if the plane orthogonal to the wave vector is at an angle to LIGO's 4-velocity, the projection of the wave vector onto that 4-velocity will be different than if the two planes were parallel.

does the fact that you have described the wave vectors and timelike 4-velocity as "null", mean that there is no impact on the spacelike plane defined by LIGO's arms at that event?

The wave vector is null, but the 4-velocity of LIGO is not null, it's timelike. Also, "null" means "lightlike", not "no impact".

Changing the "angle" of the plane orthogonal to the wave vector with respect to LIGO's 4-velocity also changes the angle of that plane with respect to the plane of LIGO's arms, because the plane of LIGO's arms is orthogonal to LIGO's 4-velocity. So the one "angle change" affects the frequency, wavelength, and amplitude that LIGO measures.

 

[RockyMarciano]

The confusion here is that, while the split of spacetime into "space" and "time" has no invariant meaning (it depends on your choice of coordinates), the property of vectors being "spacelike" or "timelike" (or null) does have invariant meaning. Also, the property of a particular spacelike vector at a given event being orthogonal to a particular timelike vector at a given event is invariant. The word "spatial" can be used to refer to the latter properties.

So a more accurate, but more cumbersome, way of saying what I was trying to say about LIGO is this: pick an event on the worldline of the laser source/interferometer at the base of LIGO. The 4-velocity of the source/interferometer at that event defines a timelike vector. There will be a particular pair of spacelike vectors which are orthogonal to that timelike vector and which point along the directions of LIGO's two arms; this pair of vectors defines a spacelike plane. Those three vectors together then pick out a unique fourth (spacelike) vector which is orthogonal to all three. These four vectors together form what is called a "tetrad".

Thanks, this is clearer. However it's not exactly what I had in mind, which I think is more in tune with the OP's and Lino's doubts.
I think it helps to compare how the Michelson interferometer basically measures phase shifts in the case of arms moving due to a passing GW as opposed to in response to any other cause.
We have that light speed is constant in both scenarios and that in the case of the GW but not in the rest of cases, we have, additionally to the presence of the change in the interferometer arms distance that produces the phase shift, a temporal change in the frequency of the laser, and a wavelength change due to the spatial change since the passing GW is a spatiotemporal perturbation, unlike any other cause that brings as a result a change of length between the interferometer's arms.

So maybe a way to understand what seems to puzzle the OP and many others judging by how frequently this comes up is that the additional frequency and wavelength changes of the laser due to the spacetime perturbation, that are absent when measuring phase shift from causes other than GWs, cancel each other out, leaving only to detect the phase shift due to the spacelike changes in the arms as explained by PeterDonis.
At least this is how I make sense of this frequently raised issue(the LIGO FAQ addresses it but it doesn't work for me), now does this makes sense to anyone else?

 

[PeterDonis]

We have that light speed is constant in both scenarios

No, we don't. We have that light moves on null worldlines. But in a curved spacetime, the coordinate speed of light, which is what you are implicitly thinking of, might not be equal to c, even though light always moves on null worldlines.

In the flat background spacetime being used to analyze the GWs, the coordinate speed of light is c, yes. But in the region affected by the GWs, it isn't.

I think a fundamental issue you are having here is that you are continuing to think of "spatial variation", "temporal variation", and "variation in the speed of light" as different things, one or more of which can be present in any given scenario. That kind of reasoning doesn't work for GWs; it can work for EM waves only because EM waves can be modeled in perfectly flat spacetime in some chosen inertial frame, where "variation in the speed of light" is absent and there is a well-defined global split between "spatial variation" and "temporal variation". But even in the case of EM waves, if you try to model them in curved spacetime, you come up against the same issues as you do for GWs: there is no longer a well-defined global split between "space" and "time", and the coordinate speed of light is no longer constant. And in the case of GWs, there is no way to model them in perfectly flat spacetime; the background spacetime can be flat, but the waves are waves of spacetime curvature, so there's no way to avoid spacetime curvature in dealing with them.

the additional frequency and wavelength changes of the laser due to the spacetime perturbation, that are absent when measuring phase shift from causes other than GWs, cancel each other out

No, they don't. The frequency and wavelength of the same GW, measured by two different LIGOs having different motion with respect to the wave vector, will be different, indicating that those changes do not "cancel each other out".

You might try thinking of it this way: what LIGO detects is tidal gravity. (Tidal gravity and spacetime curvature are just different names for the same thing in GR.) But a tidal gravity detector detects different amplitudes (and therefore different phase shifts in an interferometer, if that's the kind of detector you're using) depending on how it's oriented with respect to the source. And a GW is tidal gravity that varies sinusoidally even for a detector that is at rest, which is different from the kind we're used to, where you have to rotate the detector in order to see variation in amplitude (for example, the Earth's tides vary sinusoidally because it is rotating with respect to the Moon and Sun).

 

[RockyMarciano]

No, we don't. We have that light moves on null worldlines. But in a curved spacetime, the coordinate speed of light, which is what you are implicitly thinking of, might not be equal to c, even though light always moves on null worldlines.

In the flat background spacetime being used to analyze the GWs, the coordinate speed of light is c, yes. But in the region affected by the GWs, it isn't.

I think a fundamental issue you are having here is that you are continuing to think of "spatial variation", "temporal variation", and "variation in the speed of light" as different things, one or more of which can be present in any given scenario. That kind of reasoning doesn't work for GWs; it can work for EM waves only because EM waves can be modeled in perfectly flat spacetime in some chosen inertial frame, where "variation in the speed of light" is absent and there is a well-defined global split between "spatial variation" and "temporal variation". But even in the case of EM waves, if you try to model them in curved spacetime, you come up against the same issues as you do for GWs: there is no longer a well-defined global split between "space" and "time", and the coordinate speed of light is no longer constant. And in the case of GWs, there is no way to model them in perfectly flat spacetime; the background spacetime can be flat, but the waves are waves of spacetime curvature, so there's no way to avoid spacetime curvature in dealing with them.
No, they don't. The frequency and wavelength of the same GW, measured by two different LIGOs having different motion with respect to the wave vector, will be different, indicating that those changes do not "cancel each other out".

You might try thinking of it this way: what LIGO detects is tidal gravity. (Tidal gravity and spacetime curvature are just different names for the same thing in GR.) But a tidal gravity detector detects different amplitudes (and therefore different phase shifts in an interferometer, if that's the kind of detector you're using) depending on how it's oriented with respect to the source. And a GW is tidal gravity that varies sinusoidally even for a detector that is at rest, which is different from the kind we're used to, where you have to rotate the detector in order to see variation in amplitude (for example, the Earth's tides vary sinusoidally because it is rotating with respect to the Moon and Sun).

Thanks for the reply but I was actually referring to the speed of light measured from the laser local inertial frame, not the coordinate speed of light. And regarding wavelength versus frequency I was referring to the laser light, used as a clock or ruler depending on the pov, not to the GW. I regret the confusion, I should have been more specific.
I agree what is detected as amplitude is tidal (Weyl) curvature distortion in the way you explain.
I'm not completely sure when you say that all changes are spacetime changes if you mean all changes related to GWs or changes in general. In other words there must be something that is different between what is called ripples in spacetime from other kind of changes(perturbations, forces...). It would help to know what the exact distinction is between say electromagnetic waves and gravitational waves besides the fact the spacetime is flat in the former and a perturbative approximation to flat in the latter, but it would seem that both are "spacetime ripples".

 

[PeterDonis]

I was actually referring to the speed of light measured from the laser local inertial frame

The "laser local inertial frame" does not and cannot cover the entire LIGO device. If it did, the device would not be able to measure any GWs, because spacetime curvature would be negligible over the entire device. So your implicit reasoning here can't be right, because you are implicitly assuming that the speed of light can be assumed to be c over the length of each of LIGO's arms. It can't.

I'm not completely sure when you say that all changes are spacetime changes if you mean all changes related to GWs or changes in general.

"All changes are spacetime changes" can have two meanings.

One meaning is that all changes being considered are changes in spacetime curvature. Obviously that's only true for GWs; it's not true for everything--for example, it obviously isn't true for EM waves.

The other meaning is that there is no such thing as a "spatial change" or a "temporal change" in any invariant sense, but only a "spacetime change"--in other words, there is no such thing as a change that looks purely like a spatial change or a temporal change to all observers. That is true in general.

It would help to know what the exact distinction is between say electromagnetic waves and gravitational waves besides the fact the spacetime is flat in the former and a perturbative approximation to flat in the latter,

EM waves are waves of the electromagnetic field. GWs are waves of spacetime curvature.

Also, strictly speaking, spacetime is not flat in the case of EM waves, because EM waves have energy, and everything that has energy produces some spacetime curvature. In practice, we can ignore the spacetime curvature produced by EM waves because its effects are so much smaller than the other effects of EM waves. Obviously this can't be true of GWs, because changes in spacetime curvature are the only effects of GWs--they are the GWs.

it would seem that both are "spacetime ripples".

I think you are confusing "changes that happen in spacetime" with "changes in the curvature of spacetime". Only the latter are properly thought of as "spacetime ripples". Everything happens in spacetime, but that does not mean all changes are changes in the curvature of spacetime.

 

[RockyMarciano]

Thanks for those comments. I agree with the general reasoning behind them, but since words can be deceiving I'll try to stick closer to the actual math used in the weak-field approximation used in the detection of GWs.The following paragraph is problematic when confronting it to the actual math even if it looks quite reasonable within the explanations often found trying to facilitate understanding:

The "laser local inertial frame" does not and cannot cover the entire LIGO device. If it did, the device would not be able to measure any GWs, because spacetime curvature would be negligible over the entire device. So your implicit reasoning here can't be right, because you are implicitly assuming that the speed of light can be assumed to be c over the length of each of LIGO's arms. It can't.

You are assuming here the "ripples in spacetime" informal language that Pervect referred to in #7,(synchronous) coordinates where c varies over the arm's length imply shrinking/stretching rulers that would give the impression of a null detection, that's why the mathematical treatment usually employed to describe GWs in the weak-field approximation doesn't use coordinates where the coordinate speed of light changes, it uses harmonic coordinates instead. The gauge used models the spacetime curvature perturbation only affecting two spacelike coordinates(TT gauge) in the amplitude tensor of the plane wave solutions. The perturbative term $h_{\mu\nu}$ expressed in harmonic(quasi-minkowskian) coordinates under the usual weak-field viewpoint where $h_{\mu\nu}$ is simply a symmetric tensor field (under global Lorentz transformations) defined on a flat Minkowski background spacetime.
The spacetime cuvature is not negligible here but is clearly modeled independently of time in the linearized gravity approximation. And this demands a constant c over the length of each LIGO's arm in oder to compute a phase shift between the arms due to a difference in spacelike distances.

See also the informal treatment linked by LIGO "If light waves are stretched by gravitational waves, how can we use light as a ruler to detect gravitational waves?" by Saulson. This is the relevant paragraph with my bold:
["We are left with the question, ‘‘Are gravitational waves observable by examining the light in an interferometer?’’ It might seem that the recognition that the wave stretches with the space has in fact shown that light is unsuitable to reveal the length changes. We often say, after all, that we are usinglight as a ruler to measure the distortions of space. What good is a ruler that stretches to the same extent that space stretches?
To see why light still works perfectly well for our purpose, recall first that there is no direct sense in which we observe the wavelength of the light in the arms. Our observations are instead of the phase of the light, that is, of the arrival times of wave crests.
What happens when we observe the phase of the light wave in the x arm of our interferometer? Imagine that just before t=tau, one light wave crest had returned precisely to the beam splitter. By the same argument as we used above, that wave crest has no choice other than to remain at the beam splitter when the gravitational wave arrives. So the gravitational wave causes no phase shift at the beam splitter immediately after its arrival.
The key is the word ‘‘immediately.’’ All of the other wave crests suddenly at t=tau become farther from the beam splitter than they were before. Gravitational wave or no, light travels through the arm at the speed of light, c. The physically observable meaning of the stretching of the space is that the light in it has to cover extra distance, and so will arrive late"]

In this view the question that comes to mind is what happens to the equation $\lambda=c\nu^{-1}$ for the light inside the arm while the GW passes? I can see now that my frequency cancelling out the wavelength heuristic actually leads to a variable c over the length of the arm just like you assume so it is not in accord to the math above. In the end the Saulson paper seems to suggest we should simply ignore the wavelength/frequency variation of the laser while the GW passes since it doesn't enter the linearized gravity model that only depends on spacelike separation and constant c.
I can't say I'm satisfied with that.

 

[PeterDonis]

The following paragraph is problematic when confronting it to the actual math

No, it isn't. The harmonic coordinates you are referring to, which are the ones standardly used to describe GWs in the linearized approximation, are not the same as the coordinates of a local inertial frame. As I said, there is no local inertial frame that covers all of the LIGO device. But you can set up harmonic coordinates that cover all of the LIGO device, because the spacetime curvature on the scale of the device is small enough that the linearized approximation works.

The key is the word ‘‘immediately.’’ All of the other wave crests suddenly at t=tau become farther from the beam splitter than they were before.

This is a coordinate description of what happens, and can't be right as a description of something that is physically happening, because it involves wave crests moving "suddenly", i.e., faster than light. It is a reasonable heuristic for reasoning about what LIGO will observe, but that's all.

The physically observable meaning of the stretching of the space is that the light in it has to cover extra distance, and so will arrive late

This is a great example of why an "informal treatment" like this is not the same as the actual science, and can give a distorted understanding of the actual science. Earlier in the same quote, the author went to some pains to tell us that we don't observe the wavelength of light in the arms, i.e., we don't observe the "stretching of space", we only observe the phase of the light--the arrival of the wave crests at the beam splitter. So saying that the "physically observable meaning" is that light has to cover extra distance is contradicting the earlier part of the same quote. The only physical observable is the interference pattern at the beam splitter. The "extra distance" the light supposedly has to cover is not a physical observable; it's a model-dependent quantity that is derived from the physical observable.

I can see now that my frequency cancelling out the wavelength heuristic actually leads to a variable c over the length of the arm just like you assume so it is not in accord to the math above.

And that's fine, because the math above is not "the" unique description of what is "really happening". It's just one possible heuristic model. Yours is just another possible heuristic model, in which the variation in the speed of light caused by the GW is what causes the interference pattern. Both are heuristic models; neither is uniquely "correct". The only truly "correct" model, from a GR point of view, would be a description of the underlying spacetime geometry, including the "ripples" in spacetime that constitute the GW, purely in terms of invariants, with no coordinates at all. But such a description is hard for most people to construct and visualize. Heuristic models are a crutch to help us bridge the gap between our everyday intuitions and counterintuitive phenomena like GWs.

 

[RockyMarciano]

The only "actual science" we have in this case is the linearized approximation, it is the one used to model detection of GWs by laser interferometry. AFAIK the paragraph by Saulson is trying to put into words(awkwardly that's for sure as the apparent contradictions you underline show) how that linearization applies to the interferometer.
On the other hand I just can't figure out how the plane wave far field approximation could be compatible with the heuristic that uses variable coordinate speed of light. can you?

 

[PeterDonis]

The only "actual science" we have in this case is the linearized approximation, it is the one used to model detection of GWs by laser interferometry.

It is the model that is standardly used, yes. But it's still just an approximation. There are other possible models.

AFAIK the paragraph by Saulson is trying to put into words(awkwardly that's for sure as the apparent contradictions you underline show) how that linearization applies to the interferometer.

Yes, he is trying to describe in ordinary language how the linearized, plane wave model of GWs predicts the interference pattern detected.

I just can't figure out how the plane wave far field approximation could be compatible with the heuristic that uses variable coordinate speed of light.

It isn't, if by "the plane wave far field approximation" you mean the linearized model Saulson describes. The two heuristics are not compatible because they are different ways of trying to make sense of the underlying spacetime geometry. Neither one is "wrong" and neither one is "right". They are simply different heuristics--different models. They don't have to be compatible with each other. They only have to be compatible with the actual underlying geometry--or, more concretely, with the actual observed results, the interference pattern.

 

[RockyMarciano]

It is the model that is standardly used, yes. But it's still just an approximation. There are other possible models.

Such as?

 

[PeterDonis]

Such as?

Um, the alternate one we've been discussing, where the variable speed of light is the cause of the interference pattern.

 

[RockyMarciano]

Um, the alternate one we've been discussing, where the variable speed of light is the cause of the interference pattern.

Ah, ok, but I mean on what mathematical model is based, I'm only aware of the linearized gravity approximation to deal with GWs. Unless you are blurring the distinction between heuristics and mathematical theory.

 

[PeterDonis]

I mean on what mathematical model is based

I don't know that there is one. But you proposed a heuristic that is perfectly reasonable; it's just a different one from the standard one.

Unless you are blurring the distinction between heuristics and mathematical theory.

I suppose I am. A heuristic doesn't necessarily require a detailed mathematical theory if all you're using it for is a conceptual understanding. If you're trying to use it to make quantitative predictions, that's something different.

The point I was making is not that there is some detailed alternative model that you can use to quantitatively predict results like the LIGO observations. The point was simply that the standard linearized description of GWs is not "the way things really are"; it's just a heuristic description that, when modeled mathematically, makes predictions that are accurate enough for our purposes. So thinking about it as though "the length between the arms is changing" is what is "really happening" with a GW, and nothing else is going on, is wrong. The best way we have of saying "what is really happening", in the context of GR, is that the spacetime geometry has "ripples" in it, and how those "ripples" look to a particular detector depends on the relationship between the ripples and the detector.

 

[RockyMarciano]

I understand your view, now. Thanks.

 

[pervect]

I believe that a "local inertial frame" would cover an area considerably larger than the Ligo apparatus, though it would be finite in extent.

Here is what I get specifically.

Gravity waves can be + or x polarized, for simplicity we'll assume the wave is purely + polarized, and that it is moving in the +z direction. Then we can approximate the metric as:

$$-dt^2 + [1+2h(t-z) ]dx^2 + [1-2h(t-z)]dy^2 + dz^2$$

here h() is an arbitrary function, representing an arbitrary strain, the same dimensionless number reported by Ligo.

To first order we can approximate this metric as an orthonormal basis of one-forms

$$\omega_1 = dt \quad \omega_2 = [1+h(t-z)]dx \quad \omega_3 = [1-h(t-z)]dy \quad \omega_4 = dz$$

(GRtensor uses indices from 1 to 4, and it's too likely to induce errors for me to attempt change the notation to a more natural and readable 0-3).

One might write instead $\omega_1 = \sqrt{1+2h(t-z)}$ if one prefers, but the end result will be the same to the first order which is all that the metric is good for.

To check this, it's convenient to assume $h(u) = a \sin \Omega u$ where $u = t-z$, and $\Omega$ is just an angular frequency, which we've chosen to capitalize to avoid confusing it with the basis one forms $\omega_i$. Then the Einstein tensor G is zero to the first order - it has components of order $a^2$ which we can neglect.

We can find the Riemann

$$R_{1212} = -R_{1224} = R_{2424} = a \Omega^2 \sin \, \Omega u/ (1 + a \sin \, \Omega u ) \approx a \Omega^2 sin \, \Omega u$$
$$-R_{1313} = R_{1334} = -R_{3434} = a \Omega^2 \sin \, \Omega u / (1 - a \sin \, \Omega u ) \approx a \Omega^2 sin \, \Omega u$$

And we use the relation between the Riemann and Fermi-normal coordinates (MTW, pg 332) to find an approximate expression for the metric in Fermi-normal coordinates, coordinates which best approximate the "laser local lab frame". There are a bunch of assorted cross terms which would be tedious and error-prone to write out explicitly, the interesting term is the metric coefficient of dt^2

$$ [-1 - a \Omega^2 (x^2 - y^2) \sin \Omega u] \, dt^2$$

The gradient of this of this gives the tidal force (for small x and y). I believe that the other terms are not important for understanding the physics (I suppose I could be overlooking something, but I don't think so.)

[add]
The above was all in geometric units, a physical interpretations would be that the coefficient of "time dilation" dt^2 is proportional
$$1 + 4 \pi^2 \, a \frac{x^2 - y^2} { L^2}$$

where L is the wavelength of the gravitataional wave.

It can be seen that for large x and y the metric becomes ill behaved, and that the metric should be thought of only as being valid for "small x and y". However, due to the extremely small value of a in Ligo, on the order of $10^{-21}$, the region covered will include the whole apparatus - probably the whole solar system.

 

[PeterDonis]

I believe that a "local inertial frame" would cover an area considerably larger than the Ligo apparatus

As I said before, this can't possibly be the case as you state it, because by definition spacetime curvature in a local inertial frame is negligible, and a gravitational wave is a wave of spacetime curvature, so spacetime curvature can't possibly be negligible over the area covered by LIGO if LIGO detects a GW.

What you are constructing are basically Fermi normal coordinates centered on the LIGO detector (at the base of the arms). It's certainly true that such coordinates can cover a "world tube" that is more than wide enough to encompass the entire LIGO apparatus. But Fermi normal coordinates are not the same as a local inertial frame. In a local inertial frame the metric is $-dt^2 + dx^2 + dy^2 + dz^2$, by definition. There can't be any correction terms. If there are, it isn't a local inertial frame.

 

[RockyMarciano]

After a conversation with one of the detection paper signing authors from LIGO he has convinced me that the analysis using varaible coordinate light speed is not tenable because it is not physical(in exactly the same way the coordinate light speeds found in cosmology are not considered to break special relativity and allow superluminal signaling, a variable coordinate velocity cannot be considered a physical invariant leading to a measurable signal and if it was it would lead to a non-detection and/or to the possibility of superluminal signaling. So the only possible analysis consistent with a non-null detection is the one that considers the laser speed as constant at all times and locations in the arms whether the GW is passing or not.

 

[PeterDonis]

a variable coordinate velocity cannot be considered a physical invariant

This is true; coordinate-dependent quantities are not physical invariants. But that is equally true of the coordinate speed of light in the coordinates the LIGO authors prefer to use. The fact that that coordinate speed happens to be $c$ at all values of the coordinates doesn't mean it's an invariant.

The actual invariants in the LIGO detectors are the observables: the signals that are shown in the graphs. Basically, these are amplitudes of interference patterns in the detectors vs. time. So it is an invariant fact that, when the laser beams go down the two perpendicular arms and are reflected at the mirrors, they interfere with each other when they return. The LIGO authors' interpretation of this--that the speed of light is constant and the interference pattern is therefore due to changes in the lengths of the arms because of the passage of a GW--is certainly an easy interpretation to visualize and work with; but that doesn't make it an invariant.

in exactly the same way the coordinate light speeds found in cosmology are not considered to break special relativity and allow superluminal signaling, a variable coordinate velocity cannot be considered a physical invariant leading to a measurable signal and if it was it would lead to a non-detection and/or to the possibility of superluminal signaling

Um, what? Variable coordinate light speeds cannot allow superluminal signaling (which is true), therefore variable coordinate light speeds cannot be right because they would allow superluminal signaling? This is self-contradictory.

Actually, variable coordinate speeds of light are commonly used for interpretation in cosmology, and nobody bats an eye--precisely because they don't lead to superluminal signaling, so they're not a problem. The same is true for any coordinates in which the coordinate speed of light is variable (another common example is Schwarzschild coordinates). It just so happens that the most convenient coordinates for cosmology--comoving coordinates in FRW spacetime--lead to variable coordinate speeds of light; whereas it just so happens that the most convenient coordinates for LIGO analysis are coordinates in which the coordinate speed of light is $c$ everywhere. But that doesn't make different coordinates in which the coordinate speed of light is variable "wrong"; it just makes them less convenient for this particular problem.

 

[RockyMarciano]

The actual invariants in the LIGO detectors are the observables: the signals that are shown in the graphs. Basically, these are amplitudes of interference patterns in the detectors vs. time. So it is an invariant fact that, when the laser beams go down the two perpendicular arms and are reflected at the mirrors, they interfere with each other when they return. It just so happens that the most convenient coordinates for cosmology--comoving coordinates in FRW spacetime--lead to variable coordinate speeds of light; whereas it just so happens that the most convenient coordinates for LIGO analysis are coordinates in which the coordinate speed of light is $c$ everywhere. But that doesn't make different coordinates in which the coordinate speed of light is variable "wrong"; it just makes them less convenient for this particular problem.

I agree with this but I'm having a really hard time seeing how it applies to the LIGO detection case.
So if the observables for the detection of a GW here in the LIGO experiment are the difference in phase(different times of flight for each laser path originated in the difference in arms lengths), I would like to see a mathematical treatment using coordinates in which the coordinate speed of light is variable, even if less convenient, where the observable described in the previous paragraph is recovered, even if just in principle if the mathematical treatment in those coordinates is very involved. Can you show something like this?
I've looked for such a treatment and haven't been able to find anything. Everywhere I looked I just found the linearized gravity harmonic gauge condition(under many names:de Donder, Lorenz, Fock...) with constant c in the interferometer arms.And I myself can't figure out how to reproduce the detection observable with changing c as it seems evident that it would lead to no difference in times of flight even if there was a change in length(the change in c would cancel the change in length in both arms).

 

[PeterDonis]

if the observables for the detection of a GW here in the LIGO experiment are the difference in phase

Yes, this is a direct observable.

(different times of flight for each laser path originated in the difference in arms lengths)

But this is not. Attributing the difference in phase to "different times of flight originating in the difference in arms lengths" is already adopting a particular set of coordinates.

I would like to see a mathematical treatment using coordinates in which the coordinate speed of light is variable, even if less convenient, where the observable described in the previous paragraph is recovered, even if just in principle if the mathematical treatment in those coordinates is very involved. Can you show something like this?

No. As I think I said earlier in this discussion, I'm not aware of anyone having actually done such an analysis. But that doesn't mean there is no such analysis. Nor does it change the fact that talking about different times of flight originating in different arm lengths is adopting a particular set of coordinates.

I myself can't figure out how to reproduce the detection observable with changing c as it seems evident that it would lead to no difference in times of flight even if there was a change in length(the change in c would cancel the change in length in both arms).

Heuristically, the kind of model I'm thinking of would have no change in the coordinate lengths of the arms; the variation in the coordinate speed of light over a constant arm length would lead to a variation in coordinate times of flight, and therefore a phase shift.

 

[pervect]

After a conversation with one of the detection paper signing authors from LIGO he has convinced me that the analysis using varaible coordinate light speed is not tenable because it is not physical(in exactly the same way the coordinate light speeds found in cosmology are not considered to break special relativity and allow superluminal signaling, a variable coordinate velocity cannot be considered a physical invariant leading to a measurable signal and if it was it would lead to a non-detection and/or to the possibility of superluminal signaling. So the only possible analysis consistent with a non-null detection is the one that considers the laser speed as constant at all times and locations in the arms whether the GW is passing or not.

I would take the position that the coordinates where the laws of physics are as close to Newtonian as possible (which I regard as being the most physical because that's where my physical intuition is the strongest) are Fermi-normal coordinates, and point out that those are NOT the coordinates that the Ligo paper used.

I would also admit that using Fermi-normal coordinates would be a giant pain in the rear to use, though less of a pain if one is willing to make certain approximations.

I went through some rather detailed discussion and derivations about an approximate conversion to such coordinates and the resulting metric, but I don't think it "got through", so I don't see much point in rehashing it.

I'll quote at the end of this post a bit from the literature about the power of Fermi-normal coordinates and their "physical interpretation" from http://arxiv.org/abs/0901.4465

The point of this is this: my assumption that anyone who is searching for "physicalness" in coordinates would be well advised to consider Fermi-normal coordinates. "Physicality", a rather vague term, does seems to be the goal of the OP here, so that's why I suggested them. I hoped that the mention of the name alone would be sufficeint, but it seems not. So I'll go into some backgroud with some quotes from the literature about what these coordinates are, and their claims to "physicallity" (which are also in the literature).

http://arxiv.org/abs/0901.4465][/PLAIN]
Nowadays the Fermi normal coordinates are usually - although improperly - called Fermi coordinates. In experimental gravitation, Fermi normal coordinates are a powerful to ol used to describe various experiments: since the Fermi normal coordinates are Minkowskian to first order, the equations of physics in a Fermi normal frame are the ones of special relativity, plus corrections of higher order in the Fermi normal coordinates, therefore accounting for the gravitational field and its coupling to the inertial effects. Additionally, for small velocities v compared to light velocity c, the Fermi normal coordinates can be assimilated to the zeroth order in (v/c) to classical Galilean coordinates. They can be used to describe an apparatus in a “Newtonian” way (e.g. [1, 3, 8, 10]), or to interpret the outcome of an experiment (e.g. [11] and comment [21], [5, 6, 15, 17]). In these approaches, the Fermi normal coordinates are considered to have a physical meaning, coming from the principle of equivalence (see e.g. [18]), and an operational meaning: the Fermi normal frame can be realized with an ideal clock and a non extensible thread [29]. This justifies the fact that they are used to define an apparatus or the result of an experiment in terms of coordinate dependent quantities.

So in conclusion I think that Fermi-Normal coordinates, which are NOT the ones used in the Ligo analysis, might provide some insight into the interpretation process of the results. I'd also say that they aren't the simplest to use mathematically, and that the easiest approach is to analyze the problem in the coordinates that Ligo used, first, then convert to Fermi-normal coordinates to aid in the physical interpretation. The tools needed to convert coordinates involve knowledge of diffeomorphisms, and the tensor transformation laws.

I'll also say that when using generalized coordinates, such as GR does, coordinates are not necessarily chosen for any "physical" significance at all, and that this takes some getting used to. The metric is what converts physically non-significant coordinates into physically significant distances and times, via the mechanism of the invariant Lorentz interval. The process goes like this: coordinates go into the metric, which generates the Lorentz interval from coordinate displacements. The Lorentz interval is independent of the observer, and can be further converted into proper time intervals and proper distance intervals when one chooses an "observer".

 

[RockyMarciano]

But this is not. Attributing the difference in phase to "different times of flight originating in the difference in arms lengths" is already adopting a particular set of coordinates.

Fair enough. But this misses the point I'm trying to make in relation to gravitational radiation and linearized GR in the absence of a static spacetime situation. Here the coordinate condition acts like a gauge condition. Using the gauge terminology you are referring to coordinates sets choices in the context of this choice fixing a gauge(in this case the gauge refers to GL4 Diff(M) invariance under arbitrary coordinate transformations) like what's seen for instance in the Schwarzschild solution. But here we have to deal with a coordinate condition that is a gauge condition that doesn't fix a gauge: the Lorenz gauge condition(harmonic condition), a partial gauge, must be used due to the indeterminacy of the EFE. In other words the equation $\Box h_{\mu\nu}=0$ is only valid in the Lorenz gauge condition(i.e. harmonic coordinate condition)
Under these circumstances the gauge condition rules out the different set of coordinates you mention below, that measures variation in coordinate speed of light over arm length, which would be just a change of coordinates that is routinely performed when using interferometers for other applications (for instance when used to measure refractive changes), because we are talking about a certain gauge condition that takes advantage of the degrees of freedom left by the partial gauging to leave the coordinate t unaffected by any coordinate transformation and be able to concentrate in change of length in the arms.

Heuristically, the kind of model I'm thinking of would have no change in the coordinate lengths of the arms; the variation in the coordinate speed of light over a constant arm length would lead to a variation in coordinate times of flight, and therefore a phase shift.

Again, this could only be done in a static situation, where the existence of a timelike killing vector allows you to fix the gauge. Note that in order to measure change in coordinate speed of light you would have to have a wavelength change in the laser to keep the frequency constant, otherwise you cannot detect phase shift, and such wavelength change is not compatible with no change in arm length. This shows the usefulness of the partial Lorenz gauge in gravitational radiation detection, it allows you to ignore the change in wavelength that comes with the stretching of the interferometer's arms.

 

[PeterDonis]

Here the coordinate condition acts like a gauge condition.

Yes, it does. So what? It's still a coordinate condition, and there is still no requirement that one has to adopt any particular coordinate condition. It's a convenience, not a physical necessity.

here we have to deal with a coordinate condition that is a gauge condition that doesn't fix a gauge

All this means is that one single condition doesn't pick out a unique coordinate chart; it only picks out a family of coordinate charts (harmonic coordinates), and you have to impose a further condition to pick out one particular chart in that family. None of this changes what I was saying at all.

Under these circumstances the gauge condition rules out the different set of coordinates you mention below

No, it doesn't, because a gauge condition is not a physical condition; it's a choice of coordinates. As I said above, there is no requirement to adopt any particular coordinate condition. It's a convenience, not a physical necessity.

this could only be done in a static situation, where the existence of a timelike killing vector allows you to fix the gauge.

The existence of a timelike KVF does not require you to pick a particular coordinate chart or impose a particular gauge condition. Nor does it enable you to use a particular coordinate chart or fix a particular gauge condition that you couldn't fix in its absence. It just makes it convenient to adopt a particular choice of coordinates/gauge. But that's a convenience, not a physical necessity.

in order to measure change in coordinate speed of light you would have to have a wavelength change in the laser to keep the frequency constant, otherwise you cannot detect phase shift

The phase shift is an invariant; if it's present in one coordinate chart, it's present in any coordinate chart.

such wavelength change is not compatible with no change in arm length.

It might not be compatible with no change in arm proper length, given a particular definition of what "proper length" means. (But even that is not uniquely determined, because it requires a choice of simultaneity convention, i.e., a coordinate choice.) But I was talking about coordinate length; a wavelength change is perfectly compatible with no change in coordinate arm length.

 

[pervect]

I ran across a rather interesting paper. "Fermi-normal, optical, and wave-synchronous coordinates for spacetime with a plane gravitational wave", http://www.phys.utb.edu/downloads/0264-9381_31_8_085006.pdf

Abstract
Fermi-normal (FN) coordinates provide a standardized way to describe the
effects of gravitation from the point of view of an inertial observer. These
coordinates have always been introduced via perturbation expansions and
were usually limited to distances much less than the characteristic length
scale set by the curvature of spacetime. For a plane gravitational wave this
scale is given by its wavelength which defines the domain of validity for
these coordinates known as the long-wavelength regime. The symmetry of this
spacetime, however, allows us to extend FN coordinates far beyond the long-
wavelength regime. Here we present an explicit construction for this long-range
FN coordinate system based on the unique solution of the boundary-value
problem for spacelike geodesics. The resulting formulae amount to summation
of the infinite series for FN coordinates previously obtained with perturbation
expansions. We also consider two closely related normal-coordinate systems:
optical coordinates which are built from null geodesics and wave-synchronous
coordinates which are built from spacelike geodesics locked in phase with the
propagating gravitational wave. The wave-synchronous coordinates yield the
exact solution of Peres and Ehlers–Kundt which is globally defined. In this
case, the limitation of the long-wavelength regime is completely overcome,
and the system of wave-synchronous coordinates becomes valid for arbitrarily
large distances. Comparison of the different coordinate systems is done by
considering the motion of an inertial test mass in the field of a plane gravitational
wave.

It talked a bit about gauge conditions, but I didn't see an explicit reference to which gauge condition Fermi-normal coordinates might satisfy. That's an interesting question for the weak-field formulation of the theories, where gauge conditions are defined, that I don't know the answer to.

Gauge conditions were mentioned, for instance It was mentioned that TT gauge was useful for analyzing gravitational waves with laser interferometry.

Themost important point I think is that gauge conditions and/or coordinate choices cannot matter to experimental results. It seems to me like that basic point is getting lost somewhere in the math :(.

 

[RockyMarciano]

I ran across a rather interesting paper. "Fermi-normal, optical, and wave-synchronous coordinates for spacetime with a plane gravitational wave", http://www.phys.utb.edu/downloads/0264-9381_31_8_085006.pdf

It talked a bit about gauge conditions, but I didn't see an explicit reference to which gauge condition Fermi-normal coordinates might satisfy. That's an interesting question for the weak-field formulation of the theories, where gauge conditions are defined, that I don't know the answer to.

I've been trying to argue that the only gauge condition that the linearized gravity formulation of gravitational radiation(no other available to my knowledge) is consistent with is the harmonic (Lorenz) gauge. This is a well known fact( you can see it explicitly stated for intance in Ruffini's textbook "Basic concepts in relativistic astrophysics" pages 148-150 accesible in google books), that doesn't mean observables depend on a certain gauge any more than the fact that in the relativistic formulation of EM the Lorenz gauge being the only one compatible with a vanishing divergence of the 4-current does.

Gauge conditions were mentioned, for instance It was mentioned that TT gauge was useful for analyzing gravitational waves with laser interferometry.

The TT gauge is an additional gauge choice overimposed to the Lorenz gauge that makes the analysis easier.

Themost important point I think is that gauge conditions and/or coordinate choices cannot matter to experimental results. It seems to me like that basic point is getting lost somewhere in the math :(.

That basic point is of course unavoidable. I share the sentiment that it gets lost somewhere in the math. My posts are trying to find it back. I'm not getting much help so far.
There is an added problem in the GR case with respect to the EM case with which the analogy about gauge conditions is usually drawn. The main difference, besides the fact that Maxwell's equations are already linear so there's no need for a previous linearization of the equations like in GR, is that in EM one can formulate the wave equation without the potentials, that are the subject of the gauge freedom, it can be done just with the observable fields E, B with or without sources. This fact alone already ensures that the gauge conditions don't affect physical observables.

I would like to see exactly how this works in GR for gravitational radiation. The linearization is based in the gauge freedom of the metric (indeterminacy of the EFE on the metrics) that takes the role of the potentials in the EM analogy. In the equation $g_{\mu\nu}=η_{\mu\nu}+h_{\mu\nu}$ the decomposition between the flat background and the metric perturbation $h_{\mu\nu}$ is not uniquely determined and therefore there is not only gauge invariance in relation to the well known arbitrariness of the coordinates in which the equation is valid but also about the possible $h_{\mu\nu}$ metrics. Now it can be shown that the only way to obtain a wave equation from the linearized EFE is through the Lorenz gauge condition(that in GR also happens to be a coordinate condition). But I don't see exactly how to construct the wave equation based on the observable field, that in this case is the tidal field that is supposed to oscllate and propagate and that is represented by the Weyl curvature that appears in the absence of the Einstein tensor, with a derivation that uses the linearized EFE with Einstein tensor.

I guess it's got to do with how the Newtonian tidal force is represented in the geodesic deviation in the weak field. That in itself already demands the use of the Lorenz gauge.

 

[PeterDonis]

I don't see exactly how to construct the wave equation based on the observable field, that in this case is the tidal field that is supposed to oscllate and propagate and that is represented by the Weyl curvature that appears in the absence of the Einstein tensor, with a derivation that uses the linearized EFE with Einstein tensor.

No, the "field" that is analogous to $E$ and $B$ in electrodynamics is actually the Riemann tensor, not the Weyl tensor. The Weyl tensor is the part of the Riemann tensor that is not directly connected to the source; the Einstein tensor is the part that is (via the EFE). But you need both parts to construct the wave equation, just as you need the full EM field tensor $F$, which is the Lorentz-covariant representation of the field, to construct the EM wave equation, not just the part of $F$ that is not directly connected to the source.

(It can be difficult to see this parallel between GR and EM because we are not used to splitting up the EM field $F$ into parts analogous to the Weyl and Einstein tensors. We are used to splitting it up into $E$ and $B$, but that split is coordinate-dependent, whereas the split of the Riemann tensor into the Weyl and Einstein tensors is not. I'm not actually sure exactly what split of $F$ would correspond to the split of Riemann into Weyl and Einstein.)

 

[vanhees71]

I've not tried this, but could it be that the split of $F_{\mu \nu}$ you are thinking about for a "quasi-point-like particle" is the split in an instantaneous boosted Coulomb field (i.e., in the momentaneous rest frame of the particle you have just a Coulomb field), which contains the relation of the field to the sources (charge-current distribution four-vector) and the rest, which is a free radiation field?

 

[PeterDonis]

could it be that the split of $F_{\mu \nu}$ you are thinking about for a "quasi-point-like particle" is the split in an instantaneous boosted Coulomb field (i.e., in the momentaneous rest frame of the particle you have just a Coulomb field), which contains the relation of the field to the sources (charge-current distribution four-vector) and the rest, which is a free radiation field?

I see what you're saying: at any point the total field is a superposition of the Coulomb field due to the presence of sources at that point, and the radiation field describing propagation of the field from sources elsewhere. I'm not sure how this split would correspond to Maxwell's Equations, though; there's no way to split up $F$ itself into a part that only appears in the source-free equations, and a part that only appears in the equations with source.

 

[vanhees71]

No, what I mean is the field from a (in general accelerated) charge. Forget the notorious radiation-reaction problem. Then the four-potential is given by the Lienard-Wichert potentials and the electromagnetic field by the corresponding derivatives (see, e.g., Landau-Lifshitz vol. II, where this is derived in the most beautiful way). The electromagnetic field splits in a part that depends only on the velocity of the particle (taken at the "regarded time") and goes like $1/r^2$ with distance (for $r \rightarrow \infty$) and a part that's proportional to the acceleration that goes like $1/r$. The first term is indeed the instantaneously boosted Coulomb field (having of course both electric and magnetic components if $\vec{v} \neq 0$).

I'm not sure, however, whether this split in the electromagnetic fields in a "boosted Coulomb" and a "radiative" part is analogous to the split of the Riemann tensor in Weyl and Einstein part.

 

[pervect]

I've been trying to argue that the only gauge condition that the linearized gravity formulation of gravitational radiation(no other available to my knowledge) is consistent with is the harmonic (Lorenz) gauge. This is a well known fact( you can see it explicitly stated for intance in Ruffini's textbook "Basic concepts in relativistic astrophysics" pages 148-150 accesible in google books), that doesn't mean observables depend on a certain gauge any more than the fact that in the relativistic formulation of EM the Lorenz gauge being the only one compatible with a vanishing divergence of the 4-current does.

The TT gauge is an additional gauge choice overimposed to the Lorenz gauge that makes the analysis easier.

My position is that gauge choices are just that - a choice. When you're trying to solve the linearized equations, the Lorenz gauge is the obvious choice, because it makes the equations easier to solve and it's also the one you'll find in your textbooks. That doesn't mean that it would be incorrect and/or impossible in principle to do the analysis in some other gauge, just difficult and non-standard.

However, I also argue that the result of this gauge choice are coordinates that are not necessarily the best choice for interpreting the results, though it's obviously the best choice for solving the problem.

A related point I want to make, which I did a bit of checking my memory of this point, is that the gauge choice is equivalent to a coordinate choice. See for instance http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec10.pdf

Indeed, small deviations from $g_{\mu\nu} = \eta_{\mu\nu}$ may arise either because spacetime is perturbed from Minkowski, or because we perturbed the coordinate system (or both). So we must understand the implications of perturbing the coordinate system, or making a small gauge transformation.

.

Thus an infinitesimal change of coordinates in which the “grid” is displaced by the vector ξ changes the metric perturbation according to [[eq 17]]

$$\Delta h_{\mu\nu} = -\xi_{\mu,\nu} - \xi_{\nu,\mu}$$

Now, we already assume that $g_{\mu\nu} \approx \eta_{\mu\nu}$ to do linearized theory, but it's easiest to analyze the results when all the partial derivatives of $g_{\mu\nu}$ with respect to the coordinates $\frac {\partial g_{\mu\nu}}{ \partial x^{\rho}}$are zero, because this implies that all the Christoffel symbols are zero. We can't make this happen everywhere, but we can make it happen in a small local region, and then we can better understand the results physically in this small local region.

The issue is that the gauge choice that we made to make the linearized equations easy to solve doesn't have the property that $\frac {\partial g_{\mu\nu}}{ \partial x^{\rho}}=0$, and thus makes the physical interpretation more difficult. So we either need to deal with understanding the physical effects of the non-zero Christoffel symbols at an intuitive level (which is possible but not something suited to a brief post), or we need to convert from the coordinates implied by our gauge choice (which we did not because it was mandated, but because it was so much easier) to a more convenient set of coordinates. Because we know how tensors transform when we change coordinates, it's not all that difficult (at least in principle) to work the problem out in one set of coordinates, and change to another, easier-to-interpret set.

The set of coordinates with zero Christoffel symbols has a history and a name - it's just a special case of the Fermi-Normal coordinates I was referring to earlier. The special case is that all the Christoffel symbols are zero only for an observer with zero proper acceleration. If you have an accelerated observer, you can only make some of the Christoffel symbols zero (and you have to deal with something rather like the 'fictitious forces' you get in Newtonian physics for an accelerated frame), but when you have an observer whose proper acceleratio is zero, i.e. a geodesic observer, and you use Fermi-Normal coordinates, the Christoffel symbols all vanish - not everywhere, but in some small local region.T

 

[PeterDonis]

I'm not sure, however, whether this split in the electromagnetic fields in a "boosted Coulomb" and a "radiative" part is analogous to the split of the Riemann tensor in Weyl and Einstein part.

I'm not sure either. For one thing, the Weyl tensor is not purely "radiative"; for example, the Weyl curvature in the vacuum region around a static, spherically symmetric mass does not "propagate" anywhere, it's static; but it's still Weyl curvature.

 

[RockyMarciano]

My position is that gauge choices are just that - a choice. When you're trying to solve the linearized equations, the Lorenz gauge is the obvious choice, because it makes the equations easier to solve and it's also the one you'll find in your textbooks. That doesn't mean that it would be incorrect and/or impossible in principle to do the analysis in some other gauge, just difficult and non-standard.

Yes, in general they are just a choice by definition. But GR is not your typical gauge theory, and I already mentioned the indeterminacy problem of the EFE above that leads to GR not having a unique solution to the initial value problem(this also happens in electrodynamics with the key difference that in GR the metrics don't act on the space like the potentials in electrodynamics but define the space where the dynamics take place). In the particular case of gravitational radiation the harmonic condition acts as an initial condition constraint equation, it is mathematically unavoidable to obtain a wave equation and in which a further gauge transformation(TT gauge) allows solutions with two polarizations perpendicular between them and wrt the propagation direction of the waves. It is equivalent to the coordinate condition used in cosmology in the 3+1 formulation, that it also works as a constraint equation for the initial value problem. In as much as the gauges are used as initial conditions they are no longer a choice in that sense(cf. Straumann 2.8) though they of course exploit the gauge freedom of the theory to be able to use them.