# I Gravitational wave time variation

1. Feb 13, 2016

### 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.

2. Feb 13, 2016

### Staff: Mentor

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.

3. Feb 28, 2016

### 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?

Noel.

4. Feb 28, 2016

### Staff: Mentor

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.

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.

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.

5. Mar 1, 2016

### Lino

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.

6. Mar 1, 2016

### Staff: Mentor

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.

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.

I'm not sure what you mean by this.

7. Mar 1, 2016

### pervect

Staff Emeritus
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).

8. Mar 1, 2016

### Staff: Mentor

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.

9. Mar 1, 2016

### Ibix

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?

10. Mar 1, 2016

### Staff: Mentor

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.)

11. Mar 1, 2016

### Ibix

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.

12. Mar 1, 2016

### Staff: Mentor

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.

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.

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.

13. Mar 2, 2016

### 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.

14. Mar 2, 2016

### Staff: Mentor

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?"

Looks right to me, yes.

15. Mar 3, 2016

### RockyMarciano

I think this explanation might be confusing. If there is no invariant sense in wich 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.

16. Mar 3, 2016

### Staff: Mentor

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.

17. Mar 4, 2016

### Lino

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 seperated 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".)

Noel.

18. Mar 4, 2016

### 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.

19. Mar 4, 2016

### Staff: Mentor

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.

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.

20. Mar 4, 2016

### RockyMarciano

Thanks, this is clearer. However it's not exactly what I had in mind, wich 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?