How to measure Gravitational waves

[Vanadium 50]

Finally! A system described precisely enough to be analyzed.

Basically, you have a clock on a pedestal next to a clock hanging from a string. I don't see how a passing gravitational wave will change the reading on those clocks. Why do you think it would?

 

[timmdeeg]

Basically, you have a clock on a pedestal next to a clock hanging from a string. I don't see how a passing gravitational wave will change the reading on those clocks. Why do you think it would?

Because the hanging clock is in principle moving in the horizontal plane like a particle of the ring of particles shown here relativ to the other clock which is fastened. So there should be a frequency shift according to special relativity. I understand the arguments of @pervect but would like to discuss the idealized case.

 

[Vanadium 50]

is in principle moving in the horizontal plane

OK, work out the time dilation from this movement. Does there exist any clock this good?

 

[timmdeeg]

OK, work out the time dilation from this movement. Does there exist any clock this good?

I guess no, just trying to improve my understanding by means of such scenarios.

 

[Ibix]

Because the hanging clock is in principle moving in the horizontal plane like a particle of the ring of particles shown here relativ to the other clock which is fastened. So there should be a frequency shift according to special relativity. I understand the arguments of @perfect but would like to discuss the idealized case.

I'm struggling to convince myself that this will detect anything even in principle. The expansion and contraction of the ring is like the expansion of the universe. For any pair of the dots we can add an inertial dot halfway between, which will see the pair moving symmetrically about it. That dot will argue that there can be no difference in the clock readings between its pair, from symmetry. So I think they'll all show the same time.

This may not be the case for a ring of particles in motion with respect to this conveniently symmetric case.

 

[timmdeeg]

For any pair of the dots we can add an inertial dot halfway between, which will see the pair moving symmetrically about it. That dot will argue that there can be no difference in the clock readings between its pair, from symmetry. So I think they'll all show the same time.

But we don't have two inertial clocks. One is, the other isn't.

 

[Ibix]

But we don't have two inertial clocks. One is, the other isn't.

But @pervect pointed out in #30 that the Earth isn't rigid enough to be treated as anything significantly different from a cloud of free-floating masses.

 

[PeterDonis]

A is freely suspended like the mirrors of LIGO, B is fastened to the Earth close to A.

As @pervect pointed out, this won't make much difference in terms of the response of A and B to a gravitational wave, because for this purpose the Earth has negligible rigidity. So if we just consider the gravitational wave, without taking into account any other effects, A and B will move the same.

The reason why A is isolated in LIGO is because there are other effects--all kinds of other things that make the Earth vibrate, such as earthquake tremors and other seismic effects, vibrations due to passing cars and trucks, etc., etc. All of these things will make B move in ways that have nothing to do with a gravitational wave; but if A is isolated well enough, it won't respond to any of these things so the signal we obtain from it will be much closer to the "pure" signal we would get from a gravitational wave alone.

 

[timmdeeg]

@pervect, @Ibix and @PeterDonis , thanks for clarifying this question. I was underestimating the rigidity issue totally, so I should forget the idealization I mentioned above.

 

[pervect]

I think I've said this before, but we can rephrase the question in terms of a rigid framework rather than the Earth, if that's your intent. Our first question must be - does a rigid framework actually exist? I believe the answer is "yes, to a reasonable degree of approxiation", though I haven't seen discussed in a peer reviewed paper. When I talk about 'reasonable degree of approximation", one of the things I'm assuming is that if first order effects are on the order of 10^-20, then second order effects are on the order of 10^-40, so we can ignore them. So we're talking about a weak field. The second question we have to ask is "how big is our framework"? It turns out according to my analsyis (and this seems to match with other staetments I've read) that the ciritical dimension of the framework is how long it is in the direction of propagation of the GW. The dimension that's perpendicular to the direction of propagation can be quite large by my calculations, but not infinite. (Again, these are my personal alculations, not peer reviewed).

The techniques I used to convince myself that a rigid framework exists were to consider whether a Born-rigid congruence existed within the desired (first-order) accuracy limits. I used the criterion of vanishing shear and expansion to determine rigidity (but one could also use techniques based on the Lie derivative of the spatial projection of the metric tensor). In a plane perpendicular to the GW, we can cover a very large region before we see signficant deviations from rigidity - but if we consider a 3d spatial volume, we start to see effects of linear order in the direction of propagation of the GW, so the approach requires us to consider a "thin" plane whose thickness is negligible compared to the wavelength of the GW.

Probably the most convenient way of creating said framework is to use Fermi normal coordinates around a specified, special wordline representing a " reference" observer". I would imagine said reference observer would be following a geodesic, but one could define it in any manner one likes. The distance along a space-like geodesic to other observers in the congruence defined by constant Fermi-normal coordinates will be constant by the construction of the coordinates. The distance as measured by round-trip travel time (which is the SI definition of distance) will be approximately equal to the round-trip light light travel time when one is sufficiently close to the reference observer.

This ties in with what I've been saying before about "which coordinates".

This then reduces to a straightforwards question, what is the metric tensor for a plane GW space-time in Fermi-normal coordinates about some particular geodesic, specifically what is the $g_{00}$ term of the tensor. Unfortunately, I don't know the answer to that. There was a paper I glanced at that did discuss Fermi-normal coordinates and GW's, so there may be an answer in the literature if you care to dig for it, if the revised question is of interest. Google finds https://arxiv.org/abs/1409.4648 which is what I was recalling.

 

[Vanadium 50]

This then reduces to a straightforwards question, what is the metric tensor for a plane GW space-time in Fermi-normal coordinates about some particular geodesic, specifically what is the g00g_{00} term of the tensor.

Can you B-level-ify this?

 

[PeterDonis]

I would imagine said reference observer would be following a geodeosic

I don't think you can assume that, because a Born rigid congruence in the presence of a gravitational wave cannot possibly be a geodesic congruence; if it were, then LIGO would give a null result and would not be able to detect GWs.

 

[pervect]

I don't think you can assume that, because a Born rigid congruence in the presence of a gravitational wave cannot possibly be a geodesic congruence; if it were, then LIGO would give a null result and would not be able to detect GWs.

Most of the curves in the congruence won't be geodesics. But the reference congruence can be anything that makes sense, it represents the worldline of the "observer".

 

[pervect]

Can you B-level-ify this?

Probably not all the way to B=level, but - let's take an example of the globe. The 3-d analogy to Fermi normal coordinates would be to pick a particular point on the globe, and measure your bearing (which picks the direction you start out in) and the distance you move in that direction. Then you basically use the distance and bearing as polar coordinates on a sheet of paper (which, unlike the globe, is not curved.) The flat paper has whatever additional coordinates you like on it (say, a square grid of cartesian coordinates). The purpose of the construction is to make a map from the curved manifold of the surface of the Earth to a flat manifold (the piece of paper). The bearing/angle is an intermediate step in the contruction, not necessarily the end result. Another non-B level remark, mathematicians call this the exponential map, https://en.wikipedia.org/wiki/Exponential_map_(Riemannian_geometry), it's a mapping from the tangent space at a point to the manifold itself. The tangent space is the "flat sheet of paper", the manifold is the curved surface of the Earth.

In 2-space + 1 time, rather than a reference point, you have a reference worldline. At any instant in time (event on the worldline), you can construct an infinite number of spatial geodesics that radiate away from that point. You choose the subset of those that start out orthogonal to the reference worldline, that is a 2d space. Then you use the same principle of distance/bearing to map points on the space-time manifold to points on a flat plane. The time coordinate for any point on any of the curves reached via the radiating geodesics is the proper time reading of the clock on the observer on the reference worldline. The space coordinate are given again by the distance and bearing.

Time dilation is given by the usual procedure of the ratio of proper time to coordinate time. I'm not sure how to B-level that, but that's how time dilation in , say, Schwarzschild coordinates works.

There's several non-B level papers you can find with google, and MTW has a discussion of these coordinates.

 

[PeterDonis]

the reference congruence can be anything that makes sense

I think you mean the reference worldline. Yes, the reference worldline can be a geodesic, but no other worldlines in the congruence will be.

 

[l0st]

... So if we just consider the gravitational wave, without taking into account any other effects, A and B will move the same...

What if instead of putting A and B close to each other we set them up on the opposite sides of Earth? Would not we be able to detect a wave passing through by noticing their signals temporarily going out of sync due to different distance to the source?

 

[Ibix]

What if instead of putting A and B close to each other we set them up on the opposite sides of Earth? Would not we be able to detect a wave passing through by noticing their signals temporarily going out of sync due to different distance to the source?

Depends what you're trying to do. If you're just putting mirrors a really long way apart and bouncing signals between them then yes that would be more sensitive. It's just an upscaled LIGO, with a few minor engineering challenges to do with tunnelling through the core. You might be better to put the mirrors out in space, which is a future design I believe.

If you are planning to use the Earth as a rigid bar sensor, however, then moving the clocks further apart makes the problem worse, not better. The speed of sound is so low in rock that one side of the Earth doesn't "know" the other side is in motion until hours after the gravitational wave has passed. So your clocks would just acts like free-floating clocks and you don't get any accumulated time difference from it. Just the transients that a LIGO-style detector sees.

 

[l0st]

Transients is what I'd be looking for.
Seems like the idea is theoretically viable. Feasibility of practical application requires extensive research though.

 

[Ibix]

Transients is what I'd be looking for.
Seems like the idea is theoretically viable. Feasibility of practical application requires extensive research though.

By transients, I mean you could in principle, replace LIGO mirrors with clocks. Due to the distance change associated with the gravitational wave each clock would see the other's rate wobble. LIGO does this much more precisely with interferometry, and the practical difficulties of using clocks instead are, as V50 pointed out, way, way beyond anything you are imagining. Whatever it is you are imagining.

 

[pervect]

OK, here is my best shot at a truly B-level explanation of my remarks, sorry for the delay.

General Relativity can be understood as drawing a "map" of space-time. It's a 4 dimensional map. Sometimes this is called "the block universe". GR has a mathematical entity called the metric. This metric can be understood as representing whether or not the map is to scale - it also tells about other sorts of distortion of the map.

If you draw your map to scale, it's easier to interpret it.

Fermi-Normal coordinates are a technique for drawing maps that are to scale and not distorted in a small region around some specific observer (who is represented on the map by a worldine, the worldine is the timeline of the observer)., Because it is drawn to scale and undistorted in a small region, the "map" is easy to interpret. . There's more to it than that (and the language is rather vague) but that's the executive B-level summary and motivation for using Fermi-Normal coordinates.

Note that it's impossible to draw a map of a curved 2-d surface on a flat sheet of paper that is everywhere to scale (for instance, making a map of the surface of the Earth that is undistorted). Similar remarks apply to drawing a map of a curved 4-d space-time on a "flat" 4-d manifold.