LIGO light changes frequency not wavelength

[MeJennifer]

More generally, the point I'm trying to get across is that there is a well-known and coordinate-independent way of computing and measuring distance ...

That is true but is such "distance" a physically measurable quantity in non-stationary spacetimes?
I think not!

 

[GeorgeDishman]

No, this is not correct. If the two beams are exactly in phase when they come back after being reflected, there is no interference. This is the state of the LIGO detector when no gravitational wave is passing. Interference is only seen when a GW passes, since the passage of the GW causes the two beams to be out of phase when they return after being reflected.

Can I suggest you have a look at fig 1 in this paper:

https://arxiv.org/ftp/arxiv/papers/1411/1411.4547.pdf

and then look at paragraph 2.3:

"2.3 Gravitational wave readout
Readout of the gravitational wave signal is accomplished using an output mode cleaner in conjunction with homodyne, or DC detection. In this scheme, a local oscillator field is generated by offsetting the arm cavities slightly from their resonance (typically a few picometers), thereby pulling the Michelson slightly off the dark fringe."

 

[GeorgeDishman]

Here you are constraining the motion of the ruler, so it will not be moving on a geodesic--it will be accelerated if a GW passes. (This will be true even in a space-based experiment where there is no gravity, from Earth or anything else, to complicate things.) So you can't assume that the ruler provides an "unchanging" standard of either position or length.

I agree, but my assumption (which may not be correct) is that the test mass is free to move in the direction long the tube (though constrained in other directions) in response to the GW so if the mass follows the geodesic but the ruler does not, then there should be a resulting relative motion between them.

My thought is that, if we treat the mid-point of the tube as a fiducial location, the motion of the masses at the ends will have the opposite sense, thus both moving away from or towards the tube centre as measured against the rulers.

The alternative that might apply is that the mirrors are constrained to remain fixed relative to the rulers other than not following local seismic noise which would couple to the rulers. However, this page seems to suggest the mirrors move in response to a GW:

https://www.ligo.caltech.edu/page/vibration-isolation

"Since gravitational waves will make themselves known through vibrations in LIGO's mirrors, the only way to make gravitational wave detection possible is to isolate LIGO's components from environmental vibrations to unprecedented levels. The change in distance between LIGO's mirrors (test masses) when a gravitational wave passes will be on the order of 10^-19 m"

"The goal is to keep our hands off the masses as much as possible so they will move only due to gravitational waves."

 

[PeterDonis]

"Measuring distances in the tangent space" means that one first uses the exponential map https://en.wikipedia.org/wiki/Exponential_map_(Riemannian_geometry) to map points that are close to each other from the manifold to points in the tangent space. Then one measures the distance in the tangent space.

Yes, I understand that. The question is how close the points in the manifold have to be for this to work. The answer as I understand it is that they have to be close enough that no tidal gravity effects are measurable. As I've said several times now, that obviously can't be true if we are analyzing a gravitational wave detector, since gravitational waves are waves of tidal gravity, so if tidal gravity is not measurable, neither are GWs.

there is a well-known and coordinate-independent way of computing and measuring distance when objects are sufficiently close

Yes, but the question is what "sufficiently close" means. I'm not sure the LIGO mirrors are "sufficiently close" to the LIGO detector in the necessary sense to make the distance from mirrors to detector coordinate-independent. Certainly the distance in linearized GR is not coordinate-independent; the linearized GR distance described in MTW requires a particular choice of coordinates (harmonic gauge plus an additional constraint to make the metric correction $h_{\mu \nu}$ traceless). So if the LIGO team is using linearized GR for their analysis, as I suspect, the "distance" they are describing is dependent on that particular coordinate choice.

 

[PeterDonis]

is such "distance" a physically measurable quantity in non-stationary spacetimes?

For objects that are close enough together, yes; you can construct a local inertial frame in any spacetime whatever, whether it's stationary or not. The issue is not that GWs make the spacetime non-stationary; the issue is that GWs are made of spacetime curvature, and spacetime curvature is not observable within a single local inertial frame, by definition.

 

[PeterDonis]

my assumption (which may not be correct) is that the test mass is free to move in the direction long the tube (though constrained in other directions)

This would certainly not be true in an idealized space-based GW detector, where the apparatus as a whole is in free fall. In such a detector, the test masses would move along geodesics; their motion could not be constrained in any way, since that would mean they would be moving on non-geodesic worldlines.

In LIGO, which is not in free fall, it is true that there is a constraint imposed on the motion of the test masses; they are not in free fall, because their average position has to lie on an accelerated worldline (since the apparatus as a whole is accelerated at 1 g, not in free fall). But I don't think their motion about that average position is constrained; it can be in any direction, not just the direction along the tube.

My thought is that, if we treat the mid-point of the tube as a fiducial location, the motion of the masses at the ends will have the opposite sense

The LIGO detector is not a single long tube. It's two tubes, perpendicular to each other, in the shape of an "L", with the detector at the junction of the two tubes (the corner of the "L"), and the two mirrors at the other ends of each of the tubes. The fiducial location is the detector; the motion of each mirror due to a GW is relative to the detector, not relative to the other mirror.

this page seems to suggest the mirrors move in response to a GW

That's correct; they do.

 

[MeJennifer]

For objects that are close enough together, yes; you can construct a local inertial frame in any spacetime whatever, whether it's stationary or not.

Well obviously!
If it is so close that we consider the region flat (and thus stationary) we basically have SR.

 

[PeterDonis]

a local oscillator field is generated by offsetting the arm cavities slightly from their resonance (typically a few picometers), thereby pulling the Michelson slightly off the dark fringe."

Ah, I see; it looks like LIGO's "default" state (when no GW is detected) has an artificially induced constant phase shift between the signals from the two arms. But this is an intentionally added feature; it is not something that has to be there whenever you have beams in both arms. If the arm cavities were not offset as this paragraph describes, there would be no fringes at all at the detector (i.e., no interference) in the absence of a GW, even with beams in both arms. That is why I objected to your earlier statement that there must be interference whenever there are beams in both arms.

 

[GeorgeDishman]

Ah, I see; it looks like LIGO's "default" state (when no GW is detected) has an artificially induced constant phase shift between the signals from the two arms. But this is an intentionally added feature; it is not something that has to be there whenever you have beams in both arms. If the arm cavities were not offset as this paragraph describes, there would be no fringes at all at the detector (i.e., no interference) in the absence of a GW, even with beams in both arms. That is why I objected to your earlier statement that there must be interference whenever there are beams in both arms.

OK, it's just a slight difference in the way we look at the situation, more semantics than physics. In the "default state" without the pulling, the article notes that the photodiode would be in a "dark fringe" which for me means the two beams arrive exactly 180 degrees out of phase and hence the "interference" between them results in the null.

The article states that the pulling is of the order of a few picometres or about 10^-5 of the laser wavelength (1064nm). A path length variation of say 10fm would be 10^-8 of a wavelength and alter the photodiode current by around 0.1% and the mirror motion would be much smaller due to the Fabry-Perot design of course.

 

[GeorgeDishman]

This would certainly not be true in an idealized space-based GW detector, where the apparatus as a whole is in free fall. In such a detector, the test masses would move along geodesics; their motion could not be constrained in any way, since that would mean they would be moving on non-geodesic worldlines.

Yes, the problem is that so is the surrounding satellite so there's no local comparison. This has been mentioned before in the thread in that we can idealise the situation by thinking of the LIGO beam tube "in space" to remove the annoyance of local Earth gravity and rotation, in other words like eLISA but with a long steel tube bolted between the satellites. Again, I would put small rulers (a nanometer long!) adjacent to each mirror but fixed to the tube to get a local indication of whether the mirrors move.

In LIGO, which is not in free fall, it is true that there is a constraint imposed on the motion of the test masses; they are not in free fall, because their average position has to lie on an accelerated worldline (since the apparatus as a whole is accelerated at 1 g, not in free fall). But I don't think their motion about that average position is constrained; it can be in any direction, not just the direction along the tube.

I think they are constrained against yaw, torque and pitch as well as vertically. I'm not sure about linear motion transverse to the beam but if you think of the mirror as a flat plane, that should have negligible consequences. (In fact there aren't quite planes as the beam diameter isn't the same at the near and far ends but that's a finer detail.)

The LIGO detector is not a single long tube. It's two tubes, perpendicular to each other, in the shape of an "L", with the detector at the junction of the two tubes (the corner of the "L"), and the two mirrors at the other ends of each of the tubes. The fiducial location is the detector; the motion of each mirror due to a GW is relative to the detector, not relative to the other mirror.

Yes, you're right to an extent but this is a bit more complex in practice and will become easier in the "LIGO in space" thought experiment I want to look at. The majority of the path length in each arm is between the two "test masses" as described on this page:

https://www.ligo.caltech.edu/page/ligos-ifo

"The Fabry Perot 'cavity' actually is the full 4 km length of each arm between the beam splitter and the end of each arm. Additional mirrors placed near the beam splitter are precisely aligned to reflect each laser beam back and forth along this 4 km length about 280 times before it finally merges with the beam from the other arm. These extra reflections serve two functions:

1. It stores the laser light within the interferometer for a longer period of time, which increases LIGO's sensitivity

2. It increases the distance traveled by each laser beam from 4 km to 1120 km

With Fabry Perot cavities, LIGO's interfereometer arms are not just 4 km long, they are essentially 1120 km long, making them 144,000 times bigger than Michelson's original instrument! This bit of 'mirror magic' greatly increases LIGO's sensitivity and makes it capable of detecting changes in arm-length thousands of times smaller than a proton, while keeping the physical size of the interferometer manageable."​

this page seems to suggest the mirrors move in response to a GW

That's correct; they do.

The way I look at it, although the beams are combined at the junction of the L, the signal is primarily a measure of the "distance" (in terms of phase change) between the test masses. The effect of the short distance from the ITM to the beamsplitter has an effect 280 less than the distance variations between the test masses.

Now since both masses are free to move but the beam tube is somewhat more rigid, we can think of the masses moving in opposition within the tube. In fact there will be a common motion as well (that's where my problem really lies but more on that later) but if you think of the LIGO in space version, the rigidity of the tube would sort of average out the motion of its parts so that it would all move along roughly the geodesic of the centre plus a small strain resulting from the stress within the metal. It therefore makes more sense to put the fiducial point in the centre. However, that's really a matter of preference and I like the symmetry. All I'm going to be asking about is the motion of the test masses and the shape of their geodesics, that's what I'm trying to model, i.e a collection of independent point masses which are in free fall other than not "dropping" due to gravity.

With regard to the L shape, a bit of thought-experiment simplification can make life easier. Consider a distant binary system and our system happens to lie in the orbital plane (and ignore spins of the binary components). The signal reaching us would have "plus" polarisation. Let's set the first arm in the plane and perpendicular to the line of sight to the binary. Set the second arm aligned with the orbital axis of the binary. Now whatever signal we have in the arm in the plane, the other arm will produce an exactly equal signal but 180 degrees out of phase. In that case I think we need only consider the arm in the plane. What I want to look at is extending a series of those arms and see how they add up. (The exception to free fall mentioned above is that I'm not worrying about the equipment falling towards the distant binary star system.)

Does that make sense so far?

 

[RockyMarciano]

At a single event, yes, you can always find a timelike vector that is orthogonal to a chosen set of 3 orthogonal spacelike vectors. But your claim is much stronger: you are claiming that, in any spacetime, you can find a foliation by spacelike hypersurfaces covering the entire spacetime, and a timelike vector field covering the entire spacetime that is everywhere orthogonal to every hypersurface in the foliation. I don't think that claim is true for every possible spacetime. (For one thing, not every possible spacetime even admits a foliation to begin with; see below.)

I agree it is a much stronger claim, but again the claim is done not for any possible 4-dimensional Lorentzian manifold which is what you seem to be equating with a "spacetime". We surely agree we are only concerned here with physically plausible Lorentzian manifolds and those are the ones that I'm referring to as "spacetimes", but that's just terminology. As you say it is true of course tha not every possible Lorentzian 4-manifold even admits a foliation. But since we are talking about physics, those that are plausible as physical scenarios(leaving aside the special models of isolated objects that have timelike Killing vector fields) must admit a foliation and the foliation must be time-orthogonal if it is to have anything resembling dynamics.
But let's constrain the discussion to the GWs models, they certainly rely on linearized gravity and linearized gravity has as departure premise the decomposition of its metric tensor components in a 3+1 formulation in the strong form you mentioned above:time orthogonal spacelike hypersurfaces that covers the totality of the spacetime considered as physically relevant:Minkowski spacetime perturbed by a tidal curvature wavefield. This strong premise is actually what imposes the use of the harmonic coordinate condition in order to even have a wave equation.
When trying to understand the consequence of this particular foliation in the way the LIGO interferometer measures the tidal wavefield passing by, it is important to have means for separating the dynamics of the perturbation $h_{\mu\nu}=g_{\mu\nu}-η_{\mu\nu}$ from the dynamics of the measuring tool in GR, the inner product $g_{\mu\nu}$ because all of our other measurements are usually based in comparing with a fixed background(either because they use euclidean-Newtonian or Minkowskian-SR backgrounds or in the GR case time independent spacetimes with no dynamics) clearly separated from the dynamycs being measured. I admit that I still don't know how that is achieved mathematically, even after reading much of what Kip Thorne has written on the subject.

A foliation is not the same thing as a spacetime. Hawking & Ellis is full of examples of spacetimes that cannot be foliated by a family of spacelike hypersurfaces.

Agreed. and those examples in Hawking & Ellis are all referred(for general GR solutions without timelike KVFs) to highly pathological or physically absurd Lorentzian manifolds. IIRC there are even example of manifolds not following the Hausdorff and second countability condition, which is really pathological. Let's remember here that the majority of solutions of the EFE are not physically relevant, giving results like CTCs, or not fulfilling the minimumenergy conditions, etc.

 

[PeterDonis]

the signal is primarily a measure of the "distance" (in terms of phase change) between the test masses.

If you're viewing it in terms of distance at all, it's a measure of the difference in distance from the detector to mass A vs. the detector to mass B, magnified by a factor of 280. That difference is not the same, conceptually, as a simple fluctuation in distance between two masses lying along a single line would be. The fact that the arms are perpendicular is important because gravitational waves are spin-2, which means the oscillations along perpendicular lines are sampling different polarizations of the wave.

In that case I think we need only consider the arm in the plane.

But LIGO is not designed just for that case. The whole point is to be able to sample both polarizations with a single detector, in an arbitrary orientation. For that you need both perpendicular arms.

 

[PeterDonis]

any possible 4-dimensional Lorentzian manifold which is what you seem to be equating with a "spacetime".

Yes, because that's the standard definition in the literature. If physicists want to talk about a more restricted class of manifolds, they use a qualifier like "physically reasonable", or a more precise mathematical restriction like "globally hyperbolic". See below.

those that are plausible as physical scenarios(leaving aside the special models of isolated objects that have timelike Killing vector fields) must admit a foliation

Most physicists would agree with this, yes. The usual mathematical condition is that the spacetime be globally hyperbolic.

the foliation must be time-orthogonal if it is to have anything resembling dynamics.

This is not correct. For example, consider the foliation of Kerr spacetime (outside the static limit) by hypersurfaces of constant Boyer-Lindquist coordinate time (the analogue of Schwarzschild coordinates for Kerr spacetime). This foliation is not orthogonal to the worldlines of "static" observers (observers who are at rest relative to infinity). But those "static" observers certainly have well-defined dynamics.

linearized gravity has as departure premise the decomposition of its metric tensor components in a 3+1 formulation in the strong form you mentioned above:time orthogonal spacelike hypersurfaces

Why do you think the spacelike hypersurfaces of constant time in the linearized GR formulation are time orthogonal? What timelike vector field do you think they are always orthogonal to? Remember that the metric is not exactly Minkowski; there is a GW perturbation, which affects both the time and space part of the metric.

This strong premise is actually what imposes the use of the harmonic coordinate condition in order to even have a wave equation.

Why do you think you must have a "time orthogonal" set of hypersurfaces in order to use harmonic coordinates? Some references to back up these strong statements you are making would really help, because I have not seen anything like them in the literature I've read.

all of our other measurements are usually based in comparing with a fixed background

Why do you think this? What "fixed background" are you talking about? If you mean, for example, the "background" Minkowski spacetime in the GW case, that background metric is not observable. Only the actual physical metric $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ is.

I admit that I still don't know how that is achieved mathematically

It seems to me that this is because you are misunderstanding what is being done. See above.

Let's remember here that the majority of solutions of the EFE are not physically relevant

I'm not sure I would say "the majority", because I'm not sure how you would "count" solutions; but certainly there are many mathematically valid solutions which nobody thinks are physically relevant, yes. I'm perfectly OK with restricting discussion to the ones that are physically relevant. But since "physically relevant" is not a precise mathematical criterion (for example, even if you consider global hyperbolicity to be a necessary condition, it can't be sufficient because there are globally hyperbolic solutions, such as maximally extended Schwarzschild spacetime, which are not physically relevant), you can't point to precise mathematical theorems that only apply to physically relevant spacetimes.

 

[pervect]

But isn't the OP arguing that all those approximations you mention give an error bigger than the effect that LIGO seeks to detect?

In order to figure out the error in the approximations, we'd need something to compare the approximation to. We might have already lost the OP in the way thie discussion has developed, I don't know - I haven't seen anything from exponent137 in a while. Perhaps you're referring to someone other than the OP? If you're referring to the poster I think you might be, my impression was that communication had totally broken down, based on some rather wild misquotations expressed in a rather argumentative manner. That said, the question you ask is a perfectly reasonable one.

My general observation is that most GR textbooks will tell you how to compute an observer independent quantity called the "length" of a curve. The standard procedure for this is to divide the curve up into short segments, compute the Lorentz interval of each segments, and add the results all together. One might also describe this process as a process of integration. Furthermore, in the case of special relativity, these "lengths" can be related to proper distances (for spacelike curves) and proper times. SInce we are interested in distances, we will be interested in the lengths of space-like curves.

I have been thinking that most posters in this thread would agree with this much, at least - though on second thought, I'm not so sure. For instance, I'm not sure the OP (exponent137) would be familiar with this concept. So it's worth asking, though I'm not sure I'll have much to say if the answer is negative.

If one does accept at least this much, the second issue is just one of picking out some specific curve, and calling the length of this curve distance.

Textbooks are rather silent on this point, for instance one will see details in Wald on how to compute the length of a curve, but nothing about which curve one might pick so that it's length defines "distance". I could make some specific proposals for two general strategies on selecting a curve, and compare them. Both of the techniques I have in mind will be quite compatible in the case of Ligo. What I can't do, unfortunately, is read someone's mind as to which set of curves they think should allow one to compute the distance (assuming that they agree distance is, in fact, represented by the length of SOME curve), so as to be able to usefully compare the different ideas.

I think it would not be productive to go through more details of how one might selecting a specific curve, without an expression of some interest on the topic by someone.

 

[Ibix]

I think it would not be productive to go through more details of how one might selecting a specific curve, without an expression of some interest on the topic by someone.

Raises hand (that's what my avatar is doing with that stick thing)

 

[MeJennifer]

Perhaps someone wants to take a stab at distance in Fermi-normal coordinates?

 

[RockyMarciano]

Yes, because that's the standard definition in the literature. If physicists want to talk about a more restricted class of manifolds, they use a qualifier like "physically reasonable", or a more precise mathematical restriction like "globally hyperbolic". See below.
Most physicists would agree with this, yes. The usual mathematical condition is that the spacetime be globally hyperbolic.

You are right, so let me rephrase this condition about physically relevant spacetimes(like FRW or this thread case of linearized gravity) leaving out those with timelike Killing vector fields), by demanding their spacelike 3-surfaces to be Cauchy surfaces.

This is not correct. For example, consider the foliation of Kerr spacetime (outside the static limit) by hypersurfaces of constant Boyer-Lindquist coordinate time (the analogue of Schwarzschild coordinates for Kerr spacetime). This foliation is not orthogonal to the worldlines of "static" observers (observers who are at rest relative to infinity). But those "static" observers certainly have well-defined dynamics.

We already covered this. I'm not considering this particular case.

Why do you think the spacelike hypersurfaces of constant time in the linearized GR formulation are time orthogonal? What timelike vector field do you think they are always orthogonal to? Remember that the metric is not exactly Minkowski; there is a GW perturbation, which affects both the time and space part of the metric.

Having a Minkowskian background is enough, regardless of the GW perturbation as long as its coordinate functions must obey the wave equation, and this is guaranteed by the harmonic coordinate condition.

Why do you think you must have a "time orthogonal" set of hypersurfaces in order to use harmonic coordinates?

My phrasing was not exactly to mean this. I rather meant what I wrote inmediately above.

Why do you think this? What "fixed background" are you talking about? If you mean, for example, the "background" Minkowski spacetime in the GW case,

No, I thought it was clear by the bit between parenthesis. I meant that in the Newtonian clasical case the fixed background against which measurements were made was Euclidean absolute space, in SR and QFT it is Minkowskian spacetime, in the particular GR static case the background can be considered fixed by virtue of its special geometry, but in the GW case of GR(like in the FRW case) there is no longer a fixed background, and this introduces certain difficulties in the measurement process.

.

I'm not sure I would say "the majority", because I'm not sure how you would "count" solutions; but certainly there are many mathematically valid solutions which nobody thinks are physically relevant, yes. I'm perfectly OK with restricting discussion to the ones that are physically relevant. But since "physically relevant" is not a precise mathematical criterion (for example, even if you consider global hyperbolicity to be a necessary condition, it can't be sufficient because there are globally hyperbolic solutions, such as maximally extended Schwarzschild spacetime, which are not physically relevant), you can't point to precise mathematical theorems that only apply to physically relevant spacetimes.

I'm OK with considering global hyperbolicity a necessary but not sufficient condition when not giving any other qualification. I'm not sure I'd call the maximally extended Schwarzschild spacetime physically irrelevant, even if it cannot physical in all its features, but this discussion would be quite off topic here.

 

[PeterDonis]

let me rephrase this condition about physically relevant spacetimes(like FRW or this thread case of linearized gravity) leaving out those with timelike Killing vector fields), by demanding their spacelike 3-surfaces to be Cauchy surfaces.

This is equivalent to the spacetime being globally hyperbolic. But you also say you're not considering spacetimes with timelike KVFs; many globally hyperbolic spacetimes have timelike KVFS. I'm not clear on why you want to exclude those.

I'm not considering this particular case.

Then you shouldn't be making the general statements you are making as though they applied to every possible case. They don't even apply to every possible globally hyperbolic spacetime; Kerr spacetime is globally hyperbolic, but you say you're not considering it.

Having a Minkowskian background is enough

Can you give a proof of this statement, or a reference that gives one? It doesn't seem obvious to me.

I rather meant what I wrote inmediately above.

In other words, you think having harmonic coordinates guarantees "time orthogonality". This doesn't seem obvious to me either. Can you give a proof, or a reference that gives one?

 

[RockyMarciano]

My general observation is that most GR textbooks will tell you how to compute an observer independent quantity called the "length" of a curve. The standard procedure for this is to divide the curve up into short segments, compute the Lorentz interval of each segments, and add the results all together. One might also describe this process as a process of integration. Furthermore, in the case of special relativity, these "lengths" can be related to proper distances (for spacelike curves) and proper times. SInce we are interested in distances, we will be interested in the lengths of space-like curves.If one does accept at least this much, the second issue is just one of picking out some specific curve, and calling the length of this curve distance.

Textbooks are rather silent on this point, for instance one will see details in Wald on how to compute the length of a curve, but nothing about which curve one might pick so that it's length defines "distance". I could make some specific proposals for two general strategies on selecting a curve, and compare them. Both of the techniques I have in mind will be quite compatible in the case of Ligo. What I can't do, unfortunately, is read someone's mind as to which set of curves they think should allow one to compute the distance (assuming that they agree distance is, in fact, represented by the length of SOME curve), so as to be able to usefully compare the different ideas.

All this doesn't seem to be directly related to what is being discussed. Measurements of time and distance in relativity are ideally performed by ideal rods and clocks, just like in Newtonian physics, that are well approximated in general by our actual rods and clocks. I think the problem in the LIGO case as commented in other posts, enters when one must use the metric tensor in the 3+1 form. But if I have undertood you right you don't consider the dynamical versus fixed background an issue here, adducing that GR uses the tangent space regardless of the static or nonstatic case. The thing is that not only the tangent space is used, you also need an affine connection to connect the points so a tensor field is not the same as an algebraic tensor at a point. In GR the connexion is metric compatible and curvature is obtained using this connection, it should make a difference whether the metric is time-independent or not.

 

[MeJennifer]

The thing is that not only the tangent space is used, you also need an affine connection to connect the points so a tensor field is not the same as an algebraic tensor at a point.

I fail to see how this is even remotely relevant.

 

[PeterDonis]

adducing that GR uses the tangent space regardless of the static or nonstatic case

As I have posted several times now, I don't think this claim of pervect's is correct, because spacetime curvature effects are by definition not observable in a single tangent space (local inertial frame), and GWs are spacetime curvature effects. Your objection seems to be along similar lines, except that you are looking at the connection as an intermediate step in the reasoning.

it should make a difference whether the metric is time-independent or not.

Do you mean "shouldn't"? It is correct that the general procedure of using a metric-compatible connection and deriving the curvature tensor works for any spacetime, whether it has a "time-independent" metric or not. However, that does not mean that all of your other assumptions (such as "time orthogonality") automatically hold for any spacetime.

 

[RockyMarciano]

This is equivalent to the spacetime being globally hyperbolic. But you also say you're not considering spacetimes with timelike KVFs; many globally hyperbolic spacetimes have timelike KVFS. I'm not clear on why you want to exclude those.

Then you shouldn't be making the general statements you are making as though they applied to every possible case. They don't even apply to every possible globally hyperbolic spacetime; Kerr spacetime is globally hyperbolic, but you say you're not considering it.

The complete statement would pertain to globally hyperbolic spacetimes in the absence of timelike KVs. And the reason I'm excluding was already explained. The general case in GR doesn't have them, not when describing the universe anyway, they are used to model special scenarios with isolated objects like stars or BHs.

Can you give a proof of this statement, or a reference that gives one? It doesn't seem obvious to me.

I qualified it with "as long as the coordinate condition is fulfilled".

In other words, you think having harmonic coordinates guarantees "time orthogonality". This doesn't seem obvious to me either. Can you give a proof, or a reference that gives one?

No, let's clarify this. As you admitted in a previous post time-orthogonality is assured at a single point or event by the demand of GR spacetimes being locally Minkowskian, i.e. having a Lorentzian innerproduct(any timelike vector has a 3-dimensional subspace orthogonal to it under the Minkowski inner product. Such a subspace is spanned by vectors that are spacelike under the inner product, and all linear combinations of those vectors are spacelike.). But extending this at a point to the whole spacetime is the much stronger condition that needs the harmonic cooordinate condition.

 

[PeterDonis]

I qualified it with "as long as the coordinate condition is fulfilled".

Yes, and I don't see how that helps. See below.

As you admitted in a previous post time-orthogonality is assured at a single point or event by the demand of GR spacetimes being locally Minkowskian, i.e. having a Lorentzian innerproduct(any timelike vector has a 3-dimensional subspace orthogonal to it under the Minkowski inner product.

Yes, but as I also said, this is a very weak notion of "time orthogonality", and you appear to agree. I was talking about the much stronger notion.

extending this at a point to the whole spacetime is the much stronger condition that needs the harmonic cooordinate condition.

Once again, do you have a proof of this, or a reference that gives a proof? This is what doesn't seem obvious to me.

 

[GeorgeDishman]

If you're viewing it in terms of distance at all, it's a measure of the difference in distance from the detector to mass A vs. the detector to mass B, magnified by a factor of 280. That difference is not the same, conceptually, as a simple fluctuation in distance between two masses lying along a single line would be.

The light path can be thought of as shown in this sketch although the F-P technique doesn't have specific injection and extraction events. Obviously it's not to any kind of scale. The main distance measured is between the ITM and ETM test masses (mirrors) with around 280 passes (only a few shown) between them and only a single dependence at each of the injection and extraction on the distance to the beamsplitter (BSP). The vertical straight lines beneath the sine waves are meant to indicate the tube suspension points without implications on the distance between them, the sine waves are how I think the test mass geodesics would look relative to the suspension points.

The fact that the arms are perpendicular is important because gravitational waves are spin-2, which means the oscillations along perpendicular lines are sampling different polarizations of the wave.

It's not even that simple, each polarisation affects both arms and at an arbitrary viewing position, both are present, and in addition the instrument can have any orientation.

Plus: https://en.wikipedia.org/wiki/File:GravitationalWave_PlusPolarization.gif
Cross: https://en.wikipedia.org/wiki/File:GravitationalWave_CrossPolarization.gif

My eventual intent is to show how the mix of the polarisations depends on the inclination of the observer to the plane of the binary, the orientation of the instrument is another later beyond what I'm trying to do. I appreciate your point but I have a much bigger problem to get past first.

But LIGO is not designed just for that case. The whole point is to be able to sample both polarizations with a single detector, in an arbitrary orientation. For that you need both perpendicular arms.

Yes, but to illustrate my problem, I'm starting with the simplest configuration. Starting with the wave format in the orbital plane means the cross polarisation amplitude is zero. What I need to do though is extend it to a larger area. There was a nice illustration of the effect in the LIGO press conference for GW150914 in terms of stretching a plastic mesh. I have a video of what I think results, if there isn't a fundamental flaw in my understanding, I'll post a link to that next and explain why it perhaps isn't quite what you would expect.

 

[PeterDonis]

The main distance measured is between the ITM and ETM test masses (mirrors) with around 280 passes (only a few shown) between them

First of all, this "distance" is with respect to a particular choice of coordinates. Other choices are possible.

Second, there are two of these setups, one on each arm. And the signal at the detector is the interference between the light beams coming back from the two arms. It is not a measurement of the "length" (with respect to some coordinates) of either arm alone.

 

[GeorgeDishman]

First of all, this "distance" is with respect to a particular choice of coordinates. Other choices are possible.

Yes, I'm aware of that, that's why I said "without implications on the distance between them". Other coordinate choice might or might not have a constant distance between the resting positions and that might be the solution to my problem.

Second, there are two of these setups, one on each arm. And the signal at the detector is the interference between the light beams coming back from the two arms. It is not a measurement of the "length" (with respect to some coordinates) of either arm alone.

Again, I'm well aware of that. As I said before, if you look at the animation of the "plus" polarisation and imagine one arm horizontal and the other vertical on the Wikipedia page, then the vertical beam variation will be 180 degrees out of phase with the horizontal. The interference then just doubles the signal output. That's just the easiest example with which to explain the problem. It's what is sometimes called a "toy model", not realistic but a simplified vehicle for a complex discussion.

 

[pervect]

Raises hand (that's what my avatar is doing with that stick thing)

Well, there are two or three general approaches that come to mind for curve selection. One approach is to measure distances along geodesics curves in space-time. This ultimately leads to fermi-normal coordinates, and one can find good textbook discussions of this approach in Misner, Thorne, Wheeler's, "Gravitation". There are various posts in the PF about them as well as various papers online, many are rather technical. One needs to do a bit more work to define the coordinate system than one does to just measure distances - one might get distrated by understanding the notion of "Fermi-Walker transport", only to realize too late that it's only important for defining the coordinates and not so important for defining the notion of distance.

To define the fermi-normal distance, one starts with some point p on a reference worldline, then one considers the set of space-like geodesics that pass through p and are orthogonal to the reference worldline. Specifying the worldline (or rather the slope of the worldline) specifies the velocity of the observer.

The set of points that these geodesics reach defines a surface of "constant time". The distance from p to another point on this surface of "constant time" is given by the length of the space-like geodesic connecting p to that point.

Note that picking a different rerference worldline through the same point p will pick out a different surface of "constant time" and define a different notion of distance.

One notable point about Fermi-normal coordinates is that they don't cover all of space-time, just some local region where the geodesic curves don't cross.

Another popular approach measures distance along curves that are perpendicular to a set of time-like geodesics. (There is a name, a proper noun, for a set of curves that fill space-time. This name is a congruence, so the set of geodesics that fill space-time would be a geodesic congruence). These curves are not geodesics themselves, hence it's a different approach than the previous one. There are generally a lot of such curves, one would need additional specifications to pick out a particular member of this set, for instance the shortest such curve in this set.

There isn't much literature that I've seen on this approach (but see below) it's not even clear if it's a completely general approach. It's important, though, because this sort of approach,expressed in a coordinate-dependent manner, underlies the approach routinely used to measure distances in cosmology.

[add]If one shifts one focus from the global issues to the local issues, Eric Poisson's "A Relativists Toolkit" would be a good reference for this approach, it has a good discussion of geodesic congruences. Wald has a discussion of geodesic congruences too, but it's rather terse and hard to follow.

Another useful approach would be consider a born-rigid congruence of worldlines. This sort of construction is probably not completely general, but it ties in nicely with the notion of rigid rulers when it does exist. One starts again with a reference worldline, and attempts to construct a set of worldlines that maintain a constant distance from the reference worldline. One needs to have some pre-existing idea of how to measure the distance between worldlines when they are "close enough" to use this approach, however - which is something that I attempted to cover in previous posts.

[add]I haven't tracked down any good detailed references on this - I've seen references to papers written in German (and usually behind paywalls). Unfortunately I don't read German. One can leverage much of the theory of geodesic congruences in Poisson to other congruences, such as rigid congruences, however.

Given the complexities, limitations, and number of possibilities, I would say that the best approach by far is to first understand how distances are defined when points are sufficiently close, as the unifying notion of "distance", common to all the variants that make various compromises and trade-offs when the points are not sufficently close.

 

[Ibix]

Another useful approach would be consider a born-rigid congruence of worldlines. This sort of construction is probably not completely general, but it ties in nicely with the notion of rigid rulers when it does exist. One starts again with a reference worldline, and attempts to construct a set of worldlines that maintain a constant distance from the reference worldline. One needs to have some pre-existing idea of how to measure the distance between worldlines when they are "close enough" to use this approach, however - which is something that I attempted to cover in previous posts.

[snip]

Given the complexities, limitations, and number of possibilities, I would say that the best approach by far is to first understand how distances are defined when points are sufficiently close, as the unifying notion of "distance", common to all the variants that make various compromises and trade-offs when the points are not sufficently close.

Very interesting - thank you @pervect. The Born rigid congruence presumably works by identifying nearby worldlines with zero geodesic deviation? So if I've got a ruler in free fall near a black hole, for example, I pick the worldline of its center of mass, then find geodesics passing through the nearby parts of the ruler that have zero deviation with respect to the center of mass worldline, then rinse and repeat until I've built myself a complete set?

 

[pervect]

Consider Wald, on timelike geodesics, pg 216-217

Consider, first, a smooth congruence of timelike geodesics. Without loss of generality, we may assume that the geodesics are paramterized by proper time so that that the vector field, $\xi^a$, of the tangent is normalized to unit length, $\xi^a \xi_a = -1$.

...

We [Wald] define the "spatial metric"$h_{ab}$ by
$$h_{ab} = g_{ab} + \xi^a \xi^b$$

Thus, $h^a{}_b = g^{ac} h_{ab}$ is the projection operator onto the subspace of the tangent space perpendicular to $\xi^a$. We define the expansion, shear, and twist of the congruence by ...

Now, at the moment, we aren't directly interested in the expansion, shear, and twist. We are just interested in a how we compute spatial distance. We have a formula for it - the formula is used to calculate more advanced properties of a congruence. The formula does not require us to have a stationary metric, its a perfectly general formula. To use this formula, we don't need the connection. We don't need the Riemann tensor. All we need is the tangent to the worldline at some point p, (i.e. $\xi^a$) and the value of the 4-metric at the same point p. Then given the components of some small displacement $d^i$, we can compute the "spatial distance" via the formula $h_{ab} d^a d^b$.

It's worth thinking about why we don't need to know the Riemann tensor, or the connection, to measure distance for sufficeintly nearby points - assuming we agree on the specifics of the formula. The general discussion seems to have gotten sidetracked by making it too simple - so I'm attempting a bit more rigor this go-around. If somene thinks this is the wrong formula, I'd ideally appreciate a reference to what they think the "right" formula is, though I might settle for less.

 

[PeterDonis]

The Born rigid congruence presumably works by identifying nearby worldlines with zero geodesic deviation?

No. It identifies (if they exist) a congruence of worldlines with zero expansion and shear--intuitively, they "keep the same distance" from each other (but can possibly be rotating around each other, so vorticity can be nonzero). The worldlines do not have to be geodesics, and in fact will not be in most cases of interest. After all, we're talking about curved spacetime, and spacetime curvature is the same thing as tidal gravity, which is the same thing as geodesic deviation. So looking for zero geodesic deviation in curved spacetime is not going to work in general.

if I've got a ruler in free fall near a black hole, for example, I pick the worldline of its center of mass, then find geodesics passing through the nearby parts of the ruler that have zero deviation with respect to the center of mass worldline

No, because no such geodesics exist. A congruence of geodesics near a black hole will not be Born rigid (because of tidal gravity, i.e., geodesic deviation). A congruence describing a rigid ruler in free fall will not be a geodesic congruence; the only worldline in the congruence that will be a geodesic will be the worldline of the ruler's center of mass, the other worldlines will be non-geodesic.