I LIGO light changes frequency not wavelength

[pervect]

I don't think this is what the LIGO team is doing. I think they are simply using the usual notion of distance in their chosen frame--the "linearized GR" frame. This is almost the same as distance in an inertial frame, but not quite because of the small corrections to the metric. Those corrections make the distance as measured in their chosen coordinates fluctuate by a small amount around the "inertial" distance. Or, to put it another way, they make the 3-surfaces of simultaneity, in their chosen coordinates, slightly different from what they would be if spacetime were exactly flat.

I believe tried doing what I think you mean by "what the Ligo team is doing" in another thread. (Of course, I could be more sure if you had a reference of some sort, to make sure that what I think the Ligo team is doing is the same thing as what you think the Ligo team is doing). But rather than revisit that approach, I'll explain the approach I was using.

The essence is simple: we just apply the geodesic deviation equation. I don't think I ever saw "the Ligo team" mention the geodesic deviation equation, nor any popularization mention the geodesic equation. But I also think it's a good and reasonably simple approach - it is slightly advanced for an general audience, but it should be comprehensible to someone with some very basic knowledge of General Relativity.

In an additional effort to keep things simple, we will consider "Ligo in space". Moving Ligo into outer space gets rid of many complications that occur if we try to analyze it in situ on the Earth (such as the Earth's gravity, and the Earth's rotation), but captures the essence of the problem without introducing irrelevant details.

With that background let's proceed. The interferometer has two arms, we'll just analyze one arm. We have one fiducial test mass, where the interferometer is located, and one nearby test mass. Both our fiducial test mass and our nearby test mass are following geodesics, i.e. they are in free fall.

We let the separation between the fiducial observer and the nearby observer be represented by a vector $\xi$, as per MTW's remarks on pg 31 (and elsewhere).

We'll restrict the problem two two dimensions - time, and one spatial dimension. Via the geodesic deviation equation, we can write:

$$\frac{d^2 \xi}{d\tau^2} = R \xi$$

(See MTW 1.6).

R is some number, which is in our 2d case, the sole component of the Riemann curvature tensor.

We basically observe that in order to use the geodesic deviation equation to calculate the relative acceleration between our fiducial and our nearby observers, we needed to have some notion of the distance between our fiducial observer and our nearby observer, which we are representing by the vector $\xi$. If we didn't have some notion of distance, we couldn't calculate the second derivative of $\xi$ and call it a relative acceleration.

MTW remarks that the separation vector can be regarded as being measured in the local Lorentz frame of the fiducial observer. But the Local Lorentz frame of the fiducial observer is just the tangent space of the fiducial observer.

So there you have it, in a nutshell. The fiducial observer is following a geodesic, as is the nearby test observer. The distance between the two observers has a meaning when the fiducial observer is sufficiently close to the nearby observer, this meaning is represented by the vector $\xi$ which can be regarded as the distance in the Local Lorentz frame of the fiducial observer (i.e. the distance in the tangent space of the fiducial observer).

And - this distance is changing with time - the nearby observer is accelerating relative to the fiducial observer, due to the gravitational waves.

The only remaining issue might be to answer the question is "are the two mirrors in Ligo sufficiently close for this analysis to work". I believe the answer is yes - the residual discrepancies should be on the order of the Lorentz contraction due to velocities of nanometers per second, i.e. negligible for all practical purposes.

 

[PeterDonis]

we just apply the geodesic deviation equation.

Which cannot be done in any tangent space, because every tangent space is flat; there is no geodesic deviation.

I don't have a problem with the general strategy of analysis you are describing, at least on a first read. The only thing I have been objecting to is your use of the term "tangent space", and your claim that we can just use the usual notion of "distance" in a tangent space, when analyzing LIGO, or indeed any GW scenario. That can't be right, for the reasons I have already given (and which I just restated above).

MTW remarks that the separation vector can be regarded as being measured in the local Lorentz frame of the fiducial observer.

I'll have to look up this specific reference when I get a chance. But my previous remarks still stand: if you are analyzing geodesic deviation, you can't possibly be doing the analysis in any local Lorentz frame, because in any such frame there is no geodesic deviation at all, by definition. If your measurements are able to detect any geodesic deviation, they must cover a region of spacetime that is too large for a single local Lorentz frame.

It's possible that MTW are using the term "local Lorentz frame" somewhat inconsistently in this particular context, to mean something like Fermi normal coordinates centered on the worldline of the fiducial observer. In those coordinates, the metric is not flat and there is geodesic deviation, and I believe that deviation, expressed in terms of the separation vector $\eta$, obeys the equation you wrote down. But Fermi normal coordinates are not a tangent space, and this usage of "local Lorentz frame" would not make that term equivalent to the term "tangent space". Again, I will need to look up the specific reference to see the context.

 

[RockyMarciano]

Maybe this apparent discrepancy can be clarified(and we can recover the OP dealing with what LIGO measures since the starting point in linearized gravity is the 3+1 decomposition) by going back to a comment of PeterDonis:

it is not always possible to even find such a 3+1 foliation in a spacetime containing a given timelike congruence, such that every 3-surface in the foliation is orthogonal to every timelike curve in the congruence. It happens to be possible for the congruence of comoving observers in FRW spacetime, but you can't depend on it as a general property.

This can be further qualified in the pseudo-Riemannian case, because the Lorentzian signature introduces some caveats that don't come up in the purely Riemannian case with positive definite signature. It so happens that in the Lorentzian case the 3+1 foliation itself, by virtue of the Lorentzian inner product picks the relation of orthogonality between the spacelike and the timelike vectors at a point, and if one chooses spacelike hypersurfaces, with 3 spacelike directions tangent to it, a fourth tangent vector normal to them is necessarily timelike. So the 3+1 formalism(with simultaneity hypersurfaces, i.e. spacelike 3-surfaces) always allows to find the timelike congruence orthogonal to spacelike 3-sufaces. Another way to see this is that a manifold in GR is always locally minkowskian, so locally there is a natural 3+1 decomposition with minkowskian coordinates. Comoving observers can be found and therefore becomes a general property whenever we use the 3+1 formalism.

This issue doesn't come up in the Riemannian case where no distinction is made between timelike and spacelike vectors and coordinates, and a 3+1 foliation can't single out a special type of coordinate for the 1 in a 3+1 decomposition.

It has been debated for a long time whether the 3+1 formalism is coordinate dependent in the sense of generally covariant, i.e.:
-“The very foundation of general covariant physics is the idea that the notion of a simultaneity surface all over the universe is devoid of physical meaning”. C. Rovelli
in: H. Garcia-Compe`an (Ed.), Topics in Mathematical Physics. General Relativity and Cosmology, in honor of Jerzy Plebanski, Proceedings of 2002 International Conference (Cinvestav, Mexico City, 17-20 September, 2002);
-“the split into three spatial dimensions and one time dimension seems to be contrary to the whole spirit of relativity”, S.W. Hawking, In: S. W. Hawking and W. Israel (Eds.), General Relativity. An Einstein Centenary Survey (Cambridge University Press, Cambridge, 1979), p. 746
-“Being non-intrinsic, the 3+1 decomposition is somewhat at odds with a generally covariant formalism, and difficulties
arise for this reason” by Pon in J.M. Pons, Class. Quant. Grav. 20 (2003) 3279

 

[PeterDonis]

[Moderator's note: a subthread about expansion of the universe has been moved to the Cosmology forum. Please keep this thread focused on discussion of LIGO.]

 

[PeterDonis]

There are always two beams in the equipment so there is always interference.

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.

 

[PeterDonis]

the 3+1 formalism(with simultaneity hypersurfaces, i.e. spacelike 3-surfaces) always allows to find the timelike congruence orthogonal to spacelike 3-sufaces.

No, this is not correct. There are spacetimes in which there is no way to foliate the spacetime with any set of spacelike hypersurfaces that is everywhere orthogonal to some timelike congruence. But you do not need the orthogonality condition in order to have a foliation; you can still foliate many of these spacetimes with a set of spacelike hypersurfaces, they just won't be orthogonal everywhere to any timelike congruence.

The point you appear to be missing is that, while any vector orthogonal to a set of 3 orthogonal spacelike vectors must be timelike, it is not the case that any timelike vector must be orthogonal to a set of 3 orthogonal spacelike vectors. The latter is what would need to be true for there to always be a foliation with the properties you have specified.

 

[PeterDonis]

What I would like to visualise is that we place a ruler under each mirror with the zero point directly under the suspension. The 'string' holds the mirror from a point on the top of the tube, the ruler is bolted mounted horizontally and supported on a pillar directly under the same point. "Motion" of each mirror can then be considered relative to its adjacent ruler.

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.

 

[RockyMarciano]

No, this is not correct. There are spacetimes in which there is no way to foliate the spacetime with any set of spacelike hypersurfaces that is everywhere orthogonal to some timelike congruence. But you do not need the orthogonality condition in order to have a foliation; you can still foliate many of these spacetimes with a set of spacelike hypersurfaces, they just won't be orthogonal everywhere to any timelike congruence.

The point you appear to be missing is that, while any vector orthogonal to a set of 3 orthogonal spacelike vectors must be timelike, it is not the case that any timelike vector must be orthogonal to a set of 3 orthogonal spacelike vectors. The latter is what would need to be true for there to always be a foliation with the properties you have specified.

Remember that I'm only addressing the general case in GR, not specific spacetimes where in order to model isolated objects time independence of the metric (a timelike Killing vector field) is assumed(like it is the case for instance in the stationary but not static Kerr rotational geometry that might be the example you had in mind of spacelike 3-surface that is not orthogonal to the timelike congruence).
In the general case in GR, with no timelike Killing vectors, it is indeed the case that the choice of a spacelike hypersurface with 3 spacelike vectors, i.e. a 3+1 foliation, determines having a fourth timelike vector orthogonal to them, and if the manifold is to be locally Minkowskian, a requirement of GR spacetimes, I don't think any other foliation deserves to be called a spacetime. By all means if you have in mind a physically plausible GR spacetime with no timelike Killing vector and with a 3+1 foliation that is not time-orthogonal,bring it up, but I'd be surprised.

 

[PeterDonis]

In the general case in GR, with no timelike Killing vectors, it is indeed the case that the choice of a spacelike hypersurface with 3 spacelike vectors, i.e. a 3+1 foliation, determines having a fourth timelike vector orthogonal to them

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.)

It does happen to be true for Minkowski spacetime (obviously) and FRW spacetime. It is also true for Schwarzschild spacetime, but the "standard" foliation (using hypersurfaces of constant Schwarzschild coordinate time $t$) only covers the region outside the horizon. To cover the region inside the horizon, you need to use a non-standard foliation (such as the one implied by Kruskal coordinates), whose physical interpretation will be strained, to say the least.

I don't think any other foliation deserves to be called a spacetime.

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.

 

[pervect]

Which cannot be done in any tangent space, because every tangent space is flat; there is no geodesic deviation.

I don't have a problem with the general strategy of analysis you are describing, at least on a first read. The only thing I have been objecting to is your use of the term "tangent space", and your claim that we can just use the usual notion of "distance" in a tangent space, when analyzing LIGO, or indeed any GW scenario. That can't be right, for the reasons I have already given (and which I just restated above).

I don't necessarily have a problem with your objection to the wording, except that I don't have a better wording at the moment (and I've looked for inspiration in MTW). Perhaps the following clarification will help, though. "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.

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 when objects are sufficiently close, and that the construction does not depend on whether the underlying manifold is stationary or not.

 

[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.

 

[PeterDonis]

given the components of some small displacement $d^i$, we can compute the "spatial distance" via the formula $h_{ab} d^a d^b$.

Yes, this works--for small displacements. But what does "small" mean? It means "small enough that no tidal gravity effects are observable". In other words, we are still working in a single local inertial frame. But as I've already pointed out several times, this will not work for analyzing gravitational waves, because gravitational waves are made of tidal gravity--they are waves of spacetime curvature, and spacetime curvature is tidal gravity. So this definition of distance cannot work for analyzing GWs, because the displacements involved cannot be "small" enough.

 

[PeterDonis]

at the moment, we aren't directly interested in the expansion, shear, and twist

Yes, but only because we are limiting ourselves to "small" displacements, i.e., to a single local inertial frame. But suppose we want to go further; suppose we want to construct a family of spacelike hypersurfaces that (a) foliate spacetime, or at least the region of spacetime occupied by the congruence of worldlines we are interested in, (b) are everywhere orthogonal to the congruence of worldlines we are interested in, and (c) all have the same spatial metric at every event as the spatial metric $h_{ab}$ we see in the tangent space at that event.

The problem is that it might be impossible to meet all of these requirements at once. In particular, (b) can't be satisfied if the congruence has nonzero vorticity, and (c) can't be satisfied under conditions I'm not entirely clear about, but which at least include nonzero vorticity (heuristically, (c) can't be satisfied if the spatial metric $h_{ab}$ can only be extended beyond a single tangent space as a quotient space metric, not as the metric on an actual spacelike slice in the spacetime).

 

[Ibix]

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.

Right - the geodesics are the curved space generalisation of $F=m\ddot x=0$. They're the paths the little bits of the ruler would follow if I chopped it up.

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.

In Newtonian physics this is straightforward enough. You find the path of the COM then find the locus of points at constant distance from that. Restrict yourself to non-spinning or whatever.

But what do I do here? Find the geodesic that the COM follows - OK. Then what? If I'm understanding @pervect's #89, I define the tangent to the COM geodesic to be $\xi^a$, construct $h_{ab}=g_{ab}+\xi^a\xi^b$ (are the indices right on that? They don't seem to match up), then define a displacement field $d^a$ and require $h_{ab}d^ad^b=\mathrm{const}$. The solution to that let's me write $\xi^a+d^a$ which is the worldline of a point a constant distance from the COM.

What have I done wrong this time...?

 

[pervect]

Yes, this works--for small displacements. But what does "small" mean? It means "small enough that no tidal gravity effects are observable". In other words, we are still working in a single local inertial frame. But as I've already pointed out several times, this will not work for analyzing gravitational waves, because gravitational waves are made of tidal gravity--they are waves of spacetime curvature, and spacetime curvature is tidal gravity. So this definition of distance cannot work for analyzing GWs, because the displacements involved cannot be "small" enough.

Yes, but only because we are limiting ourselves to "small" displacements, i.e., to a single local inertial frame. But suppose we want to go further; suppose we want to construct a family of spacelike hypersurfaces that (a) foliate spacetime, or at least the region of spacetime occupied by the congruence of worldlines we are interested in, (b) are everywhere orthogonal to the congruence of worldlines we are interested in, and (c) all have the same spatial metric at every event as the spatial metric $h_{ab}$ we see in the tangent space at that event.

The problem is that it might be impossible to meet all of these requirements at once. In particular, (b) can't be satisfied if the congruence has nonzero vorticity, and (c) can't be satisfied under conditions I'm not entirely clear about, but which at least include nonzero vorticity (heuristically, (c) can't be satisfied if the spatial metric $h_{ab}$ can only be extended beyond a single tangent space as a quotient space metric, not as the metric on an actual spacelike slice in the spacetime).

I don't think I agree with your characterization, primarily because the method can be and is used to calculate the relative acceleration between geodesics. So it works better than you give it credit for.

But rather than discussing a negative, let's discuss a positive. Suppose we specify some congruence of worldlines, for a manifold with one space and one time dimension, and we have a point P on a fiducial worldline in the congruence. Then we have a 1-parameter group of worldlines that fill our 2d space-time. We let t be the time parameter along the geodesic, and s be the space parameter. We basically have a coordinate system that picks out a specific point in space-time by the values (s,t), where s picks out which worldine in our congruence, and t picks out where on the worldline a point is.

For sufficiently nearby worldines in the congruence, we can define a dispalcement "vector" d with $\Delta t=0$ and $\Delta s$ nonzero, such that $h_{ab} d^a d^b$ gives us the square of the distance of P from the worldline, and $h^a{}_b d^a$ gives us the displacement from P to P' such that (P' - P) is orthogonal to the tangent vectors of the congruence.

By leveraging this construct, we can find the distance from P to a nearby point $P_1$, and the distance from $P_1$ to $P_2$, and so on. By taking the limit with a large number of intermediate points, we can find the distance between P and an arbitrary worldline no matter how far away - given that we've picked out a congruence.

For the gravity wave case, we can examine how fast $h_{ab}$ changes. First we'd need to pick our congruence. The congruence that is easiest to pick out is the geodesic congruence. We start with the full line element for the gravity wave:

$$-dt^2 + \left( 1 + 2 f(t-z) \right) dx^2 + \left( 1 - 2 f(t-z) \right) dy^2 + dz^2 $$

(Do I need a reference here? Or do we , hopefully, have agreement).

But we set x=s, y=0, z=0 to to reduce it to our 1space-1time problem, and get:

$$-dt^2 + \left( 1 + 2 f(t) \right) ds^2 $$

f(t) is the Ligo "chirp" function, with a peak mangintude of about $10^{-21}$. Potentially confusing, I've chosen to stick with "s" as our singe spatial coordinate.

We note that $s(\tau)$ = constant is a geodesic, and that it's tangent $\xi^a = \partial_t$. We calculate $\xi_a = -dt$, so we see that $h_{ab} = g_{ab} + \xi_a \xi_b$ is simply:

$$\left | \begin{matrix} 0 & 0 \\ 0 & 1+2\,f(t) \end{matrix} \right | $$

So the end result of our elaborate discussion is that for a geodesic congruence, the distance is $\sqrt{1+2f(t)} \, \Delta s \approx (1 + f(t) ) \, \Delta s$. With no approximations needed. A result you'll see worked out in many places without the preceeding long discussion, via coordinate dependent methods. This construction leads to the "expanding space" point of view.

What if we chose a different congruence? For instance, a rigid congruence, with no expansion or shear. It's a much harder problem to work formally. I have a pretty good idea how it should work out, but I think we need to settle the other issues first.

 

[PeterDonis]

the method can be and is used to calculate the relative acceleration between geodesics.

I would say that an extension of the method can do this; but I'll defer further comment until I've had a chance to look up references.

 

[PeterDonis]

We start with the full line element for the gravity wave:

$$
-dt^2 + \left( 1 + 2 f(t-z) \right) dx^2 + \left( 1 + 2 f(t-z) \right) dy^2 + dz^2
$$

(Do I need a reference here? Or do we , hopefully, have agreement).

Shouldn't there be a $dx dy$ cross term?

Also, shouldn't the signs of the two $f(t-z)$ terms be opposite? (I'm assuming that this line element is supposed to be written in transverse traceless gauge, which means the perturbation of the metric should have zero trace, meaning its diagonal $dx^2$ and $dy^2$ components should have opposite signs.)

 

[pervect]

Shouldn't there be a $dx dy$ cross term?

Also, shouldn't the signs of the two $f(t-z)$ terms be opposite? (I'm assuming that this line element is supposed to be written in transverse traceless gauge, which means the perturbation of the metric should have zero trace, meaning its diagonal $dx^2$ and $dy^2$ components should have opposite signs.)

I don't believe the shoiuld be any cross term for a + polarizxed gravity wave.

As far as the sign of f(t) goes - yes, a typo, I'll fix it. Also another sign error, $\xi_a = -dt$, but neither affects the result. I'll fix it anyway.

 

[PeterDonis]

I don't believe the shoiuld be any cross term for a + polarizxed gravity wave.

Ah, ok, you were assuming a specific polarization. I think that's correct, yes.

 

[GeorgeDishman]

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.

I think this goes back to the picture I had in mind when talking about one arm of a "LIGO in space". The freefalling beam tube would be a congruence of worldlines held Born rigid (approximately, ignoring strain caused by the small internal stresses) by the inter-atomic forces in the material while the test masses near the ends would be following geodesics. That's why I talked of using the centre of the tube as a "fiducial point" but I realize that's a somewhat arbitrary choice. As a result, there would be a displacement of the test masses relative to the ends of the tube as shown in the first image.

Have I followed the last few posts correctly?

Going on from that, just to clarify my understanding, I would expect each photon in the laser beam between the test mass mirrors to be following light-like geodesics which would not be the usual 45 degree lines of SR but a somewhat more complex curved path from mirror to mirror, is that correct?

What I want to do is extend this picture, the first step is to imagine each test mass being mirrored on both sides and each mirror being part of the cavity of a F-P interferometer so each separation is measured as illustrated in the second image.

I think the distances would stretch and shrink virtually in phase and this visualisation matches up with this explanation given at the LIGO press conference in February (the link should go to the relevant section at 23:06):



For a beam in the binary system's orbital plane, I see each cell in the mesh as behaving like the illustration of a plus polarisation wave in Wikipedia:

https://upload.wikimedia.org/wikipedia/commons/b/b8/GravitationalWave_PlusPolarization.gif

 

[PeterDonis]

The freefalling beam tube would be a congruence of worldlines held Born rigid (approximately, ignoring strain caused by the small internal stresses)

You can't ignore the small internal stresses and the strains they cause, because those are what keep the congruence Born rigid. A geodesic congruence in the presence of spacetime curvature will not, in general, be Born rigid, because tidal gravity will cause geodesics to deviate.

the test masses near the ends would be following geodesics.

You keep on talking as if the LIGO device is a single tube with test masses at each end. It isn't; it's an L-shaped pair of tubes with the laser/detector at the corner of the L.

You have said in earlier posts that you want to consider a simpler configuration where there is a single tube with test masses at each end and the laser/detector in the middle; but that configuration is not LIGO's configuration. Even if you only consider one arm of LIGO, that arm does not have two test masses in it; it only has one, at one end of the arm, and the laser/detector at the other. Since you keep on bringing up diagrams and videos of LIGO, it is going to be a lot easier if you talk about LIGO's actual configuration.

I would expect each photon in the laser beam between the test mass mirrors to be following light-like geodesics which would not be the usual 45 degree lines of SR but a somewhat more complex curved path from mirror to mirror, is that correct?

It depends on the coordinates. As I understand it, in the preferred coordinates of the LIGO team, light still travels at $c$ and light worldlines will still be 45 degree lines.

What I want to do is extend this picture

This is probably getting beyond what we can discuss in a PF thread, unless you have a reference that analyzes this configuration.