LIGO light changes frequency not wavelength

[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.

 

[pervect]

I have some general comments about the discrepancies between the geodesic congruence and the Born rigid congruence as far as its impact on distance goes. Finding the exact expression for the congruence whose expansion is exactly zero should be theoretically possible but I didn't do it, it would be a lot of work. I settled for an expression that I believe should be accurate to the second order. To be specific, since f(t) has a magnitude of 10^-21, I counted the "order" of the discrepancy by the number of times f or any of its derivatives were multiplied together. The lowest order term in the expansion by this criterion was one proportional to $f \, df/dt$.

The geodesic congruence has an expansion of the first order in f, namely:
$$\frac{\frac{df}{dt}}{1 + 2\,f(t)}$$

Thus the expansion of the almost-rigid congruence is about 10^21 times lower than the expansion of the geodesic congruence, because it's proportional to $f df/dt$ and f is so small.

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

the almost expansion-free congruence is given by:

$$\partial_t - s \frac{df}{dt} \partial_s$$
Actually one needs to normalize this first to calculate the expansion, but it's easier to write without the normalization.

Physically, we can interpret this as saying the geodesic congruence is approximately moving with some velocity $v = -s \frac{df}{dt}$ relative to the rigid congruence. Because of this relative motion, we expect discrepancies in second order in the relative velocity $v = s\,df/dt$ due to Lorentz contraction. We are assuming f is small, and we are assuming that s is small enough that s*f is still small, and we are assuming that f is changing slowly enough that df/dt is also small.

Because $f(t) \approx 10^{-21}$ in Ligo, agreement to the second order means that the two concepts of distance agree to about 1 part in 10^20 or so. (Possibly a little less, depending on how big s is, and how fast f changes - but still an excellent approximation).

 

[GeorgeDishman]

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.

I don't intend to ignore them, quite the opposite in fact. A major part of my confusion is how we deal with the stresses in what was called the "Earth frame" earlier in the thread, but I need to explain why they're giving me a headache. The consensus here seemed to be that it was so obvious it didn't need to be discussed but I'm struggling with it.

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.

As I have said several times, I appreciate that but I think it can be dealt with fairly easily by choosing to analyse the situation where one arm is in the orbital plane of the binary system and the other is parallel to the axis. In the orbital plane, there is only plus polarisation so the signals from the arms are then the same but anti-phase.

It is my intention to extend the analysis to cover the other cases too though I'm just keeping it simple to start with.

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.

Good idea. The first attached image is the actual layout from page 3 of http://arxiv.org/abs/1411.4547:

As you can see, there are two mirrors or 'test masses' labelled ITM and ETM in each arm. The light bounces between the ITM and ETM approximately 280 times (that's an average of course, a small fraction of the photons get through on each pass). A displacement of the ITM towards the ETM by dx reduces the path length within the cavity by 560dx but only increases the path from the beamsplitter to the ITM by 2dx (once on input and once on output). Relative to a fiducial point in the middle, the ETM would move by -dx which also reduces the path length by 560dx. Overall, for displacements of each by dx and -dx, the optical path length changes by 2*560-2 = 1118dx. Obviously that is dominated by the change within the 4km cavity between the ITM and ETM. Ignoring the short distance from the beam splitter to the ITM (tens of metres at most) only introduces a small scaling error, much less than 0.1%. In other words, LIGO measures the length of the two cavities which are shown in the diagram as "4km" long and containing laser at a power level of "750kW".

The second image shows the suspension system typical of one of the four ITM and ETM mirrors. As you can see, there is nothing to stop their motion along the beam direction for either mirror so taking a fiduial point in the centre of the tube is reasonable IMHO. Of course the location we choose cancels out in the calculations anyway so it is a moot point.

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.

OK, that's interesting but I'll leave it for another day.

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

I don't. I looked around and couldn't find anything so when I found out about "PF Insights" last year, I suggested it there. There were no takers so I thought I'd try to work it out myself and as I have recently got the opportunity to start learning MATLAB, this seemed like an ideal training project. I was making good progress until I hit the problem of understanding what the "Earth frame" means.

 

[PeterDonis]

there is nothing to stop their motion along the beam direction for either mirror so taking a fiduial point in the centre of the tube is reasonable IMHO

The fiducial point for the LIGO apparatus is at the corner of the "L" (the beam splitter). It is not in the middle of either arm. If you want to analyze the actual signal output by LIGO, that is the fiducial point you need to use. If you analyze with respect to a fiducial point in the middle of an arm, you will not be analyzing the same signal that LIGO outputs.

A major part of my confusion is how we deal with the stresses in what was called the "Earth frame" earlier in the thread

The paper you linked to makes it clear that all of the key components of the LIGO detector are free to move in the "detector plane" (the plane defined by both arms, considered as two intersecting lines--we can think of this plane as being "horizontal", perpendicular to the local direction of "gravity"). So none of the components will be stressed in that plane. (The tunnel walls will be, but they play no role in determining the signal LIGO outputs.) There will obviously be stresses perpendicular to that plane, but those should not affect motion in that plane. I believe this is the basic argument used by the LIGO team to justify ignoring stresses when analyzing the signal (though of course they have to account for those stresses in the complicated apparatus that tries to keep all of the key components in the same plane and isolate them from outside disturbances).

 

[GeorgeDishman]

The fiducial point for the LIGO apparatus is at the corner of the "L" (the beam splitter). It is not in the middle of either arm. If you want to analyze the actual signal output by LIGO, that is the fiducial point you need to use. If you analyze with respect to a fiducial point in the middle of an arm, you will not be analyzing the same signal that LIGO outputs.

If you understand why the mirror on the ITM is present, it should be clear that it is the length of the Fabry-Perot cavity that is measured, not the distance to the corner of the L. In addition, the signal (i.e. the phase difference at the photodetector) should not depend on the choice of the fiducial point (or the point wouldn't be "fiducial" in the way I understand the word).

However, what you are missing is that I don't "want to analyze the actual signal output by LIGO", I want to produce a visualisation of the gravitational waves to which it is responding.

The paper you linked to makes it clear that all of the key components of the LIGO detector are free to move in the "detector plane" (the plane defined by both arms, considered as two intersecting lines--we can think of this plane as being "horizontal", perpendicular to the local direction of "gravity"). So none of the components will be stressed in that plane. (The tunnel walls will be, but they play no role in determining the signal LIGO outputs.) There will obviously be stresses perpendicular to that plane, but those should not affect motion in that plane. I believe this is the basic argument used by the LIGO team to justify ignoring stresses when analyzing the signal (though of course they have to account for those stresses in the complicated apparatus that tries to keep all of the key components in the same plane and isolate them from outside disturbances).

Right, so if you imagine a wave propagating in the direction perpendicular to the detector plane (i.e. vertical) and such that the orbital plane just happens to align with one arm, then you get the toy model I've been describing. We can consider the distortion that the GW would consider in that orientation. Then, for any realistic situation, the actual wave will have components of both plus and cross polarisation and that combination will need to be projected onto the orientation of the detector arms to take account of the effect of the suspension.

That's a separate part of the problem, all I am trying to do is the first part, to produce a visualisation of how the GW would influence a constellation of free-falling test masses whose only constraint is that they are suspend against (or we simply ignore) the basic gravitational acceleration towards the binary system. For realistic scenarios, a few dozen solar masses thousands or billions of light years away aren't going to be significant in terms of GM/r², right?

 

[PeterDonis]

it is the length of the Fabry-Perot cavity that is measured, not the distance to the corner of the L.

More precisely, because of the "magnification" effect of the cavity, the contribution to the overall distance measurement of the distance from the ITM to the beam splitter is very small (a fraction of a percent). That is because the length of the cavity is magnified by a factor of 1000 or so while the ITM-beam splitter distance is not.

the signal (i.e. the phase difference at the photodetector) should not depend on the choice of the fiducial point

It does for the LIGO team's analysis because their "fiducial" point is part of the overall construction of the coordinate chart they are using. It is true that, in principle, any choice of chart should give the same values for actual observables like the interference pattern at the photodetector, so in that sense the choice of fiducial point doesn't matter.

I don't "want to analyze the actual signal output by LIGO", I want to produce a visualisation of the gravitational waves to which it is responding.

Then that is off topic for this thread and you should start a new one.