LIGO light changes frequency not wavelength

[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.

 

[GeorgeDishman]

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.

Exactly.

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.

Thank you.

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

OK, I'll drop it at that then. However, I think the point that what is measured in practice is the phase delay resulting from the influence of the GW on the F-P cavity relates back to the original question. Thanks for your responses.