LIGO and light speed constancy

[Mister T]

not sure what was the goal of this post?

You reveal the answer in your next line.

the reality behind physical units of measurement is not arbitrary and subject to our consensus, gravity waves either stretch them or not,

The definition of physical units of measurement is subject to our consensus. Gravitational waves either stretch the arms of the LIGO detector or they don't. That was my take-away on the goal of the post.

 

[pervect]

You reveal the answer in your next line.
The definition of physical units of measurement is subject to our consensus. Gravitational waves either stretch the arms of the LIGO detector or they don't. That was my take-away on the goal of the post.

I have seen discussions of gravitational waves use both approaches. Some approaches stress the physical distance, defined ultimately by the SI definition. This changes because the round-trip light time changes, and that's how the SI system defines distance. Other approaches use a convenient set of coordinates defined by a set of free-fall observers. In this approach, the number associated with an observer on each mirror does not change, by definition.

Both approaches can provide insight and are commonly used.

I suspect many people do not use the modern SI definition of distance, and are confused about what happens according to their own views, which in some cases may not be fully thought out and hence not fully explainable. One can't hit a moving target such as an unspecified definition, but one can attempt to address the issue of what happens from a pre-relativistic definition of distance, which was based on rigid rods, physically implemented out of steel bars.

I haven't seen any discussion of this in textbooks, so the answer will necessarily be my own, not a textbook answer. My answer to this is that the SI definition in this case best approximates what a steel bar would measure. Basically, according to my take, measuring the round-trip time of light is by the principle of the constancy of the local speed of light is simply a better implementation of a "rigid ruler". The fact that this is the SI definition of distance supports the view. If you regard the SI definition of distance as being a refinement of how to accurately measure distance, then one can accept that's the defintion want to use it as the "physical" distance, and one can regard other notions of distance (such as assigning constant coordinates to inertial objects) as "non-physical". Non-physical doesn't mean useless, some of these notions can be very useful in understanding. But one needs a connection between the math, and what one can actually measure.

The complicating factor here is that rigid objects don't actually exist in a general curved space-time :(. But that seemingly alarming discussion might best belong in another thread. My personal take is that it's not vital to understanding GW's, but to justify my take requires some rather advanced mathematics.

It doesn't take much math to demonstrate a simple example to demonstrate that "rigid" two dimensional surfaces don't exist on the two dimensional surface of a 3d manifold unless the object and it's surface manifold is highly symmetrical. This won't explain why I don't think this issue isn't vital to explaining GW's, but it can be interesting in its own right, though it's a bit of a digression. I'll take the risk of explaining further, with the suggestion that followup questions probably belong in a different thread.

Let's set some ground rules. To keep things simple, we have some 3 dimensional object, and we treat it's surface as a 2d manifold. Then we address the issue of whether we can construct "rigid" objects on this 2d manifold. What do we mean by a rigid object? We could use a Born's definition, but a good treatment of that gets rather mathematical (and seems hard to find elementary treatments, a lot of the original papers aren't in English as well). I'll suggest that the notion of congruent geometric figures (such as triangles) serves as a good notion of rigidity. Basically, if all triangles are congruent, then we can move a triangle from one spot to another without changing its shape. We can break up other shapes into triangles, and then argue that if all the triangles are congruent, so are the larger figures.

But when we look at the sum of the angles of a triangle on a curved surface (such as a sphere), we start to see the issues. The sum of the angles of a triangle on a plane is always 180 degrees, but this is not true in spherical geometry. It's rather well known that the sum of angles of a spherical triangle is greater than 180 degrees, and depends on the size of the triangle. (I regard this as not being "much math", though i suppose readers unfamiliar with spherical trig might disagree. But it's certainly a lot simpler math than a full understanding of General Relativity and Riemannian geometry, though it can be a useful motivational tool for why we need Riemannian geometry).

If we only demand that triangles of the same size be congruent, we can get around this issue on a sphere. We go from "all triangles are congruent" to "all triangles of the same area are congruent". Unfortunately, when we consider more general cases, such as a flat geometry with a "bump", we see that we just can't make all triangles congruent. If we ask that all triangles be congruent, we can't have the sum of the angles be 180 degrees at one locaction, and some different number at a different location. But if we move a triangle with three equal sides from a location "on the bump" to a location "off the bump", the sum of it's angles changes (as long as the triangle has a finite area). So, we are lead to the notion that saying that "all triangles" are congruent is simply wrong, which then suggests that there are severe difficulties with "rigid objects" as a definition of distance.

Sorry, this got rather long. But I'll try to summarize my point. I believe that pre-relativistic notions of distance relied on "rigid rulers", and the lack of rigid objects on curved manifolds leads to a lot of confusion about what distance really is. Unfortunately I can't point to the literature as to what the "real" definition of distance is, because as near as I can tell there isn't much agreement on how best to formulate it. As a consequence, students wind up with little guidance on this point. I can point out that relying on "rigid objects" to define it is going to cause some issues. And I think it's fair to say that the previous incarnation of the definition of distance used "rigid rulers", so I don't think I'm too far off in suggesting that this is the source of a lot of confusion about the nature of distance.

I should stop here, but I want to add one more thing. I believe it can be fruitful to go from the notion that "all traingles are congruent" to "all sufficiently small triangles are congruent". But I haven't seen any text actually take this approach. I believe, though, that it could be used to motivate Reimannian geometry.

 

[PeterDonis]

the SI definition in this case best approximates what a steel bar would measure

I agree with this. What's more, so does GR, in the sense that GR predicts that a steel bar placed along an arm of LIGO would also contract and expand--as could be verified, relative to the current SI definition of length, by mounting a mirror on the far end of such a bar and a laser emitter/detector on the near end, and seeing the round trip light travel time change as a GW passes and the bar vibrates (i.e., as its length changes sinusoidally) in response.

In fact, the original proposal for GW detection, by Joseph Weber, was to use bars (he used alumininum, not steel, IIRC) with piezoelectric sensors to detect the vibrations induced by GWs. He even claimed, IIRC, to have detected actual GWs. But more recent calculations have made it pretty certain that he couldn't have--bar detectors simply aren't sensitive enough. One key reason why can be seen by imagining a bar laid along an arm of LIGO: the bar's length will change in response to a GW, but not by as much as the arm length between the actual LIGO mirrors, because the bar's tensile strength will resist the length change, whereas there is nothing to resist the length change between the actual LIGO mirrors. So LIGO responds fully to the GW signal, whereas a bar would only partially respond, hence would be less sensitive.

 

[vanhees71]

From the "advanced information" from the Nobel foundation:

J. Weber, Evidence for discovery of gravitational radiation, Phys. Rev. Lett. 22, 1320 (1969)
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.22.1320

 

[timmdeeg]

Aren't metallic bars sensitive in the range of their resonance frequency only?

 

[Paul Colby]

Aren't metallic bars sensitive in the range of their resonance frequency only?

Yes. Outside their band they have little sensitivity. Q values used are quit high so the effective bandwidth is often less than 1Htz.

 

[PeterDonis]

Aren't metallic bars sensitive in the range of their resonance frequency only?

Yes, as @Paul Colby has said. That's another reason why bar detectors were not pursued more: in order to detect a wide range of frequencies, you would need to build many, many bar detectors with different sizes of bars, whereas you can just build one interferometer to do the same thing.

 

[pervect]

With Ligo's arms being 4km long, and given that the speed of sound in steel is about 6000 meters/second or so, a steel bar 4km long wouldn't be terribly rigid. It'd take about 2/3 of a second for a push on one end to reach the other :(.

In contrast, it takes light only 20 microseconds to travel 6km.

 

[Paul Colby]

With Ligo's arms being 4km long, and given that the speed of sound in steel is about 6000 meters/second or so, a steel bar 4km long wouldn't be terribly rigid.

Yes. The interaction of GW with isotropic materials such as steel is through traction forces applied to the bar boundary. All the internal forces at a point interior are canceled by surrounding matter. So the speed of propagation of mechanical vibration induced on the boundary really effects the coupling to GW. This is why great care is taken to mount the LIGO mirrors so they are essentially inertial along the detector beam.

 

[PeterDonis]

With Ligo's arms being 4km long, and given that the speed of sound in steel is about 6000 meters/second or so, a steel bar 4km long wouldn't be terribly rigid.

True, but it would still be a lot more "rigid", in terms of its response to a GW, than LIGO's actual arms. The relevant criterion is not how long it would take a sound wave to travel the length of the arm, but how constrained the local motion of a given atom in the steel bar is, compared to the motion of the actual LIGO mirrors. The answer to that is that the steel bar's atoms are a lot more constrained.

 

[nitsuj]

isn't the experiment to measure distance of free space between to points, specifically to not measure the length of some rigid object. as has been pointed out the gw won't "show up" as clearly if passing through something "rigid".

 

[timmdeeg]

True, but it would still be a lot more "rigid", in terms of its response to a GW, than LIGO's actual arms. The relevant criterion is not how long it would take a sound wave to travel the length of the arm, but how constrained the local motion of a given atom in the steel bar is, compared to the motion of the actual LIGO mirrors. The answer to that is that the steel bar's atoms are a lot more constrained.

Just to understand you correctly. Does "a lot more constrained" concern the case that the steel bar isn't in resonance with the GW? So it means even though the mirrors being fixed at the ends of the steel bar aren't in free fall the distance between them would still change, however much less compared to the LIGO assembly.

 

[Mister T]

With Ligo's arms being 4km long, and given that the speed of sound in steel is about 6000 meters/second or so, a steel bar 4km long wouldn't be terribly rigid. It'd take about 2/3 of a second for a push on one end to reach the other :(.

To make the steel bar "ring" at one of its resonant frequencies you need sound waves moving in opposite directions along the bar interfering constructively, creating a standing wave.

In contrast, it takes light only 20 microseconds to travel 6km.

Is your point then that there simply isn't enough time to create a standing sound wave in the bar? I'm not that familiar with the details of the GW detection, but I had imagined that this GW is not a very long wave train, just a pulse, really. All that LIGO does is detect the peak of that wave? And the amount of time that that peak spends passing through LIGO is on the order of 20 microseconds?

I think I understand the other point being made, which is far more fundamental: A LIGO leg made of a metal bar, or indeed any material bar at all, would be way too rigid to detect a GW, which is why the basic design of LIGO involves mirrors at the leg ends that are as close to unattached as possible. That is, free to move with as little constraint as possible, given the design parameters.

 

[DanMP]

Scalar polarisation is what longitudinal waves like sound waves have. Vector polarisation is what motion-in-one-transverse-direction waves like water waves and light waves have. Tensor polarisation is what motion-in-two-transverse-directions waves have. The paper you cite is saying that the evidence is in line with theory and gravitational waves traveling in the z direction are "stretch then squish" in the x direction, "squish then stretch" in the y direction, and nothing in the z direction.

I found in wikipedia that "in longitudinal waves, such as sound waves in a liquid or gas, the displacement of the particles in the oscillation is always in the direction of propagation, so these waves do not exhibit polarization", so your above explanation (about scalar polarization, at least) appears wrong ... I need some citations, if possible. I searched 2 days and didn't find a clear explanation on this matter (scalar/vector/tensor polarization).

Maybe they didn't check at all if the signal is consistent with a longitudinal wave ... Or maybe pure tensor polarization means something else (may include longitudinal oscillations) ... I really need to know/understand what they found and meant about polarization, how https://dcc.ligo.org/public/0145/P170814/010/GW170814.pdf did stretch the interferometers arms in relation with the propagation direction.

If anyone else can help, please do. PeterDonis maybe ...

The gravity wave is propagating in a different direction from the light beams traveling down the arms. In the simplest case, the detector is perfectly transverse to the wave, so a given wave front arrives at all points on the detector (i.e., everywhere in both arms) simultaneously. In more complicated cases, the detector's plane is not perfectly transverse, but if you work out the math, that just means the amplitude of the signal detected gets reduced.

 

[PeterDonis]

Does "a lot more constrained" concern the case that the steel bar isn't in resonance with the GW?

No, it means that the amplitude of the bar's vibrations is much smaller than that of the actual LIGO arms' vibrations. And the amplitude is what determines the sensitivity of the device--the larger the amplitude for a given GW, the more sensitive the device is (i.e., the weaker the GW that the device can detect).

 

[PeterDonis]

I found in wikipedia that

That discussion isn't applicable to gravitational waves anyway, because it's talking about waves in a medium made of atoms, which spacetime isn't. (Similar remarks apply to light waves in vacuum.)

Maybe they didn't check at all if the signal is consistent with a longitudinal wave

@Ibix already answered that: per the paper you referenced, they did, and it isn't. It's consistent with tensor polarization, also called "spin-2", which is transverse and is described, for example, on p. 5 here:

http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GravitationalWaves.pdf

 

[timmdeeg]

No, it means that the amplitude of the bar's vibrations is much smaller than that of the actual LIGO arms' vibrations.

Ah, I see, thanks.

 

[timmdeeg]

Yes. Outside their band they have little sensitivity. Q values used are quit high so the effective bandwidth is often less than 1Htz.

Thanks. I just wonder why the bar has still some sensitivity, though little, "outside their band" which seems extremely narrow. What is the reason for this?

 

[Paul Colby]

What is the reason for this?

The reason to choose a narrow band is sensitivity. Basically, in any detection it's ultimately about received power versus the competing noise. Restricting the band reduces the available signal power but the corresponding noise is further reduced. The Q of a resonator, mechanical or electric, is the ratio of the power lost per cycle divided into the power stored. A high Q resonator looses energy more slowly than a low Q one because there are less loss. There is a very general relation between energy loss and noise in a detector (due I believe to Einstein but I could be wrong on this) called the fluctuation dissipation theorem. Bar detectors are designed to maximize the signal to noise in their band of operation. Received power for a bar detector is the energy transferred from the GW into mechanical vibration which can last well after the GW passes. Clearly driving a resonator on its resonant frequency imparts more energy to the resonator than driving off resonance.

 

[pervect]

To make the steel bar "ring" at one of its resonant frequencies you need sound waves moving in opposite directions along the bar interfering constructively, creating a standing wave.
Is your point then that there simply isn't enough time to create a standing sound wave in the bar? I'm not that familiar with the details of the GW detection, but I had imagined that this GW is not a very long wave train, just a pulse, really. All that LIGO does is detect the peak of that wave? And the amount of time that that peak spends passing through LIGO is on the order of 20 microseconds?

I think I understand the other point being made, which is far more fundamental: A LIGO leg made of a metal bar, or indeed any material bar at all, would be way too rigid to detect a GW, which is why the basic design of LIGO involves mirrors at the leg ends that are as close to unattached as possible. That is, free to move with as little constraint as possible, given the design parameters.

I suppose my point is that the equivalence between the SI definition of distance and the old prototype meter bar (steel ruler) definition only works when the rulers are short enough. Unfortunately, being literal and taking a 4km ruler, a steel ruler wouldn't be rigid enough for my favorite analogy to apply directly.

Working around the fact that steel rulers aren't perfectly rigid though is something that people were used to back in the days when the definition was used. It was generally understood that the "defintion" of distance in this manner did not mean that forces acting to distort the ruler distorted the concept of distance. Rather, it was intended that a force-free "ruler" be used to measure the distance.

So ideally we'd have a ruler unaffected by gravity. Unfortunately, such a thing has to be defined, as it doesn't exist physically. Gravity affects everything. I use the SI defintion of distance, and furthermore I suggest thinking of the SI concept of distance as being an idealized steel bar, that is simply "more rigid" than steel could possibly be, because the speed of sound in steel is just so low in comparison with the speed of light.

Ultimately, this approach has difficulties over large distances. But the distances involved in the case of Ligo (4km) are simply not large enough to be an issue as long as we use a ruler that's not limited by the speed of sound in steel.

 

[Ibix]

I need some citations, if possible. I searched 2 days and didn't find a clear explanation on this matter (scalar/vector/tensor polarization).

The paper you linked references (their 11 and 12) work by Eardley, Lee and Lightman in Phys Rev D. That is here:
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.8.3308
...but is behind a paywall. Searching on the title of that paper led me here:
https://ntrs.nasa.gov/search.jsp?R=19730012613
...which is by the same authors plus Wagoner and Will, is free to download, and gives an overview. General metric theories of gravity permit up to six polarisation modes for gravitational waves (including a pure longitudinal mode), but general relativity permits only two purely transverse tensor modes. The paper you linked provides experimental evidence strongly consistent with GR's pure transverse tensor modes and inconsistent with other modes.

I found in wikipedia that "in longitudinal waves, such as sound waves in a liquid or gas, the displacement of the particles in the oscillation is always in the direction of propagation, so these waves do not exhibit polarization", so your above explanation (about scalar polarization, at least) appears wrong

Terminology seems to vary a bit, but Eardley et al clearly regard the pure longitudinal mode as a polarisation state. I agree there's little point in discussing the polarisation state of a pure longitudinal wave, but for waves that permit (or may permit) both transverse and longitudinal oscillation you can break the motions down into components including ones parallel to the direction of motion. And pure parallel motion is a polarisation - at least following the terminology of your linked paper.

I may have the classification slightly wrong. The authors of your linked paper may be regarding the longitudinal mode as a vector polarisation and what Eardley et al call the $\Phi_{2,2}$ mode as a scalar polarisation. Either way, the current evidence is against longitudinal oscillation.

 

[timmdeeg]

The reason to choose a narrow band is sensitivity. Basically, in any detection it's ultimately about received power versus the competing noise. Restricting the band reduces the available signal power but the corresponding noise is further reduced...

Thanks for this explanation.
Unfortunately I addressed my question ambiguous. I was a bit surprised that you said "Outside their band they have little sensitivity". In this case with the mirrors fixed at the ends of the steel bars (preventing their free fall in the horizontal plane by this) and taking into account that the band is very narrow I wouldn't expect any sensitivity at all outside the band. Would you mind to comment on this?

Still another case. If the mirrors would be fixed to the Earth instead would the sensitivity (hereby neglecting the noise) just be markedly less compared to the LIGO assembly or almost zero? I'm asking this because in pop-science one can read quite often that the whole Earth is squashed and stretched if a gravitational wave passes by. It is even argued sometimes that the only purpose to suspend the mirrors is to avoid the enviromental noise.

 

[Paul Colby]

In this case with the mirrors fixed at the ends of the steel bars (preventing their free fall in the horizontal plane by this) and taking into account that the band is very narrow I wouldn't expect any sensitivity at all outside the band. Would you mind to comment on this?

I'll try. Mechanical vibrations of a solid obey an equation of motion subject to boundary conditions. For LIGO level GW these equations of motion are essentially,

$\rho \ddot{u_k} = T_{jk,j}$​

where indices run over just space coordinates which are Cartesian. This equation is what $F=ma$ becomes for a continuous material. The vector, $u_k(x,t)$, is the displacement of the material from it's reference (equilibrium) position. The $T_{jk}$ is the mechanical stress tensor. As usual we're using the the summation convention and $X_{,j}=\partial_j X$. In GR things are not Cartesian so one must use coordinates in which these equations are valid if one wants to get the correct answer using them. These coordinates are known as the TT (for Transverse Traceless) coordinates. In these coordinates the metric strain, $h_{jk}$, has only space components (hence GW are transverse). For very small strains, the stress is proportional to the strain in linear elastic materials like steel. For isotropic materials the applied GW stress has the property,

$T_{jk,j} = 0$​

which implies the interior material is not accelerated relative to it's reference position by a LIGO level GW. In fact, all material acceleration is due to forces that appear on the boundary of the material due to the GW. For an interior bit of matter, say an atom, to be accelerated it must be from the acoustic (shear) wave originating from the boundary, hence all the discussion of speed of sound in the material.

Now, your question was about mirrors mounted on the bar ends. The relative distance between the end points is the sum of the physical distance determined by the metric (which includes the metric strain in the TT gauge) and the change of the reference position of the endpoints. An exact answer to this is quite complicated. At very low frequencies where the period of the GW is much much longer than the acoustic propagation time between the mirrors, the reference position will quite nearly cancel the metric distance change between the mirrors. If the period of the GW is much shorter than the acoustic transit time then the distance between the mirrors would be dominated (ideally of course) by the GW. In between these cases is the resonant case where all bets are off.

I hope this covers some of your questions.

 

[timmdeeg]

At very low frequencies where the period of the GW is much much longer than the acoustic propagation time between the mirrors, the reference position will quite nearly cancel the metric distance change between the mirrors.

To put it in Layman terms could one say that if "the acoustic propagation time between the mirrors" is negligible compared to the "period of the GW" then the amplitude of length changes between the mirrors attached to the bar ends is (almost) zero? Which would mean that the sensitivity is zero.

If the period of the GW is much shorter than the acoustic transit time then the distance between the mirrors would be dominated (ideally of course) by the GW.

What does "the distance between the mirrors would be dominated (ideally of course) by the GW" mean in terms of sensitivity or amplitudes respectively?
My first impression was "dominated" means that the distance between the mirrors chances as if they were in free fall, but this makes no sense of course.

 

[Paul Colby]

My first impression was "dominated" means that the distance between the mirrors chances as if they were in free fall, but this makes no sense of course.

$x$, a real number, dominates, $y$, if $|x| >> |y|$ in this context.

I know this is confusing having struggled long and hard myself. The GW changes the distance between points as a function of time and space. This is what time and space dependent metrics do or mean. When a GW wavefront hits a hunk of matter it suddenly finds itself squished or stretched. It responds, of course, but only on a time scale commensurate with the velocity of sound in the material. In this limit the ends of the bar and the mirrors are in free fall. I say ideally because at length changes of $10^{-21} L$ or smaller a metal bar is jiggling like a bowl of jellow being kick down my driveway.

Which would mean that the sensitivity is zero.

My thesis advisor once complained about an error source in an experiment I was planning. I replied that the source was zero to which he responded "nothing is zero". The meaning, of course, is that things have relative sizes which must be understood quantitatively. The sensitivity is never zero but often negligible is a more useful way to say it.

 

[timmdeeg]

When a GW wavefront hits a hunk of matter it suddenly finds itself squished or stretched. It responds, of course, but only on a time scale commensurate with the velocity of sound in the material. In this limit the ends of the bar and the mirrors are in free fall.

Hm, interesting, but wouldn't that mean that in this case the sensitivity is comparable to LIGO's suspended mirrors? We are talking here about a non-resonant case.

 

[PeterDonis]

It responds, of course, but only on a time scale commensurate with the velocity of sound in the material.

This is true if you are talking about collective motion or vibration of the entire bar. But if you are talking about the motion of atoms at the end of the bar in response to the stretch or squeeze coming from a GW, the relevant response time is the time it takes for them to move enough away from their equilibrium positions with respect to neighboring atoms, to have those neighboring atoms exert a restoring force on them. That time scale, for a typical GW with a frequency of a few tens or hundreds of Hz, will be much faster than the time scale for the GW to stretch or squeeze the atoms. So the response of either end of the bar to the GW will be damped, compared to the response of the LIGO mirrors, which, as you note, are designed specifically to have no forces acting on them at all (to the greatest accuracy possible).

 

[Paul Colby]

Hm, interesting, but wouldn't that mean that in this case the sensitivity is comparable to LIGO's suspended mirrors? We are talking here about a non-resonant case.

Well, there are two limiting cases both non-resonant, both quite different. In case 1) the frequency of GW is well below the lowest mechanical resonance, while in case 2) the GW frequency is well above the lowest mechanical resonance. By well above and well below let's say 3 orders of magnitude in each case. I was referring to case 2, not case 1, in the quoted text. In case 1) the ends of the bar can and will move to compensate for spatial distance changes due to the slowly varying metric. In case 2 the bar ends simply can't move fast enough to compensate for the distance change due to the GW. Clearly, much depends on the size of the bar.

For measuring very high frequency GW one might well mount mirrors on a ridged mount because at 100MHz GW the mount is essentially provides an inertial mirror. This, however, is not what I would do.

 

[Paul Colby]

This is true if you are talking about collective motion or vibration of the entire bar. But if you are talking about the motion of atoms at the end of the bar in response to the stretch or squeeze coming from a GW, the relevant response time is the time it takes for them to move enough away from their equilibrium positions with respect to neighboring atoms, to have those neighboring atoms exert a restoring force on them. That time scale, for a typical GW with a frequency of a few tens or hundreds of Hz, will be much faster than the time scale for the GW to stretch or squeeze the atoms. So the response of either end of the bar to the GW will be damped, compared to the response of the LIGO mirrors, which, as you note, are designed specifically to have no forces acting on them at all (to the greatest accuracy possible).

The equations of motion given in post #58 are relevant to the discussion. For an atom in the interior of the bar the forces do to metric strain are symmetric and cancel. This is the meaning of, $T_{jk,j}=0$. There is no change in an interior atom's reference position due directly to GW. For atoms on the boundary this is not the case since there are no atoms exterior to the bar to compensate for the change in interatomic distances. For an atom on the boundary the distance change to it's neighbor is all that matters as far as the force goes. This distance change is $~10^{-21} \delta$ where $\delta$ is the interatomic distance. Typical interatomic distances are on the order of angstroms so $\delta ~ 10^{-31}$ meters which isn't large. So in case 2) as defined in #63 the distance between mirrors will be roughly $10^{-21} L$ where $L$ is the length of the bar. In case 1) it will be much smaller than this.

 

[PeterDonis]

the distance between mirrors will be roughly $10^{-21} L$

I think you mean the change in the distance between mirrors. More precisely, it's the amplitude of the oscillation in the change in the distance between mirrors. This is true for the LIGO mirrors (where the applicable $L$ is the length of the LIGO arms), but I don't think it's true for mirrors mounted on a bar, because, as I said, the inter-atomic forces on the atoms at the ends of the bar, which are the ones, as you say, that will be displaced from their equilibrium positions by the GW, can respond much faster than the time scale of the GW vibrations. That means the vibrations will be damped; the amplitude of the oscillation in the change in bar length will be smaller than $10^{-21} L$, which is the amplitude of the GW oscillations in the distance between hypothetical freely falling test particles that start out at the equilibrium positions of the atoms at opposite ends of the bar.

 

[Paul Colby]

This is true for the LIGO mirrors (where the applicable LLL is the length of the LIGO arms), but I don't think it's true for mirrors mounted on a bar, because, as I said, the inter-atomic forces on the atoms at the ends of the bar, which are the ones, as you say, that will be displaced from their equilibrium positions by the GW, can respond much faster than the time scale of the GW vibrations.

I discussed two limiting cases which are quite distinct. One in which your statement is correct and one in which it is not. If one were to magically remove all interatomic forces between atoms (and neglect their thermal motion) no change from their reference position (reference position as defined in the theory of linear elasticity) would occur independent of the frequency or time scale of the GW. The distance between atoms is changed by GW but not the reference position as defined by the equations of motion given in #58. For LIGO frequencies the answer depends on the length of the bar and the speed of sound in said bar.

 

[DanMP]

@PeterDonis and @Ibix - many thanks for your replies.

 

[PeterDonis]

The distance between atoms is changed by GW but not the reference position as defined by the equations of motion given in #58.

This is quibbling over terminology. The observed interference pattern at the LIGO detector has a given amplitude in response to a given gravitational wave. If the mirrors were mounted on bars instead, the amplitude of the observed interference pattern at the detector, for the same gravitational wave, would be smaller. That is what I am saying, and it's true regardless of how you define your "reference position" (which basically means how you choose your coordinates).

 

[timmdeeg]

In case 2 the bar ends simply can't move fast enough to compensate for the distance change due to the GW.

For measuring very high frequency GW one might well mount mirrors on a ridged mount because at 100MHz GW the mount is essentially provides an inertial mirror. This, however, is not what I would do.

Lets consider case 2: "the GW frequency is well above the lowest mechanical resonance." Which seems to fit with "very high frequency GW" (second sentence). If the mirrors are inertial here then why aren't they inertial in your case 2 description "can't move fast enough to compensate for the distance change due to the GW."?

 

[Paul Colby]

Lets consider case 2: "the GW frequency is well above the lowest mechanical resonance." Which seems to fit with "very high frequency GW" (second sentence). If the mirrors are inertial here then why aren't they inertial in your case 2 description "can't move fast enough to compensate for the distance change due to the GW."?

It all depends on how one defines the word "move". For an extended object, like a bar, one must carefully define what this means to avoid confusion. First we pick a reference frame. The reference frame I choose is the one in which the bar is at rest prior to the arrival of the GW. I further specify the coordinate system known as the Transverse Traceless one. This is a coordinate system in which inertial mass points at rest with respect to the coordinates prior to the GW arrival, remain at rest relative to the coordinates during and after the departure of the GW. Remember, the metric is time dependent so distances between points are still changing even between points at rest with respect to these coordinates.

I select this coordinate system because the equation of motion of an elastic solid, specifically what I was asked to comment on, are valid in their classical form. In this coordinate system "can't move fast enough" is referring to points within the bar which remain inertial or unaccelerated relative to the coordinates for some period of time after the arrival of the GW. By the equations of motion the surrounding matter exerts no net force on interior points of the bar. The passing GW applies a metric strain to the bar. This strain results in a stress field by Hooks law. However, this stress field has 0 divergence. By the material equations of motion, the effect of the GW on the bar is only felt through traction[1] forces which only appear on the bar boundary.

Now, let's take a bar 10m long and a GW pulse short enough in duration sound generated by the forces on the boundary only travels 1 mm. The GW will have a spatial wavelength dictated by the speed of light and so will be much greater the 10m. Such a GW pulse will act uniformly over the entire bar volume while the acoustic induced mechanical vibration barely has time to propagate 1 mm into the bar volume. In this limit the bar's overall length change will be very nearly the inertial limit. The difference between truly inertial mirrors and ones constrained by the bar will be small, on the order of 1mm/10m as the wave passes.

[1] Forces may be tangential as well as normal to the surface.