I LIGO and light speed constancy

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From LIGO website re how gravity waves are measured: "The arrival times change because when the arms of the interferometer change lengths, so too do the distances the light waves travel before exiting the interferometer. What gravitational waves do not change, however, is the speed of light. This means that a wave of light that happens to be in a longer arm during a gravitational wave has to travel farther before exiting, so it takes longer to leave than the beam that was in the shorter arm."

So does this not mean that the light speed is actually NOT constant? Will you be not measuring different speed of light when there is a gravity wave and different when there isnt?
 
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ewq said:
Will you be not measuring different speed of light when there is a gravity wave and different when there isnt?

First, a caution: the "speed of light" here means the coordinate speed of light, and that depends on your choice of coordinates. The standard coordinates that are used for analysis of LIGO experiments are such that the coordinate speed of light does not change during the passage of the gravitational wave; all that changes is the lengths of the interferometer arms. But one could also choose coordinates where the coordinate speed of light did change.

That said, in order to actually measure the speed of light, you would divide the distance a light beam covers by the time it takes. What the LIGO website is describing are changes in the distance (lengths of the interferometer arms) and time (round-trip travel time of the light beams in each arm) that happen exactly in concert so that the speed of light doesn't change. What changes at the detector is the relative travel times along each arm; that is what causes the interference fringes that are measured and give information about the gravitational wave. But this change in travel times is due to the changes in the arm lengths, not to any change in the speed of light.
 
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ewq said:
So does this not mean that the light speed is actually NOT constant?
No, the light speed is invariant. The light travel time depends on the actual length of the arm (which changes periodically). In case the arm is longer, the light travel time is longer. That's what they measure.
 
OK say I am in a tunnel, 10 my local units long. There is no gravity wave and i measure round trip of light to be 5 seconds.
Now comes the gravity wave and stretches the tunnel- I still measure it to be 10 units long because my ruler scales along with the tunnel length, but now I measure light took 6 seconds for the round trip, hence i get different speed?

Or are you saying that the gravity wave affects my clock as well? If i am watching the clock located in the "stretched" space will it seem to run slower than mine from the "unstretched" space?
 
ewq said:
I still measure it to be 10 units long because my ruler scales
The ruler doesn't scale. That is the point. Tidal gravity actually measurably deforms objects.
 
ewq said:
Now comes the gravity wave and stretches the tunnel

No, the gravity wave doesn't stretch the tunnel; it only stretches the distance between the light source/detector at one end of the tunnel, and a mirror at the other end, both of which are hung in such a way that they can move freely in the appropriate plane. In other words, they move relative to the tunnel (and to a ruler inscribed in the tunnel).

(More precisely, the gravity wave stretches the tunnel and its associated ruler so much less than it stretches the distance between the light source and the mirror, that we can consider the stretch of the tunnel to be negligible. This is because the tunnel and the ruler are constrained by internal forces between their atoms, which resist the stretch due to the changing tidal gravity of the wave. The distance from light source/detector to mirror is not so constrained, hence it can stretch freely due to the wave.)

Dale said:
Tidal gravity actually measurably deforms objects

This might be somewhat misleading, since what is actually "deformed" in this case is a distance in free space, not the size of a single object. See above.
 
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what? gravity wave affects the tunnel but not my tape measure?

of course it does, gravity wave or not i will always measure the tunnel to be 10 units long
 
PeterDonis said:
This might be somewhat misleading, since what is actually "deformed" in this case is a distance in free space, not the size of a single object. See above.

This would still mean i measure dofferent light speed with wave present and when not present?

Also does the wave affect my clock or not?
 
ewq said:
This would still mean i measure dofferent light speed with wave present and when not present?

No. The distance along the tunnel/ruler that the light has to travel changes; so does the time it takes to travel. The two change exactly in sync to keep the speed of light, measured by the ruler along the tunnel, the same.

ewq said:
does the wave affect my clock or not?

No.
 
  • #10
ok- let me see if i understand you:
my physical ruler and tunnel length stretch just a little. distance between free floating mirror and emitter stretches much more. that's the reason why I am able to notice length increase even from within "stretched frame".
my clock is unaffected

is this what you are saying?
 
  • #11
There are multiple ways of describing the situation.

What needs to be decided straight off to answer the question in terms of rulers is to decide what rulers measure and how they work. Either rulers change their length when a gravitational wave passes, or they don't. They can't do both if one wants a self-consistent explanation. One has to decide which definition applies, and have a shared understanding between the explainor and the explainee.

If one regard a ruler as something made out of something physically rigid (to the extent that's possible - nothing is perfectly rigid), one can regard rulers as not changing length ever, and gravitational waves won't change their length either. This is consistent with the SI defintion of the ruler, where one regards rulers as having a constant light round-trip travel time, and the speed of light being a defined physical constant.

Using this line of reasoning, one concludes from the fact that the ruler measures an increase in the distance between the test masses that the test masses must actually move due to the gravitational wave.

I frequently find people who do not seem to realize that the test masses to which the mirrors of the interferometer are mounted are designed to be able to move freely. The test masses are suspended in such a manner as to allow them to move freely (in two dimensions at least, the masses are supported by pendulii to support them against gravity so they can't move in the direciton in which they are suspended). This freedom to move is part of the design of LIGO.

Now, it's also perfectly consistent to regard physical rulers as stretching, which to my mind somewhat defeats the whole purpose of having a ruler in the first place, and is also not consistent with the SI definitions. You'd think that such explanations would not be popular, or at least people would mention that they're not using the standard SI defintions when presenting this sort of epxlanation, but this turns out not to be the case. There's nothing actually wrong with these explanations , however, if they are interpreted and understood correctly. Frequently, though, they cause more confusion than they prevent - in my opinoin, at least.
 
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  • #12
ewq said:
my physical ruler and tunnel length stretch just a little

In principle, they will, because the internal forces between the atoms can't respond instantaneously to the tidal gravity changes in the wave. But in practice we won't be able to measure any stretching at all. That's why the source/detector and mirrors in LIGO have to be so carefully hung and isolated from the tunnel: if they were just mounted to the tunnel, the detector would never detect a signal.

ewq said:
thats the reason why I am able to notice length increase even from within "stretched frame".

If you define the "stretched frame" as being the length measured by the ruler, yes, it's the difference between the behavior of the ruler and the behavior of the source/detector and mirrors that explains why the detector detects a signal.

However, you should be aware that this "explanation" depends on defining the "frame" the way you have. There are other ways to define a "frame" (i.e., a system of coordinates). They all make the same prediction for what the detector detects, but they don't all give the same "explanation" for why.
 
  • #13
pervect said:
There are multiple ways of describing the situation.

What needs to be decided straight off to answer the question in terms of rulers is to decide what rulers measure and how they work. Either rulers change their length when a gravitational wave passes, or they don't. They can't do both if one wants a self-consistent explanation. One has to decide which definition applies, and have a shared understanding between the explainor and the explainee.

If one regard a ruler as something made out of something physically rigid (to the extent that's possible - nothing is perfectly rigid), one can regard rulers as not changing length ever, and gravitational waves won't change their length either. This is consistent with the SI defintion of the ruler, where one regards rulers as having a constant light round-trip travel time, and the speed of light being a defined physical constant.

Using this line of reasoning, one concludes from the fact that the ruler measures an increase in the distance between the test masses that the test masses must actually move due to the gravitational wave.

I frequently find people who do not seem to realize that the test masses to which the mirrors of the interferometer are mounted are designed to be able to move freely. The test masses are suspended in such a manner as to allow them to move freely (in two dimensions at least, the masses are supported by pendulii to support them against gravity so they can't move in the direciton in which they are suspended). This freedom to move is part of the design of LIGO.

Now, it's also perfectly consistent to regard physical rulers as stretching, which to my mind somewhat defeats the whole purpose of having a ruler in the first place, and is also not consistent with the SI definitions. You'd think that such explanations would not be popular, or at least people would mention that they're not using the standard SI defintions when presenting this sort of epxlanation, but this turns out not to be the case. There's nothing actually wrong with these explanations , however, if they are interpreted and understood correctly. Frequently, though, they cause more confusion than they prevent - in my opinoin, at least.
not sure what was the goal of this post?
the reality behind physical units of measurement is not arbitrary and subject to our consensus, gravity waves either stretch them or not, let's try and stay focused
 
  • #14
ewq said:
the reality behind physical units of measurement is not arbitrary and subject to our consensus,
The SI unit of time, the second, and the SI unit of distance, the meter, are completely arbitrary and are not only subject to consensus, they are the product of a committee. It is hard to imagine something more subject to consensus than the output of a committee.
 
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  • #15
PeterDonis said:
That's why the source/detector and mirrors in LIGO have to be so carefully hung and isolated from the tunnel: if they were just mounted to the tunnel, the detector would never detect a signal.

I thought that the sole point of intricate suspension was to isolate the mirrors from external disturbances (seismic, thermal etc), didnt occur to me that it also allows it to move independently from the ceiling. So in essence the mirror is a pendulum and the grav. wave will "swing" it as it passes?
Kind of like as if there was an additional weak gravitational force acting horizontally- it wouldn't affect the walls and the ceiling of the tunnel much but it would skew the chandelier a little bit?

OK, on to the next point- why arent the clocks affected by grav wave? Isnt a gravity wave an oscillating curvature of spacetime i.e. gravity potential?
 
  • #16
ewq said:
So in essence the mirror is a pendulum and the grav. wave will "swing" it as it passes?

Sort of, yes. The motion is so small that it can be treated as being solely in the horizontal plane.

ewq said:
Kind of like as if there was an additional weak gravitational force acting horizontally

No. The concept of "gravitational force" (and "gravitational potential", see below) is only meaningful in a static spacetime, and spacetime while the gravitational wave is passing is not static. The gravitational wave is time-varying tidal gravity.

ewq said:
it wouldn't affect the walls and the ceiling of the tunnel much but it would skew the chandelier a little bit?

"Skew" in an oscillating fashion.

ewq said:
Isnt a gravity wave an oscillating curvature of spacetime

Yes.

ewq said:
i.e. gravity potential

No. Spacetime curvature is tidal gravity, not "gravitational potential". As noted above, "gravitational potential" is only a meaningful concept in a static spacetime, and spacetime while the gravitational wave is passing is not static.
 
  • #17
PeterDonis said:
This might be somewhat misleading, since what is actually "deformed" in this case is a distance in free space, not the size of a single object. See above.
You are right. My comment might have been relevant for a Weber bar type of gravitational wave detector, but not so relevant for an interferometer.
 
  • #18
PeterDonis said:
No. Spacetime curvature is tidal gravity, not "gravitational potential". As noted above, "gravitational potential" is only a meaningful concept in a static spacetime, and spacetime while the gravitational wave is passing is not static.

no matter- why wouldn't tidal gravity affect my clock? don't say that sinusoidal effect averages out to zero because the same argument could be used for distance
 
  • #19
ewq said:
why wouldn't tidal gravity affect my clock?

Does the tidal gravity of the Moon on the Earth affect Earth clocks?

Tidal gravity--at least, the kind we're talking about here--is purely spatial. It stretches and squeezes things spatially. It doesn't do anything to time.
 
  • #20
PeterDonis said:
Does the tidal gravity of the Moon on the Earth affect Earth clocks?

of course it does, but point taken -negligible when compared to what it does to oceans...
 
  • #21
ewq said:
of course it does

It does? How? What is your basis for this?
 
  • #22
every gravitational field has effect on clocks? don't know the exact math but i remember i read somewhere that time dilation on Earth because of moons influence is about 1 sec per 200000 yrs
 
  • #23
Don't know if this helps but LIGO spent a huge amount of effort designing the mirror supports which hold the mirrors. The goal of these mirror supports is to provide a near frictionless support for the mirrors along an axis parallel to local gravity and along the optical beam. The reason for this is so the mirrors that define the interferometer arm for all intents and purposes may be treated as inertial (or in free fall). The only exception to this is the radiation pressure the light beams exert on the mirrors which is central to the measurement as a whole. I bring this up because the interaction of GW with actual material like meter sticks etc is quite complex and in LIGOs case not germane.
 
  • #24
Tidal gravity--at least, the kind we're talking about here--is purely spatial. It stretches and squeezes things spatially. It doesn't do anything to time.

Exactly and it is maybe worth recalling here why this is so since you were remarking the time-varying nature of tidal gravity in a previous post and it might confuse some saying at the same time that it only affects the space part. This comes from the fact that with the linearized metric used to approximate the far source effect of GWs a gauge condition(de Donder gauge) is required to solve the EFE linearized equations so that a wave equation is obtained ##\Box h_{\mu\nu}=0##, now given this condition its solutions only affect the space components and not the time components. So mathematically it is clear enough.

One could just maybe find odd that the wavelength of the laser light is affected by the GW but not its frequency while c is constant, but this is necessary if one wishes to use the interferometer as a clock.
 
  • #25
ewq said:
every gravitational field has effect on clocks?

A "gravitational field" is not the same thing as tidal gravity. See below.

ewq said:
time dilation on Earth because of moons influence is about 1 sec per 200000 yrs

I haven't checked the math but that sounds roughly correct; but the "Moon's influence" here is not its tidal gravity. It's the gravitational potential due to the Moon. They're different things.
 
  • #26
PeterDonis said:
A "gravitational field" is not the same thing as tidal gravity. See below.
I haven't checked the math but that sounds roughly correct; but the "Moon's influence" here is not its tidal gravity. It's the gravitational potential due to the Moon. They're different things.

thanks for your patience with me guys. I though what is usually referred to as tidal gravity is just varying grav field? i will reserach this a little more but as ligo application goes this clarifies a lot

the part about laser redshifting was never an issue, that is pretty clear to me i think
 
  • #27
ewq said:
I though what is usually referred to as tidal gravity is just varying grav field?

In Newtonian physics that is the case, yes. But in Newtonian physics gravitational fields don't affect clocks.

Tidal gravity in GR is spacetime curvature, and it is a well-defined concept in any spacetime. "Gravitational field" in GR is a concept that only applies in certain spacetimes (in fact, the term is somewhat vague in GR since there is more than one concept it could refer to).
 
  • #28
PeterDonis said:
in fact, the term is somewhat vague in GR since there is more than one concept it could refer to
Because of this, I tend to think that it is best used to refer only to the Newtonian gravitational field. In GR, because of the likelihood of confusion, it should not be used at all any more.
 
  • #29
ewq said:
why wouldn't tidal gravity affect my clock?
Tidal gravity depends on the "size" of the thing being affected. That is one reason that the LIGO interferometer arms are so long. A clock is comparatively small so tidal effects are also small.
 
  • #30
PeterDonis said:
No, the gravity wave doesn't stretch the tunnel; it only stretches the distance between the light source/detector at one end of the tunnel, and a mirror at the other end, both of which are hung in such a way that they can move freely in the appropriate plane. In other words, they move relative to the tunnel (and to a ruler inscribed in the tunnel).

(More precisely, the gravity wave stretches the tunnel and its associated ruler so much less than it stretches the distance between the light source and the mirror, that we can consider the stretch of the tunnel to be negligible. This is because the tunnel and the ruler are constrained by internal forces between their atoms, which resist the stretch due to the changing tidal gravity of the wave. The distance from light source/detector to mirror is not so constrained, hence it can stretch freely due to the wave.)

If the gravity wave propagates with the same speed as light, it should affect (move) the mirror at the other end just before the light from the "displaced" source arrives there, so the distance between the light source and the mirror would not be stretched. Why we detect a stretch?
 
  • #31
DanMP said:
Why we detect a stretch?

In general relativity the metric determines distanced and time intervals between events. For a GW the metric is time dependent so the distance between points is time dependent. The length between mirrors is changing as the wave passes. This distance change doesn't depend on ones choice of coordinates. One may, for example, choose coordinates in which the mirror locations are constant and the distance between mirrors is changing because the metric is time dependent. Now, because the distance between mirrors is changing the number of wavelengths between mirrors is changing. This is what is measured.
 
  • #32
DanMP said:
If the gravity wave propagates with the same speed as light, it should affect (move) the mirror at the other end just before the light from the "displaced" source arrives there, so the distance between the light source and the mirror would not be stretched. Why we detect a stretch?
Gravitational waves are transverse. The stretch happens along the tunnel if the wave is propagating across the tunnel.
 
  • #33
DanMP said:
If the gravity wave propagates with the same speed as light, it should affect (move) the mirror at the other end just before the light from the "displaced" source arrives there

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.
 
  • #34
Ibix said:
Gravitational waves are transverse. The stretch happens along the tunnel if the wave is propagating across the tunnel.

I didn't know that "gravitational waves are transverse". I believed (and still do) that GWs are longitudinal ...

The latest detection, https://dcc.ligo.org/public/0145/P170814/010/GW170814.pdf "allowed us to probe the polarization content of the signal for the first time; we find that the data strongly favor pure tensor polarization of gravitational waves, over pure scalar or pure vector polarizations". What does this means? It implies, by any chance, longitudinal oscillations?
 
  • #35
DanMP said:
I didn't know that "gravitational waves are transverse". I believed (and still do) that GWs are longitudinal ...

The latest detection, https://dcc.ligo.org/public/0145/P170814/010/GW170814.pdf "allowed us to probe the polarization content of the signal for the first time; we find that the data strongly favor pure tensor polarization of gravitational waves, over pure scalar or pure vector polarizations". What does this means? It implies, by any chance, longitudinal oscillations?
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.
 
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  • #36
ewq said:
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.
 
  • #37
Mister T said:
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.
 
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  • #38
pervect said:
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.
 
  • #40
Aren't metallic bars sensitive in the range of their resonance frequency only?
 
  • #41
timmdeeg said:
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.
 
  • #42
timmdeeg said:
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.
 
  • #43
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.
 
  • #44
pervect said:
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.
 
  • #45
pervect said:
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.
 
  • #46
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".
 
  • #47
PeterDonis said:
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.
 
  • #48
pervect said:
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.
 
  • #49
Ibix said:
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 ...

PeterDonis said:
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.
 
  • #50
timmdeeg said:
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).
 

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