Why does a LIGO arm stretch and not move in unison?

[Henri Garcia]

LIGO is most sensitive to a GW from directly above/below. As a transverse wave hits an arm why don't the laser source and the mirror move in unison -- thereby covering up the distorted motion?

 

[PeterDonis]

As a transverse wave hits an arm why don't the laser source and the mirror move in unison

Because the wave doesn't move them in unison. It's a wave of tidal gravity; it alternately stretches and squeezes the arm.

 

[Henri Garcia]

Because the wave doesn't move them in unison. It's a wave of tidal gravity; it alternately stretches and squeezes the arm.

Thank you very much for the reply, Peter. This followup maybe inane, but please humor me and appreciate my confusion:
Is the displacement for a GW completely within the 4 spacetime axes, or are the effects in spacetime, but the displacement is in a non-existent 5th dimension?

 

[PeterDonis]

Is the displacement for a GW completely within the 4 spacetime axes, or are the effects in spacetime, but the displacement is in a non-existent 5th dimension?

I don't understand the question. How can you have a displacement in a non-existent dimension?

In any case, thinking of "displacements" can lead to misunderstandings. Spacetime is a 4-dimensional geometric object. "Gravitational waves" are just a name for a particular kind of spacetime geometry. The geometry is what it is; it doesn't "change" or get "displaced". Thinking of gravitational waves as "displacements" means we have chosen a particular way of splitting up spacetime into "space" and "time", but that's just a choice of coordinates; it doesn't change the underlying spacetime geometry.

 

[Henri Garcia]

I don't understand the question. How can you have a displacement in a non-existent dimension?.

You wouldn't. But I thought perhaps that the curvature effects in spacetime could be viewed *as if* they resulted from such a displacement. Clearly that is not the case.
My original confusion with the stretching of LIGO was that I pictured two electrons, say in an antenna, being hit simultaneously by a plane EM wave. The linearly polarized E-field would not stretch them apart, it would move them simultaneously. Similarly, why doesn't the transverse GW not move the laser and mirror together as one? I promise this is my last query on this! :-)

 

[Ibix]

. But I thought perhaps that the curvature effects in spacetime could be viewed *as if* they resulted from such a displacement. Clearly that is not the case.

All of general relativity works with four dimensions. You are thinking of "embedding" spacetime in some higher dimensional space, like a piece of paper (2d, or close enough) in 3d space. Although this kind of thing is sometimes drawn (the infamous rubber sheet model, for example) the reality is that there's no evidence of anything outside spacetime in which it could be embedded. And there's no need for it in theory.

Similarly, why doesn't the transverse GW not move the laser and mirror together as one? I promise this is my last query on this! :-)

In some senses, neither mirror nor laser is moving (if you strap an accelerometer to them, you'll find that they never accelerate). The distance between them is just changing.

 

[PeterDonis]

why doesn't the transverse GW not move the laser and mirror together as one?

Because a gravitational wave acting on objects is not the same as an EM wave acting on electrons. At the "B" level it's hard to go into much more detail than that. But @Ibix brought up one key difference, which is that objects being acted on by a gravitational wave are in free fall--unaccelerated. This is not the case for electrons being acted on by an EM wave, or indeed for any objects being acted on by any non-gravitational force.

In short, while the analogy between gravitational waves and EM waves can be helpful in some ways, it breaks down after a certain point, and the question you are asking goes beyond that point.

 

[Henri Garcia]

Thanks to you both!

 

[Mister T]

LIGO is most sensitive to a GW from directly above/below. As a transverse wave hits an arm why don't the laser source and the mirror move in unison -- thereby covering up the distorted motion?

You do understand that the mirror is mounted in such a way that it's free to move towards or away from the laser?

 

[Henri Garcia]

You do understand that the mirror is mounted in such a way that it's free to move towards or away from the laser?

Thanks for the reply, Mr. T. I am confused. Why should it matter if it is free to move or if it is rigidly held 4 km from the laser? I thought the suspension was solely to isolate the mirror. Please explain.

 

[Mister T]

Why should it matter if it is free to move or if it is rigidly held 4 km from the laser?

Because if the mirror were fixed in place the device would not be able to detect GW's for the reason you gave in your original post.

 

[Justin Hunt]

I would like to understand this as well. My understanding is that when neutron stars or black holes collide and create gravitational waves that the event would propagate in the form of expanding spheres, so that after the millions or billions it takes to get here the curvature is essentially flat. Which means they waves would be equivalent to planes moving through the earth. I would imagine anything that lines up symmetrically would look the same to both arms. Remember LIGO is only detecting differences in the arrival not the time. The symmetry between the arms, however, is in the form of a plane so it would be very unlikely for a wave to actually come in at that exact angle.

 

[PeterDonis]

My understanding is that when neutron stars or black holes collide and create gravitational waves that the event would propagate in the form of expanding spheres, so that after the millions or billions it takes to get here the curvature is essentially flat.

The question is, what is essentially flat? Just saying "the wave is" doesn't answer that question.

The first part of the answer is, the "wave fronts" are essentially flat: in other words, in the usual local coordinates used for this type of problem, whatever thing describes the wave is a function of $t$ (time) and $z$ (distance along the direction of propagation) only. (For a relativistic wave, it turns out that it can only be a function of $t - z$, but we don't need to go into that right now.)

However, that still doesn't answer the question: what is the "thing that describes the wave"? See further comments below.

Which means they waves would be equivalent to planes moving through the earth. I would imagine anything that lines up symmetrically would look the same to both arms.

You appear to have a mental picture that the "thing that describes the wave" is a scalar--a single number. But that is not the only possible kind of wave. You can also have waves where the "thing that describes the wave" is a vector or a tensor; an electromagnetic wave is an example of the first of these (vector), and a gravitational wave is an example of the second (tensor).

For a tensor wave, like a gravitational wave, the effect on a given wave front can be different in different directions. For example, see here:

https://en.wikipedia.org/wiki/Gravitational_wave#Effects_of_passing

 

[Justin Hunt]

I am sorry I wasn’t clear. It would be more of a solid passing through the earth. Not an instantaneously event.

The point is wouldn’t detectors on different parts of the earth, the moon, or even on Mars detect the same “shape”? Or are there fluctuations depending on the direction you view the event from? Shouldn’t it be symmetrical in all directions it propagates?

 

[PeterDonis]

It would be more of a solid passing through the earth.

I don't know what you mean by this, but if it means the "wave fronts" are not planes transverse to the direction of propagation, it's wrong.

Not an instantaneously event.

Of course the whole passage of the wave is not instantaneous; but a given wave front at a given point along the direction of propagation is.

wouldn’t detectors on different parts of the earth, the moon, or even on Mars detect the same “shape”?

Detectors oriented in the same direction would, yes (but at different times). Detectors oriented in different directions would not.

Shouldn’t it be symmetrical in all directions it propagates?

No. Please go back and read my previous post again, and look carefully at the images in the Wikipedia article I linked to. You can't understand how a gravitational wave works by just guessing or reasoning by analogy. You need to actually look at the math (which is what those images are based on).

 

[Justin Hunt]

Alright peter, sorry again for not using the correct scientific language. I was imagining the wave fronts being planes and it being a longitudinal wave. After reviewing the article and researching I see that gravitational waves are transverse with a transverse tensor rather than vectors and have two polarizations. But, I can’t admit to fully understanding what that means.

But, getting back to the question, you indicated the orientation of the dector will effect the “shape” of the gravity wave that it sees.

1. Does this cause an issue with multiple detectors confirming a result? If their orientations are different due to their location, wouldn’t the waves look different and make it difficult to pick out in the background noise?

2. Couldn’t a certain orientation cause both arms to be effected equally so that nothing is detected?

 

[PeterDonis]

I was imagining the wave fronts being planes and it being a longitudinal wave. After reviewing the article and researching I see that gravitational waves are transverse with a transverse tensor rather than vectors and have two polarizations.

Yes.

I can’t admit to fully understanding what that means.

"Transverse" just means that any "action" of the wave is orthogonal to the direction of propagation.

"Two polarizations" means, heuristically, that there are two "modes" in which the wave can act. The images in the Wikipedia article show the two modes. (There is a lot more math lurking beneath this, to justify why those two images are indeed the two "modes", why they are independent of each other, etc., but that is a quick heuristic description.)

you indicated the orientation of the dector will effect the “shape” of the gravity wave that it sees.

Yes, because it will affect the relative amplitude of the two polarizations as seen by the detector. In the simplest case, where the detector is oriented directly along one of the polarizations, it will see only that one, and not the other one. More generally, any single detector can only detect part of the total action of the wave; you need multiple detectors to fully detect the wave's action.

Does this cause an issue with multiple detectors confirming a result?

No; we have multiple detectors pointing in different directions precisely because one detector can only detect part of the total action of the wave.

If their orientations are different due to their location, wouldn’t the waves look different and make it difficult to pick out in the background noise?

A particular wave can be more difficult to detect in some directions as compared with others. For example, if a particular wave is entirely made of one of the two polarizations, then a detector oriented in the wrong direction for that polarization will detect nothing at all. However, if you have multiple detectors and take care to vary their orientations appropriately, at least one of them will detect any wave that is detectable at all above background noise.

Couldn’t a certain orientation cause both arms to be effected equally so that nothing is detected?

No. A certain orientation for a wave of only one polarization (the "wrong" one for that polarization) will cause both arms to not be affected at all. But if the wave affects the arms at all, it will not affect them equally. Remember that the arms, by design, are at a 90 degree angle, and look at how the images in the Wikipedia article stretch and squeeze.

 

[Justin Hunt]

@Peter two arms have a plane of symmetry. 3 arms would have a line of symmetry. You need need 4 arms to guarantee no symmetry. I had imagined you would need 4 arms to be 100% sure you didn’t miss it. Realistically speaking though, even with two arms and having a plane of symmetry, it would be very improbable for any incoming propagation ray to fall entirely inside that plane of symmetry.

But, I will take your word for it that two arms is all you need for 100% guaranteed detection possibility since I don’t completely understand the nature of the gravitational waves. Thank you very much for your information responses!

 

[PeterDonis]

two arms have a plane of symmetry.

Two arms determine a plane, period. I'm not sure why you call it a "plane of symmetry". For best detection, this plane should be perpendicular to the direction of wave propagation; in actuality it can't always be, of course, but the projection of the actual plane of the detector into the plane perpendicular to the direction of propagation turns out to be all that is relevant for predicting the detector's response.

3 arms would have a line of symmetry. You need need 4 arms to guarantee no symmetry.

I'm not quite sure what you mean by these either. But in any case, what do they have to do with what we're discussing?

even with two arms and having a plane of symmetry, it would be very improbable for any incoming propagation ray to fall entirely inside that plane of symmetry.

Huh? Any plane that is not parallel with the direction of wave propagation will have a nonzero projection into the plane perpendicular to the direction of propagation. I'm not sure what reasoning you are using here, but whatever it is, it looks wrong.

 

[Henri Garcia]

Is it possible that Justin means "detector" and not "arm"? i.e., 2 LIGOs determine a plane for source location. 3 determine a line. and 4 are needed to pin down the direction?

 

[Henri Garcia]

This discussion has helped me a lot. Thanks. And I think this next point is important.
I originally thought that the stretching of spacetime affected the 4km distance to the mirror as well as the 4km pipe that surrounds it. Both would stretch by the same amount. I thought that the delicate suspension of the mirror was only to isolate it from vibrations. Now I am led to believe that the unconstrained mirror responds more to the stretching. The solid pipe is rigid by electromagnetic forces that fight against the stretching.
thanks.

 

[PeterDonis]

I thought that the delicate suspension of the mirror was only to isolate it from vibrations.

That's correct. See below.

Now I am led to believe that the unconstrained mirror responds more to the stretching. The solid pipe is rigid by electromagnetic forces that fight against the stretching.

That's correct in principle, but in practice it might not make much of a difference. [Note: edited in response to correction from @mfb below.] In relativity there are no perfectly rigid bodies: the inter-atomic forces that keep the body rigid can only move through the body at the speed of light. The inter-atomic forces that keep the body rigid will travel through the body at the speed of sound. The time it takes light sound to travel 4 km in the Earth or the solid pipe is a lot longer than the time it takes the gravitational wave to move atoms, so in practical terms, the solid pipe is not significantly more rigid than the unconstrained mirror.

(Actually, the forces keeping the pipe rigid will move through it much slower than the speed of light, because the pipe is not made of an idealized material that has the maximum rigidity allowed by relativity. The actual speed of the forces will be the speed of sound in the pipe, which is orders of magnitude slower than the speed of light. But that just makes the argument I gave above even stronger.)

However, because the solid pipe is not isolated the way the mirror is, the solid pipe is responding to lots of other influences besides the gravitational wave. And those other influences swamp the gravitational wave's effect on the pipe. The mirror is isolated in order to remove all those other influences on its motion, that would otherwise swamp the gravitational wave's effect just as they do for the solid pipe.

 

[Henri Garcia]

You do understand that the mirror is mounted in such a way that it's free to move towards or away from the laser?

Considering only motion induced by the GW, are you implying that the unconstrained mirror moves more than if it were fixed to the Earth? (Let's neglect noise for the sake of argument)

 

[Mister T]

Considering only motion induced by the GW, are you implying that the unconstrained mirror moves more than if it were fixed to the Earth?

Yes, and for the reason you stated in Post #21. It's a necessary part of the design. Without it the device wouldn't work.

 

[PeterDonis]

It's a necessary part of the design. Without it the device wouldn't work.

No, this is not correct. See my post #22. The reason for the isolation is to remove other sources of movement besides gravitational waves. It is not because the Earth itself is any more rigid, in practical terms, than the mirror/laser/detector assemblies that form the arms of the detector, as regards their response to gravitational waves.

 

[pervect]

Well, I think the very shortest answer is that if one's understanding of GR is that the passing of a gravitational wave doesn't alter the transit time of a laser along the detection arms, that understanding is incorrect in some manner.

IT's hard to pinpoint more without knowing what that understanding is based on, the chain of logic that makes one thinks the passing of the GW shouldn't matter.

As far as concrete exposistions of a good model of GR - which may be more demanding than a popularization - I'd suggest Baez's "The Meaning of Einstien's Equations", http://math.ucr.edu/home/baez/einstein/ for the html version. There's also an arxiv version. The author, Baez, does say that an understanding of special relativity is needed to understand general relativity at the level he is writing at though, which is often an obstacle. And there are other elements of background that may be needed to appreciate this particular paper.

The paper isn't oriented specifically towards understanding gravitational waves, either - it's more a general article about GR.

The understanding of the application of Baez's paper would be that if one had a ball of coffee ground (test masses) floating in free space, and a GW passed through it, the shape of the ball would distort, but it would distort in such a manner that the appropriate derivative of the rate of change of the volume of the ball was zero. This means some coffee ground would move outwards, some inwards, deforming the ball into an ellipsoid.

Here one imagines that there is something that doesn't distort when the GW passes, which is a bit tricky actually. But for a small enough ball of coffee grounds, it's possible to find a solution for the trajectories of the grounds in the ball that keeps the distance between all pairs of grounds constant within experimental error that serves as a reference for the distortion of the ball made out of free-floating coffee grounds. I don't think this is discussed directly in Baez's paper, the critical feature is that the ball must be much smaller than the wavelength of the GW in the direction of the GW propagation to allow this reference solution where the distances between grounds don't change.

 

[mfb]

The time it takes light to travel 4 km is a lot longer than the time it takes the gravitational wave to move atoms

Huh?
Light needs about 13 microseconds for that distance. That is very short compared to the period of gravitational waves where LIGO is most sensitive (1 to 10 ms).
This is part of the design. If the arms would be long compared to the wavelength, the light couldn't travel back and forth within the period, and you wouldn't gain sensitivity from the longer arms.

 

[PeterDonis]

Light needs about 13 microseconds for that distance. That is very short compared to the period of gravitational waves where LIGO is most sensitive (1 to 10 ms).

Yes, you're right. I should have used the speed of sound instead of the speed of light--the time it takes a sound wave to travel 4 km in the Earth's material--or even in steel--is much longer than the period of the waves. I'll edit my previous post accordingly.

 

[Henri Garcia]

Thanks for the correction, PD and mfb. I am trying to ask a Physics question (i.e. spherical chickens) and I am getting engineering answers (hair in the form of vibrational noise). Assume no noise. The laser is solidly fixed. The mirror is either (a) totally unconstrained or (b) solidly fixed to the ground. As the GW passes by is there any difference in the travel times for the laser pulses for (a) vs (b)? thanks.

 

[mfb]

It doesn't matter.

The speed of sound is so low that "fixed" is meaningless. The ground next to the mirrors follows the gravitational wave in the same way the current mirrors do. Fixing the mirrors to the walls wouldn't have any effect on the signal strength, it would only make the background worse.

 

[Henri Garcia]

... it would only make the background worse.

Apparently you have never seen a spherical chicken. :-) :-) And thanks for the answer. Can you (and/or Peter) recommend reading material (text?) for a PhD in physics who wants to understand GW and LIGO?

 

[Justin Hunt]

LIGO splits a beam of photons down two perpendicular 4km arms bounce back and fourth then join back up again at the detector and the difference in their arrival is detected. With no noise or GW they would arrive simultaneously.

Now, Gravity waves are disturbances to the very fabric of space time itself just like gravity. Similar to how gravity effects the trajectory of photons, gravity waves also effect the photons traveling down the arms by distorting space time. The effect is so subtle though that they need such long arms to detect it. The noise is only relevant when the photon interacts with something like a mirror or detector, during its travel down the arms it is in a vacuum and distortions to space time is all that can effect it.

@Henry the symmetry I was talking about is the kind of symmetry we find in things such as humans. Your right side is symmetric to your right. The plane of symmetry of LIGO would be 45 degrees between the two arms. The light takes the path of two sides of a 45 45 90 degree triangle (not the hypotenuse) so the line or plane of symmetry would cut the hypotenuse in half. My question earlier was about whether this symmetry matters for GW. I did and still don’t understand the mechanics of GW themselves, but it would be feasible that gravity waves with propagation rays that lie on the plane of symmetry would have identical effects on the photons traveling down the arms and thus have a zero net effect on the detector.

 

[Ibix]

Similar to how gravity effects the trajectory of photons, gravity waves also effect the photons traveling down the arms by distorting space time.

That's not really what LIGO detects. What it detects is a difference in flight times of light down the two arms. Usually this is interpreted as a changing difference in the length of the arms. In principle it's doing nothing more complex than racing light pulses down the two arms and watching for anything other than a draw. The practice is rather more complex because the signals are so weak. (Edit: there's more than one way of describing what LIGO does, but that's one way).

I did and still don’t understand the mechanics of GW themselves, but it would be feasible that gravity waves with propagation rays that lie on the plane of symmetry would have identical effects on the photons traveling down the arms and thus have a zero net effect on the detector.

Gravitational waves (not gravity waves, by the way - those are a kind of surface wave on water) cause changes in distances perpendicular to the direction of propagation of the wave. Say the wave is propagating in the z direction. In one direction (call it x) distances stretch and then squish; in the other (y) direction distances squish then stretch. You are correct that if a wave happens to impinge on LIGO with its x and y directions exactly at 45° to the arms the wave will not be detected. That's one of the reasons for building multiple detectors with different orientations, because each detector is blind to gravitational waves from some parts of the sky. You also need multiple detectors for triangulation and speed estimates and probably other things.

 

[mfb]

You are cotrect that if a wave happens to impinge on LIGO with its x and y directions exactly at 45° to the arms the wave will not be detected.

As an example, Virgo was close to this orientation for the observed neutron star merger, and had a very small signal as result.

 

[Marcus Parker-Rhodes]

Because the wave doesn't move them in unison. It's a wave of tidal gravity; it alternately stretches and squeezes the arm.

As far as I understand a gravitational wave flexes the whole of space/time spreading across the firmament
like a ripple on a pond, but it seems to me that would not only stretch and squash the machine and the scientists operating it, but also include time and so the apparent speed of light.