High School How Does LIGO Detect Gravitational Waves Despite Changes in Spacetime?

[Heikki Tuuri]

As a gravitational wave passes an observer, that observer will find spacetime distorted by the effects of strain. Distances between objects increase and decrease rhythmically as the wave passes, at a frequency equal to that of the wave. This occurs despite such free objects never being subjected to an unbalanced force.

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

asca, the frequency of a gravitational wave in LIGO is around 300 Hz. The speed of sound in rock is roughly 5 km/s. During one cycle of the wave, sound only propagates 17 meters, which is a small fraction of the LIGO arm length of 4 km.

That is, the rock does not have time to adjust to the changed distances.

The light which is used to measure the distance, on the other hand, does adjust very quickly. We are able to measure the new distance with the light.

If the frequency of the gravitational wave would be very slow, then the stress in the rock would eventually bring the ends of the arm to the same distance as they were before the gravitational wave. Then you would no longer notice any changed spatial geometry.

A source of confusion may be that an arm of the LIGO itself is a "ruler". If that "ruler" would immediately adjust to the new geometry, so that it would no longer be under a stress, then we would not be able to observe the changes in the spatial metric.

 

[asca]

Once again, thank you all for the help you are providing. I want to ensure you the I am pretty sure I am missing something, as Pervect says, but I still didn't manage to see what.
If I go thorough all your replies, I still feel something is missing in finding the proper answer. For example, I see that Peter Donis says that I am wrong when I say the the path is curved in Eddington experiment, but then he says that it is a straight lines--geodesics (and he also uses as an example the Earth meridian), which is exactly what I said. However I think I find a common pattern in all your replies, which probably could lead to the point I am missing: it seems to me that you all say that the light wavelength does NOT get "stretched" while passing through a stretched portion of space-time. Ok, assuming this is true (I confess I do not see where Relativity states this fact but I'll try to double check), let's think about the following scenario : we setup a LIGO like detector, we run it, and it generates an interference pattern. We keep it running for months and the pattern does not change. Three months later we manage to put a mass somewhere so that one and only one of the two arms "pierces" a portion of a stretched space-time. Let's say that that arm is made up of three segments, the first one lies in the non-stretched space time, the middle one lies in the stretched space time, and the last one again in the non stretched. Now the interference pattern has changed its shape and it will stay like that until we remove the distorting mass. This change, according to what I guess you are all saying, is due to the fact that the light wavelength does NON get stretched, so if I put a detector in the middle segment, that detector should measure a shorter wavelength than the wavelength measured by two other identical detectors placed in the first and last segment of this arm. Is that what you guys are trying to tell me (please do not reply that if I put a detector I change the experiment, I really think this "quantum" effect would be negligible)?
Let me assure you I believe the signal was detected, I believe GR works, I am not one of those "flat Earth" guys , or "no moon landing" guy. I just do not manage to get around this "flaw" in my own reasoning, but if someone shows me where relativity says that wavelengths do not get affected by space-time distortion, I believe we are done. Thank you again.

 

[Heikki Tuuri]

Right, the wavelength of the laser in LIGO is constant.

It is the distance between the ends of the LIGO arm which changes.

The light which goes to the "stretched" arm does not know anything about the stretching. Its wavelength is not stretched. The light simply will think that the arm is now longer/shorter than it used to be.

Thus, the "stretching" of space is really that the distance between inertial masses (like the ends of the detector) changes. There is no "ether" which would stretch.

 

[Ibix]

I'm not sure the experiment you are describing really makes sense - you wouldn't use an interferometer to detect the presence of a static mass. You also seem to be talking about "stretched spacetime", which I suspect is the root cause of your problems.

Imagine using a radar set that emits discrete pulses of radio waves to measure the distance to a target. The radar emits a 1ns long pulse, which bounces off a target 150,000m away and returns 1ms later. Then a gravitational wave comes in. In the 1ms a radar pulse is in flight, the gravitational wave increases the distance to the target by 0.1% (a ridiculously large number, but I can't be bothered typing all the zeroes). The radar pulse gets stretched by the same fraction. So the echo returns 1.001ms later and the pulse duration is increased to 1.001ns.

But the gravitational wave is still coming in and, while the next pulse is in flight, again grows the distance by 0.1%. So the echo return time of this pulse is 1.002ms, while the pulse duration (again initially 1ns) is stretched to 1.001ns.

The gravitational wave is still coming in, and the distance grows by another 0.1%. The echo return time for the third pulse is 1.003ms and the pulse duration is 1.001ns.

Now the gravitational wave crest has passed, so the distance to the target starts to shrink. The next radar pulse then takes 1.002ms to return and has a pulse duration of 0.999ns. The next one takes 1.001ms with a pulse duration of 0.999ns, and the next one 1.000ms with a pulse duration of 0.999ns. Etcetera.

Do you see? The effect of the wave on the cavity length is cumulative because the cavity is a persistent object. But the light is moving through the cavity and isn't affected the same way.

Of course, we use an interferometer not a radar set. But this is because we only need to measure changes in flight time, not directly measure the flight time, and an interferometer can do this much more precisely than a radar set. I think the way to think of the interferometer in this context is that the light in each arm is being used as a clock to measure the return time in the other arm, while the radar set compares the flight time to an electronic clock. But it's basically the same method.

 

[A.T.]

it seems to me that you all say that the light wavelength does NOT get "stretched" while passing through a stretched...

What we say is that this doesn't really matter. It's the changing timing that matters. Watch the video again starting at 5:23. It addresses exactly your misconception at 5.42.

 

[PeterDonis]

That is, the rock does not have time to adjust to the changed distances.

The light which is used to measure the distance, on the other hand, does adjust very quickly. We are able to measure the new distance with the light.

That doesn't mean the light "adjusts" to the changed distance. It means the distance changed, and the light, which didn't change, tells us the distance changed.

The rock also tells us the distance changed, but in a different way: by the change in its internal stresses. But that's much, much harder to measure given the tiny changes involved. That's why Weber-style bar detectors for gravitational waves, which operate on the same principle--sensing the changes in internal stresses in a large solid object caused by GWs--never got to the point that LIGO has reached.

If the frequency of the gravitational wave would be very slow, then the stress in the rock would eventually bring the ends of the arm to the same distance as they were before the gravitational wave.

Not if the rock remains solid; then its length would change much less than the length between the sensor and end-of-arm mirrors in LIGO, which can move independently of each other. The atoms of the rock can't because they are bound by internal forces; so the effect of the GW shows up in the rock mostly as a change in internal stresses, rather than a change in externally measured length. There will be some small change in overall length, but again, much less than the length changes in the arms that LIGO measures.

Then you would no longer notice any changed spatial geometry.

Yes, you would, because the internal stresses in the rock will have changed. See above.

 

[PeterDonis]

it seems to me that you all say that the light wavelength does NOT get "stretched" while passing through a stretched portion of space-time.

That's not quite what I said. See my exchange with @Ibix upthread, where he ended up saying (correctly) this:

I need to measure the positions of the endpoints of the wave simultaneously, and in a dynamic spacetime there isn't a unique way to define simultaneity. So I need to specify time in terms of free-floating clocks or whatever.

In other words, as I said earlier in that subthread, there is no invariant "wavelength" of the light, so thinking in terms of whether or not this thing that isn't an invariant gets "stretched" or not is going down the wrong path.

The invariant is the presence of interference in the detector; that tells you that the physical distance between the arm ends changed. And that, as I said earlier, is the expected result of a change in tidal gravity.

Three months later we manage to put a mass somewhere so that one and only one of the two arms "pierces" a portion of a stretched space-time.

You can't. The "stretching" (not really a good term) can't be isolated like this.

 

[PeterDonis]

you wouldn't use an interferometer to detect the presence of a static mass

Actually, you could. For example, you could put it in a circular orbit around a planet like the Earth with one arm oriented radially and the other oriented tangentially. Then tidal distortion due to the Earth would make the arm lengths slightly different and a static interference pattern would show up in the detector.

 

[asca]

Ok, we probably made it. I say the latest example of RADAR pulse posted by IBIX really helps me in clarifying what was stated in the video mentioned by AT, which I liked at the beginning, but later on I was having trouble in picturing the scenario described starting minute 5.23. It also seems to me we all agree the wavelength get affected, although some of your posts seem to deny that. However the radar pulse example clears out all my troubles, Thank you all again for everything.

 

[Ibix]

It also seems to me we all agree the wavelength get affected, although some of your posts seem to deny that.

Not quite. There are two issues - one is that defining "wavelength" in non-static spacetime isn't trivial. The other is that the wavelength emitted by the laser doesn't change, and doesn't stretch by 1% just because it entered an interferometer arm that's been stretched by 1% (the idea that it does is a fairly common misconception, I think).

 

[A.T.]

... It also seems to me we all agree the wavelength get affected, although some of your posts seem to deny that. ...

As the video explains in simple terms: A wave that is present in the space while that space is being stretched will also be stretched with that space. But a wave that enters an already stretched space, will not be stretched upon entry, but will merely need longer for the passage and that's what causes the interference pattern shift.

 

[Ibix]

But a wave that enters an already stretched space,

An opinion: I'm wondering if we ought to avoid talking about "stretched space", since I think that's how we end up with the idea that things ought to stretch when they enter it. That's why I've been trying to talk always about the arm length in this thread.

 

[PeterDonis]

I'm wondering if we ought to avoid talking about "stretched space"

I think that's a good idea since it's a coordinate-dependent concept.

 

[pervect]

An opinion: I'm wondering if we ought to avoid talking about "stretched space", since I think that's how we end up with the idea that things ought to stretch when they enter it. That's why I've been trying to talk always about the arm length in this thread.

Stretched space is such a common idea, though, it's difficult to avoid. Even if it does tend to cause confusion.

The ideal way of talking about stretched space is to talk about coordinates as labels, and then introduce the metric tensor. Then we can identify the stretching of space with the metric tensor. But I don't think that's a B-level approach, it's I-level at best.

So we are left with saying that space stretches, but (physical) rulers don't. So "stretched space", whatever it may be, isn't something that's measured with physical rulers.

That may not be a good explanation of what stretched space is (the good explanations that I'm aware of are not B-level), but at least it tells us what it isn't.

The other simple point that I think needs to be made (and has been made, to some extent, but is mostly being ignored by the OP) is that it is important to consider the round trip time when light is being used as a ruler to measure distances.

The OP, though, wants to ignore the round-trip requirement, and seems to have the idea that we can talk about length from the perspective of a light beam. It seems to be leading them to incorrect conclusions about the round-trip time, so I assume they are doing something wrong. Exactly what they are doing wrong isn't entirely clear in detail.

We can certainly say that the notion that light has a wavelength (or frequency) that's independent of the observer is wrong. The OP seems to be assuming otherwise (as near as I can tell), and I'm guessing this may be the source of some of their confusion. But I could be wrong about the source of their confusion. I am sure they must be confused about something, though, because they're getting the wrong answer about the round-trip travel time.

The correct notion of wavelength (and frequencey) is that they are not the property of light, but a property of light as measured by some specific observer.

 

[PeterDonis]

The correct notion of wavelength (and frequencey) is that they are not the property of light, but a property of light as measured by some specific observer.

Strictly speaking, frequency is what a specific observer measures about a specific light ray. Wavelength is either (1) deduced from the frequency by assuming that the speed of light is $c$, or (2) deduced from measurements by multiple observers who adopt some common definition of simultaneity in order to compare their measurements "at the same time" (this is what @Ibix described in an earlier post). #1 is not a direct measurement of wavelength since it only occurs at one event in spacetime; #2 is not an invariant measurement of wavelength because it depends on the definition of simultaneity that is adopted.

 

[pervect]

Strictly speaking, frequency is what a specific observer measures about a specific light ray. Wavelength is either (1) deduced from the frequency by assuming that the speed of light is $c$, or (2) deduced from measurements by multiple observers who adopt some common definition of simultaneity in order to compare their measurements "at the same time" (this is what @Ibix described in an earlier post). #1 is not a direct measurement of wavelength since it only occurs at one event in spacetime; #2 is not an invariant measurement of wavelength because it depends on the definition of simultaneity that is adopted.

There's a third, very old, technique, with many variations. Basically, one does not have a single light beam going in one direction, but rather a pair of light beams moving in opposite directions, which generate interference fringes.

Typically, light on the return path is reflected from a mirror.

This technique requires a round trip though, so it's not really any different than the radar method in that requirement.

The interference fringes require light moving in both directions to exist, they don't exist for light moving only in one direction. One typically considers a frame where the interference fringes are stationary, and both the light source and the reflecting mirror are stationary as well. Then the distance between the fringes is said to be one wavelength in the specified frame where everything (the source, the mirror, the fringes) is stationary.

 

[Tomas Vencl]

Once again, thank you all for the help you are providing. I want to ensure you the I am pretty sure I am missing something, …..

Maybe can also help next idea (not sure if this is really the cause).
Distances in GR are physically measured by 2 ways.
By rulers (how many of them I can put along some space interval), and radiolocation (by time interval when I receive mirrored signal).
And in GR those two are not equal. The mirrored signal is also affected by time dilatation/contraction along its way. So even when I can imagine the number of rulers equal to number of wavelengths, there is still difference causing interference pattern shift.

 

[Ibix]

Stretched space is such a common idea, though, it's difficult to avoid. Even if it does tend to cause confusion.

I think a variant on Feynman's sticky beads might help. Instead of sticky beads I want frictionless beads, able to slide freely along a rod. You can then explain that the beads free-fall apart, and if the rod were sliced into a stack of thin discs then the discs would separate too. But in the actual rod, internal forces prevent that and the beads move relative to the rod. A LIGO arm isn't (conceptually) radically different from this - we've just mounted a beam splitter on one bead and a mirror on the other. The rest of the differences stem from engineering concerns driven by a desire to reduce noise, basically.

For the point about "no stretched space", imagine a really low frequency gravitational wave, with a period of minutes or hours. The rate of change of separation of the beads is undetectable on a timescale of seconds, but the cumulative change is detectable. The takeaway point would be that while the wave is coming through, there is no experiment other than "wait and see what happens" that will differentiate between the cases of the beads being distance $d+\delta$ apart because the gravitational wave has moved them $\delta$, and them having been built $d+\delta$ apart and always being that distance apart.

 

[pervect]

I think a variant on Feynman's sticky beads might help. Instead of sticky beads I want frictionless beads, able to slide freely along a rod. You can then explain that the beads free-fall apart, and if the rod were sliced into a stack of thin discs then the discs would separate too. But in the actual rod, internal forces prevent that and the beads move relative to the rod. A LIGO arm isn't (conceptually) radically different from this - we've just mounted a beam splitter on one bead and a mirror on the other. The rest of the differences stem from engineering concerns driven by a desire to reduce noise, basically.

This isn't too much different from my preference, which would basically to use Fermi-normal coordinates around a single point rather than the "stretched space" idea.

This mathematics of this approach are unwieldy, so it's not my preference for actual calculations, just my preference for visualizing what's going on.

The approach leads to the idea that gravitatioanl wave consists of tidal forces, which causes an array of test masses to actually move, as in the image from Wikipedia below. For small, planar cross sections, we can even get away with imagining that the space in which the test masses are moving is Euclidean.

I think it's less confusing than the "expanding space" idea, but it doesn't get a lot of discussion, except for the occasional diagram like the one shown above. WIthout more literature references to back it up, I'm a bit cautious about over-promoting the idea.

The idea does have some limitations that the expanding space idea does not. One of them is the issue of size. If we consider only a plane, such as in the diagram, the limitations are very modest. It's not until the diagram becomes so big that the relative velocities between the test masses start to become relativistic that we start to see the idea break down.

In three dimensions, the size limits are more severe. The basic idea of a unchanging, Euclidean space in which particles move due to "forces" breaks down when we consider a 3-d volume that's an appreciable fraction of one wavelength of the gravitational wave.

So if we had a 1khz gravitational wave , a tenth of a wavelength would be 30 kilometers, and we'd start to see some detectable issues in a volume of that size with careful enough measurements of distances between particles.

To understand most of the LIGO results, though, we don't need 3d, just 2d.

In the end, though, nothing can really replace "doing the math". But that takes more mathematics than is possible at the B level.

 

[PeterDonis]

To understand most of the LIGO results, though, we don't need 3d, just 2d.

I'm not aware of any LIGO results that cannot be understood this way.

In fact, to understand gravitational waves in general using this visualization method, 2d is sufficient, because gravitational waves are purely transverse, so all of the changing tidal effects are orthogonal to the direction of propagation.

 

[asca]

Just for the sake of completness, I found this article surfing the net :
https://pdfs.semanticscholar.org/393a/af6b1ced305ee40d175d5f3c3a2b6020348d.pdfHowever I'd just like to share the following: what happens to the clocks in the "x" arm of the example? Isn't their pace slower than the pace of clocks placed at the beam splitter? so for any observer in the x arm the time it takes for a crest to "reach the subsequent crest" (I hope you see what I mean by that) is shorter than the time measured by a clock placed at the beam splitter. But light speed is the same, so for any observer in the x arm the "distance" between two crests is shorter than the same distance measured by an observer placed at the beam splitter, in other word his ruler is longer than an identical ruler placed at the beam splitter. The solution to my puzzle maybe lies in the assumption the GW effect was and is different, actually opposite, in the two arms, and in the assumption that the observation point is somehow not (or less or more) affected by the GW passing by. So any obesrver in the x arm observing juts what happens in the x direction would not realize a GW is passing by, the same for any observer in the y arm, only the comparison by a third observer of the X and Y observations can deduct that a GW has passed by. That is probably the key.

 

[asca]

Sorry, I'm not so sure about what I just wrote: my question is what happens to the clocks in the X arm and to the clocks in the Y arm? I'm not so sure about the aswer I gave before, because the clock should not be affected by the direction of observation.

 

[Ibix]

That paper has specifically chosen a coordinate system where the passage of time is unaffected by the gravitational wave. Free floating clocks will remain synchronised throughout.

 

[asca]

Meanwhile I found an even clearer (at least to me) answer to my initial question.


Still I remain puzzled by the new question: what happens in general to the clocks placed in the X and Y arms? Why does Ibix say that their pace does not change?

 

[Ibix]

Why does Ibix say that their pace does not change?

Because that's what the maths says. For a free floating clock at rest in the coordinate system in use, $dx=dy=dz=0$, and the metric is diagonal with $g_{tt}=1$. Why do you think they should change?

 

[PeterDonis]

any obesrver in the x arm observing juts what happens in the x direction would not realize a GW is passing by, the same for any observer in the y arm, only the comparison by a third observer of the X and Y observations can deduct that a GW has passed by

That's not correct; observing just one arm can still tell you that a GW passed, because the round-trip travel time of light in the arm changes. However, it's much harder to measure that change in round-trip travel time in a single arm, then it is to measure the interference between the light in the two perpendicular arms. So an interferometer detector like LIGO is more sensitive than a single-arm round-trip travel time would be.

 

[PeterDonis]

what happens to the clocks in the "x" arm of the example?

Nothing. By the analogy the paper you linked to makes with cosmological models: as the universe expands, clocks are not affected, only distances are. Similarly, as the GW passes, clocks are not affected, only the lengths of the arms are.

 

[pervect]

Still I remain puzzled by the new question: what happens in general to the clocks placed in the X and Y arms? Why does Ibix say that their pace does not change?

Because that's an English language statement of the metric given in 2.1 in the paper you cite:

$$ds^2 = -c^2\,dt^2 + [1+h(t) ]\,dx^2 + [1-h(t)]\,dy^2 \quad [2.1]$$

In this equation, h(t) - the gravitational wave - doesn't modify the relationship between dt (coordinate time) and ds (which represents either distance or time intervals, which are unified in special relativity). So h(t) doesn't have any effect on time.

h(t) does modify the relationship between dx and dy (spatial coordinates) and ds, so the gravitatioanl wave does have an effect on space.

I'm not sure what pre-requisite knowledge you have, pessimestically I tend to assume you have little :(. Still, I'll take the risk of talking in ways that may go "over your head", as you seem to have some interest in the topic.

So, trying to thing about what you might need to know - you might start looking into what a metric is, as that's the mathematical key to answre your quesitons.

A good approach might be to start with understanding the pythagorean theoerm - the square of the hypotenuse is the sum of the square of the other two sides, then building upon this knowledge to understand what a purely spatial metric is. The translation of the pythagorean theorem in the language of metrics would be ds^2 = dx^2 + dy^2. s represents distance, x and y are coordinates, and d represents "a change in" or "differential". This would be hopefully familiar from calculus, if you have it. If you don't have calculus yet, that's another thing to add to your list of things you'd need to find out about.

At this point, you'd need to know what coordinates are. It's easy enough to state that coordinates are just lablels, but I've noticed in the past that sometimes this doesn't seem to be accepted by some PF posters, I'm not quiite sure what the difficult is.

Given that you know what coordinates are, and have some notion of what distances are, and the funamentals of calculus, then the metric allows you to take these coordinate labels (or rather their differentials), and computes differnetial distances from them.

After that, all you need to do is learn special relativity in general, and the metric approach to special relativity (the Lorentz interval), specifically.

After that, you'll have the needed bases to tackle this issue again with a firmer foundation. I've probably skipped a few steps of things you'd need to know along the way, but that's the short outline as I see it.

 

[asca]

Everything is clear now folks. When I wrote that stuff about the clocks I was momentarily carried away by the thought of a static gravitational field, but this is not the case when a wave is passing by, so I was really "off tune", sorry.

 

[pawprint]

Summary: How can an interferometer detect Gravitation waves, if the change in space time due to gravity affects all the rulers (and clocks) in that spot?

Some six or seven years ago I asked the question in here- "is detection based on (1) the length of the arm geodesics or (2) the movement of the test masses?". I received a single response who stated that it was (2). There was no further post on the thread so I have accepted this answer since. I concede that this is not the generally accepted point of view.