- #1

exponent137

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If even frequency would not be changed than spacetime is not changed, I suppose. What cannot be measured, cannot exist.

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- Thread starter exponent137
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- #1

exponent137

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If even frequency would not be changed than spacetime is not changed, I suppose. What cannot be measured, cannot exist.

- #2

PeterDonis

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Why Laser light in Ligo arms changes frequency, and not wavelength.

This is only true in one particular coordinate chart. One can choose other coordinates in which the light changes wavelength but not frequency. Or one can choose coordinates in which both change.

- #3

exponent137

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I suppose that length of static objects (arm) is proportionaly cnanged with space. In another case length of this arm is not part of this spacetime, isn't it?This is only true in one particular coordinate chart. One can choose other coordinates in which the light changes wavelength but not frequency. Or one can choose coordinates in which both change.

- #4

PeterDonis

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I suppose that length of static objects (arm) is proportionaly cnanged with space.

If you mean that the length of an object depends on your choice of coordinates, yes, that is true.

In another case length of this arm is not part of this spacetime, isn't it?

I don't understand what you mean by this.

- #5

exponent137

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I will ask differently. I suppose that arbitrary strong gravitational wave cannot stretch LIGO arm so much that it can break it. This is because i suppose that stretch of the arm follows to stretch of space. Let us neglect tidal forces.

Of course this is a theoretical question, because very strong gravitational waves probably will not happen in our history.

I think that my preffered coordinate system is in rest at LIGO, thus I do not move with a rocket, or that I am not in strong gravitational field, thus as an observer I do not feel gravitational field of gravitational wave. I think that this is the simplest situation?.

I ask because I wish to understand what is stretching of spacetime.

Of course this is a theoretical question, because very strong gravitational waves probably will not happen in our history.

I think that my preffered coordinate system is in rest at LIGO, thus I do not move with a rocket, or that I am not in strong gravitational field, thus as an observer I do not feel gravitational field of gravitational wave. I think that this is the simplest situation?.

I ask because I wish to understand what is stretching of spacetime.

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- #6

PeterDonis

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I suppose that arbitrary strong gravitational wave cannot stretch LIGO arm so much that it can break it.

The actual "arms" of LIGO are not solid structures; they are just the paths, through empty space inside underground tunnels, between the detector and the two mirrors. So there is nothing to "break".

However, a strong enough gravitational wave (far stronger than any we expect to observe on Earth) could move the mirrors enough that they would hit one of the tunnel walls. That would "break" LIGO in the sense that its measurements would no longer be accurate.

i suppose that stretch of the arm follows to stretch of space

"Stretch of space" is also coordinate-dependent.

Let us neglect tidal forces.

You can't neglect tidal forces. Tidal gravity is what gravitational waves are made of. Tidal gravity is just another name for spacetime curvature, and gravitational waves are waves of spacetime curvature.

my preffered coordinate system is in rest at LIGO

More precisely, coordinates in which the detector at the junction of the two arms of LIGO is at rest--yes, these are probably the most convenient coordinates to use for analyzing LIGO measurements, and the LIGO team uses these coordinates to describe their results. In these coordinates, the lengths of the two LIGO arms fluctuate, and the fluctuations are out of phase with each other, so the round-trip travel times of the laser beams going down the two arms are slightly different. That causes interference fringes to appear in the detector.

I wish to understand what is stretching of spacetime.

Curvature of spacetime is a better term; curvature of spacetime, as above, is just tidal gravity. So gravitational waves, which are fluctuations in the curvature of spacetime, are fluctuations in tidal gravity.

- #7

exponent137

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at the junctions of two arms,both frequencies of laser light are changed and both wavelengths are not changed.

What is/are coordinate charts where both frequencies of laser light are not changed and both wavelengths are changed or.

what is/are coordinate charts where both frequencies of laser light are changed and both wavelengths are changed.

- #8

PeterDonis

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As I understand you,

at the junctions of two arms,both frequencies of laser light are changed and both wavelengths are not changed.

Not necessarily. For interference to be present, all that needs to change is the round-trip travel time of wave crests in one arm vs. the other. That can change if the length of the LIGO arms changes, even if the frequency and wavelength of the laser light does not change. As I understand it, that is how the LIGO team's model (the one they use to analyze their results) works.

As for other coordinate charts that would show different things changing, I don't have specific ones to point to. I am simply stating a general fact about GR, that things like distance, frequency, wavelength are coordinate-dependent. The interference pattern at LIGO's detector is the only real invariant (i.e., quantity independent of coordinates) in this scenario.

- #9

exponent137

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If a gravitational wave stretches the distance between the LIGO mirrors, doesn't it also stretch the wavelength of the laser light?

A gravitational wave does stretch and squeeze the wavelength of the light in the arms. But the interference pattern doesn't come about because of the difference between the length of the arm and the wavelength of the light. Instead it's caused by the different arrival time of the light wave's "crests and troughs" from one arm with the arrival time of the light that traveled in the other arm. To get how this works, it is also important to know that gravitational waves do NOT change the speed of light.

https://www.ligo.caltech.edu/page/faq

This means, wavelength is changed, frequency is changed, lenght is changed and c is not changed,

(If c would be also changed, frequency would not be changed and intefrence pattern would not exist. But c is not changed.)

For this example I asked how it is with this.

Maybe I did not give a clear question.

- #10

PeterDonis

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If a gravitational wave stretches the distance between the LIGO mirrors, doesn't it also stretch the wavelength of the laser light?

Ah, yes, you're correct, in the LIGO team's preferred coordinates, the light wavelength does change, for the same reason the arm lengths change.

If c would be also changed, frequency would not be changed and intefrence pattern would not exist.

This is not quite right. The interference pattern is a physical observable; it does not change regardless of what coordinates you adopt. But it would be possible to choose coordinates such that the coordinate speed of light changed while the frequency of the laser light did not.

- #11

exponent137

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If a gravitational wave stretches the distance between the LIGO mirrors, and proportionally stretches wave length, does it cause elastic force of the LIGO arm? (In the opposite case, atoms are all the time in stable positions, although stretched.)

- #12

PeterDonis

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If a gravitational wave stretches the distance between the LIGO mirrors, and proportionally stretches wave length, does it cause elastic force of the LIGO arm?

The actual "arm" of LIGO is not a material thing; it's just the distance in empty space from the detector to one of the mirrors. So there isn't anything to be subjected to an elastic force.

If you consider the underground tunnel in which LIGO is placed, then yes, a passing gravitational wave will cause very tiny strains in the walls of the tunnel (and more generally in the material of the Earth). But these strains are direct observables, so they are present regardless of what coordinates you choose--including if you choose coordinates in which the "space" in which the LIGO apparatus sits is unchanging, not being "stretched" or "compressed" at all.

- #13

exponent137

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How we can choose coordinates in which the LIGO apparatus sits is unchanging, not being "stretched" or "compressed" at all?

Do you think that observer is moving with relativistic speed according to LIGO, or do you think that stronger gravitational field is present and this means another coordinates, or anything of both?

- #14

PeterDonis

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How we can choose coordinates in which the LIGO apparatus sits is unchanging, not being "stretched" or "compressed" at all?

Suppose that each LIGO arm were a single row of atoms. We could assign each atom a space coordinate (say, based on its measured distance from the reference atom at the detector at some instant of time), and then use that space coordinate to label that atom for all time. Coordinates are arbitrary; they don't have to have any kind of physical meaning.

- #15

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I will ask differently. I suppose that arbitrary strong gravitational wave cannot stretch LIGO arm so much that it can break it. This is because i suppose that stretch of the arm follows to stretch of space. Let us neglect tidal forces.

Of course this is a theoretical question, because very strong gravitational waves probably will not happen in our history.

I think that my preffered coordinate system is in rest at LIGO, thus I do not move with a rocket, or that I am not in strong gravitational field, thus as an observer I do not feel gravitational field of gravitational wave. I think that this is the simplest situation?.

I ask because I wish to understand what is stretching of spacetime.

I think that possibly you do not realize that the mirrors on Ligo are attached to test masses that are basically "hung from strings", so that the test masses, and the attached mirrors, are free to move?

(Rather than give references, I'll assume for now that the skeptical reader will look up this point in detail, and if the issue needs further clarification it will be addressed as needed.)

We'll call the thing that the test masses are suspended from "the frame". Everyone agrees that the test masses move relative to the frame. This frame is of no particular interest to the way the Ligo experiment works, so little effort is spent explaining what happens to it. If one did measure what happened to the frame , it wouldn't change measuarbly in length. The test masses, that are perfectly free to move at the slightest influence, require our most sensitive insturments to measure their motion. The frame moves even less.

I suspect this is a common misunderstanding of this point, due to the popularization of gravitational waves as "stretching and shrink space. But I'm not sure how to clear up this misunderstanding. I will try though.

Everyone agrees that the test masses move relative to the frame. If one's default viewpoint is based on the frame (which I rather suspect is the default viewpoint for nearly everyone), there is no such thing as expanding space, and no need to understand it.

The viewpoint that needs expanding space is a viewpoint that is attached, not to the frame, but to the suspended test masses. One can regard each test mass as having a constant coordinate, a coordinate that does not change with time. In this view, there are no external forces acting on these test masses, so one regards them as not moving. When the gravity wave passes by these test masses, changing their separation, but one ascribes this change in distance to "expanding and contracting space", rather than to any real force. There is no real force according to this viewpoint, the test masses are regarded isolated from any non-gravitaitonal forces, and gravity is not regarded as a real force (according to this viewpoint, which is different from the Newtonian one). One might say that the test masses are in a state of "natural motion", like a body at rest in Newtonian physics.

In this viewpoint, it's the frame that is "moving". Since the test masses are "standing still", i.e. have constant coordinates, and the frame is moving relative to the test masses, the frame must be "moving". The reason the frame moves is that internal forces generated by the interaction of the atoms that make up the frame keep the distance between atoms nearly constant. Internal forces due to the interaction of the atoms that keep the length constant (or nearly constant) are what causes the pieces of the frame to move in this viewpoint.

This viewpoint of expanding space also occurs in cosmology, and there are similar issues of (mis)understanding the popularizations in cosmology as well.

Why do people keep using these popularizations if so many people misunderstand them? I have no idea, really, it's partly a social phenomenon. It is true that a correct understanding of what the popularization are trying to say is useful, the issue as I see it is that the popularizations practically invite misunderstanding , and that there appears to be little concerted effort to address the common misunderstandings induced by the well-intentioned popularizations.

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- #16

exponent137

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This is easier to me to imagine.

If we would have one wire, which would connect two rest planets, this wire would expand and shrink, this would cause tension in the wire, One problem with understanding of expanding of the spacetime is, if we can feel some forces. I think we feel in some situation.As I understand Peter Donis, these forces are oly tidal forces.

Is it correct?

- #17

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This is only true in compving coordinates. In general, wavelength is not a property of the light itself, but also depends on the observer. Different coordinates will give you a different view of the physics, but all invariant quantities, such as the frequency a particular observer measures, remain the same.Maybe a simpler example is expanding of the universe. Because the universe is growing wee see red shift, speed of light is the same, this means that the wavelenght is not the same.

This is incorrect. The gravitational waves observed by LIGO has a frequency such that the arm length does not change significantly during the time it takes light to pass it so the light will not change its frequency ar wavelength during this time. The point is that the arm length changing with time (let us adopt a coordinate system where this is the interpretation) the interference pattern of subsequent light pulses change.If universe would expand and shrink very fast this would be a LIGO, a one arm LIGO. We would see blue and red shift is short succession.

- #18

exponent137

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I thought only rest source and rest observer, but expanding and shrinking universe.

I thought such exagerrated example that light will significantly change its frequency or wavelength during this time.

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- #19

RockyMarciano

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If you mean that the length of an object depends on your choice of coordinates, yes, that is true.

The interference pattern is a physical observable; it does not change regardless of what coordinates you adopt.

Maybe you could elaborate a little on how these two assertions can coexist, it is not evident to me. It is my understanding that the interference pattern is basically a highly magnified length measurement of the extent of the discrepance between no displacement of the fringes in the default case versus the positive case with measurable displacement. So the difference you draw between length depending on the choice of coordinates in some cases and independent of it in others looks arbitrary.

- #20

Ibix

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An interference pattern is a measure of a phase difference between two waves. That there is a phase difference and what its magnitude is are unarguable and invariant. Why there's a phase difference (change in length, change in speed of light, both, other) is coordinate dependant, if I understood Peter correctly.

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- #21

RockyMarciano

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And this phase difference measure is itself operationally performed as a measure of length(position difference in the fringes), wich is argued by PeterDonis in the quote above to be coordinate dependent.An interference pattern is a measure of a phase difference

- #22

Ibix

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- #23

RockyMarciano

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- #24

MeJennifer

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The truth is that gravitational waves are non-stationary spacetime phenomena, all (comformally flat) 3 + 1 formalisms are necessarily incapable of fully explaining them.Everyone agrees that the test masses move relative to the frame. If one's default viewpoint is based on the frame (which I rather suspect is the default viewpoint for nearly everyone), there is no such thing as expanding space, and no need to understand it.

The viewpoint that needs expanding space is a viewpoint that is attached, not to the frame, but to the suspended test masses. One can regard each test mass as having a constant coordinate, a coordinate that does not change with time. In this view, there are no external forces acting on these test masses, so one regards them as not moving. When the gravity wave passes by these test masses, changing their separation, but one ascribes this change in distance to "expanding and contracting space", rather than to any real force. There is no real force according to this viewpoint, the test masses are regarded isolated from any non-gravitaitonal forces, and gravity is not regarded as a real force (according to this viewpoint, which is different from the Newtonian one). One might say that the test masses are in a state of "natural motion", like a body at rest in Newtonian physics.

- #25

Ibix

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I see what you're getting at.

The standard is coordinate dependant. The motion of the fringes is coordinate dependant. The two dependencies cancel and the phase shift derived from the fringe shift is coordinate independent.

- #26

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The truth is that gravitational waves are non-stationary spacetime phenomena, all (comformally flat) 3 + 1 formalisms are necessarily incapable of fully explaining them.

That is certainly a true statement, however it is totally irrelevant to the points I was attempting to make. I think you're too busy talking to listen to what I have to say - and have already said. Since nobody else is commentin (at this point), it appears to be rather difficult to proceed with a useful discussiion.

- #27

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An interference pattern is a measure of a phase difference between two waves. That there is a phase difference and what its magnitude is are unarguable and invariant. Why there's a phase difference (change in length, change in speed of light, both, other) is coordinate dependant, if I understood Peter correctly.

While it's certainly possible to describe a coordinate independent result in a coordinate-independent manner, if one adopts a coordinate dependent approach, one can arrive at the coordinate-independent result in a coordinate dependent manner.

From a coordinate dependency point, the point is that any explanation based on expanding space will be coordinate dependent. "Expanding space" is a cordinate dependent idea.

To give a specific example, if we have a space-time that is empty of mass, the flat space-time of special relativity, we can regard the space-time as either being non-expanding space-time, or, if we prefer, we can regard it as an empty, but expanding, Milne universe. https://en.wikipedia.org/wiki/Milne_model.

To give a famous example, we can say that "Brooklyn is not expanding", this being a reference to Woody Allen's "Annie Hall". This is a coordinate indepenent fact. We can also say "the universe is expanding". This is also a coordinate independent fact. The ultimate goal is to understand that Brooklyn is not expanding, but the universe is expanding. "Expanding space" is an attempt at a tool to explain these facts, but sometimes I think it causes more confusion than enlightenment.

- #28

PeterDonis

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Why there's a phase difference (change in length, change in speed of light, both, other) is coordinate dependant, if I understood Peter correctly.

This is not quite what I meant, no. The phase difference, which is invariant, is caused by the gravitational wave, i.e., by a fluctuation in spacetime curvature, which is also invariant. Something that causes something else cannot be coordinate dependent. The only thing that is coordinate dependent is how you choose to interpret the fluctuation in spacetime curvature. But that is just an interpretation; it's a crutch to allow our limited cognitive abilities to try to grasp what is going on. It doesn't change the physics.

- #29

RockyMarciano

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This seems to be a simple terminology confusion, because of the dependence on sign convention of the timelike vector fields that define congruences in GR, the object referred to as expansion tensor is technically a pseudotensor.The expansion tensor is a perfectly valid tensor, both theoretically and operationally.

Right but see above.Incorrect. The definition of the expansion tensor uses the projection tensor ##h_{ab} = g_{ab} + X_a X_b## which projects out the part of an arbitrary vector or tensor field that is orthogonal to the vector field ##X##. But such projected vectors, tensors, etc. are still 4-vectors, 4-tensors, etc.

Also, could you please address #19?

In any case, would you please address #19?

- #30

PeterDonis

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the object referred to as expansion tensor is technically a pseudotensor.

No, it isn't. A pseudotensor is something that doesn't transform properly under a change of coordinates. The expansion tensor, like all genuine tensors, transforms properly regardless of which metric signature convention you are using.

would you please address #19?

I will in a separate post.

- #31

PeterDonis

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Maybe you could elaborate a little on how these two assertions can coexist

Because the interference pattern is an invariant, while the length of an object is not.

It is my understanding that the interference pattern is basically a highly magnified length measurement of the extent of the discrepance between no displacement of the fringes in the default case versus the positive case with measurable displacement.

No, it isn't. This is an interpretation, not the actual physics.

The actual physics is that an interference pattern is a pattern of wave intensities on a detector. Wave intensity is an invariant, independent of coordinates, hence the interference pattern is an invariant. The wave intensities are the result of interference between incoming waves from the two arms, hence the term "interference pattern". What you are calling the "default case" is the case of no interference--in this case the wave intensity is constant everywhere on the detector. In the case of interference, the wave intensity varies from point to point on the detector; this variation in intensity is what the term "interference fringes" describes (because the characteristic pattern of variation appears as light and dark fringes). There are no such fringes in the default case because there is no variation in intensity on the detector.

- #32

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Maybe you could elaborate a little on how these two assertions can coexist, it is not evident to me. It is my understanding that the interference pattern is basically a highly magnified length measurement of the extent of the discrepance between no displacement of the fringes in the default case versus the positive case with measurable displacement. So the difference you draw between length depending on the choice of coordinates in some cases and independent of it in others looks arbitrary.

Excuse me for butting in, but I think this is fairly easy to explain.

1) The interference fringes are a result of round-trip times. Round trip times are a measure of "proper length", via the current SI definition of the meter as "The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second."

2) Proper length is not coordinate or observer dependent quantity, it's a coordinate and observer independent quantity.

3) "Length", as opposed to proper length, can be an observer dependent quantity, as evidenced by the existence of "length contraction".

So the main confusion arises from words having multiple meanings - the interference fringes are caused by the sort of "length", proper length, that's observer independent.

- #33

MeJennifer

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I would caution everybody thinking of length in the abstract in non-stationary spacetimes (and that is what we are talking about when we deal with gravitational waves). In such spacetimes there is no such thing as an observer and coordinate independent length because there does not exist an observer and coordinate independent moment of integration.2) Proper length is not coordinate or observer dependent quantity, it's a coordinate and observer independent quantity.

- #34

PeterDonis

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Round trip times are a measure of "proper length", via the current SI definition of the meter as "The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second."

This is only true if the spacetime is stationary, at least during the travel time of the light, and if the objects between which the proper length is being measured are at rest relative to each other during the travel time of the light. If these requirements are not met, the concept of "proper length" is not well-defined.

[Edited to delete mistaken further comments.]

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- #35

RockyMarciano

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How do you figure the measure of "intensity" is performed if not with respect to a standard measurement gauge wich is an object with a certain length(that you claim is not invariant). Interferometry is actually used in defining and calibrating length standards and gauges.Because the interference pattern is an invariant, while the length of an object is not.

No, it isn't. This is an interpretation, not the actual physics.

The actual physics is that an interference pattern is a pattern of wave intensities on a detector. Wave intensity is an invariant, independent of coordinates, hence the interference pattern is an invariant. The wave intensities are the result of interference between incoming waves from the two arms, hence the term "interference pattern". What you are calling the "default case" is the case of no interference--in this case the wave intensity is constant everywhere on the detector. In the case of interference, the wave intensity varies from point to point on the detector; this variation in intensity is what the term "interference fringes" describes (because the characteristic pattern of variation appears as light and dark fringes). There are no such fringes in the default case because there is no variation in intensity on the detector.

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