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

  1. Jan 23, 2018 #1
    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?
     
  2. jcsd
  3. Jan 23, 2018 #2

    PeterDonis

    Staff: Mentor

    Because the wave doesn't move them in unison. It's a wave of tidal gravity; it alternately stretches and squeezes the arm.
     
  4. Jan 24, 2018 #3
    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?
     
  5. Jan 24, 2018 #4

    PeterDonis

    Staff: Mentor

    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.
     
  6. Jan 24, 2018 #5
    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! :-)
     
  7. Jan 24, 2018 #6

    Ibix

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    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.
    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.
     
  8. Jan 24, 2018 #7

    PeterDonis

    Staff: Mentor

    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.
     
  9. Jan 24, 2018 #8
    Thanks to you both!
     
  10. Jan 26, 2018 #9

    Mister T

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    You do understand that the mirror is mounted in such a way that it's free to move towards or away from the laser?
     
  11. Jan 26, 2018 #10
    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.
     
  12. Jan 26, 2018 #11

    Mister T

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    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.
     
  13. Jan 26, 2018 #12
    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.
     
  14. Jan 26, 2018 #13

    PeterDonis

    Staff: Mentor

    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.

    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
     
  15. Jan 26, 2018 #14
    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?
     
  16. Jan 26, 2018 #15

    PeterDonis

    Staff: Mentor

    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.

    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.

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

    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).
     
  17. Jan 26, 2018 #16
    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?
     
  18. Jan 26, 2018 #17

    PeterDonis

    Staff: Mentor

    Yes.

    "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.)

    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.

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

    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.

    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.
     
  19. Jan 26, 2018 #18
    @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!
     
  20. Jan 26, 2018 #19

    PeterDonis

    Staff: Mentor

    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.

    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?

    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.
     
  21. Jan 27, 2018 #20
    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?
     
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