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I How can gravitational waves be detected if spacetime itself warps?

  1. Jun 12, 2016 #1
    Hi,

    First: I'm pretty sure my question has been asked numerous times, so I'm absolutely happy with links to other threads. I've used search but it didn't come up with satisfying responses, probably mainly because I don't really know what search terms to use.

    So the question is: How can a Michelson-interferometer-like device detect gravitational waves based on the change of length differences between the arms? How are changes in length detectable if space itself is distorted? The length of a ruler along an arm changes as well, so why doesn't it take the light exactly the same time?
     
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  3. Jun 12, 2016 #2

    robphy

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  4. Jun 13, 2016 #3
    Relativity tells us that the speed of light is constant in every frame of reference, 'gravity' changes the length of space (and the rate of time), it is that change in length of space that is measured by instruments such as LIGO. If the length of space changes the mirrors distance apart will vary, that change in distance is measured and recorded.

    The same for things like the parthfinder satellites, they measure the variations in the length of space, the space gets longer with higher gravity and shorter with lower gravity.
     
  5. Jun 14, 2016 #4

    pervect

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    Ideal rulers don't change length when a gravity wave passes through them, one can regard rulers as defining length. Thus an idealized, perfectly rigid ruler won't change length when a gravity wave passes through it - stresses will be induced in the ruler, but it's length will remain unchanged.

    This leaves open the question of how rigid a non-ideal ruler that we can actually make actually are. In the case of Ligo gravity wave detection, the answer is "rigid enough" - the Earth can be considered as rigid as far as the gravity wave goes. The gravity wave does induce tiny stresses in the Earth, but they're insignificant.

    Note that the mirrors on Ligo are NOT attached to a rigid frame, but are attached to test masses, which are allowed to move in "natural motion" - in any direction other than the vertical direction, that is - there is a suspension system that prevents vertical motion of the test masses.

    So, from the point of view of the Earth, the test masses are free to move - and they do move. The frame doesn't move to any meaningful extent. This is a matter of convention to some extent - but one can definitely say that the test masses move relative to the rulers.

    This shouldn't be too surprising, the test masses are suspended and designed to be free to move, and they do, at least when viewed from a coordinate system in which coordinates are specified by their distances away from the origin.

    Ligo uses lasers to measure the length rather than the old platinum bar standard for length.. But there is no difference in what rigid rulers and lasers measure, except for the fact that laser system is much more precise (and also more rigid, though this doesn't matter in Ligo's case) than the older platinum bar defintion.
     
    Last edited: Jun 14, 2016
  6. Jun 21, 2016 #5
    I know it must be me, but I don't think it is a good idea to use the concept of length in non-stationary space times. For instance what would be the moment of integration in determining length?

    I would replace length with light travel time, that's more scientific!
     
  7. Jun 21, 2016 #6

    Nugatory

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    You will want to read through this thread: https://www.physicsforums.com/threads/ligo-light-changes-frequency-not-wavelength.875945/
     
  8. Jun 21, 2016 #7

    pervect

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    One does run into difficulties in cosmological contexts with defining length. However, these issues are entirely ignorable for Ligo - so I dealt with them by ignoring them, there are enough other issues to clarify. As far at the technical issues go, the "length" I'm talking about is defined in the tangent space, so we can use the SR concepts of length and ignore the issue of whether the space-time is stationary or not. I further assume that the reader is NOT necessarily familiar with the special relativivistic concept of length, so I spent a fair bit of words on discussing the issue from 1889 "prototype meter bar" perspective.
     
  9. Jun 21, 2016 #8
    So you think LIGO does not demonstrate that space time is non-stationary when it detects a wave? :confused:
     
  10. Jun 25, 2016 #9

    pervect

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    What I actually said was:
    Do you disagree with what I actually said, that non-stationary space-times have tangent spaces just like stationary space-times do? Or do you not understand the relevance of the point? Or.....what? I'm baffled , in the face of such a totally off the wall misquote of my position, as to how to keep the discussion even vaguely on-track.
     
  11. Jun 25, 2016 #10

    PeterDonis

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    Sure they do; but that's irrelevant here since you can't analyze gravitational waves using a single tangent space. See my recent post in the LIGO thread where we are discussing this point:

    LIGO light changes frequency not wavelength
     
  12. Oct 23, 2016 #11

    timmdeeg

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    This reminds me of Joseph Webers's cylinder, which should be squeezed and stretched as gravitational waves pass through. So, it seems that this cylinder isn't an ideal ruler. You say "the Earth can be considered as rigid as far as the gravity wave goes. The gravity wave does induce tiny stresses in the Earth, but they're insignificant." Are these "tiny stresses" comparable to those induced in said cylinder? Or put it another way, will the Earth be squeezed and stretched (but because of the noise not detectable) if a gravitational wave passes, thereby causing distance changes between fixed points on it which are comparable to those between the freely moving mirrors of the interferometer?

    Well, in contrast to the cylinder, the Earth doesn't consist of a homogeneous rigid material. Does this eventually make a difference regarding the distortion?
     
  13. Oct 23, 2016 #12

    pervect

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    The story of rigid objects in GR turns out to be hoarier than I thought when I wrote that post. A more careful examination of the math, which developed out of that discussion, shows that there just can't be a 3-dimensional congruence of worldlines that maintain the Born rigidity condition in a plane-wave GW space-time. So if we adopt the Born criterion for a rigid object, the mathematical solution satisfying the rigidity conditions doesn't exist. This means we can't compare the behavior of the Earth to some 3-d Born rigid model, because the 3-d Born rigid model doesn't exist.

    So I'd describe the status of rigid (Born-rigid) objects in GR this way. When they exist, they are a handy, intuitive tool - but they don't always exist.

    The good news is that we don't need to create a 3-d rigid congruence to have something useful. A 2-d rigid congruence will suffice. There doesn't appear to be any mathematical difficulty in creating such a 2 dimensional "rigid plane" that satisfies the Born rigidity conditions, assuming some reasonable constraints on the size of our "rigid plane". I'd have to dig to recall where I worked out & posted the necessary congruence, the Born rigidity condition can be described concisely (though abstractly and using highly specialized vocabulary) as saying that the Lie derivative of the spatial projection of the metric along the vector field that represents a congruence must vanish for the congruence to be rigid.

    The Ligo detector is essentially 2 dimensional. As actually implemented it has some "thickness", but this thickness isn't critical to the operation of the detector. So the existence of a 2 dimensional Born-rigid plane is basically enough to save the intuitive picture of gravitational waves using our "rigid plane" as a reference system.

    Using this rigid plane then, we can say that the lengths of rulers in our rigid plane don't change when a GW passes. What happens is that the freely floating test masses move relative to our rigid plane.

    Now is a good time to add the following caveat. We noted that our 2-d "rigid plane" wasn't infinite. The size constraints are necessary, as our model eventually would otherwise wind up with test masses exceeding the speed of light relative to our "rigid plane". This would be nonsensical, so we need auxillary assumption that the velocity of the test masses relative to the "rigid plane" is small enough not to cause significant Lorentz contraction.

    Trying to estimate the strains in the Earth itself, as a 3-d object, due to the passage of a gravitational wave becomes hard, but - we don't need to know how the Earth actually deforms. The design of Ligo makes the freely floating test masses float freely, one of the demanidng technical achivements of Ligo was to isolate the motion of the test masses from the motion of their environment to the necessary extent. Thus for our easy-to-imagine mental picture, it's sufficient to know how the masses move relative to our idealized "rigid plane".

    This leaves open the question of how to deal with situations in which rigid planes don't exist. In the gravitational wave (GW) example, the idealized plane wave is highly symmetrical, it has enough symmetry for our "rigid plane" to exist. This won't always be the case.

    The answer to this issue gets a bit complicated - basically it revolves around noting the fact that curvature tensor has only a second-order effect on distances. It causes geodesics (for example) to accelerate away from each other - a second order effect - but not to move away from each other - a first order effect.

    When we consider a small enough object, the second order effects basically don't matter, and we can apply our intuition about distances for small things. For a large object, the only good treatment I'm aware of is to do the math.
     
    Last edited: Oct 23, 2016
  14. Oct 24, 2016 #13

    timmdeeg

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    Thanks for explaining the rigidity issues.

    They (the mirrors) move relative to the rigid plane, because I think being in free fall, they follow geodesics, whereby the resulting distance changes are measured by laser technique. Now, could one argue that two fixed points on Earth (assumed to be rigid) don't follow geodesics and that their distance is constant therefore (ideally, no noise, no seismic sources, etc)? This would mean that the Earth is not deformed by gravitational waves generally. Or is this reasoning wrong, because the earth vibrates like Weber's cylinder? I don't trust the latter, because I think in contrast to the metallic cylinder the Earth hasn't a certain resonance frequency.
     
  15. Oct 24, 2016 #14

    PeterDonis

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    Not in the case of LIGO. The mirrors are at a constant altitude relative to the Earth, so they are not in free fall. One of the technical challenges in LIGO is to set things up so that the mirrors act like they are in free fall once the proper acceleration due to being at a constant altitude is factored out. Once this is done, the mirrors can be treated as though they are in "free fall" in the plane perpendicular to the direction of the proper acceleration; i.e., they do move in this plane relative to an idealized "rigid" coordinate system.

    Yes.

    The Earth does not have a single resonance frequency, but it certainly does vibrate. If it didn't, earthquakes would not be detectable far away. And it does have resonances at various frequencies, but their structure is much more complicated than the resonances of a bar made of a single substance, because of the complicated nature of the Earth's composition.
     
  16. Oct 24, 2016 #15

    timmdeeg

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    Ok, thanks for clarifying. So it seems that gravitational waves would at least in principle cause the Earth to vibrate (one of the various frequencies would fit), irrespective of their tiny amplitude, right?
    I was assuming that those vibrations are damped due to the inhomogeneous Material of which the Earth consists. But I think that's wrong, would you agree?
     
    Last edited: Oct 24, 2016
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