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I've always had a bit of fuzziness concerning relativistic contraction which I will try to put into words.

Iiuc in SR, moving objects contract in the direction of their travel, as measured by rulers at rest w.r.t. said objects. A traveling ruler when compared to the static one will appear shorter, and if we imagine a set of clocks in the moving frame spaced 1m apart in that frame, they will appear closer together in the rest frame. Thus in SR objects contract and if we take the spaced clocks as a metric then the moving frame is entirely contracted as well. To observers traveling in the moving frame however everything appears 'normal', with no contraction.

But in GR iiuc it is only space, and not objects, that contracts in the presence of a $g$-field. A ruler in the presence of (for instance) a constant $g$-field will not contract as compared to the same ruler when not in the $g$-field. But a set of meter-spaced clocks in a region of no $g$-field, will be closer together when in the presence of a $g$-field, as measured by the (non-contracting) ruler in the same g-field.

If objects were also contracted in GR, then (for instance) its hard for me to understand how LIGO could work, since the light between the mirrors would get squished just as much as the space between the mirrors was squished, and you wouldn't be able to measure any effect.

Have I got this right?

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    $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Nov 6, 2021 at 5:23
  • $\begingroup$ "A ruler in the presence of (for instance) a constant g-field will not contract as compared to the same ruler when not in the g-field." - Sure it will. The coordinate distance (e.g. between the ends of the ruler) is shorter than the same distance in the rest frame. Everything falling to a black hole becomes infinitely thin at the horizon in out coordinates, but remains intact in its rest frame. Your question is based on a wrong premise. $\endgroup$
    – safesphere
    Nov 7, 2021 at 9:54
  • $\begingroup$ there would be a slight contraction in an otherwise rigid object e.g. in earths field due to g but this amount is far exceeded by distance changes between successive (unconnected) rulers or clocks $\endgroup$ Nov 23, 2021 at 22:19
  • $\begingroup$ @jeremy_rutman No, this is incorrect. The length contraction in GR is exactly the same as in SR. The difference you are describing does not exist. Also please make sure to include the @ address on your replies. Otherwise the recipients are not notified. $\endgroup$
    – safesphere
    Jan 21, 2022 at 13:55
  • $\begingroup$ @safesphere I think you would do well to read and understand the answer provided by Steane. $\endgroup$ Jan 22, 2022 at 6:48

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You are muddling two different issues I think.

All of SR appears also in GR.

If you have a pair of parallel lines (such as the worldlines of the ends of an object of fixed proper length and constant velocity) then the distance from a point on the first straight line to a point on the second straight line parallel will depend on which points you pick. The contraction called Lorentz contraction is closely connected to this fact. The quantity called by us the "length" of the object is the distance between between events on the two worldlines that are simultaneous in the reference frame in which the length is being given. This distance comes out longest in the frame where the object is at rest. All this applies equally well to GR as to SR. The only new feature here in GR is that you must also allow for curvature effects when the distances involved are not small compared to the local radii of curvature of spacetime.

Now let's consider gravitational waves and detectors such as LIGO and VIRGO.

The mirrors in these interferometers are constrained in the vertical direction but not in horizonal direction. In the horizontal direction they are in free-fall. That is, there is no horizontal force on the mirrors except for gravity (if you want to call gravity a force, which you do not have to do).

When a gravitational wave passes by, the proper distance between the freely-falling mirrors oscillates. This is not Lorentz contraction; it is owing to spacetime curvature. An ordinary ruler, or the steel vacuum tubes which define the arms of the interferometer, will also have all its particles pushed a little together and then apart in an oscillatory motion as the wave passes by. However this gravitational squeezing and stretching involves tiny forces, so the lengths of the tubes (and of any ordinary ruler made of a stiff material such as steel) do not change except by an amount very much smaller than the amount by which the distance between the mirrors changes. Therefore the net effect is that the mirrors move relative to the steel tubes.

If there were an observer passing by Earth at high speed, then they would find that the lengths of the interferometer arms have the Lorentz contraction associated with the relative motion of observer and Earth. This contraction is purely to do with directions in spacetime and it is irrespective of whether there is any gravitational wave passing by. When a wave comes along, such an observer agrees that the mirrors move relative to the tubes. This motion is owing to the tidal force of gravity which is present when the gravitation is not uniform, and in the case of a wave the tidal force oscillates.

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  • $\begingroup$ If one calculated the Newtonian tides due to e.g. two neutron stars on a given path (like rotating from a line-of-sight and then perpendicular thereto) , would this come out to the same effect as the GR effect being seen at LIGO? $\endgroup$ Nov 6, 2021 at 4:35
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    $\begingroup$ @jeremy_rutman That's a good question! The answer is yes (to good approximation) if you are close enough to the stars that the near-field effects dominate, and no (not at all) when you are far from the stars where the gravitational wave dominates. This is because Newtonian gravity is a good first approximation to the steady field and its slow change but does not predict radiation at all. $\endgroup$ Nov 6, 2021 at 9:43
  • $\begingroup$ About whether to classify gravity as a force. Comparison: Lavoisier, Carnot etc. attributed heat to a substance 'Caloric'. It was atrributed: Steam is hot because of high Caloric content. Then came statistical mechanics; Caloric no more. Immediately 'heat' was redefined. It would be absurd to claim: "Since Caloric does not exist: steam isn't hot!". We assume/grant the existence of gravitational potential energy, and force is the gradient of potential energy. GR introduced a mediator of gravitational interaction unlike already known mediator. So: expand the scope of 'force' to accommodate. $\endgroup$
    – Cleonis
    Nov 7, 2021 at 8:34
  • $\begingroup$ @Cleonis There is no potential energy in GR. You can use it sometimes as a tool for calculations, but physically it doesn’t exist. $\endgroup$
    – safesphere
    Jan 22, 2022 at 15:53
  • $\begingroup$ @safesphere You raise a relevant point. I would love to discuss this, explaining the gamut of my considerations. Obviously comment space is totally unsuited (and totally not intended) for that. (Stackexchange chat is also a very cramped space, not suitable.) One option would be that I seek out an existing question here on physcis.SE, on the subject of how to classify gravitational interaction. I write there, I notify you, and you read that. But that would be kind of a hijack. Another option would be to agree mutually to move this to a threaded forum, such as physicsforums. Please let me know. $\endgroup$
    – Cleonis
    Jan 23, 2022 at 12:21
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Laser light is already relativistically contracted with respect to LIGO. Therefore, its contraction is not phase-locked to the apparatus reference frame sensing gravity waves of much longer wavelength.

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