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Speed of gravity?

  1. Sep 10, 2012 #1
    I understand that Newton thought that the effect of gravity was instantaneous, however Einstein believed it to 'travel' at the speed of light. I also read that in 2002 scientists concluded that gravity traveled at 1.06 times that of the speed of light but with an error of plus or minus 0.21 putting it in the range of the speed of light.

    My question that has been keeping me up at night though is, if the light from the sun is 8 minutes old, does that imply that we are in fact not orbiting the sun but orbiting where the sun was 8 minutes ago?

    Sorry if this is posted in the wrong section.
     
  2. jcsd
  3. Sep 10, 2012 #2
    Going on the basis that if the sun were to instantaneously disappear and the earth would keep on following the same orbit for 8m20s before shooting off on a tangent, then yes we are orbiting where the sun was 8m20s ago.
     
  4. Sep 10, 2012 #3

    PeterDonis

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    This is one way of looking at it, but it leaves out some key points.

    (1) Consider light first instead of gravity. The light from the Sun reaching the Earth is not only time delayed; it shows aberration as well. That is, it comes at us from the *direction* the Sun was 500 s ago, not the direction the Sun is now. This effect gives one way of estimating the speed of light.

    (2) The apparent "force" of gravity from the Sun on the Earth doesn't come from where the Sun was 500 seconds ago; it comes from where the Sun is now. (Actually that isn't exactly true--see #3 below--but it is to a very good approximation.) That is, the apparent direction of gravity from the Sun, unlike the apparent direction of light from the Sun, does *not* show aberration. So the "apparent" speed of gravity, if you try to deduce it from the "aberration" alone, is *much* faster than c.

    (3) However, the deduction of speed of an effect from observed aberration depends on the theory you use. In the case of light, our best theory (at least the one that's good enough for this purpose) is classical electromagnetism, based on Maxwell's Equations, which predict a speed of light equal to c and aberration of light from moving sources. In the case of gravity, however, our best theory is General Relativity, not Newtonian theory; but deducing the "speed of gravity" from the observed (lack of) aberration only works in Newtonian theory. In GR, gravity is not really a "force" at all, it's spacetime curvature; in cases like the Sun and the Earth, gravity in GR can still be viewed as a "force", but it's a force that depends on the Earth's velocity and acceleration as well as its position relative to the Sun, so it's not a Newtonian-type force and you can't deduce its speed from the (lack of) aberration.

    When you do the correct GR calculation of the speed of gravity from observations, it comes out to c; as the OP notes, the best current experiments are consistent with that value. So you are right that, if something were to change about the Sun that changed its effect on the Earth, it would take 500 s for the Earth to feel the effect; but that doesn't mean that the apparent "force" from the Sun on the Earth points in the direction the Sun was 500 s ago, because of the velocity and acceleration dependent effects.

    Steve Carlip wrote an excellent paper years ago on this subject, which I highly recommend reading:

    http://arxiv.org/abs/gr-qc/9909087

    This paper gives a more detailed discussion of, and also the math behind, the statements I made above.
     
  5. Sep 10, 2012 #4

    Bill_K

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    I disagree. The gravitational attraction of the sun, even in general relativity, does not exhibit retardation. The Earth is orbiting the current location of the sun. This follows immediately from solving for the orbit in the rest frame of the sun and Lorentz transforming it.
     
  6. Sep 10, 2012 #5
    Wow. So when we are leading the sun, say at midwinter (or whenever) it would follow that we are closer to the sun by something like 8 light-minutes, and when we are trailing the sun, we are further away by a like amount?

    So we don't orbit the center of mass, but rather, we orbit the place where the center of mass used to be 8 minutes ago?

    Are these correct surmises?

    And if the sun is moving very fast relative to some convenient point of reference, then is the Earth's path around it really a huge elipse?
     
  7. Sep 10, 2012 #6

    Bill_K

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    No, of course not.

    Take the solar system, and view it in a rest frame which is traveling "North" at 0.9c. According to this idea, Mercury does not know where the sun is now, it follows it, orbiting the point where the sun used to be 3 minutes ago. For the Earth it's 8 minutes, for Jupiter 40. Each planet lags a bit farther "South". Poor Pluto has to orbit a spot 4 light-hours to the South of the sun's current location. The plane of the ecliptic would not be a plane! This is nonsense.

    Newtonian gravitation does not emerge from the sun and travel outward at the speed of light. The planets react to the sun's current gravitational field. The interaction appears instantaneous. Although this seems paradoxical, and even asking for trouble, it is not. Exactly the same thing happens in electromagnetism, where the Coulomb interaction is effectively instantaneous. Nature can get away with this without violating any principles because both the Newtonian field and the Coulomb field are static. Likewise in quantum electrodynamics the Coulomb interaction is instantaneous. Changes, of course, including waves, propagate at c.
     
  8. Sep 10, 2012 #7

    Bill_K

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    This is called a wave tail. For example an outward-going wave on a Schwarzschild background will develop two kinds of tail, one an inward-going backscattered wave, the other a monopole field that reduces the Schwarzschild mass to account for the energy that's being carried away.

    Discontinuities always travel at c. The Einstein equations are hyperbolic, and their characteristics are null surfaces, bicharacteristic curves are null geodesics.
     
  9. Sep 10, 2012 #8

    A.T.

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    I like that comparison because it avoids all the curved space time complexity.

    When you travel at constant speed relative to a charged object (static E-field), that also emits light (EM-waves), you will see the object at a different direction than you will measure the Coulomb force from:
    - The seen position is outdated
    - The Coulomb force points to the correct position
    And yet when the object is suddenly accelerated, both observed directions will be affected after the same delay by the acceleration.

    I think the key problem people have is accepting that an inertially moving source can produce a static field, that moves along with the source, and doesn't stay behind it like sound waves. A lot of the wrong intuition here comes from the plane noise that comes from a direction where the plane was 15sec ago.
     
  10. Sep 10, 2012 #9

    pervect

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    My impression (which could be wrong) is that the confusion arises from the rather naive idea of thinking that the force of gravity is carried by little particles, emitted from the sun, and that the "force" occurs when these particles are absorbed.

    Unfortunately, this picture is basically wrong :-=(. But it's not clear if it's so wrong if it can't be made to work somehow, if one replaces real particles with virtual particles. However, the details of how to do this aren't particularly clear even in the much simpler electromagnetic case.

    It is also clear that conservation laws forbid the sun from disappearing suddenly (or, in the electromagnetic case, forbid charges from disappearing suddenly). So trying to work out what happens mathematically when you assume it happens anyway gives total nonsense, a set of equations that has no solution. But it seems that this observation doesn't seem to be helpful in resolving the dispute - probably because the people having the issues don't actually think about trying to solve the equations mathematically instead relying on rather simple mental pictures (like the particle carrying the force) which they are very, very attached to, while they feel that the math is somehow "beyond their grasp".
     
  11. Sep 11, 2012 #10

    Bill -

    Were you addressing me or another poster?

    By the sun's "current gravitaional field" do you mean the field which corresponds to the suns present position for a close-by body co-moving with it?

    How does any body which is at a significant distance respond to the current position, when the body "sees" the sun at a position which it occupied in the past (as determined by the close-in, comoving observer)? Why wouldn't a distant body respond to the center of mass as it appears currently to that body? For it to be otherwise, the distant body must anticipate where the sun will be in, say 8 minutes, and not where it appears currently.
     
  12. Sep 11, 2012 #11

    PeterDonis

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    [Note: I have edited this post to correct misstatements about acceleration.]

    The Carlip paper that I linked to in post #3 explains how this works; it more or less amounts to just what you say here. But as Bill_K notes, there's no paradox or violation of causality involved; what is happening is that the velocity-dependent terms [Edit: removed "acceleration dependent"] in the "force" compensate for the light-speed time delay to make the apparent "force" point in the direction the Sun is "now" instead of where it was 8 minutes ago.

    As Carlip puts it, the "force" the Earth feels from the Sun is based, not on the Sun's position 8 minutes ago, but on the Sun's position "quadratically extrapolated" from 8 minutes ago, using the Sun's velocity [Edit: removed "acceleration"] relative to the Earth, 8 minutes ago, to do the extrapolation. (As Carlip points out, in the case of electromagnetism, the extrapolation is only linear; the Coulomb force felt by a charged particle is based on the source "linearly extrapolated" over the light-speed time delay. [Edit: removed "acceleration"])

    Of course the extrapolation isn't perfect; the Earth does not "know" exactly where the Sun is "now", but only where it extrapolates the Sun to be "now" based on information from 8 minutes ago. The extra perihelion shift in planetary orbits that is predicted by GR but not by Newtonian theory (so far observed, AFAIK, only for Mercury) is a consequence of the fact that the extrapolation is not perfect (or, to put it another way, a consequence of the fact that the "speed of gravity" is the speed of light).
     
    Last edited: Sep 11, 2012
  13. Sep 11, 2012 #12

    Bill_K

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    Yes, that's exactly what does happen! And it's a good thing too. Because here's a further example: consider two equal masses in relativistic motion about each other. Normally the attraction between them is diametrical, towards the center of mass. But if the attraction between them were retarded, each would feel an attraction to a point which lags behind the other, causing the orbital motion to slow down and eventually come to a halt!

    But don't take my word for it, here's an article by John Baez, and also a published paper which addresses the question. Both discussions start with the electromagnetic case. I quote:

     
  14. Sep 11, 2012 #13

    A.T.

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    This is correct, it extrapolates the movement of the source. If the moving source was suddenly stopped, you would notice this only after a delay, and initially continue to be accelerated towards the wrong extrapolated position.

    You can think of this in two ways:

    1) We can go to the inertial rest frame of the moving source, where the field is static, and it's obvious that it always pulls the now moving object towards the correct position of the source. Unless the source accelerates.

    2) A moving source "emits" a field that points towards the source, but also "inherits" the velocity of the source. So as long the source moves at constant velocity, the distant field-lines continue to point at the current source location. When the source suddenly stops, the information about it travels through the field at c, so initially the distant parts of the field continue to move at the original velocity of the source, and now they point towards a wrong extrapolated position of the source.

    This animation shows the opposite scneriow: acceleration of a charge from rest. But stopping it is similar: The moving field-lines continue to move, and point to an extrapolated position, until the disturbance (wave) reaches their position:

    field_a.gif

    From: http://www.tapir.caltech.edu/~teviet/Waves/empulse.html
     
    Last edited by a moderator: May 3, 2017
  15. Sep 11, 2012 #14
    This is fascinating stuff.
    So is it currently thought that the graviton doesn't exist?
     
  16. Sep 11, 2012 #15

    PAllen

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    No, that is not true. A field is mediated by virtual particles, and the field of virtual spin 2, massless bosons matches the predictions of GR to a very high precision.
     
  17. Sep 12, 2012 #16
    Good, I'm glad about that! Thanks!
     
  18. Sep 17, 2012 #17

    bcrowell

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    I'm inclined to be skeptical of this claim. Logically, there are significant difficulties in defining what would be meant by the speed of propagation of a gravitational wave in the general, strong-field case. They're similar to the difficulties involved in proving geodesic motion from the field equations. You have to somehow distinguish two different metrics, one that would have existed if there had been no wave disturbance, and one in which the wave disturbance exists. Then you have to say what you mean by the speed of propagation of the wave, which presumably you want to measure in one or the other of these two metrics. So it seems likely to me that even if the second quoted sentence holds, it may not be interpretable as claimed in the first sentence, in the general strong-field case.

    I had to look up the definitions of the mathematical terms you used:
    http://en.wikipedia.org/wiki/Hyperbolic_partial_differential_equation
    http://en.wikipedia.org/wiki/Method_of_characteristics
    By the definition in the WP article, it seems clear to me that the Einstein field equations do not qualify as hyperbolic in general. This is the reason that we have to define global hyperbolicity (Hawking and Ellis, p. 206). On a spacetime that doesn't have global hyperbolicity in the Hawking and Ellis sense (e.g., a spacetime that has CTCs), we don't generally expect to have uniqueness and existence of solutions to Cauchy problems. H&E refer to a theorem by Geroch, J Math Phys 11 (1970) 437 (I haven't read it), which states that if the spacetime is globally hyperbolic in the H&E sense, then Cauchy surfaces exist. Since the existence of Cauchy surfaces is what WP is giving as the definition of a hyperbolic PDE, I think we can't generally assert that the field equations are hyperbolic -- if we could, then Geroch's 1970 theorem would have been trivially true.

    A major application of the method of characteristics seems to be to things like shock waves and fluid dynamics, where we don't have any of the GR issues I worried about above. The second WP article has this, which I think it physically suggestive: "Characteristics may fail to cover part of the domain of the PDE. This is called a rarefaction, and indicates the solution typically exists only in a weak, i.e. integral equation, sense."
     
  19. Sep 19, 2012 #18
    I was reading through, and it stuck to me that if the moving charge is suddenly stopped and we would continue to extrapolate the direction of the force(which is wrong). Does this imply that EM fields must carry momentum in order to preserve conservation of momentum ?

    And if EM fields do carry momentum, what does that even mean, when we consider NO net motion of a system carrying static fields?
     
    Last edited: Sep 19, 2012
  20. Sep 19, 2012 #19

    A.T.

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  21. Sep 19, 2012 #20

    Bill_K

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    Ben,

    I'm talking local, not global hyperbolicity, which is a much more difficult subject, hats off to Hawking and Ellis! MTW for example discuss at length the propagation of gravitational waves on a curved background in the short wavelength limit (shock waves are extremely short wavelength) They derive among other things a linear wave equation for the perturbation,

    hμν + 2Rαμβν hαβ = 0

    which is manifestly hyperbolic. (R is the background Riemann tensor.) See especially their exercise 35.15 on the "geometric optics of gravitational waves of small amplitude propagating in a curved background," which discusses propagation of various quantities along the rays (i.e. null geodesics).

    Or, if you're into Italian :biggrin: the original reference,

    ""Caratteristiche e bicaraterristiche delle equazioni gravitazionali di Einstein", T. Levi-Civita, Rend Acc Lincei Sci Fis Mat Nat 11 (1930) 3, 113.
     
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