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Questions about gravity at relativistic velocities

  1. Apr 14, 2010 #1

    nearc

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    The ‘speed’ of gravity, like everything else, is also limited to the speed of light. So if two bodies [with mass] are traveling in the same direction with one trailing directly behind the other by some distance and both bodies are traveling at .999 C how does the gravity work between them? For example: If body A is in front and body B is behind. Then A will not feel the gravitational effects for some time. In fact if A is 1 light year ahead of B then it will take 999 years before A will feel the gravitational effects from B, however B will ‘run into’ A gravitational effects in about half a year?
     
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  3. Apr 14, 2010 #2
    Let v=0.999 c, L=1ly is the distance between bodies, then:

    ct=L+vt i.e. B moves away from A by vt while the gravitational effects need to cover L+vt

    So, t=L/(c-v) which is 1/0.001=1000 years

    In the other case:

    ct'+vt'=L

    so

    t'=L/(c+v) and this is about 0.5 years.

    Now, exactly as in the MMX experiment we need to remember that L needs to be "length contracted", when viewed from the frame that measures A and B moving at v i.e. it needs to be multiplied by [tex]1/\gamma=\sqrt {1-(v/c)^2}=1/22.4[/tex]

    So, the above times become:

    [tex]1000/22.4=44.8 yr[/tex]




    Another way of solving this is to use the full-fledged Lorentz transforms. In the comoving frame, the gravitational effect cover the distance [tex]\Delta x=1ly[/tex] in the time [tex]\Delta t=1yr[/tex]
    Then, in a frame wrt the two bodies move at [tex]v[/tex] away from the observer the time separation is:

    [tex] \Delta t'= \gamma (\Delta t+v\Delta x/c^2)=\gamma*2=44.8 yrs[/tex]

    If A and B move towards the observer, the formula changes:


    [tex] \Delta t'= \gamma (\Delta t-v\Delta x/c^2)=\gamma*/1000=0.0224 yrs[/tex]

    So, we can get the correct result through two different methods. As you can see, the correct results are quite different from your initial intuition.
     
    Last edited: Apr 15, 2010
  4. Apr 14, 2010 #3

    Mentz114

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    So the two bodies have zero relative velocity ? They won't notice anything different in the gravitational interaction between them.

    You say they are travelling at .999 c or whatever but not from which frame ( which does not enter the picture in any case ) since the bodies are stationary in each others frames.
     
  5. Apr 14, 2010 #4

    bcrowell

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    I think the OP meant that in some frame K, bodies A and B are observed to be moving with the same velocity in the same direction, and the description of the time delays is meant to be in frame K.

    Yes, this is a correct description in frame K.

    But Lut is also correct -- in the frame K' in which A and B are at rest, nothing unusual happens.
     
  6. Apr 14, 2010 #5

    Matterwave

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    Can't we just see this from the point of time dilation? In the frame in which A and B don't move, the propagation is obviously 1 year. Therefore, in the frame in which A and B move, wouldn't the propagation time be t=1yr/sqrt(1-v^2/c^2)=22.36yrs? And this would then be the propagation from A to B and from B to A.

    What is wrong with my analysis?
     
  7. Apr 14, 2010 #6

    bcrowell

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    If you start from the Lorentz transformation, you can then show that [itex]\gamma[/itex] gives the time dilation between frames K and K' for events that occurred *at the same place* in one of those frames. In the example we're talking about here, the emission and receipt of the gravitational "signal" doesn't occur at the same place in either frame.
     
  8. Apr 14, 2010 #7
    You need to factor in the spatial separation as well, you can't use straight time dilation because the start point of the signal (A) is separated from the endpoint (B) by 1 lyr in the comoving frame:

    [tex] \Delta t'= \gamma (\Delta t+v\Delta x/c^2)=\gamma*2=44.8 yrs[/tex]

    See post #2 for the complete reasoning.
     
  9. Apr 14, 2010 #8

    Mentz114

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    I don't know about any of this :uhh:. Everyone seems to be answering the question as if the two bodies are receding/approaching each other at 0.999c. But they have parallel worldlines in all frames. I'm baffled :confused:.

    What am I missing ?
     
  10. Apr 14, 2010 #9
    That the gravitational effect propagates in a time interval that is frame dependent. In the comoving frame this means 1ly, in the frame that measures A and B moving away at v, this means a larger number.
     
    Last edited: Apr 15, 2010
  11. Apr 15, 2010 #10

    bcrowell

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    The long time delay in frame K is simply a time dilation.

    One thing that may be causing confusion is that since the two bodies are at rest with respect to one another, there is actually no information being propagated by gravitational waves. If we want the OP's scenario to make a little more sense, we should actually let A and B each be, say, binary neutron stars, so that each one emits gravitational waves that can be detected by the other. Unless we have some mechanism like this for radiation, the field is exactly constant everywhere in frame K'.
     
  12. Apr 15, 2010 #11

    Matterwave

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    Wait so now I'm confused. To the 2 bodies which are moving wrt me, they obviously see only a 1ly lag. To me...I see a 44.8 year lag? Or a 1000 year lag? o_O
     
  13. Apr 16, 2010 #12
    44.8, the calculations are all there.
     
  14. Apr 16, 2010 #13

    Dale

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    I agree with bcrowell, in this situation there are no gravitational waves that propagate. Without something propagating it is confusing to talk about lag.
     
  15. Apr 16, 2010 #14
    Replace the gravitational field with an em field. The problem is about calculating propagation time as a frame-dependent quantity.
     
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