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Action at a distance v curvative of space

  1. Jul 18, 2011 #1
    Einstein didn't like Newton's idea that mass causes action at a distance through gravity, so he came up with his explanation that mass causes a curvature in space time. It sounds to me like Einstein is just moving the problem around. Instead of mass causing action at a distance, now we just have mass causing space to curve at a distance. The problem is still there. We still have the problem of how mass causes action at a distance. Now the action is the curvature of space.
     
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  3. Jul 18, 2011 #2

    Mentz114

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    Re: action at a distance v curvature of space

    The action at a distance issue was dealt with by special relativity which introduces the limit of propagation speed. It wasn't a motivating factor for GR so much as the equivalence principle, which allows Ricci gravity to be transformed away in freely falling frames.

    GR has local symmetries - so energy and momentum are conserved locally under translations and rotations, for instance. No action at a distance is supposed nor required.
     
  4. Jul 18, 2011 #3

    Dale

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    Why would you think that? The EFE are local.
     
  5. Jul 18, 2011 #4
    Here is a thought experiment for you which might help you see the difference.

    Imagine that we start out with flat space along some initial spacelike hypersurface [itex]t=0[/itex]. At some later spacelike hypersurface which we will denote by time [itex]t=t_0[/itex], we deposit into this spacetime a point mass at the origin with some macroscopic mass M. What will this spacetime look like along some spacelike hypersurface [itex]t=t_0 + \epsilon[/itex] (where [itex]\epsilon[/itex] is a small parameter)?

    Next, at a second spacelike hypersurface [itex]t=t_1 > t_0[/itex], delete the point mass. What will this spacetime look like at [itex]t=t_1+\epsilon[/itex]?
     
    Last edited: Jul 18, 2011
  6. Jul 18, 2011 #5

    bcrowell

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    There's a technical problem with the scenario you suggest, which is that it doesn't conserve mass, and the EFE imply local conservation of mass. Therefore there will not be any self-consistent solution of the EFE in the scenario you're suggesting.
     
  7. Jul 18, 2011 #6
    I know. That's why I posed it more as a thought experiment rather than an actual problem.

    A well-posed version of it would be to consider what spacetime far away from a binary black hole pair would look like, but I assumed that would be harder for the OP to visualize.

    My point I guess was just to try to illustrate that the gravitational field propagates as curvature in an analogous fashion to the way the electromagnetic field propagates as photons - no spooky action at a distance required. I probably should have picked a better scenario, though.
     
    Last edited: Jul 18, 2011
  8. Jul 18, 2011 #7
    But isn't propagation action at a distance? i didn't really get your thought experiment. i'm not very good at physics.

    could you elaborate some more
     
  9. Jul 18, 2011 #8

    Dale

    Staff: Mentor

    As long as you have a differential equation you are saying that each point only affects its immediate neighbors. No action at a distance. The EFE are differential equations.
     
  10. Jul 19, 2011 #9
    Okay, let's look at E&M, since it is a much simpler theory to come to grips with.

    Imagine a (flat) spacetime containing two infinite parallel charged plates spaced some distance z apart so that the electromagnetic field is the same everywhere between these two plates. Suppose at some moment in time [itex]t_0[/itex] one of those plates flexes, causing a disturbance in the force... er... I mean a perturbation in the electromagnetic field. What happens?

    To begin with, the only regions of spacetime that will know about the perturbation are those that are within the causal future (that is, the future light cone) of the event that created the perturbation. So, the other plate won't "see" the perturbation until enough time has passed for a light ray to have traveled between the two plates; prior to then, the second plate still thinks that the field is uniform everywhere. So, the perturbation propagates outwards from the flex event at the speed of light.

    Next, let's imagine that we take z to be infinity and [itex]t_0[/itex] to be negative infinity - that is, we place the two parallel plates an infinite distance apart, and have the flex event occur an infinite time in the past. What is happening to the field?

    We still have our perturbation, propagating outwards from the initial event at the speed of light, and will keep doing so forever, since it will never reach the second plate.

    As a last step, take a snap shot of what the electromagnetic field looks like at some moment in time, and remove the two charged plates altogether. (This doesn't change anything, since the plates are infinitely far away and thus unable to affect the field in the region we are interested in.) Now, start time back up. What's going on?

    That disturbance is still there. It still propagates. Only now, it is propagating from nothing to nothing. So what is causing this?

    It can't be the two plates; the field is propagating on its own. Rather, the overall state of the field at one moment in time tells you, via Maxwell's equations, what the state of the field will be at the next moment in time. This is a purely local, purely causal event that happens independent of whatever it was that generated the field in the first place. That initial event provided the energy necessary to create the initial disturbance, but once created the disturbance propagates on its own. And, because we are talking about the electromagnetic field, we call the particle through which the field propagates the photon.

    The scenario is analogous in gravity. The gravitational field propagates as the curvature of spacetime. This is a purely local, purely causal phenomenon.
     
  11. Jul 19, 2011 #10
    No, because not only mass but also spatial curvature curves space at a distance.
     
  12. Jul 21, 2011 #11
    Aimless,

    thank you for you're careful and well-thought-out answer. I appreciate your help. I didn't understand, but that's just me. I have it bookmarked and hopefully after reading 10 popular physics books I'll understand it.
     
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