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Van Flandern's paper on "Speed of Gravity", question.

  1. Nov 27, 2014 #1
    Hello, I just recently came across this paper and I've been since reading everything I can find about Van Flandern's "credibility" and the controversies his proposals brought up. Still I'm yet unable to find a source to verify or disprove this assertion Van Flandern makes in his paper:

    "How then does the direction of Earth’s acceleration compare with the direction of the visible Sun? By direct calculation from geometric ephemerides fitted to such observations, such as those published by the U.S. Naval Observatory or the Development Ephemerides of the Jet Propulsion Laboratory, the Earth accelerates toward a point 20 arc seconds in front of the visible Sun, where the Sun will appear to be in 8.3 minutes."

    Is the above true? That the direction of Earth's acceleration goes toward the actual (not the visible) position of the Sun?
     
  2. jcsd
  3. Nov 27, 2014 #2

    Bandersnatch

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  4. Nov 27, 2014 #3
    That was great. Thanks!
     
  5. Nov 27, 2014 #4

    Vanadium 50

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    I met Mr. van Flandern. A true gentleman. Also a complete and total crackpot.

    The Carlip article is interesting, but it gets one important thing wrong: "Putting a "light travel delay" (technically called "retardation") into newtonian gravity would make orbits unstable, leading to predictions that clearly contradict Solar System observations." Consider a sun-centered coordinate system, and further consider the case where the sun is so much heavier than the other bodies that one can ignore their gravitational effect on the sun (but obviously not vice versa).

    In this frame, the effect of retardation is simple. One takes Newtonian gravity: g = -(GMm/r^3) r, where here r points towards (0,0,0) and replaces M with what M was with the retardation included. If gravity travels at c, and we are measuring g from the earth, we replace M(sun) by M(sun, 8 minutes ago). But - and here is the point that is almost universally missed - M(sun, 8 minutes ago) is the same as M(sun) now. So the size of the effect is zero.

    The is exactly analogous to trying to measure the speed of light by measuring how the brightness of the sun changes over the course of an orbit (neglecting the variability of the sun).
     
  6. Nov 27, 2014 #5
    Lol. I kinda figured he might have been from his biography. But it is true that his paper heats up the imagination of non well informed readers (like me). After reading the link above and regurgitating it on my pillow I re-realized (duh ^_^) exactly what you just said, the apparent movement of the Sun across the sky doesn't change the fact that for all accounts, the sun is static in relation to Earth.

    Another thing I wonder is if an experiment like the following could help measuring the speed of change in gravitational pull: To send an array of probes to a comet and measure it's gravitational pull, then part (nuke) the comet in two and measure how long it takes for the new configuration of it's mass to affect the orbit of the probes.
     
  7. Nov 27, 2014 #6

    Chalnoth

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    Vanadium 50: But the planet has moved, and that's where the instability comes from in this model.

    Steve Carlip also wrote a more technical version of the previously-linked post:
    http://arxiv.org/pdf/gr-qc/9909087v2.pdf

    The way I like to think of how it works is this:
    Imagine a situation where you have a star sitting all by itself in space, and a fast-moving object that comes near it, and gets its motion deflected a bit. The key insight here, I think, is to imagine gravity not as a force between two objects, but as a field that surrounds massive objects. So the fast-moving object doesn't notice or care where the star is: it just feels the gravitational field that surrounds the star.
     
  8. Nov 27, 2014 #7

    Vanadium 50

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    Yes, but it has moved from a point where the field is independent of the retardation (because it is static) to another pointy where the the field is independent of the retardation. The motion of the earth does not change the gravitational field of the sun.

    This is even easier to see working with potentials instead of fields.
     
  9. Nov 27, 2014 #8

    Chalnoth

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    Right. I think that's part of Van Flandern's confusion on this matter: he wasn't working in terms of fields, but of forces.
     
  10. Nov 27, 2014 #9

    PeterDonis

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    Don't you have to change r as well as M? Retardation may not affect the magnitude of r (if we idealize to a perfectly circular orbit), but it has to affect its direction.
     
  11. Nov 28, 2014 #10

    Vanadium 50

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    No. That's why I picked this frame to work in. The field from the sun is not affected by the motion of the earth.
     
  12. Nov 28, 2014 #11

    Doug Huffman

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  13. Nov 28, 2014 #12

    Chalnoth

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    Yes, well, that's the way the real world works. I don't think it has much to do with Van Flandern's incorrect notions about the speed of gravity.
     
  14. Nov 28, 2014 #13

    PeterDonis

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    But r does not just depend on the field of the sun. It also depends on the position of the Earth relative to the Sun, which changes during the 8 minutes it takes light to travel from the Sun to the Earth.

    Perhaps we have different ideas about what a theory of "Newtonian gravity plus retardation" actually entails. I'm assuming that it means we take each quantity in the formula for the force, and put in its retarded value instead of its instantaneous value. For M, I agree that the two values are the same, so retardation has no effect. But for r, the two values are not the same; the retarded vector r does not point in the same direction, physically, as the instantaneous vector r. That fact is independent of which frame you choose.
     
  15. Nov 28, 2014 #14

    PeterDonis

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    IIRC from Carlip's paper (the more technical one you linked to), the key problem he points out with Van Flandern's model is that it assumes the force is purely a function of position. However, in GR, in the weak field approximation, even though you can model gravity as a "force", it's a force that depends on velocity as well as position (i.e., there are "magnetic" as well as "electric" components), and the velocity-dependent components cancel out--almost--the effects of retardation. The small retardation effect that isn't canceled out appears as, for example, the perihelion shift in planetary orbits. I assume you could apply this same reasoning if you modeled gravity as a field or a potential instead of as a force in the weak field approximation.
     
  16. Nov 28, 2014 #15

    Chalnoth

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    Yes. And I think it is easiest to understand why this canceling occurs by thinking of gravity as a field.

    Carlip also explains that this kind of canceling actually must occur because without it the theory wouldn't conserve angular momentum, and we know from symmetry that it is impossible for the theory not to conserve angular momentum.

    Edit: Oh, and the perihelion shift comes from the effective [itex]1/r^3[/itex] correction term that can be added to Newtonian gravity to make it align better with General Relativity, not from any retardation effect. It might be possible to explain gravitational radiation as a result of some sort of retardation effect, but I'm not certain.
     
  17. Nov 28, 2014 #16

    Vanadium 50

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    It does in the frame I described. The sun is static here; neither its mass nor position change in time. That's why I picked it - because the motion is simplest here.

    I'm not arguing that there is no aberartion. I'm arguing the example Carlip is using is not a good one to discuss it. One cannot measure the speed of the propagation of a field using only static sources. (And if you work out the math in Carlip's paper, you will quickly come to the realization that the direction of the force towards the sun as viewed in the earth's frame is independent of the speed of propagation.)
     
    Last edited by a moderator: Nov 28, 2014
  18. Nov 28, 2014 #17

    PeterDonis

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    The sun's position doesn't change, yes, but the vector r, as I understand it, depends on both the sun's position and the earth's position, since it's defined as a vector that points from one to the other. The earth's position does change in the frame you're using.

    I do need to re-read the paper, it's been a while since I did so.

    Also, the way you phrase this--"as viewed in the earth's frame"--makes it seem like you are using a frame in which the earth is fixed, not the sun. But the comment I quoted above only makes sense in a frame in which the sun is fixed, not the earth. Are you switching frames, or are you saying Carlip used the earth-centered frame, whereas you prefer the sun-centered one?
     
  19. Nov 28, 2014 #18

    Vanadium 50

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    Something like that. You get the aberration working in the Earth frame, and you most easily see why it needs to cancel in the sun-centered frame. If you don't the argument that you can't measure the speed of propagation using only static sources.
     
  20. Dec 6, 2014 #19

    bcrowell

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    I'm not sure I buy this. Momentum is conserved in Newtonian gravity, and in GR momentum conservation is required or else the theory becomes inconsistent. In this context it may be unclear what theory we should really be discussing, but I doubt that van Flandern had in mind any theory that involves nonconservation of momentum.

    Therefore the sun has to move, and neglecting its motion is at best an approximation. If what you're describing was to be a counterexample to Carlip's claim of instability, you would have to show that the error due to neglecting the sun's motion was negligible. This is not at all obvious to me. It may be that he's right about the instability, and that the limit where the sun's mass is large is a limit in which the instability vanishes.
     
    Last edited: Dec 6, 2014
  21. Dec 6, 2014 #20

    Chalnoth

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    As I understand it, that's precisely Van Flandern's argument: that if we have a finite speed of gravity, angular momentum isn't conserved. So basically his entire argument is a straw man.
     
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