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Heated discussion about a flaw

  1. Oct 15, 2006 #1
    Dear all,

    I am having a heated discussion about this web paper: A Flaw of General Relativity ...
    I would like you to comment my opinion that this paper is wrong already in its introduction.

    I reproduce here the first lines of a figure caption found in the introduction:
    I don't discuss even the "either direction" which is totally ununderstandable to me.
    But, for me, in any situation (even near a blackhole) the traversal time could never be smaller than L/c , where L is the size of the rocket.

    Did I miss something?

    Thanks,

    Michel
     
  2. jcsd
  3. Oct 15, 2006 #2

    pervect

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    Nope. We had our own heated discussion with Zanket a while back, though his argument appears to have mutated in the time being. Don't expect to convince Zanket of anythying, though.

    There's a correct analysis of the so-called uniform gravitational field in
    http://arxiv.org/abs/physics/9810017.

    One of the most interesting things about it is that it isn't actually uniform.

    You'll find this result also in Misner, Thorne, Wheeler, "Gravitation", but the MTW reference uses a considerably higher level of math.

    Note that it actually is standard to call the gravitational field of an accelerating spaceship a "Uniform gravitational field" according to common usage in the literature, even though the proper acceleration (i.e. that which would be measured in an accelerometer) is not actually uniform.

    This is all basically a consequence of what one could losely call "gravitational time dilation", or the "relativity of simultaneity". It is not possible to synchronize clocks at the head and tail of the rocket in the accelerated frame - the two clocks appear to tick at different rates.

    Basically Zanket's argument in its current incarnation ignores this fact, and assumes that the clocks in the nose and the tail of the rocket "frame" tick at the same rate because this is what happens in inertial frames.
     
    Last edited: Oct 15, 2006
  4. Oct 16, 2006 #3
    Thanks a lot pervect.

    A quick read of this paper was already much reassuring for me.
    Equation [27] in the paper (http://arxiv.org/abs/physics/9810017) is quite clear.
    For small separations, it simply shows that acceleration has the same effect as a gravitational field.
    Like for a gravitational field, two clocks at both end of the rod must have different rates represented in metric component goo = 1-f/c², where f is the gravitational potential (or the analog for a uniform acceleration).

    Michel
     
  5. Nov 3, 2006 #4
    Nope. My text doesn't reference two clocks. Only one clock at a fixed position in the rocket is needed. The clock can be infinitesimally small.

    This thread can be locked now.
     
  6. Nov 3, 2006 #5

    JesseM

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    Zanket's argument doesn't make sense because it is based on the claim that general relativity "forbids" something which it doesn't in fact forbid, specifically the claim that an observer in a uniform gravitational field in flat spacetime (see this page for some caveats about the notion of 'uniform gravitational fields') is forbidden to see the entire universe as "observable". But this is mistaken--cosmological horizons in general relativity are based on expanding space, general relativity would not predict any sort of cosmological horizon in a flat spacetime with a uniform gravitational field, and observers in such a universe would be able to see infinitely far, according to GR. Thus the equivalence principle holds, and there is no paradox.

    edit: actually, I may be partly mistaken about that, since being at rest in a uniform gravitational field is equivalent to constant acceleration in a flat spacetime with no gravitational field, and accelerating observers do have a Rindler horizon. But a freefalling observer moving along with the ball wouldn't see any horizons, since he's equivalent to an inertial observer in flat spacetime with no gravity, and anyway I think the horizon would be "below" the tree, not "above" it (since the rindler horizon seen by an accelerating ship lies in the opposite direction that the ship is heading, so observers on the rocket perceive the horizon as 'below' them), so Zanket's argument about GR forbidding the tree to see a star arbitrarily far above it is still wrong.
     
    Last edited: Nov 3, 2006
  7. Nov 3, 2006 #6

    JesseM

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    On the other hand, I don't see a problem with the claim that the ball could appear to "fall" arbitrarily fast as seen by an accelerating observer...
    The ball can never cross the rocket faster than L/c if the rocket was moving inertially and you were talking about the rocket's inertial rest frame, but that's not what Zanket's talking about. He's talking about what is seen by a non-inertial observer at rest on the floor of the accelerating rocket. If at the moment the ball is at the ceiling, his distance from the ball is L in his own instantaneous inertial rest frame, then it is true that the proper time for the ball to reach him on the floor can be made arbitrarily short with a sufficiently high acceleration (although for small enough times the rate of acceleration would have to be so high that the G-forces would be deadly). This is equivalent to the statement that if I am at rest on earth and I see a star X light years away in the earth's rest frame, then by accelerating towards it at a high enough rate, I can make the travel time to that star as measured on my own clock arbitrarily short--see The Relativistic Rocket page for confirmation of this.
     
    Last edited: Nov 3, 2006
  8. Nov 3, 2006 #7

    pervect

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    Umm - no, I don't think so. If we were going to debate this, I'd ask to see a space-time diagram to back this claim up. Note that proper time, as measured by a single clock, can be represented as the "length" (actually the Lorentz interval) of a time-like worldline on a space-time diagram. A pair of proper time intervals can be represented by a piar of lines, etc.

    Of course, we've already debated this once, and I'm not convinced it'll do the board any good to debate it again.

    We are skating a bit on thin ice as far as PF policies go, generally we don't debate "challenges" to the standard literature here.
     
  9. Nov 4, 2006 #8

    JesseM

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    Well, what about my interpretation above that he's talking about the time-interval between the moment t1 the ball is at the ceiling, as defined by the instantaneous inertial rest frame of an observer on the floor (you might indeed need two clocks to define what time t1 the clock on the floor reads 'at the same moment' the ball is on the ceiling, but if you're using the instantaneous inertial rest frame of the clock on the floor this should be a well-defined notion), and the time t2 that the ball arrives at the floor, as measured by the clock of an observer on the floor? Isn't it true that by increasing the rate that the floor is accelerating, this time can be made arbitrarily short, just as the time for an accelerating rocket to travel from earth to a star X light years away in the earth's frame can be made arbitrarily short with sufficient acceleration?

    By the way, would the observations of an observer on a ship undergoing Born rigid acceleration be equivalent to those of an observer on a ship at rest in a uniform gravitational field? In both cases, would the G-forces felt at the front and back of the ship be the same, for example?
     
  10. Nov 4, 2006 #9

    pervect

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    I don't really have a clear idea of what today's scenario is, but my impression is that next week it will have changed. The main thing that will, however, be the same is that Zanket will still be claiming to have disproved General relativity :-(.

    If two observers are undergoing Born rigid acceleration each observer (the front and back observer) will measure a different acceleration with his or her local accelerometer.

    See for instance http://www.mathpages.com/home/kmath422/kmath422.htm

    http://arxiv.org/abs/physics/9810017 also discusses this. This is the simplest peer reviewed English language reference I've been able to find on the topic. (The mathpages article is also pretty good IMO and may be easier to follow though of course it is not peer reviewed).

    What the literature calls a "uniform gravitational field" isn't actually uniform as measured by local accelerometers.

    See for instance http://arxiv.org/PS_cache/physics/pdf/0604/0604025.pdf for an example of this usage.

    While perhaps the naming choice is unfortunate, it appears to be what the literature uses :-(.\
    The metric is defined not by setting the "felt acceleration" to a constant, but by setting the Ricci tensor to zero. I.e one is looking for a vacuum solution to Einstein's equations, this is the sort of result one gets in an accelerating spaceship, where as we've already seen the acceleration as measured with a local accelerometer depends on position (is not uniform) as long as the spaceship is "rigid".

    Note that one can apply the notion of "radar rigidity" as well as "Born rigidity" to define a "rigid spaceship" - while radar distance isn't equal to the distance as defined by the Lorentz interval, in an accelerating frame an object with a constant radar distance will also have a constant distance as measured by the Lorentz interval. Thus both distance measures will agree as far as rigidity goes, even though they are not exactly equivalent.
     
    Last edited: Nov 4, 2006
  11. Nov 4, 2006 #10
    Hi Lalbatros ;)

    Don't give Zanket the 'honour' of calling that a paper. It's an injustice to all the people who put in time and effort learning proper science and publishing proper papers.

    Even with the view of a few relativity heavy weights from these forums behind you, I doubt Zanket will listen.

    Ignorance is bliss for her.
     
  12. Nov 7, 2006 #11
    Disagreement with generally accepted viewpoints is not allowed here. If anyone wants polite discussion with me, email me by clicking on my alias on the paper, or google for an open-minded forum.
     
  13. Nov 7, 2006 #12

    Doc Al

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    Give us a break.
     
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