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Every event is simultaneous with you at some time: proof?

  1. Aug 25, 2015 #1

    bcrowell

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    Theorem: In Minkowski space, let ##r(\tau)## define a timelike, ##C^2## world-line W parametrized by proper time for ##\tau## covering the whole real line. Here ##r## is the radius four-vector extending from the origin O to a given point. Then there exists a unique time ##\tau## such that ##r(\tau)## is orthogonal to W at the point defined by ##\tau##.

    ******* This version is false. See my attempt at revision below. ********

    In other words, if you're an observer, every event is simultaneous with you at some uniquely defined time.

    How would you go about proving this? I would prefer a proof that avoids invoking any coordinates.

    At first I expected this to be trivial, but it seems a little tricky because it seems like you need a topological component to the argument. For example, I had been thinking to argue something like this. Let ##v(\tau)## be the unit tangent vector to W, and let ##f(\tau)=r(\tau)\cdot v(\tau)##. Then it seems obvious, in Minkowski space, that ##f## starts out negative and ends up positive, and since it's continuous it must be zero somewhere. But the fact that ##f## starts negative and ends positive is something that can be false if, for example, you cut out all of Minkowski space for ##t\ge 0##.

    I want to avoid making assumptions in the proof that seem reasonable but that require explicit justification to rule out counterexamples such as a flat spacetime with a nonstandard topology.
     
    Last edited: Aug 25, 2015
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  3. Aug 25, 2015 #2

    pervect

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  4. Aug 25, 2015 #3

    PAllen

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    I'm confused. There are trivial counter-examples for a non-inertial world line in perfectly normal Minkowski space. Just consider a world line with uniform proper acceleration, e.g. to the left, starting from velocity c to the right at minus infinity, going to c to the left at plus infinity. Then, there is an event on the Rindler horizon to the right, such that two or more different events on the accelerating world line would be simultaneous to this one event, by this definition.
     
    Last edited: Aug 25, 2015
  5. Aug 25, 2015 #4

    Dale

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    What is W here?
     
  6. Aug 25, 2015 #5

    Erland

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    ??? This must be wrong. For example, the birth of my mother cannot be simultaneous with any event experienced by me. All these events must be later than the birth of my mother, for all observers.
     
  7. Aug 25, 2015 #6

    PeterDonis

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    But your ##\tau## doesn't cover the whole real line; it only covers a finite range of it (you only experience a finite range of proper times). The OP is talking about idealized "observers" that exist forever, infinitely in the past and infinitely in the future.
     
  8. Aug 25, 2015 #7

    PAllen

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    You missed that the world line must extend from -infinity to +infinity of proper time on the world line. I think there may be a problem with the proposition (or with my understanding of it), but that is certainly not the problem.
     
  9. Aug 25, 2015 #8

    bcrowell

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    Oops, W was meant to be the world-line. I've edited the OP. Thanks!
     
  10. Aug 25, 2015 #9

    PAllen

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    That clarifies it, and my objection stands - the uniqueness part of the proposition is false. The existence part is true.
     
  11. Aug 25, 2015 #10

    bcrowell

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    I don't understand your counterexample. Let's say the world-line W is ##r(\tau)=(t,x)=(\sinh\tau,-\cosh\tau)## in Minkowski coordinates. Then O=(0,0) is on the Rindler horizon, and ##r(0)=(0,-1)## is orthogonal to W at ##\tau=0##. I don't see how you get another point on W that has this property...??
     
  12. Aug 25, 2015 #11

    PeterDonis

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    Every point on the worldline has this property; that is, the line ##r(\tau)## from the origin ##(0, 0)## to every point of the form ##(\sinh \tau, - \cosh \tau)##, for all real values of ##\tau##, is orthogonal to the tangent vector to the worldline at that point. That's a fundamental property of every worldline in a Rindler congruence.
     
  13. Aug 25, 2015 #12

    bcrowell

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    Aha! Very interesting. That invalidates what I thought was a nice simple proof of the 1+1-dimensional Herglotz-Noether theorem ( http://www.lightandmatter.com/sr/ , section 9.5.3).

    So maybe this is what I want:

    Theorem: In Minkowski space, let W be a timelike, ##C^2## world-line parametrized by proper time ##\tau## for ##\tau## covering the whole real line. Then (1) for any point P, there exists a point Q on W such that line PQ is orthogonal to W. Furthermore, (2) there exists a neighborhood A of W such that if P is chosen from within A, then Q is unique.

    If this version is right, then I think my proof of H-N in 1+1 dimensions works, but the theorem has to be weakened in a similar way to the weakening of the theorem above.
     
  14. Aug 25, 2015 #13

    PeterDonis

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    I think it is, since there is no constraint on how small the neighborhood can be. In fact, I believe the construction of Fermi normal coordinates depends on this theorem being true in a general spacetime, not just Minkowski spacetime; the size of the neighborhood is constrained in the general case by both the path curvature of the worldline (which is also a constraint in the Minkowski case) and the curvature of spacetime near the worldline (which of course is not present in the Minkowski case).
     
  15. Aug 25, 2015 #14

    bcrowell

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    So for H-N in 1+1, I think what this means is that a rigid motion is unique given the motion of one point on the object, but existence is not guaranteed unless the object is small enough and the proper acceleration is small enough. Given the size L and the proper acceleration a, I think we need something like ##L\lesssim c^2/a##. Basically you can't extend a rigid object beyond its own Rindler horizon.
     
  16. Aug 25, 2015 #15

    Dale

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    This seems like a plausible theorem, but difficult to prove. The proof will certainly have to involve the ##C^2## property, since a worldline with a sharp bend is a counter example.
     
  17. Aug 25, 2015 #16

    PAllen

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    Yes, I think C^2 is necessary and sufficient for a nonempty neighborhood. A Peter pointed out, the place to look in the literature for a fully rigorous proof would mathematical relativity treatments of Fermi-Normal coordinates. Physicists will take shortcuts (though, I guess, Hawking and Ellis was a rare counterexample).
     
  18. Aug 25, 2015 #17

    PAllen

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    Note that physically, this is very plausible. As acceleration approaches infinite, maximum Born rigid object size approaches zero.
     
  19. Aug 25, 2015 #18

    DrGreg

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    (1) still isn't true. Use the example of posts #11 & #12, and take P to be any point with [itex]|x| \leq |t|[/itex], i.e. on or behind the Rindler horizon (and its mirror image), e.g. (1,0).

    I believe (2) is true, but I haven't proved it.
     
  20. Aug 25, 2015 #19

    bcrowell

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    Yep. I guess I'd better stop writing revisions labeled optimistically as "theorem" :-)

    Conjecture: In Minkowski space, let W be a timelike, ##C^2## world-line parametrized by proper time ##\tau## for ##\tau## covering the whole real line. Then there exists a neighborhood A of W such that for any point P in A, there exists a unique point Q on W such that line PQ is orthogonal to W.

    This version seems to me like it really must be weak enough to hold. If we zoom in close enough on any part of a ##C^2##, timelike world-line, then it looks like a curve with constant proper acceleration, so (assuming ##a=1##, which is clearly inessential) we can make it look just like the point ##(0,-1)## on our canonical example ##(t,x)=(\sinh\tau,-\cosh\tau)##. We know that the conjecture holds in a neighborhood of this point, since the Rindler horizon lies at a finite distance from it.

    But I've been wrong so many times that I'm probably wrong again now :-)

    I originally thought that this was going to require some global knowledge of Minkowski space such as topology or causal structure, but it looks like the version of the conjecture that is actually true is much weaker and purely local, which makes proving it look much more straightforward.
     
  21. Aug 25, 2015 #20

    Nugatory

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    @Dadface may find this interesting - it's part of what I was getting at when I suggested that he look into Born rigid motion in the Bell paradox thread.
     
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