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I Why do particles move along longest proper time trajectories

  1. Jun 22, 2016 #1
    Hi,

    I am working my way thought Hartle's Gravity. In Section 5.4 he states that "The straight lines along which free particles move in spacetime are paths of longest proper time" and proceeds to proof that "in flat space time the proper time is a curve of extremal proper time".

    Can someone explain why it actually are the paths with longest proper time? Light travels along path with zero proper time. Is that a longest path (normal paths being negative)?

    Thanks
    Bas
     
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  3. Jun 22, 2016 #2

    vanhees71

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    That's the action principle. For a free particle in special relativity the only Lorentz invariant action is given by the Lagrangian
    $$L=-mc^2 \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
    This you can write in terms of an arbitrary world-line parameter as
    $$A=\int \mathrm{d} t L=-m c^2 \int \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}},$$
    where here the dot means the derivative wrt. ##\lambda##.

    Now from the principle of equivalence this translates into the generally covariant action
    $$A=-m c^2 \int \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{nu}}.$$
    The Euler-Lagrange equations leads to the equations for the geodesics in the pseudo-Riemannian manifold with pseudo-metric components ##g_{\mu \nu}##.
     
  4. Jun 22, 2016 #3

    Nugatory

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    You need an additional qualifier: it's not "a particle follows a path of maximum proper time", it is "a particle not subject to any external forces follows a path of maximum proper time". Because the paths of extremal proper time are straight lines this statement is the curved spacetime equivalent of Newton's first law - the particle is moving along a "straight" line (actually a geodesic, because that's the curved spacetime equivalent of a straight line) because there's no force acting on it to push it off that path.

    There's only one straight line between any two nearby points, and light in vacuum always travels in a straight line (any other path would imply that the light changes its speed through space, but we know that that speed is c and it doesn't change), so there's only one possible path light can follow. Thus, the zero proper-time path that light follows is extremal in a trivial sense - it is the only path so it is both the longest and the shortest path.
     
  5. Jun 22, 2016 #4

    vanhees71

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    One should stress that it's a geodesic in spacetime, not in space! E.g. the Kepler orbit of the earth around the sun is also described by a geodesic in spacetime, but for sure it's not a geodesic in space for an (asymptotic inertial) observer. It's a nice excercise to calculate this as the motion of a body in the Schwarzschild spacetime (outside the Schwarzschild radius of course ;-)).
     
  6. Jun 22, 2016 #5
    The question is how can both be true?

    What you are saying is that and a massive particle and a massless travel on a geodesic in spacetime and that (for nearby points) there is only one path.

    But a massless and a massive particle leaving the same event will never ever meet in a future event. Thus they must follow different paths.

    So that's kind of a dilemma? Right?

    Or perhaps a massive particle does not travel on a geodesic at all?
    Because perhaps it constantly couples and modulates the background?
     
  7. Jun 22, 2016 #6

    Ibix

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    If the particle isn't massless then it can't travel between the same start and end events as a photon, not in flat spacetime. So I don't see anything contradictory in what Nugatory wrote.
     
  8. Jun 22, 2016 #7

    stevendaryl

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    I think what he meant is that there is exactly one geodesic path connecting two (nearby) events. There are infinitely many geodesics that go through a single event.
     
  9. Jun 22, 2016 #8
    Between event A and B (which are nearby) how many paths are there that are geodesics?
     
  10. Jun 22, 2016 #9
    That is of course true.
     
  11. Jun 22, 2016 #10

    stevendaryl

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    Exactly one. If you use locally Minkowskian coordinates, then the geodesic can be parametrized by:

    [itex]x^\mu(s) = A^\mu + s (B^\mu - A^\mu)[/itex]
     
  12. Jun 22, 2016 #11

    vanhees71

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    Of curse in this case you have
    $$u \cdot u = (B-A) \cdot (B-A),$$
    and the geodesic cannot be both time and lightlike (null), but that's self-evident anyway :-).
     
  13. Jun 22, 2016 #12

    Ibix

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    One, as you just said to Steven.
     
  14. Jun 22, 2016 #13
    Hi vanhees71,

    I do not agree with "That's the action principle". The action principle only says you take an extreme (min or max) of the action. In optics this normally leads to the fastest path and often shortest path a photon/wave can propagate. Here it is said that the paths are those with the longest proper time. That is counter-intuitive, at least to me. I've figured out by now that proper time really is a bit special. I was wondering if there nevertheless is a easy way to understand why the extreme action always leads to the longest proper time...

    do you have any thoughts on this?

    Bas

    PS: Yes, I mean when no external forces are applied.
     
  15. Jun 22, 2016 #14
    Hi Nugatory,

    You are correct, I indeed meant a particle that is not subject to any external forces. My question however was, to put it in your own words: why are paths of extremal proper time always straight lines/the straightest lines possible?

    Bas
     
  16. Jun 22, 2016 #15

    Nugatory

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    Exactly - thx.
     
  17. Jun 22, 2016 #16

    Ibix

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    Carroll points out that you can always approximate a path in spacetime by a series of null paths (on a Minkowski diagram, approximate a curve by a connected series of 45-degree lines). By shortening the null segments you can get arbitrarily close to the curve - thus there is always a path of length zero arbitrarily near any other path. Thus the extremised path length must be a maximised one.
     
  18. Jun 22, 2016 #17

    stevendaryl

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    Well, in flat spacetime it is pretty obvious:

    Two events [itex]e_1[/itex] and [itex]e_2[/itex] can have three types of separation:
    1. Timelike, meaning that it is possible for a slower-than-light particle to move from [itex]e_1[/itex] to [itex]e_2[/itex].
    2. Lightlike, meaning that it is possible for a light signal to move from [itex]e_1[/itex] to [itex]e_2[/itex]
    3. Spacelike, meaning that neither of the above is possible.
    If [itex]e_1[/itex] and [itex]e_2[/itex] are timelike separated, then that means that there is an inertial frame in which [itex]e_1[/itex] and [itex]e_2[/itex] take place at the same spatial location, at different times. In that case, the proper time in going from [itex]e_1[/itex] to [itex]e_2[/itex] is given by:

    [itex]\tau = \int_{t_1}^{t_2} \sqrt{1-\frac{v^2}{c^2}} dt[/itex]

    It's clear that [itex]\tau \leq (t_2 - t_1)[/itex]. The largest that [itex]\tau[/itex] can be is when [itex]v=0[/itex], which gives [itex]\tau = (t_2 - t_1)[/itex]
     
  19. Jun 22, 2016 #18
    You might want to qualify what I highlighted in bold as it is a coordinate dependent quality.
    There is no absolute sense of a "same spatial location" in relativity.
     
  20. Jun 22, 2016 #19

    stevendaryl

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    I said "there is an inertial frame in which the two events take place at the same spatial location". Yes, it's frame-dependent.

    The logic is this:
    1. Pick a coordinate system such that [itex]e_1[/itex] and [itex]e_2[/itex] have the same location, but different times.
    2. In that coordinate system, convince yourself that the extremal value for [itex]\tau[/itex] is the largest, not the smallest value.
    3. If the conclusion is true in one coordinate system, then it is true in every coordinate system, since the proper time integral is invariant.
     
  21. Jun 22, 2016 #20
    Well actually that is true only in SR but not generally in GR.
    Think for instance about a ball thrown up while standing on Earth!
     
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