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Why particles follow path of extremal proper time?

  1. May 16, 2012 #1
    Hello everyone. I understand how to figure out the paths of the free particles following the principle of extremal proper time, but... where does it come from? I mean, how it's derived that particles follow a path of extremal proper time in space-time? I know that for example in flat space-time, the path of extremal proper time is a straight line in space, that agrees with Newton's laws. But how do you generalize that for all space-times?

    Thanks
     
  2. jcsd
  3. May 16, 2012 #2
    First you need to accept that particles travel along geodesics. Once you do that, you can show that stationary proper time corresponds to the geodesic equation:

    [tex]\tau = \int \sqrt{-g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} d\lambda [/tex]
    [tex]\delta \tau=0~~ \Rightarrow ~~\frac{d^2x^\sigma}{d\lambda^2}+\Gamma^\sigma_{~ \mu \nu }\dot{x}^\mu \dot{x}^\nu =0[/tex]
    where [itex]\dot{x}^\mu=dx^\mu /d\lambda[/itex].
     
  4. May 16, 2012 #3
    Au contraire mon ami. Have you ever heard of the famous Brachistochrone problem? Newton was challenged to solve the problem in 1696, and did so the very next day. The fastest path of a ball bearing rolling down an inclined slope from point A to a lower point B is an inverted cycloid. See http://mathworld.wolfram.com/BrachistochroneProblem.html.
     
  5. May 16, 2012 #4

    Mentz114

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    Gold Member

    Just being picky, but the object must slide without friction rather than roll, which would introduce angular momentum.

    In free space the extremal path is a straight line ( in Newtonian mechanics and SR).
     
  6. May 16, 2012 #5
    That's it, to get the geodesic equation, you need to assume first that the "test particles" follow the path of extremal proper time. My question is precisely about that first assumption.
     
  7. May 16, 2012 #6
    Actually, they follow the trajectory of maximal proper time!
     
  8. May 16, 2012 #7
    You only need to assume that particles follow geodesics. The fact that the path of stationary proper time corresponds to a geodesic is derived. The geodesic equation itself can be derived by other methods.

    Do you doubt that test particles follow geodesics?
     
  9. May 16, 2012 #8
    It turns out that the action for a free particle is:
    [tex]
    S = -m \, c^2 \, \int{d\tau}
    [/tex]
    where m is the rest mass of the particle.

    In classical (non-quantum) mechanics, the system evolves along those trajectories that leave the action minimal (or, extremal). This is called Hamilton's Principle of least (extremal) action.

    Notice that, because of the minus sign, if the action is minimal, then the proper time is maximal. But, you used the term 'extremal proper time'.

    In Quantum Mechanics, all trajectories are possible. Each trajectory is associated with a probability amplitude [itex]\propto \exp \left( \frac{i}{\hbar} \, S \right)[/itex]. The total transition probability is a superposition over all the possible trajectories.

    In the formal limit [itex]\hbar \rightarrow 0[/itex], which corresponds to transition from Quantum to Classical Mechanics (in classical mechanics, the Planck's constant does not enter), the complex exponential becomes wildly oscillatory, and contributions from various trajectories would cancel. The only uncompensated contribution is from the extremal trajectory, and the whole transition amplitude (path integral) can be approximated by the dominant contribution from the stationary trajectory (see Method of stationary phase, Saddle point method, Laplace method).
     
  10. May 16, 2012 #9
    Also, I should add that the principle of stationary proper time is just that - a principle. Just like the least action principle, Fermat's principle, etc., it isn't derived by definition.
     
  11. May 16, 2012 #10
    In free space in a gravitational field in Newtonian mechanics (NR), a straight line is not the fastest path. For both frictionless sliding beads and rolling ball bearings, a cycloidic path is faster than a straight line. See

    There is a similar problem with relativistic charged particles (protons) in strong focusing synchrotrons. At low energies, the higher energy protons have a faster revolution time. But above a particular energy called the transition energy, higher energy protons are slower.
     
    Last edited by a moderator: Sep 25, 2014
  12. May 16, 2012 #11
    But, is the proper time extremal? Also, you are not allowed to constrain the particle to move along a certain trajectory, because other forces, namely the forces of normal reaction act on it, so this is not a free particle in a gravitational field.
     
    Last edited by a moderator: Sep 25, 2014
  13. May 16, 2012 #12

    Mentz114

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    Gold Member

    I should have said 'in the absence of gravity or constraints' rather than 'free space'.

    The reference you gave does not deal with the rolling ball, so I'll take your word on that.
     
    Last edited by a moderator: Sep 25, 2014
  14. May 16, 2012 #13
    Well, maybe I'm so concerned about this because this principle is the key to connecting the world we see with the world of general relativity. If the space-time is curved, the metric and the line element changes, and then to know what a free particle in that space-time would do, we use the principle of extremal proper time. So still particles with respect to the big mass fall in its gravitational field due to the change in the line element and because of that, a change in the path of extremal proper time, that is the path that minimizes also the total distance S. So we could say that is a geodesic in space time.

    But what I want to see is the connection between this geodesics in space time and the way the particle really moves. I mean, why the particle follows the path that minimize an integral of an "abstract, invariant" term that we defined as dS?
     
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