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Easy question on proper time

  1. Jun 28, 2011 #1
    Hi everyone!!

    Let's take two events [itex]P_1,P_2[/itex] in a Minkowsky spacetime, and let's choose them both lying on the [itex]\omega[/itex] axis, separated by a certain distance.

    Now, I want to calculate the world line of the particle which experiences the least proper time during its trip between the two points.

    The proper time, taking the velocity of the particle costant in its modulus (but not in its direction!!) can be written [itex]d\tau=dt\sqrt{1-v^2/c^2}=dt\sqrt{dt^2-dx^2/c^2}=\frac{1}{c}\sqrt{dw^2-dx^2}[/itex]; so it's easy to set up a variational calculus, minimizing the integral[tex]\tau=\frac{1}{c}\int_{P_1}^{P_2}\sqrt{1-\left(\frac{dx}{d\omega}\right)^2}d\omega[/tex](the quantity under the square root can't be negative, you all know why:smile:).

    So by taking [itex]L=\sqrt{1-{x'}^2(\omega)}[/itex] with [itex]x'(\omega)=\frac{dx}{d\omega}[/itex] we need to calculate the euler-lagrange equation:[tex]\frac{\partial L}{\partial x(\omega)}-\frac{d}{d\omega}\frac{\partial L}{\partial x'(\omega)}=0[/tex]but since L doesn't depend on x(w) but only on x'(w) we get to[tex]\frac{d}{d\omega}\frac{\partial L}{\partial x'(\omega)}=\frac{d}{d\omega}\frac{x'(\omega)}{ \sqrt{1-{x'}^2(\omega)} }=0[/tex]and so [tex]\frac{x'(\omega)}{ \sqrt{1-{x'}^2(\omega)} }=G\Rightarrow x'=\frac{G}{\sqrt{1+G^2}}[/tex] where G is a constant, determinable by the position of the two point on the spacetime.

    So we end up with[tex]x(\omega)=\frac{G}{\sqrt{1+G^2}}\omega[/tex]

    This is supposed to be the path between the two point on the spacetime of the particle which experiences the least proper time.

    In the case considered in the beginning I took both point lying on the w axis; so the constant G turns out to be 0, so the path I'm looking for is [itex]x=0[/itex], which corresponds to a particle at rest.

    But this is impossible, since it's know that this path it's the one with the longest proper time!!

    I found various solution for this:
    1) I did not understood nothing of relativity
    2) I've done various mistake
    3) The variational method I've used gave me effectively an extremal path, but the longest, not the shortest. If so, does this means that there are not a minimum path? (I don't think so) OR I have to introduce some lagrange multiplier-style constraint on the calculation (like the fact that I cant'get back in time)??

    Can you help me to solve this riddle??
    Thanks for your disponibility!!!
     
    Last edited: Jun 28, 2011
  2. jcsd
  3. Jun 28, 2011 #2

    PAllen

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    The Euler Lagrange equation only guarantees 'stationary' variation. It does not distinguish minimum, maximum, or saddle (neither). There are other conditions you can test for that distinguish these case (these are rarely worth the bother - you can normally figure out which you have). However, in relativity, the answer is simple: there is no extremal minimum. A light like path has zero proper time but is not stationary with respect to variation. A time like path can get arbitrarily close to zero proper time. Bottom line: variation of a Minkowskian metric only gives you curves of maximum proper time (in GR, it doesn't guarantee that - there are often saddle geodesics in GR which have non-extremal paths of both longer and shorter proper time).
     
  4. Jun 28, 2011 #3
    Thanks PAllen!!

    But can you make clear to me why it can't be found any minimum path for a world line?

    The proper time can be made arbitrarily small if the world line to which it corrisponds can get close enough to the null geodesic, but if the path has to pass between fixed point there are some constraint!

    I imagine that the path which have the shortest proper time is kind of the "longest" one. So if I take some oscillating function x=Asin(aw) that meets the requirement of a timelike path and whose tangent is the closest possible to the lightlike path i guess that something has to come out; but again I realize now that the higher the frequency the longest the path, so effectively there is no minimum. Am I right?
     
  5. Jun 28, 2011 #4

    PAllen

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    There is no variational minimum. Given any time like path between two events, there is a 'nearby' path that has lower proper time. As a result, the Euler-Lagrange equation will never pick out a minimum for paths between events with time like separation.
     
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