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Infinity in Finite Proper Time

  1. Feb 16, 2006 #1

    George Jones

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    Working on pervect's "messy unsolved" problem has led me to an interesting result. Let [itex]\left( x , t \right)[/itex] be a global inertial coordinate system for Minkowski spacetime.

    Consider the worldline given by

    [tex]t \left( \tau \right) = \frac{\tau^3}{3} - \frac{1}{4 \tau}[/tex]

    [tex]x \left( \tau \right) = -\frac{\tau^3}{3} - \frac{1}{4 \tau}.[/tex]


    [tex]\frac{dt}{d \tau} = \tau^{2} + \frac{1}{4 \tau^2}[/tex]

    [tex]\frac{dx}{d \tau} =- \tau^{2} + \frac{1}{4 \tau^2}.[/tex]

    Note that [itex]dt/d\tau > 0[/itex], and that

    \left( \frac{dt}{d \tau} \right)^2 - \left( \frac{dx}{d \tau} \right)^2 &= \left( \tau^{2} + \frac{1}{4 \tau^2} \right)^2 - \left( - \tau^{2} + \frac{1}{4 \tau^2} \right)^2\\
    & = 1.

    Therefore, [itex]\tau[/itex] is the proper time for a futute-directed timelike worldline.

    Note also that when [itex]\tau = -1[/itex], both [itex]t[/itex] and [itex]x[/itex] are finite, but as [itex]\tau \rightarrow 0_-[/itex], both [itex]t[/itex] and [itex]x[/itex] wander off to positive infinity.

    The situation is unphysical because the 4-acceleration is unbounded, although there are no hyperlight speeds.


    PS I think I have found an expression for the 4-acceleration of a specific example of pervect's problem, but I have to check to see if my solution really does satisfy the necessary criteria.
    Last edited: Sep 30, 2013
  2. jcsd
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