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Map Taking Proper Time to Euclidean Length

  1. Jul 3, 2014 #1
    Is there a way to map time-like curves in Minkowski space to curves in a Euclidean space such that the length of the curve in the Euclidean space is equal to the proper time of the curve in Minkowski space?
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  3. Jul 4, 2014 #2


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    You can multiply a proper time ##\tau## by ##c## to convert it into a length ##c\tau##. What this length means is the length that light travels during that proper time.

    I don't know what specifically you're looking for. What kind of a map are you thinking about?
  4. Jul 4, 2014 #3
    What I've been trying to do is to take the tangent vectors of a curve and transform each of them so that they point in the same direction but so that their magnitude is equal to their proper time. My thinking is that this would leave the shape of the curve mostly unchanged but its length would then reflect the proper time along that curve. However, I keep running into problems with the integrals involved since they involve complex numbers.
  5. Jul 4, 2014 #4


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    I guess you mean: Embed the pseudo-Euclidean space into a Euclidean space isometrically so the Euclidean path length in the embedding space matches proper-time. This is not possible, if I remember a previous discussion here on PF correctly (search for it).

    But you can do another trick to have proper-time as a Euclidean length: Use a space-propertime diagram, where the coordiante time is the Euclidean path length, while proper-time is the time coordinate (Euclidean length along the time axis):
  6. Jul 4, 2014 #5
    What do you think of the following transformation?

    The tangent vector of a time-like curve in Minkowski space is [tex](1,{v^1})[/tex]. Now, we construct a new tangent vector w, so that

    [tex]{w^0} = \frac{{\sqrt {1 - {v^1}{v^1}} }}{{{v^1}\sqrt {1 + \frac{1}{{{v^1}{v^1}}}} }}[/tex]

    [tex]{w^1} = \frac{{\sqrt {1 - {v^1}{v^1}} }}{{\sqrt {1 + \frac{1}{{{v^1}{v^1}}}} }}[/tex]

    but so that [tex]0 \le {v^1} \le 1[/tex].

    It seems to work for the curves I've tried it on, for the most part...
  7. Jul 4, 2014 #6


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    I'm not sure this even makes sense. Why do you want to change the shape of the curve? That changes its length, which destroys the whole point of what it seems like you are trying to do.

    Also, why do you want to do this at all? The Minkowski metric already gives you the proper time along a timelike curve directly.
  8. Jul 4, 2014 #7
    I want to make a spacetime diagram in which curves with longer proper time actually appear to be longer on the diagram.
  9. Jul 4, 2014 #8


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    If you're trying to make this a unit vector, which is implied by your saying "the" tangent vector (instead of "a" tangent vector), then it's wrong. If we restrict to one spatial dimension, the unit tangent vector is ##\gamma \left( 1, v \right)##, where ##\gamma = 1 / \sqrt{1 - v^2}##.

    Which will be tangent to a *different* curve, so what's the point? What are you trying to accomplish?

    What do you mean by "work"? Can you give more specific examples? As far as I can tell, what you're doing doesn't make sense.
  10. Jul 4, 2014 #9


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    Then what you want is not a spacetime diagram. You can't represent spacetime in a diagram if you change its geometry, which is what you'd be doing if you tried the transformations you appear to be trying.

    What you want is a space-proper time diagram. Check out the link that A.T. gave you. But bear in mind one key difference between a space-proper time diagram and a spacetime diagram: in a space-proper time diagram, there is not a one-to-one mapping between points in the diagram and physical events, as there is in a spacetime diagram.

    In other words, if I designate a particular physical event, such as "lightning strikes the clock tower in the town square at exactly midnight by the town clock", this event will appear as a single point in a spacetime diagram, the point at which the worldlines of the lightning and the clock tower intersect. But it will appear as multiple points on a space-proper time diagram, because the proper times of different objects whose worldlines intersect at that event are different: for example, the town clock's proper time will be some positive number (assuming we set the "zero point" of proper time at the instant when the lightning strike is emitted up in the atmosphere), whereas the lightning's will be zero (assuming it travels at the speed of light--and bearing in mind that there are caveats to interpreting a null interval as "zero proper time", but that's what you have to do if you want to represent them on a space-proper time diagram).
  11. Jul 5, 2014 #10


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  12. Jul 5, 2014 #11
    Thank you Peter and A.T.

    I think that these Epstein diagrams actually work a lot better than what I had in mind.
  13. Jul 7, 2014 #12
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