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Special relativity time dilation question

  1. Jun 7, 2014 #1
    Out of interest I'm studying a book "A first course in general relativity" which is a great book in my opinion because it explains the subject very well. I'm a beginner though and I have a hard time understanding one particular thing mentioned quite early in the book. I'm attaching a scan of a small part of the text. I understand this part fine except for one thing. On the second page I'm attaching it says "A simple calculation shows this to be at t=sqrt((1-v²))".

    I dont understand how they arrive at this. Would someone be able to provide an explanation or some hints so that I will hopefully get it? It must be something pretty obvious because they say its simple but I just dont understand it at the moment.


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  2. jcsd
  3. Jun 7, 2014 #2


    User Avatar
    Gold Member

    Seriously? You want people to read something you've posted SIDEWAYS?
  4. Jun 7, 2014 #3
    Oops sorry! I hope this is better?

    Attached Files:

  5. Jun 7, 2014 #4
    You need to apply the Pythagorus theorem to the triangle.
  6. Jun 7, 2014 #5
    Thanks, I considered this before already but I still cannot figure it out. Perhaps you can indicate how I obtain the values of the sides of the relevant triangle?
  7. Jun 7, 2014 #6


    Staff: Mentor

    Be careful. In spacetime, the ordinary Pythagorean theorem doesn't work with triangles that have timelike sides. For example, the distance AB in the diagram on the second page that the OP posted is 1; the distance AC is ##t## (the elapsed time along O's worldline), and the distance BC is ##x = vt## (the distance an object moving at speed ##v## with respect to O travels in time ##t##). If you look at the text, you will see that it says the distance AC is ##t = 1 / \sqrt{1 - v^2}##; in other words, the "hypotenuse" of the triange, AB, is *shorter* than one of its "legs", AC.

    The correct formula, as you can see if you work it out for the distances I gave above, is ##t^2 - x^2 = 1##, where ##t## is the distance AC, ##x## is the distance BC, and ##1## is the distance AB. This is similar to the ordinary Pythagorean theorem, but it has an all-important sign change.
  8. Jun 7, 2014 #7
    Thanks a lot for the further input and explanation. That helps but I still don't understand how they arrive at the value for t at event E. Can you show me how that should follow from your reasoning above?
  9. Jun 11, 2014 #8
    Anyone that can help? Still hoping for an answer:)
  10. Jun 12, 2014 #9


    Staff: Mentor

    What triangles in the figure *can* you apply the theorem to? What distances *can* you calculate?
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