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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.

    Thanks!
     

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  2. jcsd
  3. Jun 7, 2014 #2

    phinds

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

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  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

    PeterDonis

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    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:)
    Thanks!
     
  10. Jun 12, 2014 #9

    PeterDonis

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