Special relativity time dilation question

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astrobird
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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 don't 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 don't understand it at the moment.

Thanks!
 
Attachments
  • Scanned printed page from a textbook titled "Failure of relativity?" discussing time dilation and paradoxes in special relativity, with spacetime diagram references to observers O and O-bar.
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  • Spacetime diagram with axes t and x showing worldlines of clocks O, O-bar and O-prime, with events A, B, C, E, F marked and labelled Figure 1.14, next to printed text on time dilation.
    p20.jpg
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Oops sorry! I hope this is better?
 
Attachments
  • Photo of a printed textbook page titled "Failure of relativity?" discussing the paradox of time dilation between two observers in special relativity, with the text about Figure 1.14 and clock readings.
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  • Spacetime diagram with axes t and x showing worldlines through the origin, with points A, B and C marked, and a caption reading "The proper length of AB is the time ticked by a clock at rest in O, while that of AC is the time it takes to do so as measured by O."
    p20a.jpg
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You need to apply the Pythagorus theorem to the triangle.
 
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?
 
Jilang said:
You need to apply the Pythagorus theorem to the triangle.

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.
 
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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?
 
Anyone that can help? Still hoping for an answer:)
Thanks!
 
astrobird said:
Anyone that can help? Still hoping for an answer:)
Thanks!

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