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Constantly accelerating object and its proper time

  1. Mar 24, 2009 #1
    The question is:

    A rocket has a constant acceleration in a rest frame that is 0 to 0.8c in 4s. What is its proper time?

    This question I found, I guess I'll have to use integration but don't have a clue where to start off. Could you give me some ideas please?
     
  2. jcsd
  3. Mar 24, 2009 #2

    Fredrik

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    DrGreg's posts in this thread can help you figure out what the rocket's world line looks like.
     
  4. Mar 27, 2009 #3
    Thanks for the post. By the way, I'm still struggling to find the answer for this question.

    What I've done is first find the velocity of the particle in the laboratory frame,

    [tex]a=\frac{c}{5}[/tex] so [tex]v=\frac{ct'}{5}[/tex]

    Where [tex]t'[/tex] is the time measured in the laboratory frame.

    As [tex]dt'/dt=\gamma[/tex], [tex]dt/dt'=\frac{1}{\gamma}[/tex]

    Now, integrating both sides wrt t' gives,

    [tex]t=\int\sqrt{1-\frac{v^2}{c^2}}dt'[/tex]

    with a bit of simplifying, it becomes

    [tex]t=\frac{1}{5}\int\sqrt{5^2-t'^2}dt'[/tex]

    Computing this integral with boundary t'=0...4 gives a value too large for t as t must be less than 4s.

    Please give me advice about things I've done wrong
     
  5. Mar 28, 2009 #4

    Fredrik

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    You don't have to thank me when I give you bad advice. :smile: I misread your post and thought you meant constant proper acceleration.

    What you have posted looks right, so maybe you just did something wrong when you evaluated the integral. (I haven't done that part). I think a better way to simplify the integral is to make a change of variables t''=t'/5. Then maybe a trigonometric substitution...or just look it up in an integral table.
     
  6. Mar 28, 2009 #5
    Lol.. I found what I did wrong.. I calculated the integral in degrees mode (not radians)!!
     
  7. Mar 28, 2009 #6
    There is no reason to make this so complicated. You can just use a standard relativistic rocket equation (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken])

    t' = (c/a)invsinh(at/c) = 3.665 sec.
     
    Last edited by a moderator: May 4, 2017
  8. Mar 28, 2009 #7

    Fredrik

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    You're making the same mistake as I did. This problem is about constant coordinate acceleration, not constant proper acceleration. If the problem had specified constant proper acceleration, it would have been much harder.
     
    Last edited by a moderator: May 4, 2017
  9. Mar 28, 2009 #8
    Are you sure it's about constant coordinate acceleration wrt the initial rest frame? That seems odd.
     
    Last edited by a moderator: May 4, 2017
  10. Mar 28, 2009 #9

    Fredrik

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    Why is it odd? It's just an exercise for people who have just started to learn this stuff. It would be much more difficult for them to find the proper time if the problem had specified constant proper acceleration. I mean, of course it's easy if you use a physics FAQ to find out what the world line looks like, but it's hard to do it using only the information given in the problem and the definition of proper time.
     
  11. Mar 28, 2009 #10
    Yeah, I think I missed the obvious: it looks like a homework problem. But if so, it should be more specific about whether it's proper or coordinate acceleration.
     
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