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Special relativity

  1. May 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I have the trajectories of a particle in the space-time:

    [tex]\tau[/tex]([tex]\sigma[/tex]) = [tex]\frac{1}{a}[/tex]senh([tex]\sigma[/tex])

    x([tex]\sigma[/tex]) = [tex]\frac{1}{a}[/tex]cosh([tex]\sigma[/tex])


    How can I write this equations depending on proper time t of the particle?


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 11, 2009 #2

    sylas

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    Start out by defining the proper time. If you have a trajectory for a particle, what is the proper time along that trajectory, in general?
     
  4. May 11, 2009 #3
    The proper time is :

    t = [tex]\int\sqrt{1-\frac{1}{c^{2}}(\frac{dx}{dt})^{2}}[/tex]dt
     
  5. May 11, 2009 #4

    sylas

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    Good... now how can you get dx/dt?

    By the way; you are being a bit unusual with your variable names. You have "t" being used above both as proper time (on the LHS) and as co-ordinate time (on the RHS, inside the integral).

    It's more usual to use τ (Greek tau) as the proper time. In your original post, you also use tau as the co-ordinate time.
     
  6. May 11, 2009 #5
    Maybe I can write :

    [tex]\frac{dx}{dt}[/tex] = [tex]\frac{dx}{d\sigma}[/tex] * [tex]\frac{d\sigma}{dt}[/tex]

    Where :

    t : co-ordinate time

    [tex]\tau[/tex] : proper time

    PS: you are right!!
    In my original post : [tex]\tau[/tex] [tex]\rightarrow[/tex] t
    In my second post : t [tex]\rightarrow[/tex] [tex]\tau[/tex] (in the first member)
     
  7. May 11, 2009 #6

    sylas

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    OK... there's another thing.

    In your original equations, there's no mention of "c". Were your original equations given in units where c=1? This is commonly done to keep it simple.

    To use your definition of proper time, where c is explicit, you'll have to make a small change to one of your original equations... or (easier) let c=1 in your definition of proper time and keep in mind that your units have this property.
     
  8. May 11, 2009 #7
    My prof gave me those equations without "c". I think that I must continue in units where c=1. Maybe in my definition proper time I have to put c=1.
     
  9. May 11, 2009 #8

    sylas

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    You are now well on the way to a solution. Let me give you a hint, however. When you do the integration, try changing the integration variable to σ. That is, replace "dt" with [tex]\frac{dt}{d\sigma} d\sigma[/itex]

    What this gives you, in fact, is the parametrized definition of proper time.

    It will work a bit more easily, since everything is now in terms of σ in your equations.
     
  10. May 11, 2009 #9
    Ok, so I write :

    [tex]\tau[/tex] = [tex]\int\sqrt{\frac{dt^{2}}{d\sigma^{2}}-\frac{dx^{2}}{d\sigma^{2}}}[/tex] d[tex]\sigma[/tex]

    and I can solve the integral. So I get [tex]\tau[/tex] in function of [tex]\sigma[/tex]


    But How can I write t and x in function of [tex]\tau[/tex] ?
     
  11. May 11, 2009 #10

    sylas

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    Try it. Solve the integral. It's going to be surprisingly easy.

    PS. When you use LaTeX, put the whole equation inside the {tex} {/tex} tags. It will look much nicer!
     
  12. May 11, 2009 #11
    I found :

    [tex]\tau[/tex] = [tex]\int[/tex] [tex]\sqrt{1/a^{2}}[/tex] d[tex]\sigma[/tex] = [tex]\frac{\sigma}{a}[/tex]

    So I have to replace [tex]\sigma[/tex] with a*[tex]\tau[/tex] in my original equations.
    Is it right?
     
  13. May 11, 2009 #12

    sylas

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    Yes. In fact, your original equations are almost given in terms of proper time already! They are standard parametric equations for constant proper acceleration. That is, a particle following this world line experiences the same fixed acceleration all the time.
     
  14. May 11, 2009 #13
    Yes, infact in the next part of the exercise I have to demonstrate the proper acceleration is costant and =a!
    thank you very much!!
     
  15. May 11, 2009 #14

    sylas

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    Oops. :tongue: I seem to have done too much of your exercise! But never mind; I am certain talking something like this through helps it to make better sense. Best of luck with it all.

    Let me suggest one thing worth thinking about. How would you put "c" back into these equations, so that "a" remains the proper acceleration?

    Also, you are working in a co-ordinate system with an origin (0,0). Suppose a bomb goes off at point (0,0) in space and time. When would the accelerated observer see the explosion?

    Cheers -- sylas
     
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