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Special Relativity Questions

  1. Nov 23, 2014 #1
    I was not too sure if this was the correct forum, so feel free to move if needed.

    1. The problem statement, all variables and given/known data

    A spaceship is measured to be exactly 1/3 of its proper length.

    (a) What is the speed parameter β of the spaceship relative to the observer's frame?

    (b) By what integer factor do the spaceship's clocks run slow, compared to clocks in
    the observer's frame?


    2. Relevant equations
    [itex]
    L=\frac{L_0}{\gamma} \\
    t=\frac{t_0}{\gamma} \\
    \gamma = \frac{1}{\sqrt{1-\beta^2}} \\
    \beta = \frac{v}{c}
    [/itex]

    3. The attempt at a solution
    For A i did:
    [tex]
    L=L_0 \sqrt{1-\beta^2} \\
    \frac{L_0}{3}=L_0 \sqrt{1-\beta^2} \\
    \frac{1}{3}= \sqrt{1-\beta^2} \\
    \frac{1}{9}=1-\beta^2 \\
    -\frac{8}{9}=- \beta^2 \\
    \frac{8}{9}=\beta^2 \\
    \beta = \sqrt{\frac{8}{9}}
    [/tex]

    I am not to sure that is correct. But for part B I was stuck but during typing this up managed to get an integer answer so hopefully it is correct.
    [tex]
    t=\frac{t_0}{\sqrt{1-\beta^2}} \\
    \frac{t}{t_0}=\frac{1}{\sqrt{1-\beta^2}} \\
    \frac{t}{t_0}=\frac{1}{\sqrt{1-\frac{8}{9}}} \\
    \frac{t}{t_0}=\frac{1}{\frac{1}{3}} =3 \\
    [/tex]

    Would appreciate any help/advice/feedback, thanks :)
     
  2. jcsd
  3. Nov 23, 2014 #2

    Simon Bridge

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    Your answers are fine, but it sounds like you don't understand the equations.

    ##L_0=\gamma L\\
    T=\gamma T_0##
    Where the 0's index the proper time, for this case.

    You were told: ##L=\frac{1}{3}L_0##. This means that ##\gamma = 3##.
    ... so you have automatically answered the second question without any further working out: ##T=3T_0##

    (Pretty much the first thing you want to know in any SR problem is ##\gamma##.)

    ... for the first question, you want ##\beta##: $$\gamma = \frac{1}{\sqrt{1-\beta^2}} \implies \beta = \sqrt{1-\frac{1}{\gamma^2}}$$ ... it is best practice to do the algebra with the symbols before putting numbers in.
    $$\beta = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} \implies v= 0.9428c $$ ... can you see how much easier that was that what you did?
    No worries though everyone does it the hard way at first ;)
     
  4. Nov 24, 2014 #3
    Many
    Many thanks for your help and feedback, much appreciated :)
     
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