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Homework - time dilation problem (Lorenz factor is unknown)

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Alpha Centauri is ##4.4## light years away from Earth. What speed ##u##
    would a spaceship headed towards Alpha centauri had to have in order
    to last ##t' = 10## years for a passanger onboard?

    2. Relevant equations
    I know equations for time dilation, length contraction:
    \begin{align}
    \Delta t &= \gamma \Delta t' \xleftarrow{\text{time dilation}}\\
    \Delta x' &= \gamma \Delta x \xleftarrow{\text{length contraction}}
    \end{align}
    and Lorenz transformations:
    \begin{align}
    \Delta x &= \gamma(\Delta x' + u \Delta t')\\
    \Delta x' &= \gamma(\Delta x - u \Delta t)\\
    \Delta t&= \gamma\left(\Delta t' - \Delta x' \frac{u}{c^2}\right)\\
    \Delta t'&= \gamma\left(\Delta t + \Delta x \frac{u}{c^2}\right)
    \end{align}

    3. The attempt at a solution
    The only thing i know how to do is to determine the proper time which is the one measured on a spaceship...
    \begin{aligned}
    \boxed{t'\equiv \tau}
    \end{aligned}
    By using the equations above i simply can't calculate any other variable because ##\gamma## is unknown and is present in all 6 equations...
     
  2. jcsd
  3. Jun 22, 2013 #2
    Right.
    And the Lorentz-γ solemnly depends on your "u", and you've given data for Δt', Δx, and Δt = Δx/u to be matched.
    So simply substitute and solve for u.
     
    Last edited: Jun 22, 2013
  4. Jun 22, 2013 #3
    Ok so i am sure that:

    $$
    u = \frac{\Delta x}{\Delta t}
    $$

    and this holds for an observer in a coordinate system ##xy##. In this system only the observers clock isn't moving so this means that his clock is measuring the proper time ##\Delta t = \tau## from this it follows that all other times are longer (holds also for the time measured by the clock in a moving ##x'y'## frame): ##\Delta t' = \gamma \Delta t##. If i use this in the equation above i get:

    \begin{align}
    u = \frac{\Delta x}{\Delta t} = \frac{\Delta x \gamma}{\Delta t'} \xrightarrow{\text{i use only the RHS}} u &= \frac{\Delta x \gamma}{\Delta t'}\\
    u&=\frac{\Delta x}{\Delta t' \sqrt{1 - u^2 / c^2}}\\
    u^2&=\frac{\Delta x^2}{(\Delta t')^2 (1 - u^2/c^2)}\\
    u^2 - \frac{u^4}{c^2} &= \frac{\Delta x^2}{(\Delta t')^2}
    \end{align}

    How can i now solve this?
     
  5. Jun 22, 2013 #4
    Sorry, it should by a minus,
    [tex]\Delta t'= \gamma\left(\Delta t - \Delta x \frac{u}{c^2}\right)[/tex]
    not a plus in the Δt' expression; I had missed your typo in my first answer.

    Now do yourself a favor, set c:=1, and continue
    [tex]\Delta t'= \gamma\left( \Delta x/u - \Delta x\, u\right) = \gamma\left( \Delta x \frac{1-u^2}{u}\right) [/tex]

    Now expand γ and see what cancels out.
     
  6. Jun 22, 2013 #5
    Your first Lorentz Transformation equation will give you your answer:

    [tex]\Delta x = \gamma(\Delta x' + u \Delta t')[/tex]

    The x' coordinate of the spaceship is constant, so Δx' = 0. Δx = 4.4 light years, and Δt' = 10 years. And, you already noted that γ is a function of u/c.
     
  7. Jun 23, 2013 #6
    Thank you. If i set point in a origin this is easier :)

    Can anyone tell me which time here is the proper time and why.
     
  8. Jun 23, 2013 #7
    The proper time is the time measured by the guy on the spaceship who is physically present at both events (spaceship leaves, spaceship arrives), and whose coordinate is not changing during the trip.
     
  9. Jun 24, 2013 #8
    Thanks.
     
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