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Special relativity - sending a light signal from the rear of spaceship

  1. Jun 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Spaceship is ##\scriptsize100m## long in its proper coordinate system and is moving relative to Earth. Astronaut sends a light signal from the rear part of the space ship towards the front part. It seems to an observer on Earth, that a light signal needed ##\scriptsize 0.57\mu s##. What is the relative speed ##\scriptsize u## of the spaceship compared to Earth.

    2. Relevant equations
    • Lorentz transformations,
    • time dilation,
    • length contraction.

    3. The attempt at a solution
    I set spaceship's rear end in an origin of ##\scriptsize x'y'## and the observer on Earth in an origin of ##\scriptsize xy##. The problem states that length ##\scriptsize 100m## is in the proper coordinate system, so i know that ##\scriptsize \Delta x' = 100m \equiv \ell## (i ll denote proper length using ##\ell##). I also know that observer on Earth measured ##\scriptsize \Delta t = 0.56\mu s##.

    Time events do not happen on the same place for neither of the observers (astronaut nor the observer on Earth), so i don't know if the time dilation eq. is reliable enough to calculate ##\scriptsize \gamma## and ##\scriptsize u ## afterwards. Can you confirm this please?

    The speed of light is constant so:
    \begin{align}
    c=const.\quad
    \left\{
    \begin{aligned}
    c&=\frac{\Delta x}{\Delta t} \xrightarrow{\text{i calculate $\Delta x$}} \Delta x = c \Delta t = 2.99\cdot10^{8}\tfrac{m}{s} \cdot 0.56 \cdot 10^{-6} s = 173.42m\\
    c&=\frac{\Delta x'}{\Delta t'} \xrightarrow{\text{i calculate $\Delta t'$}} \Delta t' = \frac{\Delta x'}{c} = \frac{100m}{2.99\cdot 10^8 \tfrac{m}{s}} = 3.34\cdot10^{-7}m
    \end{aligned}
    \right.
    \end{align}

    1st it is weird to me that the length ##\scriptsize \Delta x > \Delta x'##. Especially because the problem states that ##\scriptsize \Delta x' \equiv \ell## which indicates that it must hold ##\scriptsize \Delta x' = \gamma \Delta x##. But it doesnt:

    \begin{align}
    \Delta x' &= \gamma \Delta x \xleftarrow{\text{because $\gamma \geq 1 \longrightarrow \Delta x'>\Delta x$}}\\
    \gamma &= \frac{\Delta x'}{\Delta x}\\
    \gamma &= \frac{100m}{173.42m}\\
    \gamma &= 0.57
    \end{align}

    The ##\scriptsize \gamma## above is weird and i don't know how to continue to calculete the right ##\scriptsize u##. Please I need some explaination on this. Where did i do anything wrong? I think using the invariant interval ##\scriptsize \Delta x^2 - (c\Delta t)^2## could solve the problem fast, but please show me how to solve this using the Lorentz transformations or time dilation / length contraction equations.
     
    Last edited: Jun 24, 2013
  2. jcsd
  3. Jun 24, 2013 #2

    TSny

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    You need to interpret the meaning of the Δx that you calculated. Can you see why it's not the length of the spaceship as measured in the earth frame?
     
  4. Jun 24, 2013 #3
    Well i know only that in the Earth frame (##\scriptsize xy##) the ship is mooving. If it stood still in the ##\scriptsize xy## it would be ##\scriptsize 100m## long (like it is for an observer in the system ##\scriptsize x'y'## mooving with the ship). But it is moving and it is bizarre to me that i get a length dilation instead of a length contraction... I can't explain this to myself.

    Maybe the problem gives wrong data and the time ##\scriptsize \Delta t = 0.58\mu s## is just too large... Mistake? From my perspective it may be, but i am not an expert and am therefore not sure...
     
    Last edited: Jun 24, 2013
  5. Jun 24, 2013 #4

    TSny

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    Δx is the distance measured between two events in the earth frame. What are those two events? Draw a sketch showing the location of the two events in the earth frame and the location of the spaceship for each of the two events.

    Note that in Newtonian mechanics Δx would also be expected to be greater Δx'.
     
    Last edited: Jun 24, 2013
  6. Jun 25, 2013 #5
    Are you trying to say that ##\Delta x=u\Delta t + \gamma \Delta x'## ? If this is the case then i get the equation below:

    \begin{align}
    c&=\frac{\Delta x}{\Delta t}\\
    c&=\frac{u\Delta t + \gamma \Delta x'}{\Delta t}\\
    c&= u + \gamma \frac{\Delta x'}{\Delta t}
    \end{align}

    It is true that ##\gamma## is a function of ##u##, but i don't know how to bust it out of this equation. Do i need another eq?
     
  7. Jun 25, 2013 #6
    I did solve this case using 4 lorentz transformations. From all of the transformations i substracted ##\gamma u## which is a mixed part and is hard to calculate with. If i combined only Lorentz transformations only i could only get ##\Delta x## out of it. But if i combined Lorentz transformations with reverse Lorentz transformations i could get ##\gamma##. Then i could calculate ##u=0.49c##.
     
  8. Jun 25, 2013 #7

    TSny

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    Almost. In the earth frame, the ship is contracted. So, did you get the gamma factor in the right place in the last term?
     
  9. Jun 25, 2013 #8

    TSny

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    I think .49c is the right answer.
     
  10. Jun 25, 2013 #9
    What exactly should i get? Please post an equation.
     
  11. Jun 25, 2013 #10

    TSny

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    How did you get the last term in ##\Delta x=u\Delta t + \gamma \Delta x'##? The first term is how far the spaceship travels relative to the earth. What does the last term represent?
     
  12. Jun 25, 2013 #11
    Well tha last term ##\Delta x'## represents the length of a moving ship in coordinate system ##xy##. Should there be no ##\gamma##?
     
  13. Jun 25, 2013 #12
    The last term is supposed to be the length of a mooving spaceship which is mooving in an ##xy##. I think i know what u mean... Should it be like this:

    \begin{align}
    \Delta x &= u \Delta t + \Delta x'\\
    \Delta x &= u \Delta t + \frac{\Delta x}{\gamma}
    \end{align}

    ?
     
  14. Jun 25, 2013 #13

    TSny

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    Yes. That's [STRIKE]correct[/STRIKE]. See if you can solve this for u.

    EDIT: Whoops, I didn't notice that there is still an error in what you wrote for the last term on the right. Δx on the left is how far the light signal travels relative to the earth frame. That should equal the distance the ship travels relative to the earth (uΔt) plus the length of the ship as measured in the earth frame. So, Δx = uΔt + L where L is length of ship as measured in the earth frame. How can you express L in terms of the length of the ship as measured in the ship frame?
     
    Last edited: Jun 25, 2013
  15. Jun 26, 2013 #14
    Well length of the ship in the ships frame is equal to the proper length ##\scriptsize \ell=100m## if we measure this distance from the earth while the ship is mooving we can write down:

    $$\Delta x = u\Delta t + \frac{\ell}{\gamma}$$

    Is this now ok? I think i was confused about ##\scriptsize \Delta x## as i thought it is a length, but it is NOT it is distance between events! Even in the Lorentz transformations this is the case! And i wasn't even using ##\ell## so far so i better start...
     
  16. Jun 26, 2013 #15

    TSny

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    That looks good.
     
  17. Jun 27, 2013 #16
    Thank you.
     
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