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When to use lorentz transformations vs time dilation?

  1. Oct 20, 2013 #1
    I don't understand when I should use the Lorentz transformation versus time dilation or length contraction.

    I found this: http://www.phas.ubc.ca/~mav/p200/lttips.html

    but it's still unclear to me...

    "Length contraction applies when you are talking about a distance that is independent of time (e.g. the distance between two objects that are fixed relative to one another or the distance between two ends of a single object)."
    I *think* I understand that...
    - length contraction would be the distance between two planets to a moving observer
    - lorentz transformation would be the distance between a planet and a moving observer after some time t
    Is that correct?

    "Time dilation applies when you are talking about two events that are at the same place in some frame (e.g. two ticks on the same clock)."
    this one I'm just utterly lost on. can someone give me an example for time dilation and then a slightly tweaked example for a lorentz transformation? (like the ones I made for length contraction above)
     
  2. jcsd
  3. Oct 20, 2013 #2

    UltrafastPED

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    The Lorentz transformation allows you to shift the point of view between two (or more) coordinate systems - moving observer vs laboratory, etc.

    Or if you are only interested in a particular time or length you can just directly use the Lorentz factor to extract that information ... based on one set of coordinates you obtain that single result as seen by the other coordinate system - but without converting everything.
     
  4. Oct 20, 2013 #3
    If you have a single observer with a clock in one frame of reference sweeping past a sequence of clocks in another frame of reference, he will observe that the time exhibited on his clock is changing more slowly than the times he observes on the clocks that he sweeps past. If there are also a sequence of observers at the locations of the clocks in the other frame of reference, and they note down the time on his clock and on their own clocks as he sweeps past them, when they get together later and compare notes, they will conclude that the time on his clock was changing more slowly than the time on their clocks. Both these sets of observations describe time dilation.

    In many calculations, if you have some experience and really know what you are doing, then you can confidently take time dilation into consideration and get the right answer. However, if you are less confident and experienced, you can always use the Lorentz Transformation and have a better chance of getting the right answer. I recommend doing problems both ways until you have more experience and confidence.
     
  5. Oct 20, 2013 #4
    ...wouldn't the observer in this case see the OTHER clock moving more slowly, since in his frame it's the other frame that's moving, not him?
     
  6. Oct 20, 2013 #5
    ultrafast - That didn't help me, tweaked examples as referred to please.
     
  7. Oct 21, 2013 #6

    vela

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    You can derive time dilation from the Lorentz transformations. Say you have two events separated by ##\Delta t## and ##\Delta x## in one frame and ##\Delta t'## and ##\Delta x'## in another. The Lorentz transformations tell us that
    \begin{align*}
    c\Delta t' &= \gamma(c\Delta t - \beta \Delta x) \\
    \Delta x' &= \gamma(\Delta x - \beta c\Delta t')
    \end{align*} You recover the time dilation formula ##\Delta t' = \gamma \Delta t## when ##\Delta x=0##.

    Say there's a moving clock. In its rest frame, two ticks occur at the same place in space. In any other frame, the time interval ##\Delta t'## will be greater than the time ##\Delta t## because ##\gamma\ge 1##. In other words, observers will see the moving clock ticking more slowly than their own clocks.
     
  8. Oct 21, 2013 #7
    No. He's not looking at ONE OTHER clock. He is looking at the times shown on a sequence of clocks in the other frame as he sweeps past each of them. As he goes by, if he writes down the time on his own clock and the time on the clock directly opposite him in the other frame, and does this for each of the clocks he goes past, he will find that the times displayed on the clocks in the other frame (as captured on his data sheet) will show intervals that are greater than the intervals displayed on his own clock.

    Chet
     
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