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Homework Help: Relativistic Time for spaceship movement

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A certain star is 76.6 light-years away. How many years would it take a spacecraft traveling 0.946c to reach that star from Earth, as measured by observers on Earth? How many years would it take to reach that star from Earth, as measured by observers on the spacecraft?

    2. Relevant equations
    Relativistic Time = Original Change in Time / sqrt(1 - v/c)^2

    3. The attempt at a solution
    For the first question, 1/.946 = 1.057
    1.057 * 76.6 light years = 81 years

    For the second question, I can't get the right answer. What I did was:
    81 / sqrt[1- (.946)^2] and got 250, but that is wrong. not sure what I'm doing wrong...
  2. jcsd
  3. Apr 21, 2010 #2

    Filip Larsen

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    The factor is the inverse, that is with your names it should be "Relativistic Time = Original Change in Time * sqrt(1-(v/c)^2)". Usually one writes it as

    [tex]\tau = \frac{t}{\gamma}[/tex]

    where [itex]\tau[/itex] is the proper time for the spacecraft and

    [tex]\gamma = \frac{1}{\sqrt{1-(v/c)^2}}[/tex]
  4. Apr 21, 2010 #3
    Good to know. Our professor gave us the wrong equation in notes. Thanks for your help!
  5. Apr 22, 2010 #4

    Filip Larsen

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    It could also be that your professor gave the correct equation, but somewhere along the way in your notes the meaning of the two times got swapped, especially since the first equation I gave above very often is written with gamma on the other side, i.e.

    [tex]\tau \gamma = t[/tex]

    or, using unprimed time for the spacecraft and primed for the earth time,

    [tex]\Delta t' = \Delta t \gamma[/tex]

    As you can probably gather it is easy to get these times mixed up, and one way to keep them straight is to remember, that proper time, i.e. the time interval as measured in a frame at rest relative to the object in question which is the spacecraft in this case, will always be the smallest time interval and since gamma always is equal or bigger than one you can easily deduce with time interval is which in the last equation above.
    Last edited: Apr 22, 2010
  6. Apr 22, 2010 #5
    I did know that relativistic time (where observer is moving) should be shorter than proper time (where observer is stationary), but our professor gave us this:
    proper time = t0
    relativistic time = t
    relativistic time = t0 / sqrt[1 - (v/c)^2], which doesn't give the right relationship....
    so I can just change my equation to t = t0*sqrt[1 - (v/c)^2]
    t = t0 / 1 / sqrt[1 - (v/c)^2]
    but i guess the biggest clue is that unprimed time or relativistic time should be less than primed time or proper time?
  7. Apr 22, 2010 #6

    Filip Larsen

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    This is the correct relationship. Notice, that t = t0 / sqrt(...) = t0 * 1/sqrt(..) = t0*gamma, and that t > t0 for v > 0.

    In the context of your original question, the time you find in the first part is what is called relativistic time in your original post, i.e. time t, and you then need to find the proper time t0 (called Original Change in Time in your original post). I guess you just got the times swapped at some point along the way.
  8. Apr 22, 2010 #7
    Ah, I see. I was confused about the wording, because I thought "proper time" (t0) was the time measured by the person on earth, but "proper time" is actually the time that is measured by the person that is moving on the spaceship. Thank you so much for your help!
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