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Homework Help: Time Dilation-AHH

  1. Apr 21, 2008 #1
    [SOLVED] Time Dilation-AHH!!

    So, I feel like this should be a simple problem, but according to my online homework, none of the answers are correct.

    Here's the problem:
    A spaceship is moving at 0.66 c (0.66 times the speed of light) with respect to the earth. An observer on the spaceship measures the time interval between two events on earth as 35 hrs. If the spaceship had been moving with a speed of 0.9 c with respect to the earth, what would the time interval between the events have been?

    So, I used the equation T= gamma*To, and To as 35 hours, but then I thought maybe I was doing it backwards and used the 35 hours as T, and still couldn't get the right answer!! Is the time interval of 35 hours the proper time interval? From what I read in the book, proper time is time as measured by an observer who sees the events occur at the same position, and is this is a time dilation problem, then the answer must be longer than 35 hours. I'm just a little frustrated, I don't know what I'm doing wrong!!
  2. jcsd
  3. Apr 21, 2008 #2


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    This seems like a homework problem. Can you please show your work? I can't know what you have done wrong, if you don't show me.
  4. Apr 21, 2008 #3
    Using the equation delta T= gamma*delta(Tproper)....

    delta T= 1/(sqrt(1-.9^2))*35 hours, getting about 81 hours.
  5. Apr 21, 2008 #4
    (But this is wrong, but I don't know why)
  6. Apr 21, 2008 #5
    Omg, relativity.

    Can I have the link to your online homework, I want to start learning it also
  7. Apr 22, 2008 #6
    Hint: Convert the time interval to the Earth's frame first, and then convert to the case for 0.9c.
  8. Apr 22, 2008 #7
    If book describes proper time is exactly the way you worded it I do not think too highly of it.

    Proper time is the local elapsed time you measure when you are for instance on earth or on a spaceship, not when you measure something away from you that is in motion with respect to you.

    It does not state if the spaceship is moving towards or away from the earth which obviously makes a difference in the measurement of the interval. Possibly the creator of the problem did not even think about it or wanted the student to ignore this part, you might want to bring this up.
    Last edited: Apr 22, 2008
  9. Apr 22, 2008 #8
    Actually, that definition is quite common. Proper time is the interval as measured in the reference frame where events occur in the same spatial coordinates. Draw a Minkowski diagram and you will see this also corresponds to the shortest period of time possible.

    I think you can disregard the propagation of light signals from the Earth to the ship for this problem. In that case, it's irrelevant.
  10. Apr 22, 2008 #9

    Doc Al

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    Staff: Mentor

    When the problem talks about an observer "measuring" a time interval one can assume that they are not talking about raw observations and that light travel time has already been corrected for. (It would be quite a different problem if that's not what's meant.)
  11. Apr 22, 2008 #10
    Okay, some of you guys seem to contradict one another.

    So...T(earth)=T(spaceship)*gamma, correct? I get 46.588 for this, but this isn't the final answer. I get a big fat "NO" for this one.

    so then I used T(.9c)=T(earth)*gamma, using .9c here and using .66c above, getting 106.880 hours. Still not right.

    WHAT THE H!!!
  12. Apr 22, 2008 #11
    This isn't correct. The time measured in the Earth is the proper time, so it should also be the shortest interval. Think about it and check your equation.
  13. Apr 22, 2008 #12
    Just to check if I'm understanding the problem correctly, is the answer supposed to be 64.2 hours?
  14. Apr 22, 2008 #13
    No, 64.2 hours is wrong as well.
  15. Apr 22, 2008 #14
    I GOT IT FINALLY!! Alrighty, so, T(earth) I found using 35/gamma, getting 26.294 (and using .66c for velocity). I then found T(.9c) by dividing 26.294 by gamma using .9c and getting 60.323 hours.
  16. Apr 22, 2008 #15
    Simple calculation errors made a huge difference, I was dividing 35 by sqrt(1-.66^2) rather than 35/[1/(sqrt(1-.66^2))]. Thanks!
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