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Find t in the inertial frame

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A rocket travels in a straight line with speed 0.6c, where c is the speed of light in a vacuum. A wrist-watch is on board the rocket. The rocket moves past a clock tower on the earth.

    a. In the inertial frame where the tower is at rest, how long does it take the watch (on board the rocket) to register 10 s?

    b. In the inertial frame where the rocket is at rest, how long does it take the clock in the tower to register 10s?

    2. Relevant equations

    I don't remember


    3. The attempt at a solution

    To solve the time for the inertial frame
     
  2. jcsd
  3. Oct 22, 2009 #2

    sylas

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    Do you have a text book?
     
  4. Oct 22, 2009 #3
    yes, I do, but I don't have the text book with me right now.
    However, I remember one of the formula, but not too sure about it.

    t=t0*sqaure root(1-v^2/c^2)

    I am not sure if the t or the t0 should be switch on the place or not, which the t0 is the t knot as the way how my professor said it.

    Also, I have question is how do I know if I'm looking for the "t0", which is the t knot, or the "t" itself. My lab professor told me that if the inertial frame is on the tower and is asking for the time on the tower, then I would be looking for the t0, which is the t knot; if the inertial frame is on the tower, but is asking the time on the rocket, then it would be t? I don't know how to distinguish the difference on this.

    Please help!
     
    Last edited: Oct 22, 2009
  5. Oct 22, 2009 #4

    sylas

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    OK. You have the correct formula.
    [tex]t = t_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]​

    Both the tower and the rocket are inertial. The rocket is also inertial, because it is moving at a constant velocity.

    The formula works for any inertial observer. t0 is the duration of time for the inertial observer that is required for a clock moving at constant speed v to record a time difference of t.

    For example. The rocket is moving at speed 0.6v relative to the tower, and the moving rocket clock records 10s. So you can use t0 for the time measured on the tower for how long it takes the rocket clock to record 10 seconds.

    What confuses everyone at first is that the situation is completely symmetrical. You can use the same formula with t0 as the time measured on the rocket for the tower clock to record 10 seconds.

    Try it... what answers do you get for the two questions?

    Cheers -- sylas
     
  6. Oct 22, 2009 #5
    How do I determine when the question is asking about the t knot or the t? This was something that I was trying to make sense of. I know I can solve it, but for the 10 s that was given, how do I know if the 10 s was the time for the "t" or for the "t knot"?
    Please help explaining this to me.
     
  7. Oct 22, 2009 #6

    sylas

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    t0 is the actual time in the frame for the observer making a measurement or determination. t is the time recorded on a moving clock over the duration t0.
     
  8. Oct 23, 2009 #7
    a. In the inertial frame where the tower is at rest, how long does it take the watch (on board the rocket) to register 10 s?

    So, in this question, the t0 is the tower, and the t is on the rocket?

    b. In the inertial frame where the rocket is at rest, how long does it take the clock in the tower to register 10s?

    In this one, t0 is the rocket? and the t is clock in the tower?
     
  9. Oct 23, 2009 #8
    a. In the inertial frame where the tower is at rest, how long does it take the watch (on board the rocket) to register 10 s?

    T=?

    [tex]
    t = t_0 \sqrt{1-\frac{v^2}{c^2}}
    [/tex]
    [tex]
    t= 10s \sqrt {1-0.6}
    [/tex]
    [tex]
    t= 10s \sqrt {0.4}
    [/tex]
    [tex]
    t= 10s \sqrt {4*10^(-1)}
    [/tex]
    [tex]
    t= 10s * 2 \sqrt {10}
    [/tex]

    b. In the inertial frame where the rocket is at rest, how long does it take the clock in the tower to register 10s?
    [tex]
    t_0= ?
    [/tex]
    [tex]
    \frac{t}{\sqrt{1- \frac{v^2}{c^2}} = t_0?
    [/tex]
    Is this the way how I solve for question b?

    I tried to type out the equation for finding
    [tex]
    t_0
    [/tex]
    just can't fix it.

    So, what i was trying to say is that, to find out t0, I would need to divide t over square root of 1-((v^2)/(c^2))
    is that right?
     
    Last edited: Oct 23, 2009
  10. Oct 23, 2009 #9

    sylas

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    Yes.

    At this point, you have two problems. You have forgotten to square the 0.6. And you have mixed up the t and t0. In msg #7 above, you noted correctly that for the first question where the rocket is moving, we have "the t0 is the tower, and the t is on the rocket". So the moving rocket, which registers 10 seconds, is t.

    With substitution for a clock on the rocket measuring ten seconds, and a duration of time t0 for the tower, the equation should be
    [tex]
    10s = t_0 \sqrt {1-0.6^2}
    [/tex]

    Carry on from there. You won't need a calculator; the square root is an exact decimal fraction, and t0 will be the length of time passing on the tower for the rocket clock to register 10 seconds.

    Cheers -- sylas
     
  11. Oct 23, 2009 #10
    a. In the inertial frame where the tower is at rest, how long does it take the watch (on board the rocket) to register 10 s?

    [tex] 10s = t_0\sqrt {1-0.6^2} [/tex]
    [tex] 10s = t_0\sqrt {1-0.36} [/tex]
    [tex] 10s = t_0\sqrt {.64} [/tex]
    [tex] 10s = t_0(.8) [/tex]
    [tex] \frac {10s} {.8} = t_0 [/tex]
    [tex] t_0 = 12.5s [/tex]

    b. In the inertial frame where the rocket is at rest, how long does it take the clock in the tower to register 10s?

    [tex] t= 10s \sqrt {1-0.6^2} [/tex]
    [tex] t= 10 s (.8) [/tex]
    [tex] t= 8s [/tex]

    So, is this right now?
     
    Last edited: Oct 23, 2009
  12. Oct 23, 2009 #11

    sylas

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    Part a., correct.

    Part b, not correct. In this case, you are calculating in the rocket frame, so t0 is what you are calculating, using t = 10s for the moving tower. You should get the same result as in the first problem.

    Most people find this really confusing. But the situation of the rocket and the tower is symmetrical. Each one concludes that the clock of the other is running slow. This is not a contradiction.

    If you have two events that are separated from each other in time and space, different observers will have different notions of the distance between them, and of the time between them.

    For example, in the tower frame, the tower clock reads 12.5 in the same instant that the rocket clock reads 10. So in the tower frame, the tower clock reads 10 well before the rocket clock reads 10.

    In the rocket frame, the order of these events in time is reversed. In the rocket frame, the rocket clock reads 10 and then a while later the tower clock reads 10.

    Cheers -- sylas
     
  13. Oct 23, 2009 #12
    thank you sylas. Even though I'm still a little bit confuse, but I believe I will find a way to understand it. Thanks for your time. I really appreciate your time. Thanks again.
     
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