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Special relativity -- Proper time

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Alice is driving a race car around an essentially circular track at a constant speed of 60 m/s. Brian, who is sitting at a fixed position at the edge of the track, measures the time that Alice takes to complete a lap by starting his watch when Alice passes by his position (call this event E) and stopping it when Alice passes his position again (call this event F). This situation is also observed by Cara and Dave, who are passengers in a train that passes very close to Brian. Cara happens to be passing Brian just when Alice passes Brian the second time (see Figure). Assume that the clocks used by Alice, Brian and Cara are close enough together that we can consider them all to be “present” at event E; similarly, that those used by Alice, Brian and Dave are “present” at event F. Assume that the ground frame is an inertial reference frame.

    (a) Who measures the shortest time between these events? Who measures the longest?
    (b) If Brian measures 100 s between the events, how much less time does Alice measure between the events?
    (c) If the train carrying Cara and Dave moves at a speed of 30 m/s, how much larger or smaller is the time that they measure compared to Brian’s time?
    2q9ypol.jpg

    2. Relevant equations

    Coordinate time ≥ Space-time interval ≥ Proper time
    $$ Δt≥Δs≥Δr $$

    Measuring proper time in an inertial reference frame between events A and B, and V is constant:
    $$\Delta r_{AB} = \sqrt{1-v^2}\Delta t_{AB} \approx (1-0.5 v^2) \Delta t_{AB}$$ (Using Binomial approx.)

    3. The attempt at a solution
    my answers:
    A- Alice, B- Brian, CD- Cara and Dave

    (a)
    The Shortest time-
    A and B measures Δr (which is defferent in each RF)
    Because B measures Δs also, and A is moving (therefore time moves slower for her) – Alice measures the shortest time.

    The longest time- B and CD measures Δt
    B measures Δs which is the shortest possible Δt.
    Since CD don’t measure Δs they measure the longest time interval.

    *** Question: Is it correct to state that CD don't measure Δs ? and therefore they measure the longest time interval? ***

    (b)
    Converting the speed the SR units:
    $$ 60 m/s = 20*10^{-8} (SR) $$
    $$ Δt_{AB-Brian} = 100_s $$
    Manipulating the equation by subtracting Δt in order to get the answer:
    $$\Delta r_{AB- Alice} - Δt_{AB} = (1-0.5 v^2) \Delta t_{AB} - Δt_{AB} = (1-0.5 v^2-1) \Delta t_{AB} = -0.5 Δt_{AB} $$
    $$ -0.5 * (20*10^{-8})^2 *100_s = -2*10^{-12}_s $$
    *** No questions about this section ***

    (c)
    Train speed $$ 30 m/s = 10^{-7} (SR) $$
    Meaning that in CD reference frame B is moving in a constant speed of 10^{-7}

    *** Question: I thought I need to use the same equation like in section b, but CD don't measure proper time, or maybe they are indeed measuring proper time ?
    I'm not sure how to combine the given data into the equation ****


    Thanks so much in advance !
     
  2. jcsd
  3. Nov 22, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Sure, they have different points on the train for the two events.

    Why do you write seconds as index?

    You can find the time C/D measure by considering the reference frame of the train, for example. B is moving from C to D with a known velocity and for a known proper time.
     
  4. Nov 23, 2015 #3
    mfb thank you so much !
    The second's index was a mistake.

    I made the train RF:
    qrhroi.jpg

    so is it correct to use the equation as follows:

    $$ Δr_{EF-Brian} = (1-0.5 v^2) \Delta t_{ED-CD}$$
    $$ \Delta t_{ED-CD} = \frac{100 }{(1-0.5*(10^{-7})^2)} = \frac {100}{(1-5*10^{-15})} $$

    And therefore CD Coordinate time is bigger.

    Or maybe should I use the metric equation (but im not sure its possible due to the data) ?
     
  5. Nov 23, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Correct.
     
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