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Homework Help: Rockets approach: time they meet w.r.t. rocket A?

  1. Dec 16, 2017 #1
    This is an IB problem from November 2017. Two rockets are separated by 6E8m (2 sec x c, or two light-seconds), w.r.t. Earth, and are approaching Earth from opposite directions. Rocket A approaches from left at 0.6c. Rocket B approaches form right at 0.4c.
    According to Earth, when do they meet? I get 2sec. All seems good.

    According to A, how fast is B? Using Lorentz transformation for velocity, I get 0.81c. All seems good.

    According to A, how long until A and B meet? So I treat the rockets as if in a train car moving right, with rocket A at the far left of the car (stationary w.r.t. the car) and B moving leftward (at 0.81c w.r.t. the car) from the right end of the car. The car is two light-seconds long, w.r.t. Earth.

    I use v=0.6c in gamma to change the time of the event (rockets coming together to meet). I use gamma to transform Earth's perception of an event of 2sec to an event w.r.t. A of 1.6sec.

    But if I do this a different way, finding the initial distance between rockets w.r.t. A and then using rate x time = distance (all w.r.t. A), I get 0.81c x time = 2sec x c/gamma (gamma = 5/4, using v=0.6c). Therefore, time=1.98sec.

    I tried using the Lorentz transformation for time, to get a tie-breaker, but had no luck. I got t'=gamma(using v=0.6c) (t-vx/c^2) = 5/4 (2sec - 0.6c (2 sec c)/c^2) = 1sec.

    Thanks for any help in pointing out where I'm making my mistakes.
     
  2. jcsd
  3. Dec 16, 2017 #2

    mfb

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    Staff: Mentor

    I moved the thread to our homework section.

    In the reference frame of A the rocket B is not 2 light seconds away when the scenario starts - and not 2 light seconds divided by gamma either. Relativity of simultaneity makes the view of A and B more complicated. Calculating everything in the frame of Earth is much easier here.
     
  4. Jan 1, 2018 #3
    Thank you.

    Okay, so we can't just divide time (2 seconds) by γ. A and B should agree on rate (w.r.t. each other) and distance from each other, so they should agree on how long until they meet (rate ⋅ time = dist). But dividing by two different γ (using v or u) gives two different lengths, which shouldn't be their observation: the asymmetry in their speeds (0.4c vs. 0.6c) arises only from Earth's perception.

    S' is A's frame of reference. View B as moving left in A's FOR. View A's FOR as a train car moving right at 0.6c w.r.t. Earth. So use the formula
    Δt'=γA w.r.t. Earth(Δt-vA w.r.t EarthΔx/c2)
    (=1.25(2 sec - 0.6c 2 sec ⋅ c / c2)= 1 sec)​

    I suspect we can do this problem another way, finding Δx' and then using rate ⋅ time = distance.
    Δx'=γA w.r.t. Earth(Δx-vA w.r.t EarthΔt)
    (=1.25(2 sec ⋅ c - 0.6c ⋅ 2 sec)= 1 sec ⋅ c)​
    Using rate ⋅ time = distance gives 1 sec ⋅ c/0.806c = 1.24s. (No match. Hmm.)

    We can further try to confirm our answer by verifying that A and B agree on how much time they'll take to meet. After all, their perception of each other's speed should match, being independent of Earth perception.
    Δt'=γB w.r.t. Earth(Δt-vB w.r.t EarthΔx/c2)
    (=1.19(2 sec - 0.4c ⋅ 2 sec ⋅ c / c2)= 1.43 sec)​

    Three approaches and none match, when I expected them all to match. What went wrong?
     
  5. Jan 1, 2018 #4

    mfb

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    Distance when? They can't even agree on this "when" because events that happen at the same time for A do not happen at the same time for B.
     
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