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Special relativity question with lots of frames

  1. May 11, 2016 #1
    1. The problem statement, all variables and given/known data
    At exactly 00:00:00 hours, a group of convicts escape from a planet in a space-ship that travels at speed
    ##v=\frac{4}{5}c##.
    After 11 min, a patrol spaceship goes after them with ##v_P=\frac{24}{25}c##.
    Ignore all acceleration periods.
    (i) The convicts immediately notice the patrol spaceship taking off, and release a stealth missile to destroy it. The missile has speed
    ##v_M=\frac{40}{41}c##
    with respect to the convicts. Show that, when the missile hits the patrol ship, the time on a watch worn by the pilot of the patrol ship is 00 : 23 : 36.

    2. Relevant equations
    ##u_x' = \frac{u_x-v}{1-\frac{u_xv}{c^2}}## (1)
    ##t' = \gamma (t - \frac{vx}{c^2})## (2)
    3. The attempt at a solution
    I should say this is from a past paper, so it won't get marked. There are way too many frames! I'm going to assume that all the speeds given are 'proper' speeds.

    ##u_x=\frac{40}{41}##, I think v is the speed of one frame relative to the other so that might be ##\frac{24}{25}-(-\frac{40}{41}) = \frac{1984}{1025}##. Then I realised that ##u_x'## might not be useful anyway, I was going to just use speed = distance/time, but can't do that because I don't know distances. And (2) probably isn't useful either. Do I try and work out at what time in the ship's frame the missile hits, and transform that? But then do I need to transform the 40/41 to see what the missile's speed is in the ship's frame? Don't I need distances for this approach as well?
     
  2. jcsd
  3. May 11, 2016 #2

    Orodruin

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    There is no such thing. It is very well stated in the problem what the speeds are relative to.

    What is not clear from the problem is what this means:
    Now, there is only one physically sound interpretation of this, but I am not sure that this is what the problem writer intended.

    Also, you do not need to consider more than one frame, you simply need to track what happens in one frame and possibly use concepts of relative simultaneity and so on.
     
  4. May 11, 2016 #3
    Oops, didn't read it properly, clearly. All speeds relative to the planet. OK, I'll try sticking to the frame of the planet. I thought proper times, length etc were in the rest frame of an object?
     
  5. May 11, 2016 #4

    Orodruin

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    No, all velocities are not given in the planet rest frame, but the velocities in the planet rest frame can be easily figured out using relativistic velocity addition.

    Yes, but these are invariant quantities. There is no meaningful way of defining a proper velocity. If you take the velocity of an object in its rest frame it is zero by definition of the rest frame.
     
  6. May 11, 2016 #5
    OK. So are the convict ship and spaceship velocities given in the planet's rest frame then? Just the missile one that isn't? If not I really don't understand the question.
     
  7. May 11, 2016 #6
    My suggestion would be to calculate the speed of the missile relative to the patrol ship -
    at what speed does the patrol ship see the approaching missile?
    Then you would need the distance between the two ships after 11 minutes -
    this would be relativistically contracted when the patrol ship takes off (you know how far the convict's ship
    has traveled relative to earth).
    Now if you have this information, speed and distance you should be able
    to calculate the time for the missile to reach the patrol ship.
     
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