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New to STR, trying to confirm my understanding

  1. Jan 28, 2009 #1
    I am trying to learn STR on my own, from various real text books (not the ones for layman). I would like to check my understanding with the following thought experiment. Can someone please confirm if I am correct, or if not, what am I doing wrong? Thanks…

    Consider the following:

    An stationery observer shoots a particle to the left at a velocity of 0.9c. Immediately following this, independent of the first particle, he shoots another particle to the right at a velocity of 0.9c. So for the observer, the two particles are separating at a velocity of 1.8c. This, I think, is OK and there is no violation of the max speed postulate of STR. The two particles are simply separating apart at a speed of 1.8c and there is no information exchange between them. There is no information being sent at a velocity greater than c. The left and right particles which are carrying information from the observer, themselves are going at velocities of 0.9c, which is still less than c.

    Now consider the left particle. What is the speed of the right particle wrt left? Remember they were shot out independent of each other.

    I think from the left particle viewpoint also, the right particle is moving at a velocity of 1.8c. We do not have to use the relativistic addition of velocities because the right particle did not jump off of the left particles – the two particles are independent of each other. Again, since the right particle was shot out independent of the left, there is no information exchange between the left and right particles. So the right particle can indeed move at 1.8c wrt the left particle without violating the max speed postulate of STR.

    Am I right or did I totally make a fool of myself?

    If I am wrong, and I had to use the relativistic addition of velocities, then the right particle will be moving at a velocity of 0.99c wrt left. But now, there will be a paradox:

    The stationery observer will calculate the distance between the left and right particle in 1 sec to be 1.8c. But the left particle, in 1 sec, will calculate that the right particle is at a distance of 0.99c.
     
  2. jcsd
  3. Jan 28, 2009 #2

    JesseM

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    Science Advisor

    Yes, that's right, 1.8c is sometimes called the "closing speed" or "closing velocity" (because if the two particles were moving towards each other rather than away from each other, the observer in the middle would measure the distance between them to be closing at a rate of 1.8c)
    Relativistic velocity addition doesn't have anything to do with any physical details about the object aside from their velocities, it doesn't matter whether one jumped off the other or what. It's just a matter of transforming the coordinates of each object's path from one inertial frame to another (the velocity addition formula can be derived from the Lorentz transformation which tells you how two different frames assign space and time coordinates to the same event).
    Suppose in the stationary observer's frame, both particles start at position x=0 at time t=0. Then one second later in this frame, the left particle is at x=-0.9 light-second at t=1 second, and the right particle is at x=0.9 l.s. at t=1 s. Suppose at this moment each particle changes color, so then we can ask where and when the events of the two particles changing color happened in other frames. If we want to look at the coordinates of any event in the frame of the right particle, and we know the coordinates of this event x,t in the frame of the stationary observer, then the coordinates x',t' in the right particle's frame are given by the Lorentz transform:

    x' = gamma*(x - vt)
    t' = gamma*(t - vx/c^2)
    where gamma = [tex]1/\sqrt{1 - v^2/c^2}[/tex] and v is the velocity of the right particle in the stationary observer's frame. Since v=0.9c, gamma = 2.294

    So, for the event of the left particle changing color, which has coordinatex x=-0.9 and t=1 in the stationary frame (in units where c=1), we have:

    x' = 2.294*(-0.9 - 0.9*1) = 2.294*(-1.8) = -4.13
    t' = 2.294*(1 - 0.9*-0.9) = 2.294*(1.81) = 4.15

    And for the event of the right particle changing color, which has coordinates x=0.9 and t=1 in the stationary frame, we have:

    x' = 2.294*(0.9 - 0.9*1) = 0
    t' = 2.294*(1 - 0.9*0.9) = 2.294*(0.19) = 0.436

    So you can see that although the events of the two particles changing color happened simultaneously at t=1 in the stationary frame when the two particles were 1.8 ls apart in this frame, in the right particle's frame these events were not simultaneous, a feature of known as the relativity of simultaneity. Also, if you think in terms of time dilation, it makes sense that these events happened when they did in the right particle's frame. In the stationary frame, both particles were moving at 0.9c so clocks attached to these particles would be slowed down by a factor of [tex]\sqrt{1 - 0.9^2}[/tex] = 0.436. So, after 1 second of time in the stationary frame, each particle's clock has only ticked 0.436 seconds, so each particle's clock should read 0.436 seconds at the time it changes color. In the right particle's frame, the right particle is at rest, so its clock is ticking normally in this frame, meaning it should read 0.436 seconds at coordinate time t'=0.436 which is what was found above using the Lorentz transformation. And in this frame the left particle is moving at 1.8c/1.81 = 0.9945c, so in this frame its clock is slowed down by [tex]\sqrt{1 - 0.9945^2}[/tex] = 0.105, meaning it should take 0.436/0.105 = 4.15 seconds for its clock to tick 0.436 seconds in this frame. So, the left particle should change color at t'=4.15 seconds in this frame, which is also what was found above using the Lorentz transformation.
     
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