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Events and how they are percieved.

  1. Apr 2, 2008 #1
    Hey guys, to further my understanding of special relativity, could you help me out with this problem:

    Two spaceships A and B fly directly towards each other. Each has a velocity of magnitude c/4 relative to the Earth, in a direction perpendicular to the line from the Earth to the spaceship. Each is capable of firing a powerful laser beam with frequency, f.

    According to an observer on Earth (who receives radio notification of the event), both spaceships fire their lasers directly at each other at the same time.

    Who fired first according to an observer in:

    a) Spaceship A?

    b) Spaceship B?

    c) A third spaceship (C), which is travelling in the same direction and with the same speed as A, but which has already passed B?

    For a) I said that in A's rest frame, B is coming towards it with speed v=8c/17. So the laser pulse from A will reach B in a time,t1, where B comes to meet the pulse. t1 is shorter than the time B's pulse takes to reach A, since A is not moving relative to the pulse.

    Is this logic correct? Any maths to back it up/corrections/help with other parts would help a lot.
    Thanks in advance :-)
  2. jcsd
  3. Apr 2, 2008 #2


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    The question was who fired first in A's frame, not who will be hit by the other's pulse first. You seem to assume in your answer that both of them fire at the same time in A's frame just as they did in the Earth's frame, but that's not correct--are you familiar with the relativity of simultaneity? If the events of both ships firing are simultaneous in the Earth's frame, they cannot be simultaneous in A's frame. Imagine looking at things in the Earth's frame, and picking the point in space that's at the exact midpoint of the ships at the moment they fire (the problem doesn't say whether they're equal distances from the Earth when they fire, so this point could be out in space somewhere). Suppose there's a buoy sitting at that position, at rest relative to the Earth; since both beams travel at the same speed in the Earth's frame they must pass the buoy simultaneously. And since this all happens at a single location in space, the two beams must pass the buoy simultaneously in other frames as well.

    So in A's frame, the buoy is moving towards A at c/4 while B is moving towards A at 8c/17. Since both ships are at equal distances from the buoy in the Earth's frame when they fire, it must be true that they both fire at equal distances from the buoy in A's frame too (imagine there were rods sticking out from either side of the buoy, of the right length so that each ship fired at the same moment it passed the end of the rod; if the rods are the same length in the Earth's frame, and they are both shrunk by the same length contraction factor in A's frame, then they'll still be the same length in A's frame). But in A's frame the buoy is moving towards the position A was when it fired, while it is moving away from the position B was when it fired, so if both beams move at c and start at the same distance from the buoy, the only way they can both pass it at the same moment is if B fired earlier in A's frame so there was more time for the beam from B to catch up to the buoy.

    In general if two clocks are at rest relative to one another and synchronized in their own rest frame, and the distance between them in their rest frame is L, then in a frame where they are moving at speed v along the axis that joins them, the clock in the lead will show a time-reading that's behind the reading on the clock in the rear, by a constant amount vL/c^2. So, imagine two clocks at rest relative to Earth, and sitting at exactly the positions that the rockets are when they fire their beams. If these clocks are synchronized in Earth's frame, they must both show the same reading at the moment the ships are next to them and firing. In A's frame, the two clocks are rushing towards A at 0.25c, so the clock on A's side is in the lead and so its time must be behind the time of the clock at B's location at any given moment, including the moment when B fires (which means when that clock reaches the time when B is next to it and firing, the clock on A's side has not yet reached the same time, which is the time when A is next to it and firing). So this is another way of showing that in A's frame, B fired first.

    Of course the logic is exactly the same in B's frame, so in this frame it must be A who fired first. And if C has the same speed as A then it shares the same rest frame with A, it doesn't matter that it's at a different location, except in terms of when it sees the events (but hopefully you understand that when events happen in a given frame is not the same as when an observer in that frame sees the light from the events--see the links I gave on this thread).
    Last edited: Apr 2, 2008
  4. Apr 2, 2008 #3
    Spaceships A and B cannot be flying "directly towards each other" in directions that are simultaneously orthogonal to radial vectors from the center of the Earth unless they have collided head-on at a single point.


  5. Apr 9, 2008 #4
    Hey thanks for replies, I understand it fully now :D. ps. I don't understand your point Antenna Guy, do you mean they will have a point at which they will collide, ie. in between them ?
  6. Apr 9, 2008 #5
    If the velocity of each spaceship is perpendicular to a radial vector from the center of the Earth, each is tangent to a sphere about the Earth. As such, neither velocity can point "directly toward" another point on the same sphere (i.e. where the other spaceship is).


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