Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A paradox of my own

  1. Oct 11, 2008 #1
    In trying to come up with an example in S.R., I came up with the following: 2 rockets, a distance d apart as measured in the earth frame, each have a speed of 0.8c and are on a head on collision with the other. What is the time to impact as measured in the earth frame and on one of the rockets. As first this did not seem like a difficult example, but when I tried to change the speed of one of the rockets to 0.4c, that's when the conflicts in my mind began. Here is how my original solution follows and then why it bothers me:

    In the earth frame, the point of collision is at the origin IF you start each rocket d/2 on either side. Thus, an observer on earth would measure the time to impact as t=(d/2)/0.8c, the time it takes either rocket to travel a distance of d/2 at a speed 0.8c. This yields t=0.625(d/c). To find the collision time in the rocket frame, it seems to me that I have 3 possible methods:

    1) simple time dilation
    2) lorentz transformations
    3) t=d/v in the rocket frame

    1) Using method 1, since the rocket is measuring proper time, I use t=(1/gamma)*t(in earth frame), with gamma calculated with v=0.8c. This yields a time to collision of 0.375(d/c).

    2) Using method 2, with the event coordinates as:
    S (0,0,0,0) initial
    S (d/2,0,0,0.625(d/c)) final
    S' (0,0,0,0) initial
    S' (0,0,0,t') final
    the t' L.T. equation yields 0.375(d/c) - same as method 1

    3) In finding t=d/v in the rocket frame, I used the velocity transformation equation for Vx and got -(40/41)c. Using the rocket on the left (at -d/2 earth frame), the left rocket sees the distance between the rockets as being contracted by gamma*d. Again using v=0.8c in gamma, I get a contracted length of 0.6d. Calculating d/v yields 0.615(d/c), which is not the same as before. If I use (40/41)c for gamma which I think is correct, the result is 0.225(d/c) - still not equal.

    It seems to me that all these methods should yield the same collision time for either rocket. I have not had S.R. in almost 15 years, so I am really rusty on this and hope someone more gifted in this area can see my flaw and/or mis-understanding.

    This example was modified from a homework problem I came across in which a fixed time to evacuate a ship is 1.5 hrs. By calculating the collision time in each rocket frame (one at 0.8c the other at -0.6c), is there enough time to get every one off each rocket before the collision? The book had an earth collision time of 1.67hrs, rocket A at 1.47 hrs and rocket B at 1.97 hrs. Thus, the crew of rocket A don't make it.

    However, my problem with this approach is that I would think that both rockets clocks should have collisions times slower than that of the earths? If you view this result in terms of tick rates of the clocks (R=1/t), rocket A would have a faster tick rate than the earth clock - which goes against the relativity statement that "moving clocks tick slower". It was at this point that the meltdown occurred. If anyone has a clue to the resolution of my paradox, I would greatly appreciate it. Thanks in advance.
  2. jcsd
  3. Oct 11, 2008 #2


    User Avatar
    Science Advisor

    PhysicsProf, are you familiar with the relativity of simultaneity? Suppose there are two buoys in space and at rest relative to the Earth, at a distance of d apart from one another with Earth at the midpoint. Do you understand that if in the Earth frame the events of the two rockets passing the two buoys are simultaneous, in the rocket frame these events are not simultaneous?
  4. Oct 12, 2008 #3
    Yes, I am aware of the issue of simultaneity. That does not appear to be the issue here. I am NOT trying or expecting to see the events at the same time in both frames. The fact is the rockets WILL/MUST collide at some point (but the time will not be measured as the same from the different frames). You can't have one rocket appear to explode before hitting the other. So my thinking is, since they must collide, how much time runs off each frames clock before they do?
  5. Oct 12, 2008 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Simultaneity is a an issue. You say the rockets are a distance d apart in Earth's frame. I think you means that there are two events A and B with A on one rocket's worldline and B on the other's such that in the Earths frame: A and B have a spatial separation of d; A and B have no time separation.

    Events A and B are not simultaneous in either of the rockets' frames, so Lorentz contraction cannot be applied so easily.

    Are you familiar with the equations for Lorentz transformations?
  6. Oct 12, 2008 #5


    User Avatar
    Science Advisor

    "How much time runs off each frames clock" between what event and the collision? If each rocket sets its clock to zero as it passes the buoy, for example (i.e. both read zero when they are a distance d apart in the Earth's frame) then clearly simultaneity is an issue since in the right rocket's frame it will judge the left rocket to have passed its buoy much earlier than the right rocket did. If you want each rocket to set its clock to zero at some other event on its worldline, you need to specify which one.
  7. Oct 12, 2008 #6
    I will not lie that I do not fully understand George's reply. Sorry. However, it seems JesseM's comment along with George's is starting to help unravel the problem. The original problem is that they both started at the same time (event 1) at different locations (separated by a distance d in the earth frame) and start heading toward each other and eventually collide (event 2). I was assuming from the way the original problem was stated that all 3 clocks were sync'd at t=0. If that is not the case, it was not clearly stated and I would never have assumed otherwise. It does make me think, can problems involving collisions of two or more moving objects not be solved by method 1 or 2, but only by method 3? In all the solutions of this nature in the text, they use method 3, perhaps this is why. If so, this was NOT clearly stated at all in the text nor in any of the example within the chapter. Could be the text I was told to use is not so good.

    Again, thanks for your replies. They are helping, its just slow to come back.
  8. Oct 12, 2008 #7
    Whenever you talk about "at the same time at different locations" or "all 3 clocks were synched," you must specify the reference frame you're talking about. They might both start at the same time in Earth's frame, but if so, they will NOT be starting at the same time in the rocket's frame. The clocks might all be initially synched in Earth's frame, but if so, they will NOT be synched in the rocket's frame.

    As usual, the solution to this "paradox" is the solution to all SR "paradoxes:" you have neglected to include failure of simultaneity in your analysis. A correct analysis of the situation:

    In the Earth's reference frame, each of the rockets sets its clock to 0 when passing its buoy, stationary in Earth's frame. Earth synchs its own clocks (including those of the buoys) at this same time. This is set up to occur simultaneously for both rockets. The rockets slam into one another at t=0.625d/c. The Earth observer notes the ships' clocks each slowed by a factor of gamma=5/3. So the time read out on each ship's clock is 0.375d/c.

    In the reference frame of the left-hand ship, when he passes the buoy, he sets his own clock to 0, the buoy reads 0, but no other clocks read zero. The distance to the Earth (where the collision will take place) is contracted to 3/5 (1/2 d) = 3/10 d. Thus the time for him to reach it is t' = (3/10 d)/(8/10 c) = 3/8 d/c, and this is what his clock reads when he gets there. During the trip, the time-dilated clock at the Earth has progressed by an amount (gamma)t' = 3/5 (.375 d/c) = 0.225d/c. However, the Earth's clock in fact reads 0.625d/c when he gets there, because it was not initially synched with the buoy. In fact, at the moment that the rocket pilot passed the buoy, and the buoy was reading time 0, Earth's clock was already reading time vl/c^2=(.8c)(.5d)/c^2=.4d/c. So by progressing another 0.225d/c, it will read the correct 0.625d/c when the collision occurs.

    Remaining in the reference frame of the left-hand ship (call it L), when he passes his buoy and sets his clock to 0, his buoy reads 0, but the clock on the buoy on the other side of Earth reads 0.8d/c. But the other ship was there when it read 0, which was (according to L) a time 5/3*0.8d/c= 4/3 d/c ago. The other ship is approaching at speed 4/5 [itex]\oplus[/itex] 4/5 = 40/41 c, so the observed relative velocity of the other ship and earth is [itex]v_s-v_E=\frac{40}{41} - \frac{4}{5}=\frac{36}{205}c[/itex], so it reaches Earth at a time (L's clock) of t'= t'_0 + d'/v' = -4/3 d/c + (3/10 d)/(36/205 c) = 3/8 d/c, and the ships do collide there because they are there at the same time. The other ship's clock read zero at t'=-4/3 d/c, and the ships collide at t'=3/8 d/c. So the other ship's clock, which runs slow by a gamma factor of 41/9, reads (4/3 + 3/8)(9/41) d/c = 3/8 d/c when it reaches the collision. Perfect.

    This is your problem. You are taking the Earth's reference frame and thinking of it as an absolute rest frame. There is no absolute sense in which rocket A is moving, and thus it is improper to categorically state that A's clocks are slower than the Earth's. In some reference frames they are, and in others, they are not. In rocket A's frame, in particular, Rocket A is at rest, and the Earth is moving and it is the Earth's clocks, not the rocket's, which are slow.

    Hope this helps.
  9. Oct 13, 2008 #8
    Hmm.... This will require some thinking/processing... I appreciate your detailed response, ZikZak. I do find it distressing that most texts don't do enough to explain the issue of simultaneity with examples such as this. Perhaps some text book writers don't fully understand what they are writing about. As far as assuming the earth is an absolute ref. frame, i plead guilt to that. Thanks for pointing that out.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook