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

Relativistic problem ?

  1. Jun 28, 2011 #1
    I was thinking,

    Suppose there are three boys in one room. A B C. They time there watches perfectly with each other. In each of their watches the time is 6:00. Boy A starts flying around with velocity {[3^(1/2)]/2}.c.........he flies and boys B sees that the time on A's watch is ticking slowly. But boy A thinks that it is B who is moving and so when he looks at B's clock he thinks that it is ticking slowly. The boy B and A come to rest after 30 mins.(on B's clock)...... Now they talk with each other, A and B... A says that since B was moving with high speed( As he thinks), his clock ticked slower and thats why B's clock is at 6:15.....and he is looking at B's clock ticking at 6:15. B says that "no".....U moved with high speed, my clock is at 7:00, urs is at 6:30.....B claims that since A was moving his clock slowed down.......and he is looking at A's clock and in it is 7:00 pm.........

    Mr. C comes, he observes both watches...........what is this??? what will he see??? Why is it that A , and B, aftter B's flight; are looking at the same watches and observing different times????
  2. jcsd
  3. Jun 29, 2011 #2


    User Avatar
    Science Advisor
    Gold Member

    You're only complicating things by having a third observer C who remains stationary with respect to B during the entire scenario. In Special Relativity, we select any single inertial (non-accelerating) frame of reference from which to describe, analyze and calculate what happens. The easiest frame of reference to do this with your scenario is the one in which all three boys start and end at rest. During the time that A is flying around at high speed, his clock will run slower than time as defined by the frame of reference. The clocks for B and C will also run at the normal time defined by the frame of reference. At the end when A stops near B and C, everyone will observe and agree that A's clock has accumulated less time the clocks for B and C (which read the same time). This is really a very simple problem. Why do you think it is any more complicated than what I have described?
  4. Jun 29, 2011 #3
    Here is the complication. How do u tell that it was "A" who moved with the speed i mentioned. What if "A" claims that it was the surrounding (containing the "B" and "C") that moved. Doesn't "A" have the right to say that? Isnt Motion relative?....If "A" does have the right to say that, then he also has the right to say, that it is actually the time of the surrounding which dilated, and hence, "A" has the authority to claim that the clocks of "B" & "C" have accumulated less time, and so what he sees is if his clock is at 6:30, their clocks would be at 6:15........ Whose clock is showing more time??....isnt "A" 's argument correct?
  5. Jun 29, 2011 #4


    User Avatar
    Science Advisor

    In order that there be any problem, A and B must be stationary, and at the same place, with respect to each other at both beginning and ending but in relative motion in between. That is only possible if at least one accelerated and was not in inertial motion. According to you, that was A. That breaks the symmetry.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook