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Sketched solution

  1. Aug 17, 2011 #1
    I would like to get some help on how to calculate a solution, I was only able to sketch a solution.

    Suppose there's two persons. One person in travelling at 0.6c and one is at fixed position. They agree to signal after 1min.

    I could only come up with a solution sketching a diagram. First [itex]t_v = \gamma t = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}= 75s[/itex]

    My diagram gave me.
    Fixed person gets signal at 120s.
    Moving person gets signal at 150s.

    I think it's right, but I like to calculate it in a proper way. If anyone could nudge me in the right direction...
     
  2. jcsd
  3. Aug 17, 2011 #2

    ghwellsjr

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    If they both agree to send the signal after one minute according to their own clock traveling with them (how else could they do it?), then they both will receive the signal from the other person at the same time (according to their own clock).

    This is an application of Relativistic Doppler which yields an answer of two minutes. So both would get the same answer that you got for the first person.

    I think you may be overlooking the fact that the traveler measures the reception of the signal from the stationary person with his time dilated clock (just like he sends it according to his time dilated clock) and so he sees 120 seconds on his clock when the signal is received.
     
  4. Aug 17, 2011 #3

    mathman

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    There is a symmetry that is being overlooked. If person A is stationary and B is moving at .6c, then the problem is being described in A's reference frame. However, in B's reference frame, B is stationary, while A is moving at .6c. Therefore in their own frames A and B must get the same answer.
     
  5. Aug 17, 2011 #4

    phinds

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    Nicely put.
     
  6. Aug 17, 2011 #5

    ghwellsjr

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    It doesn't matter which frame the problem is being described in, they both will get the same answer, as long as you do the analysis correctly, in this case, realizing that only the traveler is experiencing time dilation. The OP apparently was aware of this when calculating the delayed reaction of the traveler sending the signal back to the stationary person, but overlooked this when calculating the traveler's measurement of the stationary person's signal.

    And you don't have to limit the analysis to a frame of reference in which one of the people is at rest. For example, you could pick a FoR in which both people travel in opposite directions from a common starting point. This would be the ultimate in symmetry because both would be experiencing the same time dilation, but a correct analysis will show again that they both will measure two minutes from the time they depart until they see the signal from the other person.
     
    Last edited: Aug 18, 2011
  7. Aug 18, 2011 #6
    Well, thanks for the vigorous explanation of the different reference systems in this case. My idea was that the reference system would be the stationary. I should have mention that.
     
  8. Aug 18, 2011 #7

    ghwellsjr

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    In Special Relativity, we consider every inertial Frame of Reference to be stationary and define the positions and motions of all the observers and objects with respect to that FoR. You stated that the first person was "fixed" and the second person was "moving" with respect to the assumed stationary FoR. That was very clear. You didn't need to mention anything else.

    Do you now understand why the moving person also measures 120s?

    Have you looked up Relativistic Doppler to see how that confirms your analysis?
     
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