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A question regarding Special Relativity.

  1. Oct 29, 2011 #1
    Suppose there is a rocket that is traveling at a constant .9c starting at an initial position x1 and fires a photon every second to observers on earth. (Observers on earth do not know the velocity of the rocket)

    If 1s is the proper time [itex]\Delta[/itex][itex]\tau[/itex] then the time passed on earth between each photon being fired would be [itex]\gamma[/itex] or 2.29416s. Then dependent on the initial position x1 the time between each measured photon on earth would be the time for each photon to travel the distance d from each position x at every 2.29416s? From measuring the time between each photon that reaches earth you could calculate the positions x2...xn. But you would only be able to know the positions of where the rocket was not where it is presently.

    Assuming I understand the above, how would someone go about by calculating the positions if the rocket is accelerating?
     
  2. jcsd
  3. Oct 29, 2011 #2
    Wait a minute, if the rocket is accelerating then [itex]\Delta[/itex]t and [itex]\Delta\tau[/itex] would be changing but the observers on earth wouldn't be able to determine the relativistic time because it would be an element of the time it takes the photon to travel to earth.
     
  4. Oct 29, 2011 #3

    Dale

    Staff: Mentor

    Yes. That is always true.
     
  5. Oct 30, 2011 #4

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    You are right, the time observed from earth of the 1 photon per second coming from the rocket will take even longer than the 2.29416 seconds. It's easy to calculate using the Relativistic Doppler Factor which for comparing intervals for objects moving directly apart from each other is:

    D = √[(1+β)/(1-β)]

    For β = 0.9
    D = √[(1+0.9)/(1-0.9)]
    D = √[(1.9)/(0.1)]
    D = √19
    D = 4.3589

    That means as long as the rocket travels at the constant speed of 0.9c away from earth and continues to send out a photon every second, earth will receive them every 4.3589 seconds.

    Now with a little algebra, you can turn the above formula around to calculate β if you know D:

    β = |(1 - D2) / (1 + D2)|

    So let's see if this works:

    D = 4.3589
    β = |(1 - D2) / (1 + D2)|
    β = |(1 - 4.35892) / (1 + 4.35892)|
    β = |(1 - 19) / (1 + 19)|
    β = |-18/20|
    β = |-0.9|
    β = 0.9

    So now we know how to determine the delayed speed of the rocket as it accelerates. If we wanted to keep track of the change in position just by looking at the Doppler Shift Periods, we could integrate the calculated speed measured as each photon was received.
     
    Last edited: Oct 30, 2011
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