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Doubt on FRAME of REF(Its killing me !Help)

  1. Nov 26, 2007 #1
    Doubt on FRAME of REF(Its killing me!!!!!!Help)

    Hello Guys
    I am really fascinated by Theory of Relativity and although its not in my educational syllabus(not a phy scholar) , I really study a lot about it. Now coming to the point , I have a doubt on FoR and it may be basic, so I hope you fellas can clear it out.

    Talking about TWIN paradox, lets assume Mr.STILL remains on earth while his brother ,Mr.MOVE moves in a spaceship at very high speed(say 87% of light). From my understanding this is what will happen from Mr.STILL's point of view(or Frame of Ref):
    Mr.MOVE is moving at 87% of Light's speed so time will slow down for Mr.MOVE. Mr. STILL waits and waits and after 6 days his brother ,Mr.MOVE returns after taking a trip at .87c. Mr.STILL will predict this:Since I am stationary , and my brother is moving at .87c, time will slow down for Mr.MOVE(my brother). At 87% of light , time slows by a factor of 2. Now my clock says that I have spent 6 days ,waiting. So when he will return ,Mr.MOVE won't feel as if he spent 6 days, infact he will think that he spent just 3 days up there.
    From Mr.MOVE's point of view:

    Now assume that Mr.MOVE don't know that he is Motion and thinks he is still, so he will see Mr.STILL moving at .87c in opposite direction(along with earth). After waiting for 3 days,as his clock will tell him(or will it be 3 days? I am confused) he will see Mr.STILL and earth returning.So this is what Mr.MOVE will predict:Since I am stationary , and my brother (Mr.STILL)is moving at .87c time will slow down for Mr.STILL(my brother). At 87% of light , time slows by a factor of 2. Now my clock says that I have spent 3 days ,waiting. So when he will return ,Mr.STILL won't feel as if he spent 3 days, infact he will think that he spent just 1.5days down there on earth.

    From my understanding, both are right since laws of physics is same in every frame of reference.And both are relativly in motion with each other so both can assume taht they are stationary. Now both Mr.STILL and Mr.MOVE thinks that Mr.MOVE has spent 3 days in ship(Mr.STILL does it by calculation while Mr.MOVE is told by his clock). But....... According to Mr.STILL he has spent 6 days on earth(clock will say) while Mr.MOVE tell by his calculations that Mr.STILL will spend only 1.5 days in earth and not 6.

    But both can't be true, Mr.STILL has spent a definite time up there.Where is the assumption wrong?. Please tell me.Its killing me.
  2. jcsd
  3. Nov 26, 2007 #2

    D H

    Staff: Mentor

    Special relativity only pertains to observers in inertial frames. That is why it is called "special", meaning "of restricted applicability". Your Mr. STILL is in a nearly inertial frame (he is accelerating due to the Earth's orbit around the Sun, let's ignore that minor detail) while you Mr. MOVE is not. He has to accelerate four times: once to get started, a second time to come to a stop at the target (this is still acceleration to a physicist), a third time to get started back home, and one last time to come to a stop when he reaches home. Those periods of acceleration make the problem non-symmetric and remove the apparent paradox.
  4. Nov 26, 2007 #3


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    No, the usual equations of SR, including the equation for time dilation, only work in inertial frames of reference, i.e. the frame of reference of someone who moves with constant speed and direction (you'll know if you change speed or direction, i.e. accelerate, because you'll feel G-forces). Mr.MOVE has to turn around in order to reunite with Mr.STILL, so he does not remain in a single inertial frame, and cannot apply the laws of SR in the same way. You are free to analyze the situation from the point of view of a different inertial frame where Mr.STILL is moving rather than still, but all inertial frames end up predicting that Mr.MOVE ages less over the course of the entire journey.

    For example, you could pick a frame where Mr.MOVE was at rest during the outbound leg of the journey before turning around, and Mr.MOVE was traveling at 0.87c; but then after Mr.MOVE turns around, Mr.STILL continues to travel at 0.87c in this frame while Mr.MOVE has an even higher speed (it works out to 0.9897c in this frame, using the formula for addition of relativistic velocities). So on the inbound leg after Mr.MOVE turns around, he is aging more slowly than Mr.STILL in this frame, and the inbound leg lasts for more time than the outbound leg in this frame--the outbound leg lasts 1.5 days in this frame while the inbound leg lasts 10.5 days, and Mr.STILL's aging is slowed down by a factor of 2 throughout the journey in this frame, while Mr.MOVE ages normally during the outbound leg since he's at rest in this frame, then ages at [tex]\sqrt{1 - 0.9897^2}[/tex] = 0.143 the normal rate during the inbound leg, meaning Mr.MOVE is predicted to age a total of (1)*(1.5) + (0.143)*(10.5) = 1.5 + 1.5 = 3 days in this frame, the same amount he's predicted to age in Mr.STILL's rest frame.

    For more details on the twin paradox I recommend this site:

  5. Nov 27, 2007 #4
    In a nutshell, because the SPACE-SHIP is accelerated (4 times) it would be wrong to apply SR from SPACE-SHIP. Is it?
  6. Nov 27, 2007 #5
    Can you please clarify what is INBOUND and OUT-BOUND leg of the journey?

    I'm new to Relativity , so forgive my noobness.
  7. Nov 27, 2007 #6


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    They're not really physics terms, by "outbound" I just mean the segment of the trip where Mr.MOVE is moving out away from the Earth at constant velocity (from the moment of departure until the turnaround), and by "inbound" I mean the segment of the trip where he's moving back in towards the Earth at constant velocity (from the turnaround until the arrival on Earth). Sorry if that was unclear, "inbound" and "outbound" are often used for things like airline flights (taking off from an airport vs. landing there) and subways (heading towards the center of the city or away from it).

    In the frame where Mr.MOVE is at rest during the outbound leg of the trip, during the outound leg Mr.STILL is moving at 0.866c, so the distance between them is increasing at a rate of 0.866c, and it takes 1.5 days for the distance to increase to (0.866 light-days/day)*(1.5 days) = 1.299 light-days, the distance between them at the moment of the turnaround in this frame (I'm assuming the turnaround itself is very brief so the distance doesn't have time to change noticeably during the turnaround). Then on the inbound leg Mr.STILL continues to move at 0.866c in this frame, while Mr.MOVE is now moving in the same direction at 0.9897c, so the distance between them is only decreasing at a rate of 0.9897c - 0.866c = 0.124c, therefore it takes (1.299 light-days)/(0.124 light-days/day) = 10.5 days for the distance between them to shrink to zero in this frame, a considerably longer span of time. Of course, in the rest frame of Mr.STILL, both the inbound leg and the outbound leg take the same amount of time, 3 days each.
    Last edited: Nov 27, 2007
  8. Nov 27, 2007 #7

    I would like to thank a lot to you. You cleared my doubts completly. Thanks a lot.
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