New to special relativity, I am trying to understand and compute an example of the paradox of symmetrical time dilation explained in http://en.wikipedia.org/wiki/Minkowski_diagram, that "A second observer, having moved together with the clock from O to B, will argue that the other clock has reached only C until this moment and therefore this clock runs slower." I've tried to check for this already being answered, but couldn't find quite the same problem. Sorry if I missed it...

If two observers move in opposite directions at the same speed (say 0.5c) from a resting observer, how will the clock of the symmetrical system seem to tick for the opposite observer ?

All observers start at t0=0, x=0. I apply c=1 At times resp t1=0,5 and t2=1, the positions of the two moving obs are + and - 0,25 and 0,5 in the resting frame of ref. When I compute their position and time in their own frames, I logically obtain: x'=0 at all times (observers at rest in own ref), t'1=0,433 ; t'2=0,86. Their clocks run slower than the clock in rest frame. All fine (i think).

So now I proceed to compute the relative speed of one of the moving observers with respect to the other one. I get 0,8c, which seems to be correct.

Subsequently, I apply Lorentz transform once more to compute the x''s and t''s of the observer moving at 0,8c in the frame of ref of its companion, using [x'1,t'1], and [x'2,t'2]. I obtain x''1 = 0,577; t''1 = 0,72 x''2 = 1,154; t''2 = 1,44

Being new to relativity, I would have expected the clock of the observer moving away at 0.8c to be slower, but it doesn't seem to work that way...

Moreover, I checked the Minkowsky invariants, and they are invariant, so I surmise that my calculations could be right, but that my brains don't interpret the result correctly.

Where is my mistake? In the way I apply the Lorentz transform, or in my interpretation of the result?


Use the velocity addition formula $$ v'~=~\frac{v~+~u}{1~+~\frac{uv}{c^2}}. $$ which is for $u~=~v~=~.5c$ gives $v'~=~.8c$.

You can derive this for $$ dx'~=~\gamma(dx~+~udt),~dt'~=~\gamma(dt~+~(u/c^2)dx), $$ and compute $v'~=~dx'/dt'$ $$ \frac{dx'}{dt'}~=~\frac{dx~+~udt}{dt~+~(u/c^2)dx} $$ $$ =~\frac{dx/dt~+~v}{1~+~(u/c^2)dx/dt}. $$ you get the result for $v~=~dx/dt$.


Your calculations are correct. Now think about what they mean.

Suppose there are mileposts located (according to the earthbound observer) at half-mile intervals, starting at the origin.

$E$ is the event "Traveler A reaches the first milepost to the right of the origin."

$F$ is the event "Traveler B reaches the first milepost to the left of the origin".

According to the earthbound observer, both events take place at time $.5$.

According to traveler A, event $E$ takes place at time $.433$. According to traveler A, event F takes place at time $.722$.

By the same calculations, according to traveler B, event $F$ takes place at time $.433$ --- which is what his clock says when he passes that milepost.

Now how does traveler $A$ describe event $F$? He says "Event $F$ takes place at time $.722$. At that event, traveler B's clock reads only $.433.$! His clock seems to be running at about 6/10 of normal speed!

  • $\begingroup$ Dear WillO. Thanks to you I realized my mistake. According to what I thought, .722 should have been event F on B's clock since I had applied LT to A's data about B! And that couldn't be, since in A's ref frame, B is travelling and A is resting... so t1'' should be <t1'=.433. My mistake: I used incorrect xB' . Since the relative speed of B in A's frame is -.8, xB1'=-0.8*.443=-.346 . When applying LT to [t1';xB1']=[.433; .346], [t1'';xB1'']=[.260; 0] which seems correct now, since B is at rest in her own ref, and her clock seems slower to A. You helped me a lot to dig deeper ! $\endgroup$ – Michel Sougi Aug 11 '16 at 0:08
  • $\begingroup$ I'm glad this was helpful and that the question was not closed (as four people seem to think it should have been). There are a great many questions about elementary relativity from people who have made no serious effort, and sometimes people jump to the conclusion that all questions about SR are in that category. You had obviously thought hard, and you were confused on a very specific point which has probably been addressed elsewhere, but I don't see how you could have successfully searched for the question that addresses it. $\endgroup$ – WillO Aug 11 '16 at 1:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.