What is the apparent time of a clock moving x-wise at 0.5c for an observer moving -x-wise at 0.5c?

Question

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?

Answer 1

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 $v~=~dx/dt$.

Answer 2

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!