I am trying to understand this apparent "paradoxes" but probably i am missing something important.(adsbygoogle = window.adsbygoogle || []).push({});

Imagine that accourding to stationary observer on earth two twin cats are moving in the opposite directions with speed [itex]-v[/itex] and [itex]v[/itex]. When the two cats meet the stationary observer at the begining [itex]O[/itex] of his coordinate system [itex]K_{tx}[/itex] their clocks are synchronized. Left cat has coordinate system [itex]K^{\prime}_{t^{\prime}x^{\prime}}[/itex] right cat [itex]K^{\prime\prime}_{t^{\prime\prime}x^{\prime\prime}}[/itex].So at the center [itex]t_O=t^{\prime}_O=t^{\prime\prime}_O=0[/itex] and [itex]x_O=x^{\prime}_O=x^{\prime\prime}_O=0[/itex].

Now accourding to the stationary observer the two twin cats both travel [itex]T[/itex] until they reach points [itex]x_{-S}=-vT[/itex] and [itex]x_{S}=vT[/itex] and then they go back and meet at the center.

Let [tex]\gamma_x=\frac{1}{\sqrt{1-\frac{x^2}{c^2}}}[/tex]

Now lets take the viewpoint of the left cat.To find the time at witch the point [itex]x_{-s}[/itex] reaches it we use the Lorentz transformation:

[tex]t^\prime_{-S}-t^\prime_{O}=\gamma_v(t_{-S}-t_{O}+\frac{vx_{-S}}{c^2}-\frac{vx_{-S}}{c^2})=\gamma_vT [/tex]

Since the point [itex]x_{-S}[/itex] stays stationary accourding to the unprimed frame.

Using the same calculation for the rigth cat the point [itex]x_S[/itex] reaches it at time [itex] t^{\prime\prime}_S=\gamma_vT[/itex]

Now accourding to the left cat the right cat is moving with speed [itex]w=\frac{2v}{1+\frac{v^2}{c^2}}[/itex]. Now using the lorentz transformations again we can find that the right cat moves from from [itex]x_O[/itex] to [itex]x_S[/itex] accourding to the left cat for time [itex]t_S^\prime-t^\prime_O=t_S^\prime=\gamma_wt^{\prime\prime}_{S}=\gamma_w\gamma_vT[/itex].

And doing the same thing the left cat moves from [itex]x_{O}[/itex] to [itex]x_{-S}[/itex] accourding to the right cat for time [itex]t_{-S}^{\prime\prime}-t^{\prime\prime}_O=t_{-S}^{\prime\prime}=\gamma_w\gamma_vT[/itex].

Now if we do the same thing for the reverse direction at the end we will find the exactly same thing. Each cat thinks that the other is younger at the end of their path. But their situation is symmetric and they actualy did exactly the same thing. Shouldnt they age exactly the same at the end?

How can I resolve mathematicaly this disagreement on which cat is younger when from symmetry viewpoint they should be the same age?

I will appreciate any help. Thanks.

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# Symmetric movement with respect to "stationary" observer.

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