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The three clocks $A, B, C$ at the same position in spacetime at the beginning of the measurement. $A$ starts moving to the right with the velocity $v$ with respect to $B$ and $B$ starts tracing $A$ with the velocity $u$ with respect to $C$, $v>u$. By time dilation relations can be written $$t_A=t_B\sqrt{1-\frac{v^2}{c^2}}, t_B=t_C\sqrt{1-\frac{u^2}{c^2}}$$. Then one can find $$t_A=t_C\sqrt{1-\frac{v^2}{c^2}}\sqrt{1-\frac{u^2}{c^2}}$$. Using the rule of the addition of velocities, the velocity of $A$ with respect to $C$ is defined as $$w=\frac{u+v}{1+\frac{uv}{c^2}}$$ Then $$t_A=t_C\sqrt{1-\frac{w^2}{c^2}}=\frac{t_C\sqrt{1-\frac{v^2}{c^2}}\sqrt{1-\frac{u^2}{c^2}}}{1+\frac{uv}{c^2}}$$ which is different from the one obtained by chaining up the relations. What is wrong here?

I tried to understand this in the following way. But there the question is still unanswered.

We can describe the motion of $A$ in $C$(stationary) and $B$(moving) frames using the Lorentz transformations. $$t_C=\frac{t_B+\frac{ux_B}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}$$ (1). $t_B$ is dilated time seemed to a stationary observer on $C$. $x_B=vt_B$ the position of $A$ in $B$ frame. Standing on $B$ one can write $$t_B=\frac{t_A}{\sqrt{1-\frac{v^2}{c^2}}}$$ (2) and replacing $t_B$ in (1) relation is derived the expected relation between $t_C$ and $t_A$

$$t_C=\frac{t_A(1+\frac{uv}{c^2})}{\sqrt{1-\frac{u^2}{c^2}}\sqrt{1-\frac{v^2}{c^2}}}$$

The question is if $t_B$ in equations (1) and (2) are equivalent? In relation (1) it is dilated time seemed to a stationary observer on $C$ frame. In equation (2) it is the proper time of $B$ frame?

Constantin
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    These seems like a duplicate of https://physics.stackexchange.com/q/392515/ . See my reply https://physics.stackexchange.com/a/392656/148184 – robphy Nov 28 '19 at 16:34

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Your proposed chaining method is inappropriate. To understand why, consider a simple analogy. Suppose I am holding a metre rule. You are standing at some distance from me, so that the rule appears to be half its length, ie 50cm. Suppose your friend is standing the same distance from you, so that a metre rule you are holding appears 50cm to her. You cannot 'chain' these effects to conclude that my metre rule seems 25cm to your friend. If she were standing next to me, for example, my rule would seem a full metre long.

Marco Ocram
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  • If the $A, B, C$ clocks are located at the origin of the reference frames $A, B, C$ respectively. Can I chain up the time relations using the full machine of the Lorentz transformations? $t_A=\frac{t_B-\frac{vx_B}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$, $t_B=\frac{t_C-\frac{ux_C}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}$. Then question comes up to compare the readings of the clocks at the origins of their own reference frames and derive the relation between $t_A$, and $t_C$. – Constantin Nov 29 '19 at 02:52
  • I see the mistake $t_B$ measured by $A$ and $C$ are not the same. It is fascinating that in every inertial frame time seems going normally for observers of that frame and seems passing differently than their clocks when they are moving – Constantin Nov 29 '19 at 05:04