This seemed at first glance very easy. But there appeared some confusion.

$A$ is moving to the right with velocity $v$ with respect to $B$. The proper time for $A$ is $$t_a=t_b\sqrt{1-v^2/c^2}$$

and $B$ is moving to the right with velocity $u$ with respect to $C$. Proper time for $B$ is


Now, $t_a$ can be found by $t_c$


Further, by using the law of addition of relativistic velocities one can find the relative velocity of $A$ with respect to $C$


And defining the proper time for $A$ by $w$ I found


which is different from the previous one.

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 $t_A$


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?


$t_a=t_c\sqrt{1-u^2/c^2}\sqrt{1-v^2/c^2}$ is incorrect. The proper time formula can't be chained up like that. The reason is a little subtle - it's because $A$ isn't in the same position at the start than when $t_B$ has passed in the $B$ frame. Because of that, you need the full machinery of the Lorentz transformation to go from $B$ to $C$. If you chain up two Lorentz transformations and look at the constant of proportionality for relating $t_A$ to $t_C$ you'll get what you derived using addition of velocities.

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  • $\begingroup$ @Constantin See sentence 3 of my answer. $\endgroup$ – Sean E. Lake Nov 27 '19 at 12:28
  • $\begingroup$ Is $t_b=t_c\sqrt{1-u^2/c^2}$ correct in general if the clocks start ticking instantaneously in both frames? $\endgroup$ – Constantin Nov 27 '19 at 12:39
  • $\begingroup$ @Constantin But the clock that measures proper time in $A$ isn't in $B$, it's in $A$. Because of that, it is moving through $B$, so no single clock in $B$ can measure any passage of time in $A$ without some notion of events at different locations being simultaneous in an absolute sense, and relativity throws that right out the window. That formula only applies to transforming the ticks of a clock that is stationary in one frame to a frame where it's moving. $\endgroup$ – Sean E. Lake Nov 28 '19 at 11:47
  • $\begingroup$ @Constantin I recommend watching this playlist, especially the Lorentz transform video and the length contraction/time dilation video. Bottom line: the formula you want to depend on is a special case that only works sometimes. $\endgroup$ – Sean E. Lake Nov 28 '19 at 11:51
  • $\begingroup$ $t_B=\frac{t_A+x_A\frac{v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$ and $t_C=\frac{t_B+x_B\frac{v}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}$ assuming three inertial frames at the same position at the beginning and $v>u$. $A$ and $B$ start moving. Both of them(clocks) at the origin of their inertial frames. Then $x_A=0$ , $t_B=\frac{t_A}{\sqrt{1-\frac{v^2}{c^2}}}$. The Clock $B$ at the origin of $B$ frame $x_B=0$, $t_C=\frac{t_B}{\sqrt{1-\frac{u^2}{c^2}}}$. $\endgroup$ – Constantin Nov 28 '19 at 12:06

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