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I'm trying to derive the relativistic velocity addition equation using the time dilation equations but I get a wrong result. Assume the widely used scenario in which there are 3 clocks: O, A and B. A is moving relative to O at velocity $v$. B is moving relative to A at velocity $w$. Find the velocity of B relative to O (in O's frame of reference). The time dilation equations says:

$$t_A = t_O \sqrt{1-v^2/c^2}$$

$$t_B = t_A \sqrt{1-w^2/c^2}$$

$$t_B = t_O \sqrt{1-u^2/c^2}$$

From these 3 equations we want to derive $u$. Then we get:

$$t_O \sqrt{1-u^2/c^2} = t_O \sqrt{1-v^2/c^2} \sqrt{1-w^2/c^2}$$

$$1-u^2/c^2 = (1-v^2/c^2) (1-w^2/c^2)$$

$$1-u^2/c^2 = 1-v^2/c^2-w^2/c^2+v^2 w^2/c^4$$

$$u^2 = v^2 + w^2 - v^2 w^2/c^2$$

This is different from the equation

$$u = \frac{v + w}{1+v w/c^2}$$

Why doesn't the derivation of velocity using the time dilation equations alone work?

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  • $\begingroup$ Welcome to Physics! Note that we have an equation editor built into the site which helps for readability of your post. $\endgroup$ – Kyle Kanos Mar 15 '18 at 21:15
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    $\begingroup$ I think the problem here is remembering exactly what it meant by these time dilation equations. How about thinking about this little puzzler?: If a time $t_0$ goes by in O's frame then a time $t_A=t_0 \sqrt{1-v^2/c^2}$ has gone by in A's frame, right? But from A's perspective when that same time $t_A$ has gone by on his clock, then according to his use of the time dilation equation a time $t_0=t_A \sqrt{1-v^2/c^2}$ should have gone by on O's clock. But then if you look at these equations and try to combine them, they're incompatible with each other for any v except v=0 ! What went wrong? $\endgroup$ – Samuel Weir Mar 15 '18 at 22:28
  • $\begingroup$ yes, they are incompatible, there should be asymetry in order for them to work; and actually the only case which i see for the equations to work, is when let's say observer A accelerated related to observer O and somehow his state changed; if both accelerates then equations are not working anymore $\endgroup$ – Andrei Mar 16 '18 at 21:34
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The clocks on ships A and B are not just running slower compared to clock O. They are also offset from each other due to their different locations. If you look at the Lorentz transform for the time coordinate $$t' = \gamma\left(t - \frac{xv}{c^2}\right),$$ you can see that the distance between the clocks is also a factor in how much the clocks differ in their readings. Since clocks A and B are moving at difference speeds, they must be at different locations, so the difference in time is not just due to a different running speed, but a difference in starting time (when the clocks would read zero). This is related to the relativity of simultaneity.

In short, relativity means you can't treat space and time separately, as a change in motion through one affects your motion through the other.

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    $\begingroup$ Thanks, this means that I used incomplete equations, since they are valid only in the case the comparison of time is performed at the same location in space (x = 0) $\endgroup$ – Andrei Mar 16 '18 at 22:01
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Your first derived equation implies that $\gamma_u=\gamma_v\gamma_w$, which is not correct (time-dilation factors are not multiplicative). Instead, the correct relation [dropping the $c$'s] is $$\gamma_u=\gamma_v\gamma_w(1+vw).$$ This is most easily seen using rapidities [Minkowski-angles] $\theta$, where $v=\tanh\theta$.
I will modify your notation for clarity.
$v=\tanh\theta_{AO}$ and $\gamma_v=\cosh\theta_{AO}$ (where $\theta_{AO}$ is the rapidity from $O$ to $A$).
$w=\tanh\theta_{BA}$ and $\gamma_w=\cosh\theta_{BA}$
$u=\tanh\theta_{BO}$ and $\gamma_u=\cosh\theta_{BO}$
Since rapidities add, we have $\theta_{BO}=\theta_{BA}+\theta_{AO}$.
Thus,
$\begin{align}\gamma_u=\cosh\theta_{BO}&=\cosh(\theta_{BA}+\theta_{AO})\\ &=\cosh\theta_{BA}\cosh\theta_{AO}+\sinh\theta_{BA}\sinh\theta_{AO}\\ &=\cosh\theta_{BA}\cosh\theta_{AO}(1+\tanh\theta_{BA}\tanh\theta_{AO})\\ &=\gamma_w\gamma_v(1+wv)\\ \end{align}$


So, your first derived equation is incorrect. To see why that is, it would be good to draw a spacetime diagram. I'll draw one using rotated graph paper so that we can visualize the ticks along the segments.

Velocity Composition on Rotated Graph Paper

Your first time-dilation equation can be written as $t_O=t_A \gamma_v=t_A \cosh\theta_{AO}$,
which says that
the adjacent leg $t_O$ is equal to the hypotenuse $t_A$ times $\cosh\theta_{AO}$ in a Minkowski-right-triangle. Furthermore, the event at $t_O$ along O's worldline is what he says is "simultaneous with the distant event at $t_A$ along Alice's worldline"... so maybe call it $t_{OsimA}$].
Similarly,
the adjacent leg $t_A$ is equal to the hypotenuse $t_B$ times $\cosh\theta_{BA}$, and
the adjacent leg $t_O$ is equal to the hypotenuse $t_B$ times $\cosh\theta_{BO}$.
But, from the diagram, if the first two are true, the last one can't be true since the event at $t_O$ on O's worldline can't be simultaneous with both $t_A$ on A's worldline and $t_B$ on B's worldline. Simultaneity isn't generally transitive.

So, you need to do some work to get the correct third time-dilation equation. The spacetime diagram might help.


By the way, since the rapidities add, we have
$\begin{align}u=\tanh\theta_{BO}&=\tanh(\theta_{BA}+\theta_{AO})\\ &=\frac{\tanh\theta_{BA}+\tanh\theta_{AO}}{1+\tanh\theta_{BA}\tanh\theta_{AO}}\\ &=\frac{w+v}{1+wv}\\ \end{align}$
which is the velocity-composition formula you seek.
In the diagram shown, $v=(3/5)$, $w=(5/13)$, and $u=(16/20)=(4/5)$.
so, the time-dilation factors are [adjacent/hypotenuse]:
$\gamma_v=(10/8)=(5/4)$, $\gamma_w=(13/12)$, and $\gamma_u=(20/12)=(5/3)$.

You can verify the velocity composition formula:
$\begin{align}u&=\frac{w+v}{1+wv}\\ \left(\frac{4}{5}\right)&=\frac{\left(\frac{5}{13}\right)+\left(\frac{3}{5}\right)}{1+\left(\frac{5}{13}\right)\left(\frac{3}{5}\right)}\\ \end{align}$
and the time-dilation factor relation:
$\begin{align}\gamma_u&=\gamma_w\gamma_v(1+wv)\\ \left(\frac{5}{3}\right)&=\left(\frac{13}{12}\right)\left(\frac{5}{4}\right)\left(1+\left(\frac{5}{13}\right)\left(\frac{3}{5}\right)\right) \end{align}$


In addition, a multiplicative relation does hold for the doppler factors $k_v=\sqrt{\frac{1+v}{1-v}}$, which is equal to the exponential function of the rapidity:
$\begin{align}k_u=\exp\theta_{BO}&=\exp(\theta_{BA}+\theta_{AO})\\ &=\exp\theta_{BA}\exp\theta_{AO}\\ &=k_w k_v\\ \end{align}$
you can express this in terms of the velocities... do some algebra and get the velocity-composition formula. One develops this formula using a light ray (as shown) and relating the elapsed times from the common meeting event. (This is in the spirit of the Bondi k-calculus.)

Doppler factors on Rotated Graph Paper

In terms of the velocities in the diagram, the doppler factors are [reception-leg/emission-leg]:
$k_v=(8/4)=2$, $k_w=(12/8)=(3/2)$, and $k_u=(12/4)=(3)$.

You can verify the Doppler-factor relation:
$\begin{align}k_u&=k_w k_v\\ (3)&=\left(\frac{3}{2}\right)(2) \end{align}$

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