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Situation: We consider two inertial systems $1$ and $2$ in standard configuration, i.e. system $2$ is moving into the direction $x_1$ with a speed of $v$. We only consider one spatial axis. An object in uniform motion is observed, from system $1$ and $2$.

System $1$ sees the object passing through a spatial distance of $s_1$ in a time difference of $t_1$ and concludes that it has a speed of $v_1 = s_1/t_1$.

System $2$ sees the object passing through a spatial distance of $s_2$ in a time difference of $t_2$ and concludes that it has a speed of $v_2 = s_2/t_2$.

Now in system $2$ spatial distances are Lorentz contracted and temporal distances are Lorentz dilatated: $\displaystyle s_2 = s_1/\gamma$ and $t_2 = \gamma t_1$ where $$\gamma = {1 \over \sqrt{1 - v^2/c^2}}$$ is the Lorentz factor. Thus:

$$v_2 = {s_2\over t_2} = {1\over \gamma^2} {s_1\over t_1} = \left(1 - {v^2 \over c^2}\right) v_1 \quad {\rm and\ thus}\quad v_1 = {v_2 \over 1 - v^2/c^2}$$

On the other hand, textbook velocity addition tells us that $$v_1 = v \oplus v_2 = {v + v_2 \over 1 + v\,v_2/c^2}$$ and, quite obviously, my derivation is wrong. This can be seen also from the fact that my "velocity formula" leads to an expression which is linear in $v_2$.

My question: Where does my analysis go wrong? Why do I get a different formula than the textbook velocity addition? I want to know what I am doing wrong here.

Note: I do understand that my analysis (obviously) is wrong and I am aware of the textbook derivation of velocity addition. I am interested in the exact point where my reasoning is wrong.

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    $\begingroup$ You need to re-learn (or properly learn) what Lorentz contraction and time dilation is actually saying. Those formulas cannot be applied here. $\endgroup$
    – Prahar
    Commented Apr 10, 2022 at 16:49
  • $\begingroup$ Just to add: time dilation and length contraction are themselves derived quantities, they are not suitable to use as a starting point because to define them in the first place you need to use the Lorentz Transform! $\endgroup$
    – m4r35n357
    Commented Apr 10, 2022 at 17:33
  • $\begingroup$ I added a note since it looks like the focus of my question was not clear enough. $\endgroup$ Commented Apr 10, 2022 at 20:53
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    $\begingroup$ Your reasoning is wrong as you substituted s2 by s1/gamma while this transformation is not valid. $\endgroup$ Commented Apr 10, 2022 at 21:05
  • $\begingroup$ Actually, this should be the formula for length contraction. It does not apply here because...?!?...which...requirement...?!? is not met? $\endgroup$ Commented Apr 10, 2022 at 21:11

2 Answers 2

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Spatial distances are not contracted and times as well. The transformation rule is: $$x_1=\gamma (x_2+vt_2)$$ $$t_1=\gamma(t_2+vx_2/c^2)$$

A velocity $v_2$ measured by the second frame system will be measured by the first frame as: $$\frac{x_1}{t_1}=\frac{x_2+vt_2}{t_2+vx_2/c^2}=\frac{v_2t_2+vt_2}{t_2+vv_2t_2/c^2}=\frac{v_2+v}{1+vv_2/c^2}$$ supposing that the particle with $v_2$ start at the origin and both systems' origin coincide. These calculations have been carried out knowing that $x_2=v_2t_2$ and then dividing the numerator and denominator by $t_2$.

The first two equations can give the length contraction and time dilation. if you think about it Look: $$x_1=\gamma (x_2)$$ if $t_2$ is zero, and $$t_1=\gamma(t_2)$$ if $x_2$ is zero. But, why does $x_1$ depend on both $x_2$ and $t_2$? In special relativity, time is also a measurable coordinate as space. Indeed if you move respect a system $S_1$, you will not "see" just everything contracted, you will "see" the future of $S_1$ in front of you and the past of $S_1$ behind you. That means, your present will be his future and past. The implications of this have many important consequences, like the fact that what an observer $S_1$ measures as simultaneous (like two bombs at different places exploding at t=0) you may not (you might measure one explosion at t=1 and another at t=4). Minkowski diagrams will help you see it a lot better. So, that means that an interval of space for $S_1$, that means $\Delta x_1\neq 0$ and $\Delta t_1 =0$ may be an interval of space $\Delta x_2 \neq 0$ and also an interval of time $\Delta t_2 \neq 0$ for $S_2$. I insist on the fact that Minkowski diagrams will help you much better.

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(Ok. Comment editing time and length is limited. For the sake of completeness, I will put down what I see as the final answer thus far in a bit broader wording. May it help the next person passing by this place.)

When substituting $s_2$ by $s_1/\gamma$ I do not claim to "invoke a transformation" but I apply a proposition stating that a measured length of an object is shorter than its proper length by the Lorentz factor.

This is where I go wrong: $s_1$ is not a proper length in system 1 but just a (normal) length in system 1. The proposition does not apply here.

The reason of the problem being that - trained as a mathematician - I expect all preconditions of a proposition to be stated precisely in that proposition and not being dispersed somewhere in the explanation. All the texts I consulted on this actually did have that restriction somewhere in the same chapter but not as part of the "proposition". That is the reason I missed the limitation in the applicability of the formula.

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