Where does this derivation of the velocity addition theorem go wrong?

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.

• You need to re-learn (or properly learn) what Lorentz contraction and time dilation is actually saying. Those formulas cannot be applied here. Commented Apr 10, 2022 at 16:49
• 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! Commented Apr 10, 2022 at 17:33
• I added a note since it looks like the focus of my question was not clear enough. Commented Apr 10, 2022 at 20:53
• Your reasoning is wrong as you substituted s2 by s1/gamma while this transformation is not valid. Commented Apr 10, 2022 at 21:05
• Actually, this should be the formula for length contraction. It does not apply here because...?!?...which...requirement...?!? is not met? Commented Apr 10, 2022 at 21:11

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.

(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.