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.
Best Answer
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.