Relativity, What actually happens if the Lorentz transformation. Lorentz boost [closed]

Question

If we look the relations of Lorentz Boost; $$t'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} (t-\frac{vx}{c^2})$$ $$x'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} (x-vt)$$ $$y'=y$$ $$z'=z$$

we can concentrate on 2D case, and as $$v=\frac{ax}{bt}$$ and $$v'=\frac{ax'}{bt'}$$ Here $a$ is some certain numeric amount of length, and $b$ some certain numeric amount of time, so that $0<v=\frac{a}{b}<c$, ofcource the $x$ and $t$ could have been directly such amounts. By using $a$ and $b$ it can be prevented that something in the calculation becomes "Geometrized".

then we can write; $$t'=\frac{1}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} (t-\frac{(\frac{ax}{bt})x}{c^2})$$ $$x'=\frac{1}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} (x-(\frac{ax}{bt})t)$$

And these Reduce a bit to; $$t'=\frac{1}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} (t-\frac{\frac{ax^2}{bt}}{c^2})$$ $$x'=\frac{1}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} (x-\frac{ax}{b})$$

Wo when these are combined to $v'$ then $$v'=\frac{\frac{a}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} (x-\frac{ax}{b})}{\frac{b}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} (t-\frac{\frac{ax^2}{bt}}{c^2})}$$

Which can be written; $$v'=\frac{a(x-\frac{ax}{b})}{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}} \frac{\sqrt{1-\frac{(\frac{ax}{bt})^2}{c^2}}}{b(t-\frac{\frac{ax^2}{bt}}{c^2}) }$$

And this reduces quickly just to; $$v'=\frac{a(x-\frac{ax}{b})}{1} \frac{1}{b(t-\frac{\frac{ax^2}{bt}}{c^2}) }=\frac{a(x-\frac{ax}{b})}{b(t-\frac{ax^2}{btc^2})}$$

And $$v'=\frac{ax-\frac{a^2x}{b}}{bt-\frac{bax^2}{btc^2}}=\frac{ax-\frac{a^2x}{b}}{bt-\frac{ax^2}{tc^2}}$$

So, this would be the Lorentz -Transformation for velocity; $$v'=\frac{ax-\frac{a^2x}{b}}{bt-\frac{ax^2}{tc^2}}$$

Question; Are these kind of higher level Lorentz Transformation been properly analysed before? (Source)
1. Velocity
2. Acceleration
3. Jerk
4. Momentum

Answer 1

The problem with your derivation starts before you attempt to generalize by defining speed as $$v = \frac{x}{t}.$$ This is incorrect because $x$ and $t$ are coordinates, not lengths or durations. In your final expression, the velocity is a function of time, which shouldn't be the case if the velocities are constant. The proper definition of velocity is $$v = \frac{x_2 - x_1}{t_2 - t_1}$$ or, for instantaneous velocity, $$v = \frac{dx}{dt}.$$

So, the proper procedure to find the velocity transformation (well, one procedure), is to define an object in the primed frame traveling at a constant velocity $v'$ with respect to that frame. $$v' = \frac{dx'}{dt'}.$$ Then, you can solve for the speed in the unprimed frame by solving the Lorentz boost equations for the unprimed $x$ and $t$ in terms of their primed coordinates and calculating $$v = \frac{dx}{dt}.$$ Note that, due to the chain rule, you'll need the expression for $$\frac{dt'}{dt},$$ which is the time dilation factor.