Enovik said:
Sorry, this reference doesn't contain any formulas
There are 29 numbered formulas in the paper.
Enovik said:
My computer doesn't open it (for safety).
Download the TeX source if you don't trust PDFs.
The rest of your post is simply nonsense. You pull random formulae out of nowhere with no justification and claim that they somehow mean something. Show what (you think is) an actual derivation, and then we can explain whatever you are misunderstanding that leads to the nonsense you keep getting.
Enovik said:
If you know another result write, please, formulas, not a row of references
In an inertial frame, an object moving inertially has equation of motion ##x=ut+x_0##. In another inertial frame it has ##x'=u't'+x'_0##. This is a definition of an inertial frame. Therefore, if there exist global inertial frames any transform between them must map straight lines to straight lines. The most general transform that does this is an affine transform. The transforms can therefore be written in matrix form$$\begin{eqnarray*}
\vec x'&=&\Lambda\vec x\\
\left(\begin{array}{c}x'\\t'\end{array}\right)
&=&
\left(\begin{array}{cc}A&B\\C&D\end{array}\right)
\left(\begin{array}{c}x\\t\end{array}\right)
\end{eqnarray*}$$where the capital letters are unknowns that may be functions of the velocity between the frames but cannot be functions of the coordinates (or the transform would not have the required properties).
Now note the following things.
1: if the velocity of the primed frame is ##v##, an object moving at velocity ##v## in the unprimed frame must have speed zero under transform. That is, ##(x,t)^T=(vT_1,T_1)^T##, where ##T_1## is an arbitrary time, must map to ##(x',t')^T=(0,T'_1)^T##.
2: Similarly, the inverse transform must map ##(x',t')^T=(-vT'_2,T'_2)^T##, where ##T'_2## is an arbitrary time, to ##(x,t)^T=(0,T_2)^T##.
3: If you set ##T_1=T'_2## then ##T'_1## must equal ##T_2##.
The first of these is the definition of "moving at velocity ##v##" for an inertial frame. The second and third are applications of the principle of relativity (Einstein's first postulate). If you work through the maths, you will find that you can express ##B##, ##C## and ##D## in terms of ##A## and ##v##.
Now add a new requirement that a composition of these transforms must be a transform too. Add a subscript ##v## to get ##\Lambda_v## whose elements are all functions of ##A_v## and ##v##, and define a transform ##\Lambda_u## whose elements are obtained from ##\Lambda_v## by changing all ##v##s to ##u##s. This is the same type of transform but to a different velocity, ##u##, Write the composition, ##\Lambda_w=\Lambda_u\Lambda_v##, which is a boost to a frame with some velocity ##w## relative to the unprimed frame. Multiply that out. You will have found that the leading diagonal components of ##\Lambda_v## are equal, so insist on this for ##\Lambda_w##. You will get a messy equality that you can tidy up so that all the ##u## and ##A_u## are on one side and all the ##v## and ##A_v## are on the other. It must be true for arbitrary ##u## and ##v##, so pick ##A## to eliminate the dependency.
You will find three ways to do this: ##A=0##, ##A=1##, and ##A=(1-Ev^2)^{-1/2}##, where ##E## is an unknown constant. The first is a spurious solution that makes the matrix singular. The second and third produce the Galilean and Lorentz transforms (albeit with an unknown constant). The sole application of the second postulate is here: the Galilean transforms do not respect it so we reject them, and the Lorentz transforms only do so if ##E=1/c^2##, so we require this.
That concludes the derivation of the Lorentz transforms, reasoning from first principles. Note that none of your formulas 1-4 will appear (because they have nothing to do with the Lorentz transforms), although 5 and 6 are one option for deducing the value of ##E##.
