Vector transformation in special relativity Please note that I am very new on this website so have some difficulties in writings as required here but trying really hard to learn quickly. La-Tex is the main problem but please understand me that I am serious.
How do Vectors transform from one inertial reference frame to another inertial reference frame in [special relativity].
A bound vector in an inertial reference frame ($x$,$ct$) has its line of action as one of the space  axis in that frame and is described by $x$*i*,then what would it be in form of new base vectors (a) and (b) in a different inertial system ($x`$,$ct`$) moving with respect to the former inertial system with $v$*i* velocity.Let (i) and (j) be the two bounded unit vectors with the line of action as co-ordinate axis($x$) and($ct$) respectively and senses in the positive side of co-ordinates and similarly (a) and (b) are defined for co-ordinates ($x`$) and ($ct`$) respectively.
 A: Well, vectors (3D vectors) don't really transform linearly. Unlike Galilean transformations, you need now know anything "extra" when transforming a vector. Here, due to the "mixing" of space and time, you do. To transform displacement, you need to know time, and vice versa. Same with energy and momentum.
Four-vectors, on the other hand, transform linearly. These are four-dimensional vectors which transform linearly via the Lorentz matrix ($\beta=\frac{v}{c},\gamma=\frac{1}{\sqrt{1-\beta^2}}$): 
$$L=\begin{bmatrix}
\gamma&-\beta \gamma&0&0\\
-\beta \gamma&\gamma&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{bmatrix}$$
For example, if you want to transform position and/or time, you use the four-position
$$X=\begin{bmatrix}
c t \\ x \\ y \\ z
\end{bmatrix}$$
this can be compactly written as $(ct,\vec x)$--this just means that you can expand the second "vector" term to get the next three four-vector components.
Anyway, the four vector transforms as:
$$X'=L\times X$$
(matrix product)
This, expanded, is:
$$\begin{bmatrix}
c t' \\ x' \\ y' \\ z'
\end{bmatrix}
=
\begin{bmatrix}
\gamma&-\beta \gamma&0&0\\
-\beta \gamma&\gamma&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{bmatrix}
\begin{bmatrix}
c\,t \\ x \\ y \\ z
\end{bmatrix} ,$$
which are your normal lorentz transformations. A property of four vectors is that if we are talking about the same four vector $(a,\vec b)$ in two frames, the value of $a^2-|\vec b|^2$ is the same in both. For four-position, you get $c^2t^2-x^2-y^2-z^2=c^2t'^2-x'^2-y'^2-z'^2$
Other four vectors are:


*

*Four-velocity: $(\gamma c,\gamma\vec u)$

*Four-momentum: $(E/c,\vec p)$

*Four-current density: $(\gamma\rho,\vec J)$

*Four-potential: $(\frac\phi{c},\vec A)$


(And a few more which I can't remember)
