 # 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 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. easternerjx 2022-09-17 Answered
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 ($\mathbf{a}$) and ($\mathbf{b}$) 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 ($\mathbf{a}$) and ($\mathbf{b}$) are defined for co-ordinates ($x‘$) and ($ct‘$) respectively.
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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 ($\beta =\frac{v}{c},\gamma =\frac{1}{\sqrt{1-{\beta }^{2}}}$):
$L=\left[\begin{array}{cccc}\gamma & -\beta \gamma & 0& 0\\ -\beta \gamma & \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$
For example, if you want to transform position and/or time, you use the four-position
$X=\left[\begin{array}{c}ct\\ x\\ y\\ z\end{array}\right]$
this can be compactly written as $\left(ct,\stackrel{\to }{x}\right)$-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:
${X}^{\prime }=L×X$
(matrix product)
This, expanded, is:
$\left[\begin{array}{c}c{t}^{\prime }\\ {x}^{\prime }\\ {y}^{\prime }\\ {z}^{\prime }\end{array}\right]=\left[\begin{array}{cccc}\gamma & -\beta \gamma & 0& 0\\ -\beta \gamma & \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}c\phantom{\rule{thinmathspace}{0ex}}t\\ x\\ y\\ z\end{array}\right],$
which are your normal lorentz transformations. A property of four vectors is that if we are talking about the same four vector $\left(a,\stackrel{\to }{b}\right)$ in two frames, the value of ${a}^{2}-|\stackrel{\to }{b}{|}^{2}$ is the same in both. For four-position, you get ${c}^{2}{t}^{2}-{x}^{2}-{y}^{2}-{z}^{2}={c}^{2}{t}^{\prime 2}-{x}^{\prime 2}-{y}^{\prime 2}-{z}^{\prime 2}$
Other four vectors are:
Four-velocity: $\left(\gamma c,\gamma \stackrel{\to }{u}\right)$
Four-momentum: $\left(E/c,\stackrel{\to }{p}\right)$
Four-current density: $\left(\gamma \rho ,\stackrel{\to }{J}\right)$
Four-potential: $\left(\frac{\varphi }{c},\stackrel{\to }{A}\right)$

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