# Vector transformation in special relativity

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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.

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By a Lorentz transformation, $V\to \Lambda\cdot V$ where dot is the matrix product. Vectors transform in the same way as $x^\mu$ which is one example of a vector. –  Luboš Motl Nov 27 '12 at 14:04
I'm really not clear what you mean when you say "line of action." Do you mean $\mathbf i$ is tangent (parallel) to the $x$-axis, and so on? –  Muphrid Nov 27 '12 at 16:33
First of all $i$ is a bound vector and it is not only parallel but in the same $x$ axis. –  Abhinav Anand Nov 27 '12 at 16:55
You're saying $\mathbf i$ is not only parallel to the $x$-axis, but also it lies along the $x$-axis (as opposed to being parallel but located somewhere other than along the axis)? –  Muphrid Nov 27 '12 at 17:36
Did not understand you at all. –  Abhinav Anand Nov 27 '12 at 17:40

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)

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I think it might be useful to clarify your first sentence. I think by "vectors" you were referring to "three-vectors", but it's common usage to call four-vectors just "vectors" in relativity. Vectors are just things living in a tangent space and they transform linearly. –  twistor59 Nov 27 '12 at 14:55
@twistor59: Clarified, thanks :) –  Manishearth Nov 27 '12 at 14:56
I have edited the post.Please pay attention on it once more.You helped me to understand it more closely. –  Abhinav Anand Nov 27 '12 at 16:08
@AbhinavAnand: I'm afraid I don't really understand what the new question in your update is. If you're looking at it from a linear algebra point of view, I'm afraid I can't help you :/ –  Manishearth Nov 27 '12 at 16:12
A blunder.If you have enough tolerance for an exhausted mind accidentally? –  Abhinav Anand Nov 27 '12 at 20:10