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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Luboš Motl Nov 27 '12 at 14:04
  • $\begingroup$ 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? $\endgroup$ – Muphrid Nov 27 '12 at 16:33
  • $\begingroup$ First of all $i$ is a bound vector and it is not only parallel but in the same $x$ axis. $\endgroup$ – Abhinav Anand Nov 27 '12 at 16:55
  • $\begingroup$ 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)? $\endgroup$ – Muphrid Nov 27 '12 at 17:36
  • $\begingroup$ Did not understand you at all. $\endgroup$ – 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)

  • $\begingroup$ 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. $\endgroup$ – twistor59 Nov 27 '12 at 14:55
  • $\begingroup$ I have edited the post.Please pay attention on it once more.You helped me to understand it more closely. $\endgroup$ – Abhinav Anand Nov 27 '12 at 16:08
  • $\begingroup$ @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 :/ $\endgroup$ – Manishearth Nov 27 '12 at 16:12
  • $\begingroup$ A blunder.If you have enough tolerance for an exhausted mind accidentally? $\endgroup$ – Abhinav Anand Nov 27 '12 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.