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I've been self studying some special relativity by reading some textbooks, and I'm somewhat confused about the process of computing velocities. The book I'm reading primarily identifies computation of velocities by just carrying out the boost operators, but the general case remains unclear.

If you have a reference frame $\pmatrix{x & t}$ at rest, and another reference frame $\pmatrix{x' & t'}$ moving with respect to the other frame, and you have coordinate equations $x'(x,t)$ and $t'(x,t)$ relating the one frame to the other, how do you compute the relative velocity of the $\pmatrix{x'& t'}$ frame with respect to the $\pmatrix{x & t}$ frame?

I'm struggling with some of the semantics, so maybe some elaboration on intuition might be appreciated too. Thanks.

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    $\begingroup$ have you practiced using the spacetime chart method? $\endgroup$
    – niels nielsen
    Jan 20, 2021 at 6:10
  • $\begingroup$ I've heard of it, but I'm hoping to find an analytical approach/method without having to go to graphs to visualize it $\endgroup$
    – stanfordude
    Jan 20, 2021 at 6:12
  • $\begingroup$ Click on this link. $\endgroup$
    – joseph h
    Jan 20, 2021 at 6:25
  • $\begingroup$ @Drjh thx, I read that but was kind of confused. In the page, they calculate $\frac{dx}{dt}$ in terms of $\frac{dx'}{dt'}$. Does that give the velocity of the moving frame with respect to the rest frame? In my context, would I have to invert $x'(x,t)$ to a function $x(x', t')$ in order to calculate the correct velocity then? $\endgroup$
    – stanfordude
    Jan 20, 2021 at 6:34
  • $\begingroup$ The value of the spacetime chart is not in allowing you to compute graphically, but in allowing you to keep track of all the things that need to be kept track of. Don't ignore it, it is very useful and it remains useful even when velocities are not constant and the frame approach no longer works properly (following @nielsnielsen 's comment and your reply) $\endgroup$
    – Cryo
    Jan 20, 2021 at 23:08

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Since for a boost $$ \left( \begin{array}{c} t'\\ x'\\ \end{array} \right) = \left( \begin{array}{cc} \gamma & \gamma v\\ \gamma v & \gamma\\ \end{array} \right) \left( \begin{array}{c} t\\ x\\ \end{array} \right), $$ given $$ \left( \begin{array}{c} t'\\ x'\\ \end{array} \right) = \left( \begin{array}{cc} A & B\\ B & A\\ \end{array} \right) \left( \begin{array}{c} t\\ x\\ \end{array} \right), $$ where $\det(M)=A^2-B^2=1$,
the relative velocity is $$\frac{B}{A}=\frac{\gamma v}{\gamma}=v.$$

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  • $\begingroup$ Does this block matrix form hold for vector coordinates $x'$? $\endgroup$
    – stanfordude
    Jan 20, 2021 at 20:31
  • $\begingroup$ In the special situation when all relative motions are along the x-directions (that is the relative-velocity vector is coplanar with the tx and t'x' planes), we could write $\left( \begin{array}{c} t'\\ x'\\ y' \\ z' \end{array} \right) = \left( \begin{array}{cc} \gamma & \gamma v & 0 &0 \\ \gamma v & \gamma &0 &0 \\ 0&0&1&0\\ 0&0&0&1 \end{array} \right) \left( \begin{array}{c} t\\ x\\y\\z \end{array} \right),$ Then the block form method would still apply. For something more general, the transformation matrix is more complicated and may be hard to solve for $\vec v$. $\endgroup$
    – robphy
    Jan 20, 2021 at 22:31

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