The lorentz transformation matrix (for all 3 spatial axes, not just a single dimension boost) appears to be commonly defined as the following: $$ \begin{bmatrix} \gamma &-\gamma v_x/c &-\gamma v_y/c &-\gamma v_z/c \\ -\gamma v_x/c&1+(\gamma-1)\dfrac{v_x^2} {v^2}& (\gamma-1)\dfrac{v_x v_y}{v^2}& (\gamma-1)\dfrac{v_x v_z}{v^2} \\ -\gamma v_y/c& (\gamma-1)\dfrac{v_y v_x}{v^2}&1+(\gamma-1)\dfrac{v_y^2} {v^2}& (\gamma-1)\dfrac{v_y v_z}{v^2} \\ -\gamma v_z/c& (\gamma-1)\dfrac{v_z v_x}{v^2}& (\gamma-1)\dfrac{v_z v_y}{v^2}&1+(\gamma-1)\dfrac{v_z^2} {v^2} \end{bmatrix} $$ I tried to derive it myself by combining the matrices for the individual boost directions and making $v=|\vec{v}|$and ended up at $$ \begin{bmatrix} ct' \\ x' \\ y' \\ z' \\ \end{bmatrix} = \begin{bmatrix} \gamma & -\beta_x\gamma& -\beta_y\gamma & -\beta_z\gamma \\ -\frac{\beta_y} {\gamma_{v_x}} & \frac{1}{\gamma_{v_x}} & 0 & 0 \\ -\frac{\beta_y}{\gamma_{v_y}} & 0 & \frac{1}{\gamma_{v_y}} & 0\\-\frac{\beta_z}{\gamma_{v_z}} & 0 & 0 & \frac{1}{\gamma_{v_z}} \\ \end{bmatrix} \begin{bmatrix} ct \\ x \\ y \\ z \\ \end{bmatrix} $$ Where $\gamma = \displaystyle\frac{1}{\sqrt{1-\displaystyle\frac{|\vec{v}|^2}{c^2}}} $ and $\gamma_{v_x} = \displaystyle\frac{1}{\sqrt{1-\displaystyle\frac{v_x^2}{c^2}}}$

2 questions. Where do the bottom right 9 terms come from in the common definition and why is the top $\gamma$ and not $\frac{1}{\gamma}$ given that $l′=\frac{l}{\gamma}$ but $t′=t\gamma$

  • 1
    $\begingroup$ Your matrix doesn’t reduce to the simple case for motion in the $x$-direction when you set $\beta_y$ and $\beta_z$ to zero. So it can’t be right. $\endgroup$
    – G. Smith
    Jul 15 '20 at 22:13
  • $\begingroup$ It seems to to me. I misunderstand $$y'= -ct\beta_y\gamma_{v_y} + y\gamma_{v_y}, ~~ \text{let}~~ v_y =0$$ $$=-ct*0 + y$$ $$=y$$ same with z $\endgroup$ Jul 16 '20 at 1:45
  • $\begingroup$ I think you could just use the rotational transformation equations on the Lorentz transformations, i.e. apply them one by one $\endgroup$
    – SK Dash
    Jul 16 '20 at 2:31
  • $\begingroup$ I changed it so the length is reciprocal of gamma and not time I made a mistake there $\endgroup$ Jul 16 '20 at 10:55
  • $\begingroup$ @LewisKelsey it appears your question has been answered. Should this question be closed ? $\endgroup$
    – Lelouch
    Jul 16 '20 at 11:30

$$\Delta x =x_f-x_i=\gamma(\Delta x'+v\Delta t')=\gamma(x'_f-x'_i+v(t'_f-t'_i))$$ $$ \Delta t'=t'_f-t'_i=0$$ $$\Delta t'=t'_f-t'_i=\gamma(\Delta t-\frac{v(x_f-x_i)}{c^2})$$ $$\Delta x=x_f-x_i=0$$

We deduce these facts: $\Delta x=\gamma \Delta x'=l=\gamma \Delta l'$ and $\Delta t'= \gamma \Delta t$ not $t'=\gamma t$.The reason why is that we don't talk about a time of an event in spacetime. What we're interested in is time difference between two events.


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.