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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 $t′=t\gamma$

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    $\begingroup$ 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, 2020 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, 2020 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, 2020 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, 2020 at 10:55
  • $\begingroup$ @LewisKelsey it appears your question has been answered. Should this question be closed ? $\endgroup$
    – Lelouch
    Jul 16, 2020 at 11:30

2 Answers 2

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Your main problem is that boosts are not closed under composition in Relativity - except when they are in the same or opposite directions. The composition of two boosts in different directions is a combination of a boost and a rotation of the axes. So, you can't derive the resulting boost by applying them one at a time along each axis ... without also reorienting the axes.

Velocity-space in Relativity is not flat, but curved.

The situation is the same as if you were to try to take a grid on the Earth's surface on the equator at 90 degrees longitude west, with X pointing east Y pointing north, then "boost" it up to the north pole, with X not pointing south along the prime meridian and Y pointing south along the 90 degrees east longitude; then "boost" it south back to the equator at 180 degrees longitude, with X now pointing north and Y pointing west, and then "boost" it back to 90 degrees west, with X still pointing north and Y pointing west. The result is a 90 degree rotation of the axes counter-clockwise.

A similar thing would happen with 3 boosts in different directions in velocity-space that bring you back to 0: there will be a rotation of axes - clockwise ... the other way, because the velocity space has negative curvature.

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

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.

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