Lorentz transformation matrix for all 3 spatial axes

Question

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 $t′=t\gamma$

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. — G. Smith, Jul 15 '20 at 22:13
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 — Lewis Kelsey, Jul 16 '20 at 01:45
I think you could just use the rotational transformation equations on the Lorentz transformations, i.e. apply them one by one — SK Dash, Jul 16 '20 at 02:31
I changed it so the length is reciprocal of gamma and not time I made a mistake there — Lewis Kelsey, Jul 16 '20 at 10:55
@LewisKelsey it appears your question has been answered. Should this question be closed ? — Lelouch, Jul 16 '20 at 11:30
@Lelouch No, I still don't know why all the matrices I see have a $-\beta \gamma$ term rather than $-\frac{\beta}{\gamma}$ term on the bottom 3 rows given that given that $t′=t\gamma$ but $l′=\frac{l}{\gamma}$ — Lewis Kelsey, Jul 17 '20 at 23:56
The composition of two Lorentz boosts along two different axes is not a boost along a third axis : Combining two Lorentz boosts. — Frobenius, Aug 19 '21 at 10:01
For the Lorentz boost along an arbitrary direction : Deriving Λij components of the Lorentz transformation matrix. — Frobenius, Aug 19 '21 at 10:05

score 1 · Answer 1 · answered Jan 02 '23 at 02:47

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.

score 0 · Answer 2 · answered Jul 24 '20 at 20:41

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

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.

Lorentz transformation matrix for all 3 spatial axes

2 Answers2