# Lorentz transformation matrix for all 3 spatial axes

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. Jul 15, 2020 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 Jul 16, 2020 at 1:45
• I think you could just use the rotational transformation equations on the Lorentz transformations, i.e. apply them one by one Jul 16, 2020 at 2:31
• I changed it so the length is reciprocal of gamma and not time I made a mistake there Jul 16, 2020 at 10:55
• @LewisKelsey it appears your question has been answered. Should this question be closed ? Jul 16, 2020 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.