Lorentz transformation matrix for all 3 spatial axes

Physics Asked on January 4, 2022

The lorentz transformation matrix (for all 3 spatial axes, not just a single dimension boost) appears to be commonly defined as the following:
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}

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_xgamma& -beta_ygamma & -beta_zgamma \ -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 = displaystylefrac{1}{sqrt{1-displaystylefrac{|vec{v}|^2}{c^2}}} $ and
$gamma_{v_x} = displaystylefrac{1}{sqrt{1-displaystylefrac{v_x^2}{c^2}}}$

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 $t′=tgamma$

One Answer

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

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.

Answered by Fatma Keskin on January 4, 2022

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