D.5 Lorentz group prop­erty de­riva­tion

This note ver­i­fies the group prop­erty of the Lorentz trans­for­ma­tion. It is not rec­om­mended un­less you have had a solid course in lin­ear al­ge­bra.

Note first that a much more sim­ple ar­gu­ment can be given by defin­ing the Lorentz trans­for­ma­tion more ab­stractly, {A.4} (A.13). But that is cheat­ing. Then you have to prove that these Lorentz trans­form are al­ways the same as the phys­i­cal ones.

For sim­plic­ity it will be as­sumed that the ob­servers still use a com­mon ori­gin of space and time co­or­di­nates.

The group prop­erty is easy to ver­ify if the ob­servers B and C are go­ing in the same di­rec­tion com­pared to A. Just mul­ti­ply two ma­tri­ces of the form (1.13) to­gether and ap­ply the con­di­tion that $\gamma^2-\beta^2\gamma^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for each.

It gets much messier if the ob­servers move in dif­fer­ent di­rec­tions. In that case the only im­me­di­ate sim­pli­fi­ca­tion that can be made is to align the co­or­di­nate sys­tems so that both rel­a­tive ve­loc­i­ties are in the $x,y$ planes. Then the trans­for­ma­tions only in­volve $z$ in a triv­ial way and the com­bined trans­for­ma­tion takes the generic form

$$\Lambda_{C\leftarrow A} = \left( \begin{array}{cccc} \l... ..._1 & \lambda^2{}_2 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

It needs to be shown that this is a Lorentz trans­for­ma­tion from A di­rectly to C.

Now the spa­tial, $x,y$, co­or­di­nate sys­tem of ob­server C can be ro­tated to elim­i­nate $\lambda^2{}_0$ and the spa­tial co­or­di­nate sys­tem of ob­server A can be ro­tated to elim­i­nate $\lambda^0{}_2$. Next both Lorentz trans­for­ma­tions pre­serve the in­ner prod­ucts. There­fore the dot prod­uct be­tween the four-vec­tors $(1,0,0,0)$ and $(0,0,1,0)$ in the A sys­tem must be the same as the dot prod­uct be­tween columns 1 and 3 in the ma­trix above. And that means that $\lambda^1{}_2$ must be zero, be­cause $\lambda^1{}_0$ will not be zero ex­cept in the triv­ial case that sys­tems A and C are at rest com­pared to each other. Next since the proper length of the vec­tor $(0,0,1,0)$ equals one in the A sys­tem, it does so in the C sys­tem, so $\lambda^2{}_2$ must be one. (Or mi­nus one, but a 180$\POW9,{\circ}$ ro­ta­tion of the spa­tial co­or­di­nate sys­tem around the $z$-​axis can take care of that.) Next, since the dot prod­uct of the vec­tors $(0,1,0,0)$ and $(0,0,1,0)$ is zero, so is $\lambda^2{}_1$.

That leaves the four val­ues re­lat­ing the time and $x$ com­po­nents. From the fact that the dot prod­uct of the vec­tors $(1,0,0,0)$ and $(0,1,0,0)$ is zero,

$$- \lambda^0{}_0 \lambda^0{}_1 + \lambda^1{}_0 \lambda^1{}_1... ...bda^1{}_1} = \frac{\lambda^1{}_0}{\lambda^0{}_0} \equiv \beta$$

where $\beta$ is some con­stant. Also, since the proper lengths of these vec­tors are mi­nus one, re­spec­tively one,

$$- \lambda^0{}_0^2 + \lambda^1{}_0^2 = -1 \qquad - \lambda^0{}_1^2 + \lambda^1{}_1 = 1$$

or sub­sti­tut­ing in for $\lambda^0{}_1$ and $\lambda^1{}_0$ from the above

$$- \lambda^0{}_0^2 + \beta^2 \lambda^0{}_0^2 = -1 \qquad - \beta^2 \lambda^1{}_1^2 + \lambda^1{}_1 = 1$$

It fol­lows that $\lambda^0{}_0$ and $\lambda^1{}_1$ must be equal, (or op­po­site, but since both Lorentz trans­for­ma­tions have unit de­ter­mi­nant, so must their com­bi­na­tion), so call them $\gamma$. The trans­for­ma­tion is then a Lorentz trans­for­ma­tion of the usual form (1.13). (Since the spa­tial co­or­di­nate sys­tem can­not just flip over from left handed to right handed at some point, $\gamma$ will have to be pos­i­tive.) Ex­am­in­ing the trans­for­ma­tion of the ori­gin $x_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 iden­ti­fies $\beta$ as $V$$\raisebox{.5pt}{$/$}$​$c$, with $V$ the rel­a­tive ve­loc­ity of sys­tem A com­pared to B, and then the above two equa­tions iden­tify $\gamma$ as the Lorentz fac­tor.

Ob­vi­ously, if any two Lorentz trans­for­ma­tions are equiv­a­lent to a sin­gle one, then by re­peated ap­pli­ca­tion any ar­bi­trary num­ber of them are equiv­a­lent to a sin­gle one.