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}
...\lambda^2{}_2 & 0 \\
0 & 0 & 0 & 1

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{}_...
...a^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
- \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
- \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.