D.4 Lorentz transformation derivation

This note derives the Lorentz transformation as discussed in chapter 1.2. The question is what is the relationship between the time and spatial coordinates $t_{\rm {A}},x_{\rm {A}},y_{\rm {A}},z_{\rm {A}}$ that an observer A attaches to an arbitrary event versus the coordinates $t_{\rm {B}},x_{\rm {B}},y_{\rm {B}},z_{\rm {B}}$ that an observer B attaches to them.

Note that since the choices what to define as time zero and as the origin are quite arbitrary, it can be arranged that $x_{\rm {B}},y_{\rm {B}},z_{\rm {B}},t_{\rm {B}}$ are all zero when $x_{\rm {A}},y_{\rm {A}},z_{\rm {A}},t_{\rm {A}}$ are all zero. That simplifies the mathematics, so it will be assumed. It will also be assumed that the axis systems of the two observers are taken to be parallel and that the $x$ axes are along the direction of relative motion between the observers, figure 1.2.

It will further be assumed that the relationship between the coordinates is linear;

\begin{displaymath}
\begin{array}{ccc}
t_{\rm {B}} =
a_{tx} x_{\rm {A}} + ...
... {A}} + a_{zz} z_{\rm {A}} + a_{zt} t_{\rm {A}}
\end{array}
\end{displaymath}

where the $a_{..}$ are constants still to be found.

The biggest reason to assume that the transformation should be linear is that if space is populated with observers A and B, rather than just have a single one sitting at the origin of that coordinate system, then a linear transformation assures that all pairs of observers A and B see the exact same transformation. In addition, the transformation from $x_{\rm {B}},y_{\rm {B}},z_{\rm {B}},t_{\rm {B}}$ back to $x_{\rm {A}},y_{\rm {A}},z_{\rm {A}},t_{\rm {A}}$ should be of the same form as the one the other way, since the principle of relativity asserts that the two coordinate systems are equivalent. A linear transformation has a back transformation that is also linear.

Another way to look at it is to say that the spatial and temporal scales seen by normal observers are miniscule compared to the scales of the universe. Based on that idea you would expect that the relation between their coordinates would be a linearized Taylor series.

A lot of additional constraints can be put in because of physical symmetries that surely still apply even allowing for relativity. For example, the transformation to $x_{\rm {B}},t_{\rm {B}}$ should not depend on the arbitrarily chosen positive directions of the $y$ and $z$ axes, so throw out the $y$ and $z$ terms in those equations. Seen in a mirror along the $xy$-​plane, the $y$ transformation should look the same, even if $z$ changes sign, so throw out $z_{\rm {A}}$ from the equation for $y_{\rm {B}}$. Similarly, there goes $y_{\rm {A}}$ in the equation for $z_{\rm {B}}$. Since the choice of $y$ and $z$ axes is arbitrary, the remaining $a_{z.}$ coefficients must equal the corresponding $a_{y.}$ ones. Since the basic premise of relativity is that the coordinate systems A and B are equivalent, the $y$ difference between tracks parallel to the direction of motion cannot get longer for B and shorter for A, nor vice-versa, so $a_{yy}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Finally, by the very definition of the relative velocity $v$ of coordinate system B with respect to system A, $x_{\rm {B}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y_{\rm {B}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z_{\rm {B}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 should correspond to $x_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $vt_{\rm {A}}$. And by the principle of relativity, $x_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z_{\rm {A}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 should correspond to $x_{\rm {B}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-vt_{\rm {B}}$.

You might be able to think up some more constraints, but this will do. Put it all together to get

\begin{displaymath}
\begin{array}{ccc}
t_{\rm {B}} = a_{tx} x_{\rm {A}} + a_...
... x_{\rm {A}} + z_{\rm {A}} + a_{yt} t_{\rm {A}}
\end{array}
\end{displaymath}

Next the trick is to consider the wave front emitted by some light source that flashes at time zero at the then coinciding origins. Since according to the principle of relativity the two coordinate systems are fully equivalent, in both coordinate systems the wave front forms an expanding spherical shell with radius $ct$:

\begin{displaymath}
x_{\rm {A}}^2 + y_{\rm {A}}^2 + z_{\rm {A}}^2 = c^2 t_{\rm...
...rm {B}}^2 + y_{\rm {B}}^2 + z_{\rm {B}}^2 = c^2 t_{\rm {B}}^2
\end{displaymath}

Plug the linearized expressions for $x_{\rm {B}},y_{\rm {B}},z_{\rm {B}},t_{\rm {B}}$ in terms of $x_{\rm {A}},y_{\rm {A}},z_{\rm {A}},t_{\rm {A}}$ into the second equation and demand that it is consistent with the first equation, and you obtain the Lorentz transformation. To get the back transformation giving $x_{\rm {A}},y_{\rm {A}},z_{\rm {A}},t_{\rm {A}}$ in terms of $x_{\rm {B}},y_{\rm {B}},z_{\rm {B}},t_{\rm {B}}$, solve the Lorentz equations for $x_{\rm {A}}$, $y_{\rm {A}}$, $z_{\rm {A}}$, and $t_{\rm {A}}$.

To derive the given transformations between the velocities seen in the two systems, take differentials of the Lorentz transformation formulae. Then take ratios of the corresponding infinitesimal position increments over the corresponding time increments.