The Lorentz transformations can be derived without any specific reference to light just by assuming that the laws of electromagnetism are seen as the same by all inertial observers.

My favourite version is the one which came to me (as I am sure it also has to others) while doing an experiment in the second year physics lab. The experiment was to observe the magnetic force between two parallel current-carrying wires, and the thought experiment that leads to SR is just to ask what happens if the wires are eliminated and the current is in the form of electron beams instead. (I first wondered what happens if I travel with the electrons in the wires, but then the protons are travelling in the opposite direction and the magnetic force is the same in both cases. But if the protons are removed then we do have a current, and so a magnetic force, in one frame but not the other.)

My first thought was that the laws of electromagnetism may only apply exactly in one particular “aether” frame and that there should be a correction for frames moving relative to the aether. But of course we see from many experiments both that there are no seasonal variations and even objects moving relative to the earth continue to obey exactly the same laws.

IF the laws of electromagnetism are the same for a moving as for a stationary observer, then the attractive magnetic force between the currents of two electron beams is not apparent to an observer moving along with the electrons. So how does the change to relatively moving lab coordinates lead to the appearance of a magnetic force?

In the case of an observer $#O#$ moving with the electrons we have the following picture:

So the acceleration of each electron is solely due to the electrostatic repulsion and is given by $$a=\frac{F_{E}}{m}=\frac{k_{E}}{m}\frac{qQ_{eff}}{d^{2}}$$ where $$Q_{eff}=\int{(q/s)cos(\theta)}$$

And for the lab observer $#O’#$ moving at speed $#v#$ along the lines of electrons we have:

$$\begin{align}a’ &=\frac{(F_{E}-F_{M})}{m’}=\frac{k_{E}}{m’}\frac{q’Q’_{eff}}{d’^{2}}-\frac{k_{M}}{m’}\frac{q’Q’_{eff}v}{d’^{2}}\\ &=\frac{q’}{m’d’^{2}}Q’_{eff}(k_{E}-k_{M}v)\end{align}$$

Now the distance $#d#$ is measured perpendicular to the direction of motion and so should also be unchanged and so also $#\frac{Q’_{eff}}{Q_{eff}}=\frac{q’/s’}{q/s}=\frac{q’s}{qs’}#$

So we get $$\frac{a’}{a}=\frac{(k_{E}-k_{M}v)q’Q’_{eff}/m’}{k_{E}qQ_{eff}/m}=\frac{(k_{E}-k_{M}v)q’^{2}sm}{k_{E}q^{2}s’m’}$$

or $$\frac{a’s’m’/q’^{2}}{asm/q^{2}}=\frac{(k_{E}-k_{M}v)}{k_{E}}=1-\frac{k_{M}}{k_{E}}v$$

But from the first principles argument we know that ….