# Derivation of Lorentz transformation from Maxwell's equations

Some books on special theory of relativity (or books that cover STR among other things) say that the Lorentz transformation can be derived from Maxwell's equations directly (or rather from Maxwell's equations and the requirement that they work in all inertial systems). However, I couldn't find any textbook that actually shows such a derivation. The only textbook that did something in that direction is Feynman lectures on physics volume 2, but that is more of a plausibility argument (and also it doesn't even mention how t is transformed). After some research, I also found those papers:

https://www.tandfonline.com/doi/abs/10.1163/1569393053303884?journalCode=tewa20 But I didn't find that very convincing since for this derivation the permanence principle is required which, as far as I know, is more of a heuristics and not really a rigorous mathematical principle or something you can argue to be true from basic physical principles (like causality or something like that).

https://royalsocietypublishing.org/doi/abs/10.1098/rsta.2007.2195 A central step there is that if $$\partial_x(a {H_z^{'}} - a v \epsilon_0 {E_y^{'}}) = \epsilon_0 \partial_t (a {E_y^{'}} + \frac{(1-a^2)}{a v \epsilon_0} \cdot {H_z^{'}})$$ and $$\partial_x{H_z} = \epsilon_0 \partial_t{E_y}$$ , then $$H_z = a {H_z^{'}} - a v \epsilon_0 {E_y^{'}}$$ and $$E_y = a {E_y^{'}} + \frac{(1-a^2)}{a v \epsilon_0} \cdot {H_z^{'}}$$, which I don't see. The paper argues that that follows from the principle of relativity in some way, but that is just stated without any reasoning and I don't see why that necessarily has to be true.

So, my question would be: Is it possible to derive the Lorentz transformation directly from Maxwell's equations and if yes, could you give a source to such a derivation? Or explain to me why the derivation given in at least one of the papers I cited is actually correct?

• Lorentz transformations are not simply derived from Maxwell's equations. If there is a relationship between them, namely that the Lorentz transformations form a group of transformations that leave Maxwell's equations invariant (but it is possible that there are other transformations that also satisfy this condition). It is easy to see that Maxwell's equations in a vacuum imply that the field propagates in the form of waves and that the wave equation is NOT invariant with respect to the Galileo group but with respect to the Lorentz group. Commented May 21, 2023 at 14:30
• So you're saying that actually there are several coordinate transformations that leave Maxwell's equations invariant? I am aware that the Galilei transformations (which form the Galilei group) don't leave them invariant and the Lorentz transformations (forming the Lorentz group) do. I had just heard and read several times that actually, the Lorentz transformations are the only ones that leave Maxwell's equations invariant. Commented May 21, 2023 at 14:35
• Tying Lorentz transformations to Maxwell's equations is a bad idea because it misrepresents physics IMHO. Special relativity follows trivially from the fact that space is metric, locally homogeneous and isotropic and that all physics is relative (we are also assuming that we have something like a time and distance measurement handy and that we define velocity in the usual way). Maxwell's equations and the entire slew of relativistic quantum field theories then follow from the resulting symmetry group. Commented May 21, 2023 at 16:10
• I know and agree, my question is more out of scientific curiosity. Obviously, it is much more efficient and much more general to derive the Lorentz transformations from Einstein's postulates than from Maxwell's equations. But since I've often heard that it is possible to derive STR from Maxwell's equations, I would like to see that derivation. But if I had to explain to aliens how we understand physics, I obviously wouldn't try to derive STR from Maxwell's equations. Commented May 21, 2023 at 17:16

The Lorentz transformations are defined by the condition that they leave c$$^2$$dt$$^2$$-dr$$^2$$=0. This condition comes from the Maxwell equations combined with the Michelson-Morley result.