So I have been trying to derive the Lorentz transormation via the equation of a photon:

$$\frac {\partial^2 \psi}{\partial t^2}-c^2 \frac{\partial^2 \psi}{\partial x^2}=0$$

The premise is that this should be invariant under any transformation so it can be shown that Galileian transformations don't work. Then I have tried to find a new one (which I know should be Lorentz's ), but it is turning to be quite difficult to operate.

Can anyone give me some help with this? I know it's probably not the best method to derive this, but I'm quite curious and Google is not giving me answers.

  • $\begingroup$ On the website mathpages.com there is an essay discussing theoretical exploration, published in 1887, by a physicist called Woldemar Voigt, who (in the course of exploring Doppler effects) sought to find whether there is any linear transformation that leaves an ordinary wave equation intact. The essay recounts Voigts explorations, noting that he came ever so close to stating Lorentz transformation. That is, the essay demonstrates that there is a path from 'a time-dependent scalar field f(x,t) in one dimension' equation to Lorentz transformation. $\endgroup$
    – Cleonis
    Commented Mar 4, 2018 at 10:13

1 Answer 1


The procedure is easy if you assume two postulates:

  1. $\psi$ is a scalar under change of reference frame;

  2. the transformations you consider are linear (non-homogeneous in general since you are always free, in a given reference frame, to move the origin of space and time).

The second postulate can be proved assuming other principles, but I will not enter into the details of a deeper formulation.

Your equation can be re-written

$$\frac {\partial^2 \psi}{\partial \tau^2}- \frac{\partial^2 \psi}{\partial x^2}=0\:,\tag{1}$$

where I introduced the new variable $\tau = ct$. If $\tau =x^0$ and $x=x^1$, (1) can be rewritten into an apparently more abstract form $$\eta^{ab}\frac{\partial^2 \psi(x)}{\partial x^a \partial x^b}=0\:,\tag{2}$$ where I use the standard convention of summation over repeated indices ($a$ and $b$) and the $2\times 2$ real matrix $\eta$ of coefficients $\eta^{ab}$ is diagonal $$\eta = diag(-1,1)\:.$$ Now, we change reference frame, passing to coordinates $y^0,y^1$ connected to $x^0,x^1$ by a linear non-homogeneous transformation according to postulate 2.

$$y^a = \Lambda^a\:_b x^b + c^a\:.$$

Evidently, this map must be invertible since the role of reference frame must be swapped in view of their physical equivalence. It is easy to prove that this is equivalent to requiring that the matrix $\Lambda$ is invertible. So, we also have the inverse transform:

$$x^a = \Lambda'^a\:_b y^b + c'^a\:,$$

where $\Lambda' = \Lambda^{-1}$ and $c'^d = -\Lambda'^d\:_e c^e$. Equation (2), changing coordinates and taking postulate 1 into account, gives $$\eta^{ab} \frac{\partial y^c}{\partial x^a}\frac{\partial y^d}{\partial x^b}\frac{\partial^2 \psi(x(y))}{\partial y^c \partial y^d}=0\:,$$ where I also used postulate 1 to cancel a second derivative including only coordinates otherwise appearing (at this stage, one may require that this derivative vanishes in order to preserve the form of the wave equation and thus proving 2 form more basic assumptions). In other words, if $\psi'(y) := \psi(x(y))$, $$\eta^{ab} \Lambda^c\:_a\Lambda^d\:_b \frac{\partial^2 \psi'(y)}{\partial y^c \partial y^d}=0 \tag{3}\:.$$ Comparing (2) and (3), you see that the form of the equation is preserved provided $$\left(\eta^{ab} \Lambda^c\:_a\Lambda^d\:_b - \eta^{cd}\right)\frac{\partial^2 \psi'(y)}{\partial y^c \partial y^d}=0 \:.\tag{4}$$ It is evident that the form of the wave equation is preserved if $$\eta^{ab} \Lambda^c\:_a\Lambda^d\:_b = \eta^{cd}\:.\tag{5}$$ The converse is true under some mild physical assumption. If, for any given event $y$ of the spacetime and for any choice of $\{c_0,d_0\}$ where $c_0,d_0$ take values in $\{0,1\}$, we are able to produce a wave such that $\frac{\partial^2 \psi'(y)}{\partial y^c \partial y^d}=0$ if $\{c,d\} \neq \{c_0,d_0\}$ and $\frac{\partial^2 \psi'(y)}{\partial y^{c_0} \partial y^{d_0}} = \frac{\partial^2 \psi'(y)}{\partial y^{d_0} \partial y^{c_0}} \neq 0$, then (4) implies (5).

Notice that the transformations ${\mathbb R^2} \ni x \mapsto y \in {\mathbb R}^2$ $$y^a = \Lambda^a\:_b x^b + c^a$$ where the matrices $\Lambda$ satisfy (5) define the 2D Poincaré group and thus can be written into the standard form almost everybody knows.

If $c^0=c^1=0$ and $t'= y^0/c$, $x'= y^1$,

$$t' = \gamma\left(t- \frac{vx}{c^2}\right)\:, \qquad x' = \gamma(x-vt)$$

where $\gamma = (1- \frac{v^2}{c^2})^{-1/2}$ and $v$ is the speed of a point fixed in the reference frame at rest with $x'$ measured in the non-primed reference frame.

Writing down the wave equation in coordinates $t'= y^0/c$, $x'= y^1$, the invariant form of the equation is

$$\frac{\partial^2 \psi'}{\partial t'^2}- c^2\frac{\partial^2 \psi'}{\partial x'^2}=0\tag{1'}$$

Interpreting $t$ and $x'$ as the temporal and spatial coordinate of the other reference frame, the standard theory of D'Alembert equation proves that the speed of propagation of waves is $c$ in both reference frames.


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