Is it possible to derive Lorentz transformation equation without Einstein's postulates?

At some time, you need physical postulates. For instance, suppose these two possible transformations applying to the infinitesimal space-time components $dt$ and $dz$, between a frame $R$, and an other frame $R'$ moving at a velocity $v_z=v$ relatively to $R$ :

$$\begin{pmatrix} dt'\\dz'\end{pmatrix} = \begin{pmatrix} \cos \lambda & \sin \lambda\\-\sin \lambda&\cos \lambda\end{pmatrix} \begin{pmatrix} dt\\dz\end{pmatrix} \tag{1}$$ with $\tan \lambda = \dfrac{v}{c}$

$$\begin{pmatrix} dt'\\dz'\end{pmatrix} = \begin{pmatrix} \cosh \lambda & -\sinh \lambda\\-\sinh \lambda&\cosh \lambda\end{pmatrix} \begin{pmatrix} dt\\dz\end{pmatrix} \tag{2}$$ with $\tanh \lambda = \dfrac{v}{c}$

These two possible transformations verify some basic postulates : They form a one-dimensional subgroup (of $SL(2, \mathbb R))$, in particular for $\lambda=0(v=0)$, you recover the identity matrix, you have $A(\lambda)A(-\lambda)=A(0)= Id$ (calling $A(\lambda)$ the transformation matrix).

Finally for $dz= vdt$, you get $dz'=0$, which means that a object moving at speed $v$ relatively to $R$, is moving at speed zero relatively to $R'$.

However, how to choose between these two kind of transformations ?

Let us take $dz=0, dt >0$, the first transformation leads to $dt' = \cos\lambda \quad dt$, while the second transformation gives $dt' = \cosh\lambda \quad dt$.

We see that the first transformation does not guarantee that $dt' >0$, while the second guarantees the positivity of $dt'$

Now, we are ready to make an interpretation of $dz=0, dt >0$. We could interpret this as a possible causal relation between two process corresonding to space-time events $t,z$ and $t+dt,z$ (the former being the cause of the latter).

We now make the postulate, that in the frame $R'$, this causal relation appears as $dt'>0$, that is, in the $R'$ chronology ($t'$), the cause must not precede the consequence.

If you admit this postulate, then you can eliminate the first kind of transformations $(1)$, and you get the correct special Lorentz transformation (boost)$(2)$


Preliminary remarks: if a book say that invariance of $c$ is a direct consequence of M-M experiment, stop reading it.

Answer to your question: yes, we can derive relativistic kinematics from different postulates than the one of invariance of $c$. Consider these 4 assumptions

1) speed of light is isotropic only in one inertial frame (that we can call absolute space, presumably anchored to cosmic masses).

2) when an object is in absolute motion, its dimensions are contracted along the directions of motion by $\gamma$ factor (Fitzgerald idea).

3) observers in different inertial frames synchronize clock using Einstein convention.

4) when an object is in absolute motion, all his processes are affected by a slowdown by a $\gamma$ factor (Lorentz idea, I suppose).

The first 2 assumption explain the Michelson Morley experiment. If we take also the last one we explain the Kennedy-Thorndike experiment too. By the way, the last assumption is not indispensable to explain this experiment: by principle we know nothing about absolute motion of the solar system, that could be orthogonal to the ecliptic, so that module of absolute speed doesn't vary. (I suppose non circularity of the orbit or rotation, have effects too small to be detected).

We can say more: taking these assumption we can derive Lorentz transformation (that's why I said to you "the answer is yes"). So the hypothesis of equivalence of inertial frame in measuring speed of light is a way to relativistic kinematics, but not the only one (the problem is that the absence of an operative procedure that allows to locate the absolute space push to the simplicity of the "invariant $c$" point of view, and the experimentally tested impossibility to send superluminal signal too).


I found the natural extension to dynamics of this point of view: consider these other three assumptions:

5) For observers in absolute rest, classical physics works.

6) When a point mass is in absoulte motion, its mass is increased by a $\gamma$ factor, while charge remain unchanged.

7) If a force $\mathbf{F}$ is applied on a body then, in virtue of its absoulte motion, is exerted an additive force $- ( \mathbf{F} \cdot \mathbf{u} ) \frac{\mathbf{u}}{c^2}$, where $\mathbf{u}$ is the absolute speed.

If these seven assumptions works, then every inertial observer, doing electrodynamics experiment, has no way of detecting the absolute motion: acceleration of particles measured in different inertial frames are correctly connected in the way showed by Lorentz trasformations, and this agree with the fact that we know that first four assumptions gives relativistic kinematic: all sound right (things go in a different way if you include gravitation: charge doesn't change when absolute speed varies while mass does). There is nothing of strange if a force depend on speed or on exerted forced (think to the force of a spring, or to magnetic force). I was forced to take into consideration this three furher hypotesys by reflecting on $\mathbf{F}=\frac{d}{dt}(\gamma m \mathbf{u})$, wich as showed in Barone book can be rearranged in this way $$ \gamma m \mathbf{a} = \mathbf{F} - (\mathbf{F} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} $$ (to da that, you can start for example from problem 12.36 of Griffiths ED book III ed: $$ \mathbf{F}= \frac{m}{\sqrt{1-u^2/c^2}} \left[ \mathbf{a} + \frac{\mathbf{u} (\mathbf{u}\cdot \mathbf{a})}{c^2-u^2} \right] $$ and as intermediate step multiply by $\mathbf{u}$ showing that $\mathbf{a} \cdot \mathbf{u} = \frac{\mathbf{F} \cdot \mathbf{u}}{\gamma^3 m}$). Of course sometimes the seven assumptions give results very far from classical physics (after all they are based on Lorentz transformations and relativistic 2nd Newton law: they are equivalent to relativity) but I like express myself as in assumption 5, because here we have not a paradigm change. I add that sometimes using this assumption solving problems is faster (try to find acceleration of a point charge in a generic electromagnetic field to see what I mean).