# Overview

Einstein's proof for the Lorentz transformation is given here:

From $O$'s view point, $x^2+y^2+z^2 = (ct)^2$.

Form $O'$'s view point, $x'^2+y'^2+z'^2 = (ct')^2$.

We find that Einstein has imparted the information that the velocity of light observed by $O$ and $O'$ is constant (Einstein's first postulate).

# Problem statement

All this has made me wonder why should a postulate derive his result? Could his results be derived by the existing laws of physics? Every textbook that I have referred says that his postulates are a direct application of Michelson-Morley experiment's results. But what actually causes this time dilation and length contraction?

# Update

I haven't asked about why we define postulates. I understand that at one point we must have postulates. But to elaborate, my question is about: is it possible to derive Einstein's postulates form works of Maxwell?

Note: I know that a postulate does not have a derivation, but how do you think this postulate works?

To show you that Maxwell's laws have something to do with Einstein's postulates:

Consider 2 point charges moving with the same velocity parallel to each other. We observe that force acts on both the particles due to magnetic field. But, the observer moving along with the charge must observe no force at all! Don't you think Einstein's postulates can be derived from Maxwell's work?

• Actually one could derive the constant speed of light postulate from the relativity postulate provided the formulation encompasses electromagnetic phenomena. As Maxwell equations, consequently, must have the same form in every inertial frame, the speed of light in vacuum, $c= 1/\sqrt{\mu_0\epsilon_0}$ has to be the same in all inertial frames. – Valter Moretti Jun 28 '14 at 10:45
• @V.Moretti that simply replaces the speed of light postulate with the postulate that the permittivity and permeability being the same in all inertial frames. Since free-space EM can be equivalently described with constants $(c_0,Z_0)$ as with constants $(\epsilon_0,\mu_0)$, we're effectively trading invariance of lightspeed with invariance of lightspeed and invariance of impedance. That's not an improvement, and arguably, a step back, since it postulates more things. – Stan Liou Jun 28 '14 at 11:42
• This wikipedia says (en.wikipedia.org/wiki/Postulates_of_special_relativity) that Minkowski showed speed of light to be a space-time constant. The English translation of Minkowski's paper can be found here (en.wikisource.org/wiki/Translation:Space_and_Time) – user7757 Jun 28 '14 at 12:25
• Finally, though we're all big fans of neat mathematical treatments (they're usually the springboard for deep understanding), physics is an empirical business and our postulates (unlike those of the mathematicians) have to give results consistent with experiment. Or else. It is that agreement that justifies the postulates and issues of "beauty" and "elegance" that are used to chose which formulation is "most fundamental". – dmckee --- ex-moderator kitten Jun 28 '14 at 18:38
• @V.Moretti Wow really? I would imagine there must be some other set of equivalent assumptions one uses? – joshphysics Jun 29 '14 at 20:00

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).

• Some of the hypotheses you listed are discussed here: web.ist.utl.pt/ist13264/publicat/TP1-v3.pdf – Marco Disce Aug 19 '16 at 17:48
• Thank for the pdf, I think that relativity is usually introduced in a very fast and abstract way that I find unaesthetic (for my liking). I agree with author: "special relativity should be taught starting from the idea of a preferred frame" and then show that this frame has to be abandoned (obviously I strongly agree with Born: no place in physics for what is not observable by principle!). Anyway pdf speaks only about kinematics, but my problem is that I feel it must exists a natural extension of this point of view to dynamics and I don't find it. – Fausto Vezzaro Aug 21 '16 at 15:56
• @MarcoDisce in my home page I added a pdf (click on motoassoluto) where I wrote why I don't see any irremediable incompatibility between newtonian mechanics and maxwellian electromagnetism (I develop calculus that I speak about here above). We gain in viewability but of course we have to pay by increasing the number of assumption that hold the theory. It is in italian. – Fausto Vezzaro Nov 1 '16 at 17:52

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)$(2)$

• One of the minus signs on the Euclidean "boost" matrix shouldn't be there. Also, what about Galilean boost? (One can do $e^{\epsilon\lambda} = \mathrm{cosp}\,\lambda + \epsilon\,\mathrm{sinp}\,\lambda$ with $\epsilon^2 = 0$ to give $dt' + \epsilon\,dx' = e^{\epsilon v}(dt + \epsilon\,dx)$, but it's probably not worth to stretch the trig analogy too far.) – Stan Liou Jun 28 '14 at 10:52
• @StanLiou : Thanks for the typo ! Yes, you are right about galilean boosts, but I tried to make a simple choice . In fact, one has to consider the five possible $1$-dimensional subgroups of $SL(2,R)$, which are structured in $3$ families (as exponentials of a matrix which trace may be positive, zero or negative), and galilean boost transformations are part of them. – Trimok Jun 28 '14 at 11:02
• I know what a postulate is, but I don't know how Einstein's postulate works, if we derive his postulate from Maxwell's equations. I believe that constancy of speed of light in all inertial frames of reference does have something to do with Maxwell's laws. – Sreram Jun 30 '14 at 4:25
• @user3633270 : In fact, Maxwell's equation (in the vacuum, without charges and currents) is invariant by Lorentz transformations. This means that Maxwell's equation is a relativistic equation. – Trimok Jun 30 '14 at 8:03
• @rob I defined the "parabolic" trig functions by the equation given. That makes them completely trivial because the the power series cuts off after the linear term, $\mathrm{sinp}\,\lambda = \lambda$, etc., but I phrased it in this way to carry the analogy to Euclidean rotations ($i^2 = -1$ with $e^{i\phi} = \cos\phi + i\sin\phi$) and Lorentz boosts ($j^2 = +1$ with $e^{j\alpha} = \cosh\alpha + j\sinh\alpha$). – Stan Liou Aug 14 '14 at 3:07