# LOG#005. Lorentz transformations(I).

For physicists working with objects approaching the light speed, e.g., handling with electromagnetic waves, the use of special relativity is essential.

The special theory of relativity is based on two single postulates:

1st) Covariance or invariance of all the physical laws (Mechanics, Electromagnetism,…) for all the inertial observers ( i.e. those moving with constant velocity relative to each other).

It means that there is no preferent frame or observer, only “relative” motion is meaningful when speaking of motion with respect to certain observer or frame. Indeed, unfortunately, it generated a misnomer and a misconception in “popular” Physics when talking about relativity (“Everything is relative”). What is relative then? The relative motion between inertial observers and its description using certain “coordinates” or “reference frames”. However, the true “relativity” theory introduces just about the opposite view. Physical laws and principles are “invariant” and “universal” (not relative!).  Einstein himself was aware of this, in spite he contributed to the initial spreading of the name “special relativity”, understood as a more general galilean invariance that contains itself the electromagnetic phenomena derived from Maxwell’s equations.

2nd) The speed of light is independent of the source motion or the observers. Equivalently, the speed of light is constant everywhere in the Universe.

No matter how much you can run, speed of light is universal and invariant. Massive particles can never move at speed of light. Two beams of light approaching to each other does not exceed the speed of light either. Then, the usual rule for the addition of velocities is not exact. Special relativity provides the new rule for adding velocities.

In this post, the first of a whole thread devoted to special relativity, I will review one of the easiest ways to derive the Lorentz transformations. There are many ways to “guess” them, but I think it is very important to keep the mathematics as simple as possible. And here simple means basic (undergraduate) Algebra and some basic Physics concepts from electromagnetism, galilean physics and the use of reference frames. Also, we will limite here to 1D motion in the x-direction.

Let me begin! We assume we have two different observers and frames, denoted by S and S’. The observer in S is at rest while the observer in S’ is moving at speed $v$ with respect to S. Classical Physics laws are invariant under the galilean group of transformations:

$x'=x-vt$

We know that Maxwell equations for electromagnetic waves are not invariant under Galileo transformations, so we have to search for some deformation and generalization of the above galilean invariance. This general and “special” transformation will reduce to galilean transformations whenever the velocity is tiny compared with the speed of light (electromagnetic waves). Mathematically speaking, we are searching for transformations:

$x'=\gamma (x-vt)$

and

$x=\gamma (x'+vt')$

for the inverse transformation. There $\gamma=\gamma(c,v)$ is a function of the speed of light (denoted as c, and constant in every frame!) and the relative velocity $v$ of the moving object in S’ with respect to S. The small velocity limit of special relativity to galilean relativity imposes the condition:

$\displaystyle{\lim_{v \to 0} \gamma (c,v) =1}$

By the other hand, according to special relativity second postulate, light speed is constant in every reference frame. Therefore, the distance a light beam ( or wave packet) travels in every frame is:

$x=ct$ in S, or equivalently $x^2=c^2t^2$

and

$x'=ct'$ in S’, or equivalently $x'^2=c^2t'^2$

Then, the squared spacial  separation between the moving light-like object at S’ with respect to S will be

$x^2-x'^2=c^2(t^2-t'^2)$

Squaring the modified galilean transformations, we obtain:

$x'^2=\gamma ^2(x-vt)^2 \rightarrow x'^2=\gamma ^2 (x^2+v^2t^2-2xvt) \rightarrow x'^2-\gamma ^2x^2+2\gamma ^2xvt=\gamma ^2 v^2t^2$

$x^2=\gamma ^2 (x'+vt')^2 \rightarrow x^2-\gamma ^2x'^2-2\gamma ^2x'vt'=\gamma ^2v^2t'^2$

The only “weird” term in the above last two equations are the mixed term with “xvt” (or the x’vt’ term). So, we have to make some tricky algebraic thing to change it. Fortunately for us, we do know that $x'=\gamma(x-vt)$, so

$x'=\gamma x -\gamma vt \rightarrow \gamma x'=\gamma ^2 x-\gamma ^2 vt \rightarrow \gamma xx'=\gamma ^2 x^2-\gamma ^2 xvt$

and thus

$2\gamma xx'=2 \gamma ^2 x^2-2\gamma ^2 xvt \rightarrow 2\gamma ^2 xvt =2\gamma ^2x^2-2\gamma xx'$

In the same way,  we proceed with the inverse transformations:

$x=\gamma x'+\gamma vt' \rightarrow \gamma x=\gamma ^2x'+\gamma ^2vt' \rightarrow \gamma xx'=\gamma ^2x'^2-\gamma ^2x'vt'$

and thus

$2\gamma xx'=2\gamma^2x'^2+2\gamma^2x'vt' \rightarrow 2\gamma^2x'vt'=2\gamma^2x'^2-2\gamma^2xx'$

We got it! We can know susbtitute the mixed x-v-t and x’-v-t’ triple terms in terms of the last expressions. In this way, we get the following equations:

$x'^2=\gamma ^2(x-vt)^2 \rightarrow x'^2=\gamma ^2(x^2+v^2t^2-2xvt) \rightarrow x'^2-\gamma ^2x^2+2\gamma ^2x^2-2\gamma ^2xx'=\gamma ^2v^2t^2 \rightarrow x'^2+\gamma ^2x^2-2\gamma ^2xx'=\gamma ^2v^2t^2$

$x'^2=\gamma ^2(x'+vt')^2 \rightarrow x^2=\gamma ^2(x'^2+v^2t^2+2x'vt') \rightarrow x^2-\gamma ^2x'^2+2\gamma ^2x'^2-2\gamma ^2xx'=\gamma ^2v^2t'^2 \rightarrow x^2+\gamma ^2x'^2-2\gamma ^2xx'=\gamma ^2v^2t'^2$

And now, the final stage! We substract the first equation to the second one in the above last equations:

$x^2-x'^2+\gamma ^2(x'^2-x^2)=\gamma ^2v^2(t'^2-t^2) \rightarrow (x'^2-x^2)(\gamma ^2-1)= \gamma ^2v^2(t'^2-t^2)$

But we know that $x^2-x'^2=c^2(t^2-t'^2)$, and so

$(x'^2-x^2)(\gamma ^2-1)= \gamma ^2v^2(t'^2-t^2) \rightarrow c^2(x'^2-x^2)(\gamma ^2-1)= \gamma ^2v^2(x'^2-x^2)$

then

$c^2(\gamma ^2-1)= \gamma ^2v^2 \rightarrow -c^2= -\gamma ^2c^2+\gamma ^2v^2 \rightarrow \gamma ^2=\dfrac{c^2}{c^2-v^2}$

or, more commonly we write:

$\gamma ^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

and therefore

$\gamma =\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Moreover, we usually define the beta (or boost) parameter to be

$\beta = \dfrac{v}{c}$

To obtain the time transformation we only remember that $x'=ct'$ and $x=ct$ for light signals, so then, for time we  obtain:

$x'=\gamma (x-vt) \rightarrow t' =x' /c= \gamma (x/c-vt/c)=\gamma ( t- vx/c^2)$

Finally, we put everything together to define the Lorentz transformations and their inverse for 1D motion along the x-axis:

$x'=\gamma (x-vt)$

$y'=y$

$z'=z$

$t'=\gamma \left( t-\dfrac{vx}{c^2}\right)$

and for the inverse transformations

$x=\gamma (x'+vt)$

$y=y'$

$z=z'$

$t=\gamma \left( t'+\dfrac{vx'}{c^2}\right)$

ADDENDUM: THE EASIEST, FASTEST AND SIMPLEST DEDUCTION  of $\gamma$ (that I do know).

If you don’t like those long calculations, there is a trick to simplify the “derivation” above.  The principle of Galilean relativity enlarged for electromagnetic phenomena implies the structure:

$x'=\gamma (x-vt)$ and $x=\gamma (x'+vt')$ for the inverse.

Now, the second postulate of special relativity says that light signals travel in such a way light speed in vacuum is constant, so $t=x/c$ and $t'=x'/c$. Inserting these times in the last two equations:

$x'=\gamma (1-v/c)x$ and $x=\gamma (1+v/c)x'$

Multiplying these two equations, we get:

$x'x =\gamma ^2(1+v/c)(1-v/c)xx'$.

If we consider any event beyond the initial tic-tac, i.e., if we suppose $t\neq 0$ and $t'\neq 0$, the product $xx'$ will be different from zero, and we can cancel the factors on both sides to get what we know and expect: $\gamma^2(1-v^2/c^2)=1$ i.e.

$\gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$