# This Quantum World/Appendix/Relativity/Lorentz transformations

### Lorentz transformations (general form)

We want to express the coordinates $t$ and $\mathbf{r}=(x,y,z)$ of an inertial frame $\mathcal{F}_1$ in terms of the coordinates $t'$ and $\mathbf{r}'=(x',y',z')$ of another inertial frame $\mathcal{F}_2.$ We will assume that the two frames meet the following conditions:

1. their spacetime coordinate origins coincide ($t'{=}0,\mathbf{r}'{=}0$ mark the same spacetime location as $t{=}0,\mathbf{r}{=}0$),
2. their space axes are parallel, and
3. $\mathcal{F}_2$ moves with a constant velocity $\mathbf{w}$ relative to $\mathcal{F}_1.$

What we know at this point is that whatever moves with a constant velocity in $\mathcal{F}_1$ will do so in $\mathcal{F}_2.$ It follows that the transformation $t,\mathbf{r}\rightarrow t',\mathbf{r}'$ maps straight lines in $\mathcal{F}_1$ onto straight lines in $\mathcal{F}_2.$ Coordinate lines of $\mathcal{F}_1,$ in particular, will be mapped onto straight lines in $\mathcal{F}_2.$ This tells us that the dashed coordinates are linear combinations of the undashed ones,

$t'=A\,t+\mathbf{B}\cdot\mathbf{r},\qquad \mathbf{r}'=C\,\mathbf{r}+(\mathbf{D}\cdot\mathbf{r})\mathbf{w}+\,t.$

We also know that the transformation from $\mathcal{F}_1$ to $\mathcal{F}_2$ can only depend on $\mathbf{w},$ so $A,$ $\mathbf{B},$ $C,$ and $\mathbf{D}$ are functions of $\mathbf{w}.$ Our task is to find these functions. The real-valued functions $A$ and $C$ actually can depend only on $w=|\mathbf{w}|={}_+\sqrt{\mathbf{w}\cdot\mathbf{w}},$ so $A=a(w)$ and $C=c(w).$ A vector function depending only on $\mathbf{w}$ must be parallel (or antiparallel) to $\mathbf{w},$ and its magnitude must be a function of $w.$ We can therefore write $\mathbf{B}=b(w)\,\mathbf{w},$ $\mathbf{D}=[d(w)/w^2]\mathbf{w},$ and $=e(w)\,\mathbf{w}.$ (It will become clear in a moment why the factor $w^{-2}$ is included in the definition of $\mathbf{D}.$) So,

$t'=a(w)\,t+b(w)\,\mathbf{w}\cdot\mathbf{r},\qquad \mathbf{r}'=\displaystyle c(w)\,\mathbf{r}+ d(w){\mathbf{w}\cdot\mathbf{r}\over w^2}\mathbf{w}+e(w)\,\mathbf{w}\,t.$

Let's set $\mathbf{r}$ equal to $\mathbf{w} t.$ This implies that $\mathbf{r}'=(c+d+e)\mathbf{w} t.$ As we are looking at the trajectory of an object at rest in $\mathcal{F}_2,$ $\mathbf{r}'$ must be constant. Hence,

$c+d+e=0.$

Let's write down the inverse transformation. Since $\mathcal{F}_1$ moves with velocity $-\mathbf{w}$ relative to $\mathcal{F}_2,$ it is

$t=a(w)\,t'-b(w)\,\mathbf{w}\cdot\mathbf{r}',\qquad \mathbf{r}=\displaystyle c(w)\,\mathbf{r}'+ d(w){\mathbf{w}\cdot\mathbf{r}'\over w^2}\mathbf{w}-e(w)\,\mathbf{w}\,t'.$

To make life easier for us, we now chose the space axes so that $\mathbf{w}=(w,0,0).$ Then the above two (mutually inverse) transformations simplify to

$t'=at+bwx,\quad x'=cx+dx+ewt,\quad y'=cy,\quad z'=cz,$
$t=at'-bwx',\quad x=cx'+dx'-ewt',\quad y=cy',\quad z=cz'.$

Plugging the first transformation into the second, we obtain

$t=a(at+bwx)-bw(cx+dx+ewt)=(a^2-bew^2)t+(abw-bcw-bdw)x,$
$x=c(cx+dx+ewt)+d(cx+dx+ewt)-ew(at+bwx)$
$=(c^2+2cd+d^2-bew^2)x+(cew+dew-aew)t,$
$y=c^2y,$
$z=c^2z.$

The first of these equations tells us that

$a^2-bew^2=1$  and  $abw-bcw-bdw=0.$

The second tells us that

$c^2+2cd+d^2-bew^2=1$  and  $cew+dew-aew=0.$

Combining $abw-bcw-bdw=0$ with $c+d+e=0$ (and taking into account that $w\neq0$), we obtain $b(a+e)=0.$

Using $c+d+e=0$ to eliminate $d,$ we obtain $e^2-bew^2=1$ and $e(a+e)=0.$

Since the first of the last two equations implies that $e\neq0,$ we gather from the second that $e=-a.$

$y=c^2y$ tells us that $c^2=1.$ $c$ must, in fact, be equal to 1, since we have assumed that the space axes of the two frames a parallel (rather than antiparallel).

With $c=1$ and $e=-a,$ $c+d+e=0$ yields $d=a-1.$ Upon solving $e^2-bew^2=1$ for $b,$ we are left with expressions for $b, c, d,$ and $e$ depending solely on $a$:

$b={1-a^2\over aw^2},\quad c=1,\quad d=a-1,\quad e=-a.$

Quite an improvement!

To find the remaining function $a(w),$ we consider a third inertial frame $\mathcal{F}_3,$ which moves with velocity $\mathbf{v}=(v,0,0)$ relative to $\mathcal{F}_2.$ Combining the transformation from $\mathcal{F}_1$ to $\mathcal{F}_2,$

$t'=a(w)\,t+{1-a^2(w)\over a(w)\,w}x,\qquad x'=a(w)\,x-a(w)\,wt,$

with the transformation from $\mathcal{F}_2$ to $\mathcal{F}_3,$

$t''=a(v)\,t'+\frac{1-a^2(v)}{a(v)\,v}x',\qquad x''=a(v)\,x'-a(v)\,vt',$

we obtain the transformation from $\mathcal{F}_1$ to $\mathcal{F}_3$:

$t''=a(v)\left[a(w)\,t+{1-a^2(w)\over a(w)\,w}x\right]+{1-a^2(v)\over a(v)\,v} \Bigl[a(w)\,x-a(w)\,wt\Bigr]$
$=\underbrace{\left[a(v)\,a(w)-{1-a^2(v)\over a(v)\,v}a(w)\,w\right]}_{\textstyle\star}t+ \Bigl[\dots\Bigr]\,x,$
$x''=a(v)\Bigl[a(w)\,x-a(w)\,wt\Bigr]-a(v)\,v\left[a(w)\,t+{1-a^2(w)\over a(w)\,w}x\right]$
$=\underbrace{\left[a(v)\,a(w)-a(v)\,v{1-a^2(w)\over a(w)\,w}\right]}_{\textstyle\star\,\star}x-\Bigl[\dots\Bigr]\,t.$

The direct transformation from $\mathcal{F}_1$ to $\mathcal{F}_3$ must have the same form as the transformations from $\mathcal{F}_1$ to $\mathcal{F}_2$ and from $\mathcal{F}_2$ to $\mathcal{F}_3$, namely

$t''=\underbrace{a(u)}_{\textstyle\star}t+{1-a^2(u)\over a(u)\,u}\,x,\qquad x''=\underbrace{a(u)}_{\textstyle\star\,\star}x-a(u)\,ut,$

where $u$ is the speed of $\mathcal{F}_3$ relative to $\mathcal{F}_1.$ Comparison of the coefficients marked with stars yields two expressions for $a(u),$ which of course must be equal:

$a(v)\,a(w)-{1-a^2(v)\over a(v)\,v}a(w)\,w=a(v)\,a(w)-a(v)\,v{1-a^2(w)\over a(w)\,w}.$

It follows that $\bigl[1-a^2(v)\bigr]\,a^2(w)w^2=\bigl[1-a^2(w)\bigr]\,a^2(v)v^2,$ and this tells us that

$K={1-a^2(w)\over a^2(w)\,w^2}={1-a^2(v)\over a^2(v)\,v^2}$

is a universal constant. Solving the first equality for $a(w),$ we obtain

$a(w)=1/\sqrt{1+Kw^2}.$

This allows us to cast the transformation

$t'=at+bwx,\quad x'=cx+dx+ewt,\quad y'=cy,\quad z'=cz,$

into the form

$t'={t+Kwx\over\sqrt{1+Kw^2}},\quad x'={x-wt\over\sqrt{1+Kw^2}},\quad y'=y,\quad z'=z.$

Trumpets, please! We have managed to reduce five unknown functions to a single constant.