# 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

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

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 $w\neq 0$ ), 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 $e\neq 0,$  we gather from the second that $e=-a.$

$y=c^{2}y$  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.