# This Quantum World/Appendix/Relativity/Lorentz transformations

### Lorentz transformations (general form)

We want to express the coordinates ${\displaystyle t}$ and ${\displaystyle \mathbf {r} =(x,y,z)}$ of an inertial frame ${\displaystyle {\mathcal {F}}_{1}}$ in terms of the coordinates ${\displaystyle t'}$ and ${\displaystyle \mathbf {r} '=(x',y',z')}$ of another inertial frame ${\displaystyle {\mathcal {F}}_{2}.}$ We will assume that the two frames meet the following conditions:

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

What we know at this point is that whatever moves with a constant velocity in ${\displaystyle {\mathcal {F}}_{1}}$ will do so in ${\displaystyle {\mathcal {F}}_{2}.}$ It follows that the transformation ${\displaystyle t,\mathbf {r} \rightarrow t',\mathbf {r} '}$ maps straight lines in ${\displaystyle {\mathcal {F}}_{1}}$ onto straight lines in ${\displaystyle {\mathcal {F}}_{2}.}$ Coordinate lines of ${\displaystyle {\mathcal {F}}_{1},}$ in particular, will be mapped onto straight lines in ${\displaystyle {\mathcal {F}}_{2}.}$ This tells us that the dashed coordinates are linear combinations of the undashed ones,

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

We also know that the transformation from ${\displaystyle {\mathcal {F}}_{1}}$ to ${\displaystyle {\mathcal {F}}_{2}}$ can only depend on ${\displaystyle \mathbf {w} ,}$ so ${\displaystyle A,}$ ${\displaystyle \mathbf {B} ,}$ ${\displaystyle C,}$ and ${\displaystyle \mathbf {D} }$ are functions of ${\displaystyle \mathbf {w} .}$ Our task is to find these functions. The real-valued functions ${\displaystyle A}$ and ${\displaystyle C}$ actually can depend only on ${\displaystyle w=|\mathbf {w} |={}_{+}{\sqrt {\mathbf {w} \cdot \mathbf {w} }},}$ so ${\displaystyle A=a(w)}$ and ${\displaystyle C=c(w).}$ A vector function depending only on ${\displaystyle \mathbf {w} }$ must be parallel (or antiparallel) to ${\displaystyle \mathbf {w} ,}$ and its magnitude must be a function of ${\displaystyle w.}$ We can therefore write ${\displaystyle \mathbf {B} =b(w)\,\mathbf {w} ,}$ ${\displaystyle \mathbf {D} =[d(w)/w^{2}]\mathbf {w} ,}$ and ${\displaystyle =e(w)\,\mathbf {w} .}$ (It will become clear in a moment why the factor ${\displaystyle w^{-2}}$ is included in the definition of ${\displaystyle \mathbf {D} .}$) So,

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

Let's set ${\displaystyle \mathbf {r} }$ equal to ${\displaystyle \mathbf {w} t.}$ This implies that ${\displaystyle \mathbf {r} '=(c+d+e)\mathbf {w} t.}$ As we are looking at the trajectory of an object at rest in ${\displaystyle {\mathcal {F}}_{2},}$ ${\displaystyle \mathbf {r} '}$ must be constant. Hence,

${\displaystyle c+d+e=0.}$

Let's write down the inverse transformation. Since ${\displaystyle {\mathcal {F}}_{1}}$ moves with velocity ${\displaystyle -\mathbf {w} }$ relative to ${\displaystyle {\mathcal {F}}_{2},}$ it is

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

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

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

Plugging the first transformation into the second, we obtain

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

The first of these equations tells us that

${\displaystyle a^{2}-bew^{2}=1}$  and  ${\displaystyle abw-bcw-bdw=0.}$

The second tells us that

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

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

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

Since the first of the last two equations implies that ${\displaystyle e\neq 0,}$ we gather from the second that ${\displaystyle e=-a.}$

${\displaystyle y=c^{2}y}$ tells us that ${\displaystyle c^{2}=1.}$ ${\displaystyle c}$ 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 ${\displaystyle c=1}$ and ${\displaystyle e=-a,}$ ${\displaystyle c+d+e=0}$ yields ${\displaystyle d=a-1.}$ Upon solving ${\displaystyle e^{2}-bew^{2}=1}$ for ${\displaystyle b,}$ we are left with expressions for ${\displaystyle b,c,d,}$ and ${\displaystyle e}$ depending solely on ${\displaystyle a}$:

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

Quite an improvement!

To find the remaining function ${\displaystyle a(w),}$ we consider a third inertial frame ${\displaystyle {\mathcal {F}}_{3},}$ which moves with velocity ${\displaystyle \mathbf {v} =(v,0,0)}$ relative to ${\displaystyle {\mathcal {F}}_{2}.}$ Combining the transformation from ${\displaystyle {\mathcal {F}}_{1}}$ to ${\displaystyle {\mathcal {F}}_{2},}$

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

with the transformation from ${\displaystyle {\mathcal {F}}_{2}}$ to ${\displaystyle {\mathcal {F}}_{3},}$

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

we obtain the transformation from ${\displaystyle {\mathcal {F}}_{1}}$ to ${\displaystyle {\mathcal {F}}_{3}}$:

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

The direct transformation from ${\displaystyle {\mathcal {F}}_{1}}$ to ${\displaystyle {\mathcal {F}}_{3}}$ must have the same form as the transformations from ${\displaystyle {\mathcal {F}}_{1}}$ to ${\displaystyle {\mathcal {F}}_{2}}$ and from ${\displaystyle {\mathcal {F}}_{2}}$ to ${\displaystyle {\mathcal {F}}_{3}}$, namely

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

where ${\displaystyle u}$ is the speed of ${\displaystyle {\mathcal {F}}_{3}}$ relative to ${\displaystyle {\mathcal {F}}_{1}.}$ Comparison of the coefficients marked with stars yields two expressions for ${\displaystyle a(u),}$ which of course must be equal:

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

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

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

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

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

This allows us to cast the transformation

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

into the form

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

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