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:

their spacetime coordinate origins coincide ( $t'{=}0,\mathbf {r} '{=}0$ mark the same spacetime location as $t{=}0,\mathbf {r} {=}0$ ),
their space axes are parallel, and
${\mathcal {F}}_{2}$ 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: