This Quantum World/Appendix/Relativity/Lorentz transformations

Lorentz transformations (general form)

We want to express the coordinates $t$ and $𝐫=(x,y,z)$ of an inertial frame ${ℱ}_1$ in terms of the coordinates $t^{\prime }$ and ${𝐫}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ of another inertial frame ${ℱ}_2.$ We will assume that the two frames meet the following conditions:

1. their spacetime coordinate origins coincide ($t^{\prime }=0,{𝐫}^{\prime }=0$ mark the same spacetime location as $t=0,𝐫=0$),
2. their space axes are parallel, and
3. ${ℱ}_2$ moves with a constant velocity $𝐰$ relative to ${ℱ}_1.$

What we know at this point is that whatever moves with a constant velocity in ${ℱ}_1$ will do so in ${ℱ}_2.$ It follows that the transformation $t,𝐫\to t^{\prime },{𝐫}^{\prime }$ maps straight lines in ${ℱ}_1$ onto straight lines in ${ℱ}_2.$ Coordinate lines of ${ℱ}_1,$ in particular, will be mapped onto straight lines in ${ℱ}_2.$ This tells us that the dashed coordinates are linear combinations of the undashed ones,

$t^{\prime }=At+𝐁\cdot 𝐫,{𝐫}^{\prime }=C𝐫+(𝐃\cdot 𝐫)𝐰+t.$

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

$t^{\prime }=a(w)t+b(w)𝐰\cdot 𝐫,{𝐫}^{\prime }=c(w)𝐫+d(w)\frac{𝐰\cdot 𝐫}{w^2}𝐰+e(w)𝐰t.$

Let's set $𝐫$ equal to $𝐰t.$ This implies that ${𝐫}^{\prime }=(c+d+e)𝐰t.$ As we are looking at the trajectory of an object at rest in ${ℱ}_2,$ ${𝐫}^{\prime }$ must be constant. Hence,

$c+d+e=0.$

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

$t=a(w)t^{\prime }-b(w)𝐰\cdot {𝐫}^{\prime },𝐫=c(w){𝐫}^{\prime }+d(w)\frac{𝐰\cdot {𝐫}^{\prime }}{w^2}𝐰-e(w)𝐰t^{\prime }.$

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

$t^{\prime }=at+bwx,x^{\prime }=cx+dx+ewt,y^{\prime }=cy,z^{\prime }=cz,$

$t=a{t}^{\prime }-bw{x}^{\prime },x=c{x}^{\prime }+d{x}^{\prime }-ew{t}^{\prime },y=c{y}^{\prime },z=c{z}^{\prime }.$

Plugging the first transformation into the second, we obtain

$t=a(at+bwx)-bw(cx+dx+ewt)=(a^2-be{w}^2)t+(abw-bcw-bdw)x,$

$x=c(cx+dx+ewt)+d(cx+dx+ewt)-ew(at+bwx)$

     $=(c^2+2cd+d^2-be{w}^2)x+(cew+dew-aew)t,$

$y=c^{2}y,$

$z=c^{2}z.$

The first of these equations tells us that

$a^2-be{w}^2=1$  and  $abw-bcw-bdw=0.$

The second tells us that

$c^2+2cd+d^2-be{w}^2=1$  and  $cew+dew-aew=0.$

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

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

Since the first of the last two equations implies that $e\ne 0,$ 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 $e^2-be{w}^2=1$ for $b,$ we are left with expressions for $b,c,d,$ and $e$ depending solely on $a$:

$b=\frac{1-a^2}{a{w}^2},c=1,d=a-1,e=-a.$

Quite an improvement!

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

$t^{\prime }=a(w)t+\frac{1-a^2(w)}{a(w)w}x,x^{\prime }=a(w)x-a(w)wt,$

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

$t^{″}=a(v)t^{\prime }+\frac{1-a^2(v)}{a(v)v}{x}^{\prime },x^{″}=a(v)x^{\prime }-a(v)v{t}^{\prime },$

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

$t^{″}=a(v)[a(w)t+\frac{1-a^2(w)}{a(w)w}x]+\frac{1-a^2(v)}{a(v)v}[a(w)x-a(w)wt]$

     $=\underset{\star }{\underbrace{[a(v)a(w)-\frac{1-a^2(v)}{a(v)v}a(w)w]}}t+[\dots ]x,$

$x^{″}=a(v)[a(w)x-a(w)wt]-a(v)v[a(w)t+\frac{1-a^2(w)}{a(w)w}x]$

     $=\underset{\star \star }{\underbrace{[a(v)a(w)-a(v)v\frac{1-a^2(w)}{a(w)w}]}}x-[\dots ]t.$

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

$t^{″}=\underset{\star }{\underbrace{a(u)}}t+\frac{1-a^2(u)}{a(u)u}x,x^{″}=\underset{\star \star }{\underbrace{a(u)}}x-a(u)ut,$

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

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

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

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

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

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

This allows us to cast the transformation

$t^{\prime }=at+bwx,x^{\prime }=cx+dx+ewt,y^{\prime }=cy,z^{\prime }=cz,$

into the form

$t^{\prime }=\frac{t+Kwx}{\sqrt{1+K{w}^2}},x^{\prime }=\frac{x-wt}{\sqrt{1+K{w}^2}},y^{\prime }=y,z^{\prime }=z.$

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