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.