This Quantum World/Appendix/Relativity/Lorentz transformations

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Lorentz transformations (general form)[edit]

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,


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





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


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.