This Quantum World/Appendix/Relativity/Lorentz transformations

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[edit] 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, \mathbf{D}, and  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) = (a2 − bew2)t + (abw − bcw − bdw)x,
x = c(cx + dx + ewt) + d(cx + dx + ewt) − ew(at + bwx)
      = (c2 + 2cd + d2 − bew2)x + (cew + dew − aew)t,
y = c2y,
z = c2z.


The first of these equations tells us that

a2 − bew2 = 1  and  abw − bcw − bdw = 0.

The second tells us that

c2 + 2cd + d2 − bew2 = 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 e2 − bew2 = 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 = c2y tells us that c2 = 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 e2 − bew2 = 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.

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