View source for This quantum world/Appendix/Relativity/Lorentz transformations

==Lorentz transformations (general form)==
We want to express the coordinates <math>t</math> and <math>\mathbf{r}=(x,y,z)</math> of an inertial frame&nbsp;<math>\mathcal{F}_1</math> in terms of the coordinates <math>t'</math> and <math>\mathbf{r}'=(x',y',z')</math> of another inertial frame&nbsp;<math>\mathcal{F}_2.</math> We will assume that the two frames meet the following conditions:

<br>
# their spacetime coordinate origins coincide (<math>t'{=}0,\mathbf{r}'{=}0</math> mark the same spacetime location as <math>t{=}0,\mathbf{r}{=}0</math>),
# their space axes are parallel, and
# <math>\mathcal{F}_2</math>&nbsp;moves with a constant velocity&nbsp;<math>\mathbf{w}</math> relative to&nbsp;<math>\mathcal{F}_1.</math>

<br>What we know at this point is that whatever moves with a constant velocity in&nbsp;<math>\mathcal{F}_1</math> will do so in&nbsp;<math>\mathcal{F}_2.</math> It follows that the transformation <math>t,\mathbf{r}\rightarrow t',\mathbf{r}'</math> maps straight lines in&nbsp;<math>\mathcal{F}_1</math> onto straight lines in&nbsp;<math>\mathcal{F}_2.</math> Coordinate lines of&nbsp;<math>\mathcal{F}_1,</math> in particular, will be mapped onto straight lines in&nbsp;<math>\mathcal{F}_2.</math> This tells us that the dashed coordinates are linear combinations of the undashed ones,

<br>
:<math>
t'=A\,t+\mathbf{B}\cdot\mathbf{r},\qquad \mathbf{r}'=C\,\mathbf{r}+(\mathbf{D}\cdot\mathbf{r})\mathbf{w}+\,t.</math>

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

<br>
:<math>
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.</math>

<br>Let's set <math>\mathbf{r}</math> equal to <math>\mathbf{w} t.</math> This implies that <math>\mathbf{r}'=(c+d+e)\mathbf{w} t.</math> As we are looking at the trajectory of an object at rest in&nbsp;<math>\mathcal{F}_2,</math> <math>\mathbf{r}'</math> must be constant. Hence,

:<math>
c+d+e=0.</math>

Let's write down the inverse transformation. Since <math>\mathcal{F}_1</math> moves with velocity <math>-\mathbf{w}</math> relative to <math>\mathcal{F}_2,</math> it is

<br>
:<math>
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'.</math>

<br>To make life easier for us, we now chose the space axes so that <math>\mathbf{w}=(w,0,0).</math> Then the above two (mutually inverse) transformations simplify to

<br>
:<math>
t'=at+bwx,\quad x'=cx+dx+ewt,\quad y'=cy,\quad z'=cz,</math>
:<math>
t=at'-bwx',\quad x=cx'+dx'-ewt',\quad y=cy',\quad z=cz'.</math>

<br>Plugging the first transformation into the second, we obtain

<br>
:<math>
t=a(at+bwx)-bw(cx+dx+ewt)=(a^2-bew^2)t+(abw-bcw-bdw)x,</math>
:<math>
x=c(cx+dx+ewt)+d(cx+dx+ewt)-ew(at+bwx)</math>
:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>=(c^2+2cd+d^2-bew^2)x+(cew+dew-aew)t,</math>
:<math>
y=c^2y,</math>
:<math>
z=c^2z.</math>

<br>The first of these equations tells us that

:<math>
a^2-bew^2=1</math> &nbsp;and&nbsp; <math>abw-bcw-bdw=0.</math>

The second tells us that

:<math>
c^2+2cd+d^2-bew^2=1</math> &nbsp;and&nbsp; <math>cew+dew-aew=0.</math>

Combining <math>abw-bcw-bdw=0</math> with <math>c+d+e=0</math> (and taking into account that <math>w\neq0</math>), we obtain <math>b(a+e)=0.</math>

Using <math>c+d+e=0</math> to eliminate&nbsp;<math>d,</math> we obtain <math>e^2-bew^2=1</math> and <math>e(a+e)=0.</math>

Since the first of the last two equations implies that <math>e\neq0,</math> we gather from the second that <math>e=-a.</math>

<math>y=c^2y</math>&nbsp;tells us that <math>c^2=1.</math> <math>c</math>&nbsp;must, in fact, be equal to&nbsp;1, since we have assumed that the space axes of the two frames a parallel (rather than antiparallel).

With <math>c=1</math> and <math>e=-a,</math> <math>c+d+e=0</math> yields <math>d=a-1.</math> Upon solving <math>e^2-bew^2=1</math> for&nbsp;<math>b,</math> we are left with expressions for <math>b, c, d,</math> and <math>e</math> depending solely on&nbsp;<math>a</math>:

:<math>
b={1-a^2\over aw^2},\quad c=1,\quad d=a-1,\quad e=-a.</math>

Quite an improvement!

To find the remaining function <math>a(w),</math> we consider a third inertial frame&nbsp;<math>\mathcal{F}_3,</math> which moves with velocity <math>\mathbf{v}=(v,0,0)</math> relative to&nbsp;<math>\mathcal{F}_2. </math> Combining the transformation from <math>\mathcal{F}_1</math> to&nbsp;<math>\mathcal{F}_2,</math>

<br>
:<math>
t'=a(w)\,t+{1-a^2(w)\over a(w)\,w}x,\qquad x'=a(w)\,x-a(w)\,wt,</math>

<br>with the transformation from <math>\mathcal{F}_2</math> to <math>\mathcal{F}_3,</math>

<br>
:<math>
t''=a(v)\,t'+\frac{1-a^2(v)}{a(v)\,v}x',\qquad x''=a(v)\,x'-a(v)\,vt',</math>

<br>we obtain the transformation from <math>\mathcal{F}_1</math> to <math>\mathcal{F}_3</math>:

<br>
:<math>
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]</math>
:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>
=\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,</math>
:<math>
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]</math>
:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>
=\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.</math>

<br>The direct transformation from <math>\mathcal{F}_1</math> to <math>\mathcal{F}_3</math> must have the same form as the transformations from <math>\mathcal{F}_1</math> to <math>\mathcal{F}_2</math> and from <math>\mathcal{F}_2</math> to <math>\mathcal{F}_3</math>, namely

<br>
:<math>
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,</math>

<br>where <math>u</math> is the speed of <math>\mathcal{F}_3</math> relative to&nbsp;<math>\mathcal{F}_1.</math> Comparison of the coefficients marked with stars yields two expressions for&nbsp;<math>a(u),</math> which of course must be equal:

<br>
:<math>
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}.</math>

<br>It follows that <math>\bigl[1-a^2(v)\bigr]\,a^2(w)w^2=\bigl[1-a^2(w)\bigr]\,a^2(v)v^2,</math> and this tells us that

<br>
:<math>
K={1-a^2(w)\over a^2(w)\,w^2}={1-a^2(v)\over a^2(v)\,v^2}</math>

<br>is a universal constant. Solving the first equality for <math>a(w),</math> we obtain

:<math>
a(w)=1/\sqrt{1+Kw^2}.</math>

This allows us to cast the transformation

:<math>
t'=at+bwx,\quad x'=cx+dx+ewt,\quad y'=cy,\quad z'=cz,</math>

into the form

<br>
:<math>
t'={t+Kwx\over\sqrt{1+Kw^2}},\quad x'={x-wt\over\sqrt{1+Kw^2}},\quad y'=y,\quad z'=z.</math>

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