# This Quantum World/Appendix/Relativity/Lorentz

### The case against $K>0$ In a hypothetical world with $K>0$  we can define $k\equiv 1/{\sqrt {K}}$  (a universal constant with the dimension of a velocity), and we can cast $u=v+w/(1-Kvw)$  into the form

$u={v+w \over 1-vw/k^{2}}.$

If we plug in $v=w=k/2,$  then instead of the Galilean $u=v+w=k$  we have $u={4 \over 3}k>k.$  Worse, if we plug in $v=w=k,$  we obtain $u=\infty$ : if object $O$  travels with speed $k$  relative to ${\mathcal {F}}_{2},$  and if ${\mathcal {F}}_{2}$  travels with speed $k$  relative to ${\mathcal {F}}_{1}$  (in the same direction), then $O$  travels with an infinite speed relative to ${\mathcal {F}}_{1}$ ! And if $O$  travels with $2k$  relative to ${\mathcal {F}}_{2}$  and ${\mathcal {F}}_{2}$  travels with $2k$  relative to ${\mathcal {F}}_{1},$  $O$ 's speed relative to ${\mathcal {F}}_{1}$  is negative: $u=-{4 \over 3}k.$

If we use units in which $K=k=1,$  then the invariant proper time associated with an infinitesimal path segment is related to the segment's inertial components via

$ds^{2}=dt^{2}+dx^{2}+dy^{2}+dz^{2}.$

This is the 4-dimensional version of the 3-scalar $dx^{2}+dy^{2}+dz^{2},$  which is invariant under rotations in space. Hence if $K$  is positive, the transformations between inertial systems are rotations in spacetime. I guess you now see why in this hypothetical world the composition of two positive speeds can be a negative speed.

Let us confirm this conclusion by deriving the composition theorem (for $k{=}1$ ) from the assumption that the $x'$  and $t'$  axes are rotated relative to the $x$  and $t$  axes.

The speed of an object $O$  following the dotted line is $w=\cot(\alpha +ta)$  relative to ${\mathcal {F}}',$  the speed of ${\mathcal {F}}'$  relative to ${\mathcal {F}}$  is $v=\tan \alpha ,$  and the speed of $O$  relative to ${\mathcal {F}}$  is $u=\cot \beta .$  Invoking the trigonometric relation

$\tan(\alpha +\beta )={\tan \alpha +\tan \beta \over 1-\tan \alpha \tan \beta },$

we conclude that ${1 \over w}={v+1/u \over 1-v/u}.$  Solving for $u,$  we obtain $u={v+w \over 1-vw}.$

How can we rule out the a priori possibility that $K>0$ ? As shown in the body of the book, the stability of matter — to be precise, the existence of stable objects that (i) have spatial extent (they "occupy" space) and (ii) are composed of a finite number of objects that lack spatial extent (they don't "occupy" space) — rests on the existence of relative positions that are (a) more or less fuzzy and (b) independent of time. Such relative positions are described by probability distributions that are (a) inhomogeneous in space and (b) homogeneous in time. Their objective existence thus requires an objective difference between spactime's temporal dimension and its spatial dimensions. This rules out the possibility that $K>0.$

How? If $K<0,$  and if we use natural units, in which $c=1,$  we have that

$ds^{2}=+\,dt^{2}-dx^{2}-dy^{2}-dz^{2}.$

As far as physics is concerned, the difference between the positive sign in front of $dt$  and the negative signs in front of $dx,$  $dy,$  and $dz$  is the only objective difference between time and the spatial dimensions of spacetime. If $K$  were positive, not even this difference would exist.

### The case against zero K

And what argues against the possibility that $K=0$ ?

Recall the propagator for a free and stable particle:

$\langle B|A\rangle =\int {\mathcal {DC}}e^{-ibs[{\mathcal {C}}]}.$

If $K$  were to vanish, we would have $ds^{2}=dt^{2}.$  There would be no difference between inertial time and proper time, and every spacetime path leading from $A$  to $B$  would contribute the same amplitude $e^{-ib(t_{B}-t_{A})}$  to the propagator $\langle B|A\rangle ,$  which would be hopelessly divergent as a result. Worse, $\langle B|A\rangle$  would be independent of the distance between $A$  and $B.$  To obtain well-defined, finite probabilities, cancellations ("destructive interference") must occur, and this rules out that $K=0.$

### The actual Lorentz transformations

In the real world, therefore, the Lorentz transformations take the form

$t'={t-wx/c^{2} \over {\sqrt {1-w^{2}/c^{2}}}},\quad x'={x-wt \over {\sqrt {1-w^{2}/c^{2}}}},\quad y'=y,\quad z'=z.$

Let's explore them diagrammatically, using natural units ($c=1$ ). Setting $t'=0,$  we have $t=wx.$  This tells us that the slope of the $x'$  axis relative to the undashed frame is $w=\tan \alpha .$  Setting $x'=0,$  we have $t=x/w.$  This tells us that the slope of the $t'$  axis is $1/w.$  The dashed axes are thus rotated by the same angle in opposite directions; if the $t'$  axis is rotated clockwise relative to the $t$  axis, then the $x'$  axis is rotated counterclockwise relative to the $x$  axis.

We arrive at the same conclusion if we think about the synchronization of clocks in motion. Consider three clocks (1,2,3) that travel with the same speed $w=\tan \alpha$  relative to ${\mathcal {F}}.$  To synchronize them, we must send signals from one clock to another. What kind of signals? If we want our synchronization procedure to be independent of the language we use (that is, independent of the reference frame), then we must use signals that travel with the invariant speed $c.$

Here is how it's done:

Light signals are sent from clock 2 (event $A$ ) and are reflected by clocks 1 and 3 (events $B$  and $C,$  respectively). The distances between the clocks are adjusted so that the reflected signals arrive simultaneously at clock 2 (event $D$ ). This ensures that the distance between clocks 1 and 2 equals the distance between clocks 2 and 3, regardless of the inertial frame in which they are compared. In ${\mathcal {F}}',$  where the clocks are at rest, the signals from $A$  have traveled equal distances when they reach the first and the third clock, respectively. Since they also have traveled with the same speed $c,$  they have traveled for equal times. Therefore the clocks must be synchronized so that $B$  and $C$  are simultaneous. We may use the worldline of clock 1 as the $t'$  axis and the straight line through $B$  and $C$  as the $x'$  axis. It is readily seen that the three angles $ta$  in the above diagram are equal. From this and the fact that the slope of the signal from $B$  to $D$  equals 1 (given that $c{=}1$ ), the equality of the two angles $\alpha$  follows.

Simultaneity thus depends on the language — the inertial frame — that we use to describe a physical situation. If two events $E_{1},E_{2}$  are simultaneous in one frame, then there are frames in which $E_{1}$  hapens after $E_{2}$  as well as frames in which $E_{1}$  hapens before $E_{2}.$

Where do we place the unit points on the space and time axes? The unit point of the time axis of ${\mathcal {F}}'$  has the coordinates $t'=1,x'=0$  and satisfies $t^{2}-x^{2}=1,$  as we gather from the version $(t')^{2}-(x')^{2}=t^{2}-x^{2}$  of (\ref{ds2}). The unit point of the $x'$  axis has the coordinates $t'=0,x'=1$  and satisfies $x^{2}-t^{2}=1.$  The loci of the unit points of the space and time axes are the hyperbolas that are defined by these equations: