### The case against

In a hypothetical world with

${displaystyle K>0}$${displaystyle kequiv 1/{sqrt {K}}}$

(a universal constant with the dimension of a velocity), and we can cast

${displaystyle u=v+w/(1-Kvw)}$into the form

If we plug in

${displaystyle v=w=k/2,}$then instead of the Galilean

${displaystyle u=v+w=k}$we have

${displaystyle u={4 over 3}k>k.}$${displaystyle v=w=k,}$

we obtain

${displaystyle u=infty }$: if object

${displaystyle O}$travels with speed

${displaystyle k}$relative to

${displaystyle {mathcal {F}}_{2},}$and if

${displaystyle {mathcal {F}}_{2}}$travels with speed

${displaystyle k}$relative to

${displaystyle {mathcal {F}}_{1}}$(in the same direction), then

${displaystyle O}$travels with an infinite speed relative to

${displaystyle {mathcal {F}}_{1}}$! And if

${displaystyle O}$travels with

${displaystyle 2k}$relative to

${displaystyle {mathcal {F}}_{2}}$and

${displaystyle {mathcal {F}}_{2}}$travels with

${displaystyle 2k}$relative to

${displaystyle {mathcal {F}}_{1},}$${displaystyle O}$

‘s speed relative to

${displaystyle {mathcal {F}}_{1}}$is negative:

${displaystyle u=-{4 over 3}k.}$

If we use units in which

${displaystyle K=k=1,}$then the invariant proper time associated with an infinitesimal path segment is related to the segment’s inertial components via

This is the 4-dimensional version of the 3-scalar

${displaystyle dx^{2}+dy^{2}+dz^{2},}$which is invariant under rotations in space. Hence if

${displaystyle K}$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

${displaystyle k{=}1}$) from the assumption that the

${displaystyle x’}$and

${displaystyle t’}$axes are rotated relative to the

${displaystyle x}$and

${displaystyle t}$axes.

The speed of an object

${displaystyle O}$following the dotted line is

${displaystyle w=cot(alpha +ta)}$relative to

${displaystyle {mathcal {F}}’,}$the speed of

${displaystyle {mathcal {F}}’}$relative to

${displaystyle {mathcal {F}}}$is

${displaystyle v=tan alpha ,}$and the speed of

${displaystyle O}$relative to

${displaystyle {mathcal {F}}}$is

${displaystyle u=cot beta .}$Invoking the trigonometric relation

we conclude that

${displaystyle {1 over w}={v+1/u over 1-v/u}.}$Solving for

${displaystyle u,}$we obtain

${displaystyle u={v+w over 1-vw}.}$

How can we rule out the *a priori* possibility that

${displaystyle K>0.}$

${displaystyle K<0,}$

and if we use natural units, in which

${displaystyle c=1,}$we have that

As far as physics is concerned, the difference between the positive sign in front of

${displaystyle dt}$and the negative signs in front of

${displaystyle dx,}$${displaystyle dy,}$

and

${displaystyle dz}$ is the *only* objective difference between time and the spatial dimensions of spacetime. If

were positive, not even this difference would exist.

### The case against zero K[edit]

And what argues against the possibility that

${displaystyle K=0}$?

Recall the propagator for a free and stable particle:

If

${displaystyle K}$were to vanish, we would have

${displaystyle ds^{2}=dt^{2}.}$There would be no difference between inertial time and proper time, and every spacetime path leading from

${displaystyle A}$to

${displaystyle B}$would contribute the same amplitude

${displaystyle e^{-ib(t_{B}-t_{A})}}$to the propagator

${displaystyle langle B|Arangle ,}$which would be hopelessly divergent as a result. Worse,

${displaystyle langle B|Arangle }$would be independent of the distance between

${displaystyle A}$and

${displaystyle B.}$To obtain well-defined, finite probabilities, cancellations (“destructive interference”) must occur, and this rules out that

${displaystyle K=0.}$

### The actual Lorentz transformations[edit]

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

Let’s explore them diagrammatically, using natural units (

${displaystyle c=1}$). Setting

${displaystyle t’=0,}$we have

${displaystyle t=wx.}$This tells us that the slope of the

${displaystyle x’}$axis relative to the undashed frame is

${displaystyle w=tan alpha .}$Setting

${displaystyle x’=0,}$we have

${displaystyle t=x/w.}$This tells us that the slope of the

${displaystyle t’}$axis is

${displaystyle 1/w.}$ The dashed axes are thus rotated by the same angle in *opposite* directions; if the

axis is rotated clockwise relative to the

${displaystyle t}$axis, then the

${displaystyle x’}$axis is rotated counterclockwise relative to the

${displaystyle x}$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

${displaystyle w=tan alpha }$relative to

${displaystyle {mathcal {F}}.}$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

${displaystyle c.}$

Here is how it’s done:

Light signals are sent from clock 2 (event

${displaystyle A}$) and are reflected by clocks 1 and 3 (events

${displaystyle B}$and

${displaystyle C,}$respectively). The distances between the clocks are adjusted so that the reflected signals arrive simultaneously at clock 2 (event

${displaystyle D}$). 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

${displaystyle {mathcal {F}}’,}$where the clocks are at rest, the signals from

${displaystyle A}$have traveled equal distances when they reach the first and the third clock, respectively. Since they also have traveled with the same speed

${displaystyle c,}$they have traveled for equal times. Therefore the clocks must be synchronized so that

${displaystyle B}$and

${displaystyle C}$ are simultaneous. We may use the *worldline* of clock 1 as the

axis and the straight line through

${displaystyle B}$and

${displaystyle C}$as the

${displaystyle x’}$axis. It is readily seen that the three angles

${displaystyle ta}$in the above diagram are equal. From this and the fact that the slope of the signal from

${displaystyle B}$to

${displaystyle D}$equals 1 (given that

${displaystyle c{=}1}$), the equality of the two angles

${displaystyle alpha }$follows.

Simultaneity thus depends on the language — the inertial frame — that we use to describe a physical situation. If two events

${displaystyle E_{1},E_{2}}$are simultaneous in one frame, then there are frames in which

${displaystyle E_{1}}$hapens after

${displaystyle E_{2}}$as well as frames in which

${displaystyle E_{1}}$hapens before

${displaystyle E_{2}.}$

Where do we place the unit points on the space and time axes? The unit point of the time axis of

${displaystyle {mathcal {F}}’}$has the coordinates

${displaystyle t’=1,x’=0}$and satisfies

${displaystyle t^{2}-x^{2}=1,}$as we gather from the version

${displaystyle (t’)^{2}-(x’)^{2}=t^{2}-x^{2}}$of (ref{ds2}). The unit point of the

${displaystyle x’}$axis has the coordinates

${displaystyle t’=0,x’=1}$and satisfies

${displaystyle x^{2}-t^{2}=1.}$The loci of the unit points of the space and time axes are the hyperbolas that are defined by these equations: