16 Why special relativity? (II)

Let dx, dy, dz, and dt be the coor­di­nate inter­vals asso­ci­ated with an infin­i­tes­imal path seg­ment dC. Another impor­tant con­se­quence of the trans­for­ma­tion (2.14.1) is that the expression

(2.15.1)   ds2 = dt2 + K(dx2 + dy2 + dz2)

is invariant under it: the value of ds2 does not change if dx, dy, dz, and dt are replaced by dx’, dy’, dz’, and dt’. If K=0, this simply means that the same time coor­di­nate can be used for all iner­tial frames; t is the “absolute time” of New­tonian physics.

To dis­cover the meaning of ds if K<0, we con­sider a clock that fol­lows a space­time path C. During any infin­i­tes­imal time interval, the incre­ment of time it shows equals the incre­ment of the time coor­di­nate of any iner­tial frame rel­a­tive to which it is “momen­tarily” (that is: during that infin­i­tes­imal time interval) at rest. Since dx=0, dy=0, and dz=0 in such a frame, ds2 equals dt2 in such a frame. If the clock travels with vari­able speed, the time that passes “on” it — its proper time — dif­fers from iner­tial time regard­less of which iner­tial frame is used.

Observe that if K is pos­i­tive, Eq. (2.15.1) con­tains nothing that makes the time coor­di­nate dis­tinct from any space coor­di­nate. (To make this com­pletely obvious, choose units in which K=1.) This con­firms our pre­vious con­clu­sion that no objec­tive dif­fer­ence exists between a space axis and the time axis if K>0.

If K≤0, there exists an invariant speed. (What­ever moves with an invariant speed rel­a­tive to one iner­tial frame, moves with exactly the same speed rel­a­tive to any other iner­tial frame.)

If K=0, the invariant speed is infi­nite: what travels with infi­nite speed rel­a­tive to one iner­tial frame, does so rel­a­tive to every iner­tial frame. Dif­fer­ently put, if the depar­ture of an object in one place and its arrival in another are simul­ta­neous with respect to one iner­tial frame, they are so with respect to every iner­tial frame. Simul­taneity is absolute.

If K<0, the invariant speed is finite and is given by c = 1/√(−K). Infi­nite or finite, the invariant speed cannot be attained by any object that can be at rest, inas­much as this would require an infi­nite supply of energy. And since one has to attain the invariant speed before one can make a U–turn in a space­time plane con­taining the time axis, it is the exis­tence of an invariant speed that makes it impos­sible to go back in time by per­forming such a U–turn.

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The case against K = 0

The invariant interval of proper time ds is the “length” of the path seg­ment dC — the func­tion of dx, dy, dz, and dt that we were looking for. With it, the prop­a­gator for a free and stable par­ticle (2.13.2) takes the form

(2.15.2)   <B,tB|A,tA> = ∫DC [1: −b∫C ds] = ∫DC [1: −b∫C dt² + K(dx² + dy² + dz²) ].

What jumps out at us straight­away is that if K were zero, every space­time path from (A,tA) to (B,tB) would con­tribute the same ampli­tude [1: −b∫C dt] = [1: −b(tB−tA)] to the prop­a­gator (2.15.2), which would be infi­nite and there­fore phys­i­cally mean­ing­less as a result — as well as inde­pen­dent of the dis­tance between A and B! If it is to yield finite prob­a­bil­i­ties, can­cel­la­tions (“destruc­tive inter­fer­ence”) must occur. The com­plex num­bers [1: −bs(C)] = [1: −b∫C ds] must not be the same for every path C from (A,tA) to (B,tB). Con­sid­ered as vec­tors in a plane, they must not all point in the same direc­tion. (While this does not quite rule out the option K=0 — after all, there is such a thing as non-​​relativistic quantum mechanics — the manner in which the non-​​relativistic theory is obtained from the rel­a­tivistic one makes it clear that the former can only be an approx­i­ma­tion. It is useful as long as the particle’s speed v is so small com­pared to the invariant speed c that all powers of v2/​c2 but the first can be ignored.)

With K = −1/​c2 < 0, the trans­for­ma­tions (2.14.1) become the famous Lorentz trans­for­ma­tions, which lie at the heart of the spe­cial theory of relativity:

(2.15.3)   t’ = (t − wx/c²)/√1 − w²/c²,   x’ = (x − wt)/√1 − w²/c²,   y’ = y,   z’ = z.

It should be stressed that we have derived this trans­for­ma­tion law without invoking the con­stancy of the speed of light. Instead, we found that a finite invariant speed must exist. This turns out to be the speed with which light prop­a­gates (in vacuum).

It may not have escaped your notice that the proper time interval ds deter­mined by Eq. (2.15.1) may be pos­i­tive as well as neg­a­tive. As it hap­pens, it is both. There are two types of par­ti­cles: the “ordi­nary” par­ti­cles, for which it is pos­i­tive, and their so-​​called antipar­ti­cles, for which it is neg­a­tive.[1]

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1. [↑] Costella, J.P., McKellar, B.H.J., and Rawl­inson, A.A. (1997). Clas­sical antipar­ti­cles, Amer­ican Journal of Physics 65 (9), 835–841.