This Quantum World/Appendix/Relativity/Composition theorem and proper time

Composition of velocities
In fact, there are only three physically distinct possibilities. (If $$K\neq0,$$ the magnitude of $$K$$ depends on the choice of units, and this tells us something about us rather than anything about the physical world.)

The possibility $$K=0$$ yields the Galilean transformations of Newtonian ("non-relativistic") mechanics:



t'=t,\quad \mathbf{r}'=\mathbf{r}-\mathbf{w} t,\quad u=v+w,\quad ds=dt.$$

(The common practice of calling theories with this transformation law "non-relativistic" is inappropriate, inasmuch as they too satisfy the principle of relativity.) In the remainder of this section we assume that $$K\neq0.$$

Suppose that object $$C$$ moves with speed $$v$$ relative to object $$B,$$ and that this moves with speed $$w$$ relative to object $$A.$$ If $$B$$ and $$C$$ move in the same direction, what is the speed $$u$$ of $$C$$ relative to $$A$$? In the previous section we found that


 * $$a(u)=a(v)\,a(w)-{1-a^2(v)\over a(v)\,v}a(w)\,w,$$

and that


 * $$K={1-a^2(v)\over a^2(v)\,v^2}.$$

This allows us to write



a(u)=a(v)\,a(w)-{1-a^2(v)\over a^2(v)\,v^2}a(v)\,v\,a(w)\,w= a(v)\,a(w)(1-Kvw).$$

Expressing $$a$$ in terms of $$K$$ and the respective velocities, we obtain



{1\over\sqrt{1+Ku^2}}={1-Kvw\over \sqrt{1+Kv^2}\sqrt{1+Kw^2}},$$

which implies that



1+Ku^2={(1+Kv^2)(1+Kw^2)\over(1-Kvw)^2}.$$

We massage this into



Ku^2={(1+Kv^2)(1+Kw^2)-(1-Kvw)^2\over(1-Kvw)^2}={K(v+w)^2\over(1-Kvw)^2},$$

divide by $$K,$$ and end up with:



u={v+w\over1-Kvw}.$$

Thus, unless $$K=0,$$ we don't get the speed of $$C$$ relative to $$A$$ by simply adding the speed of $$C$$ relative to $$B$$ to the speed of $$B$$ relative to $$A$$.

Proper time
Consider an infinitesimal segment $$d\mathcal{C}$$ of a spacetime path $$\mathcal{C}.$$ In $$\mathcal{F}_1$$ it has the components $$(dt,dx,dy,dz),$$ in $$\mathcal{F}_2$$ it has the components $$(dt',dx',dy',dz').$$ Using the Lorentz transformation in its general form,



t'={t+Kwx\over\sqrt{1+Kw^2}},\quad x'={x-wt\over\sqrt{1+Kw^2}},\quad y'=y,\quad z'=z,$$

it is readily shown that



(dt')^2+K\,d\mathbf{r}'\cdot d\mathbf{r}'=dt^2+K\,d\mathbf{r}\cdot d\mathbf{r}.$$

We conclude that the expression



ds^2=dt^2+K\,d\mathbf{r}\cdot d\mathbf{r}=dt^2+K(dx^2+dy^2+dz^2)$$

is invariant under this transformation. It is also invariant under rotations of the spatial axes (why?) and translations of the spacetime coordinate origin. This makes $$ds$$ a 4-scalar.

What is the physical significance of $$ds$$?

A clock that travels along $$d\mathcal{C}$$ is at rest in any frame in which $$d\mathcal{C}$$ lacks spatial components. In such a frame, $$ds^2=dt^2.$$ Hence $$ds$$ is the time it takes to travel along $$d\mathcal{C}$$ as measured by a clock that travels along $$d\mathcal{C}.$$ $$ds$$ is the proper time (or proper duration) of $$d\mathcal{C}.$$ The proper time (or proper duration) of a finite spacetime path $$\mathcal{C},$$ accordingly, is



\int_\mathcal{C} ds=\int_\mathcal{C}\sqrt{dt^2+K\,d\mathbf{r}\cdot d\mathbf{r}}=\int_\mathcal{C} dt\sqrt{1+Kv^2}.$$

An invariant speed
If $$K<0,$$ then there is a universal constant $$c\equiv1/\sqrt{-K}$$ with the dimension of a velocity, and we can cast $$u=v+w/(1-Kvw)$$ into the form



u={v+w\over1+vw/c^2}.$$

If we plug in $$v=w=c/2,$$ then instead of the Galilean $$u=v+w=c,$$ we have $$u={4\over5}c<c.$$ More intriguingly, if object $$O$$ moves with speed $$c$$ relative to $$\mathcal{F}_2,$$ and if $$\mathcal{F}_2$$ moves with speed $$w$$ relative to $$\mathcal{F}_1,$$ then $$O$$ moves with the same speed $$c$$ relative to $$\mathcal{F}_1$$: $$(w+c)/(1+wc/c^2)=c.$$ The speed of light $$c$$ thus is an invariant speed: whatever travels with it in one inertial frame, travels with the same speed in every inertial frame.

Starting from



ds^2=(dt')^2-d\mathbf{r}'\cdot d\mathbf{r}'/c^2=dt^2-d\mathbf{r}\cdot d\mathbf{r}/c^2,$$

we arrive at the same conclusion: if $$O$$ travels with $$c$$ relative to $$\mathcal{F}_1,$$ then it travels the distance $$dr=c\,dt$$ in the time $$dt.$$ Therefore $$ds^2=dt^2-dr^2/c^2=0.$$ But then $$(dt')^2-(dr')^2/c^2=0,$$ and this implies $$dr'=c\,dt'.$$ It follows that $$O$$ travels with the same speed $$c$$ relative to $$\mathcal{F}_2.$$

An invariant speed also exists if $$K=0,$$ but in this case it is infinite: whatever travels with infinite speed in one inertial frame — it takes no time to get from one place to another — does so in every inertial frame.

The existence of an invariant speed prevents objects from making U-turns in spacetime. If $$K=0,$$ it obviously takes an infinite amount of energy to reach $$v=\infty.$$ Since an infinite amount of energy isn't at our disposal, we cannot start vertically in a spacetime diagram and then make a U-turn (that is, we cannot reach, let alone "exceed", a horizontal slope. ("Exceeding" a horizontal slope here means changing from a positive to a negative slope, or from going forward to going backward in time.)

If $$K<0,$$ it takes an infinite amount of energy to reach even the finite speed of light. Imagine you spent a finite amount of fuel accelerating from 0 to $$0.1\,c.$$ In the frame in which you are now at rest, your speed is not a whit closer to the speed of light. And this remains true no matter how many times you repeat the procedure. Thus no finite amount of energy can make you reach, let alone "exceed", a slope equal to $$1/c.$$ ("Exceeding" a slope equal to $$1/c$$ means attaining a smaller slope. As we will see, if we were to travel faster than light in any one frame, then there would be frames in which we travel backward in time.)