Transformation of Velocity

Consider a particle in some inertial reference frame, $S$, that moves at the fixed (subluminal) velocity ${\bf u}= (u_x,\,u_y,\,u_z)$. Suppose that the particle is located at the origin at time $t=0$. It follows that, at time $t$, the particle is located at point ${\bf r}= (u_x\,t,\,u_y\,t,\,u_z\,t)$. Let us observe the motion of the particle in a second inertial frame, $S'$, that moves with (subluminal) velocity ${\bf v} = v\,{\bf e}_x$ with respect to $S$, and is in a standard configuration with respect to $S$. It follows from Equations (3.100)–(3.103) that the particle is located at the origin of $S'$ at $t'=0$. The location of the particle in $S'$, at time $t'$, can be written ${\bf r}'= (u_x'\,t',\,u_y'\,t',\,u_z'\,t')$, where ${\bf u}'= (u_x',\,u_y',\,u_z')$ is the particle's velocity in $S'$. It follows from Equations (3.100)–(3.103) that

$$u_x'\,t' = \gamma\left(u_x\,t-v\,t\right),$$ (3.118)

$$u_y'\,t' =u_y\,t,$$ (3.119)

$$u_z'\,t' =u_z\,t,$$ (3.120)

$$t' = \gamma\left(t-\frac{v\,u_x\,t}{c^2}\right),$$ (3.121)

which yields

$$u_x' = \frac{u_x-v}{1-u_x\,v/c^2},$$ (3.122)

$$u_y' = \frac{u_y}{\gamma\,(1-u_x\,v/c^2)},$$ (3.123)

$$u_z' = \frac{u_z}{\gamma\,(1-u_x\,v/c^2)}.$$ (3.124)

This result is known as the transformation of velocity.

Let $u=(u_x^{\,2}+u_y^{\,2}+u_z^{\,2})^{1/2}$ and $u'=(u_x'^{\,2}+u_y'^{\,2}+u_z'^{\,2})^{1/2}$ be the speeds of the particle in frames $S$ and $S'$, respectively. It is easily demonstrated, from the transformation of velocity, that

$$c^2-u'^{\,2} = \frac{c^2\,(c^2-u^2)\,(c^2-v^2)}{(c^2+ u_x\,v)^2}.$$ (3.125)

If $\vert u\vert<c$ and $\vert v\vert<c$ then the right-hand side is positive, implying that $\vert u'\vert<c$. In other words, the resultant of two subluminal velocities is another subluminal velocity. It is evident that a particle can never attain the velocity of light relative to a given inertial frame, no matter how many subluminal velocity increments it is given. It follows that no inertial frame can ever appear to propagate with a superluminal velocity with respect to any other inertial frame (because we can track a given inertial frame in terms of a particle that remains at rest at the origin of that frame). Of course, if $\vert u\vert=c$ then $\vert u'\vert=c$. In other words, a particle traveling at the speed of light in one inertial frame does so in all inertial frames.

It is evident from Equation (3.125) that there is only a single speed—namely, $u=c$—that is the same in all inertial frames of reference. Now, according to Einstein's first postulate, any wave that propagates in the absence of a physical medium must propagate at the same speed in all inertial frames of reference, otherwise the different wave speeds in different reference frames could be used to distinguish between the frames. Hence, we deduce that all waves that propagate in the absence of a physical medium (e.g., a gas, liquid, or solid) must propagate at the common speed $c$ in all inertial reference frames. Thus, gravitational waves, which are ripples in the fabric of spacetime, must travel at the same speed, $c$, as electromagnetic waves, because both waves propagate in the absence of media. Thus, we could just as well designate $c$ as the speed of gravitational waves.

Note, finally, that the Lorentz transformation is the only (linear) transformation of coordinates that preserves the speed $c$, and morphs into the tried and tested Galilean transformation in the limit that $v/c\ll 1$. In fact, it is possible to guess the form of the Lorentz transformation by searching for a (linear) coordinate transformation that has these two properties.