Spacetime Interval

Consider two events, $1$ and $2$, whose spacetime coordinates in some inertial frame $S$ are ($x_1$, $y_1$, $z_1$, $t_1$) and ($x_2$, $y_2$, $z_2$, $t_2$), respectively. Let us form the differences between these coordinates, ${\mit\Delta}x = x_2-x_1$, ${\mit\Delta}y=y_2-y_1$, ${\mit\Delta} z = z_2-z_1$, and ${\mit\Delta} t = t_2-t_1$. The spatial distance, ${\mit\Delta}d$, between the two events is written

$$({\mit\Delta} d)^2 = ({\mit\Delta} x)^2 + ({\mit\Delta} y)^2 + ({\mit\Delta} z)^2.$$ (3.108)

Suppose that we shift the origin of our coordinate system in $S$. It is obvious that this process does not change the distance between our two events. In other words, ${\mit\Delta}d$ is invariant under a shift of the origin of the coordinate system, as is easily verified. Suppose that we rotate our coordinate axes in $S$. (See Section A.5.) Such a process is length-preserving. In other words, ${\mit\Delta}d$ is invariant under a rotation of the coordinate axes, as is easily verified. However, it is evident, by inspection, that ${\mit\Delta}d$ is not invariant under a Lorentz transformation. Let us try to find a quantity that is invariant.

Consider a second inertial frame, $S'$, that moves with velocity ${\bf v} = v\,{\bf e}_x$ with respect to $S$, and is also in a standard configuration with respect to $S$. Let events $1$ and $2$ have spacetime coordinates ($x_1'$, $y_1'$, $z_1'$, $t_1'$) and ($x_2'$, $y_2'$, $z_2'$, $t_2'$), respectively, in $S'$. Let us again form the differences between these coordinates, ${\mit\Delta}x' = x_2'-x_1'$, ${\mit\Delta}y'=y_2'-y_1'$, ${\mit\Delta} z' = z_2'-z_1'$, and ${\mit\Delta} t' = t_2'-t_1'$. The spacetime interval between events 1 and 2 in $S$ is defined

$$({\mit\Delta} s)^2 = c^2\,({\mit\Delta} t)^2 - ({\mit\Delta} x)^2 - ({\mit\Delta} y)^2 - ({\mit\Delta} z)^2.$$ (3.109)

The Lorentz transformation, (3.100)–(3.103), yields

$${\mit\Delta}x' = \gamma\,({\mit\Delta}x-v\,{\mit\Delta}t),$$ (3.110)

$${\mit\Delta}y' ={\mit\Delta}y,$$ (3.111)

$${\mit\Delta}z' ={\mit\Delta}z,$$ (3.112)

$${\mit\Delta}t' = \gamma\left({\mit\Delta}t-\frac{v\,{\mit\Delta}x}{c^2}\right).$$ (3.113)

Hence, the spacetime interval between the two events in $S'$ is

$$({\mit\Delta} s')^2 = c^2\,({\mit\Delta} t')^2 - ({\mit\Delta} x')^2 - ({\mit\Delta} y')^2 - ({\mit\Delta} z')^2$$

$$= \gamma^2\left[c^2\,({\mit\Delta}t)^2-2\,v\,{\mit\Delta x}\,{\mi... ...Delta}x)^2-2\,v\,{\mit\Delta x}\,{\mit\Delta t}+ v^2\,({\mit\Delta t})^2\right]$$

$$\phantom{=} - ({\mit\Delta} y)^2 - ({\mit\Delta} z)^2$$

$$=c^2\,({\mit\Delta} t)^2\,\gamma^2\left(1-\frac{v^2}{c^2}\right) ... ...\gamma^2\left(1-\frac{v^2}{c^2}\right)- ({\mit\Delta} y)^2 - ({\mit\Delta} z)^2$$

$$= c^2\,({\mit\Delta} t)^2 - ({\mit\Delta} x)^2 - ({\mit\Delta} y)^2 - ({\mit\Delta} z)^2$$

$$= ({\mit\Delta} s)^2,$$ (3.114)

where use has been made of Equation (3.72). Thus, we conclude that the spacetime interval between two events is invariant under a Lorentz transformation. In other words, the interval is the same in all inertial reference frames. What is the significance of this result? Suppose that a light ray travels in a straight-line from event 1 to event 2. The speed of the light ray in $S$ is

$$\frac{{\mit\Delta}d}{{\mit\Delta t}}= c\left[1-\frac{({\mit\Delta} s)^2}{(c\,{\mit\Delta} t)^2}\right]^{1/2},$$ (3.115)

where use has been made of Equations (3.108) and (3.109). However, we know, from Einstein's second postulate, that this speed is equal to $c$. Hence, we deduce that

$${\mit\Delta} s = 0.$$ (3.116)

In other words, if a light ray travels between two events in spacetime then the interval between these two events is zero. This implies that the interval between the two events is zero in all inertial frames of reference. In particular, the interval in $S'$ is ${\mit\Delta} s' =0$. Now, the speed of the light ray in $S'$ is

$$\frac{{\mit\Delta}d'}{{\mit\Delta t'}}= c\left[1-\frac{({\mit\Delta} s')^2}{(c\,{\mit\Delta} t')^2}\right]^{1/2} = c.$$ (3.117)

Hence, we deduce that the invariance of the spacetime interval under Lorentz transformation guarantees that light travels through a vacuum at the speed $c$ in all inertial frames of reference.