We noted above that t is the time the event actually happens. This is an important point, one that bears further discussion. Light waves take time to travel. Messages, whether they’re transmitted by light pulses, telephone, or courier on horseback, take time to be delivered. An experimenter observes an event, such as an exploding firecracker, only at a later time when light waves reach his or her eyes. But our interest is in the event itself, not the experimenter’s observation of the event. The time at which the experimenter sees the event or receives information about the event is not when the event actually occurred.

Suppose at $t=0\text{s}$ a firecracker explodes at $x=300\text{m}$. The flash of light from the firecracker will reach an experimenter at the origin at $t_1=1.0\mu \text{s}$. The sound of the explosion will reach a sightless experimenter at the origin at $t_2=0.88\text{s}$. Neither of these is the time $t_{\text{event}}$ of the explosion, although the experimenter can work backward from these times, using known wave speeds, to determine $t_{\text{event}}$. In this example, the spacetime coordinates of the event—the explosion—are $(300\text{m,}$ 0 m, 0 m, $0\text{s})$.

EXAMPLE 36.2 Finding the time of an event

Experimenter A in reference frame S stands at the origin looking in the positive x-direction. Experimenter B stands at $x=900\text{m}$ looking in the negative x-direction. A firecracker explodes somewhere between them. Experimenter B sees the light flash at $t=3.0\mu \text{s}$. Experimenter A sees the light flash at $t=4.0\mu \text{s}$. What are the spacetime coordinates of the explosion?

MODEL Experimenters A and B are in the same reference frame and have synchronized clocks.

VISUALIZE FIGURE 36.14 shows the two experimenters and the explosion at unknown position x.

FIGURE 36.14 The light wave reaches the experimenters at different times. Neither of these is the time at which the event actually happened.

FIGURE 36.14 Full Alternative Text

SOLVE The two experimenters observe light flashes at two different instants, but there’s only one event. Light travels at $300\text{m}/\mu \text{s}$, so the additional $1.0\mu \text{s}$ needed for the light to reach experimenter A implies that distance $(x-0\text{m})$ is 300 m longer than distance $(900\text{m}-x)$. That is,

$$(x-0\text{m})=(900\text{m}-x)+300\text{m}$$

This is easily solved to give $x=600\text{m}$ as the position coordinate of the explosion. The light takes $1.0\mu \text{s}$ to travel 300 m to experimenter B, $2.0\mu \text{s}$ to travel 600 m to experimenter A. The light is received at $3.0\mu \text{s}$ and $4.0\mu \text{s,}$ respectively; hence it was emitted by the explosion at $t=2.0\mu \text{s}$. The spacetime coordinates of the explosion are $(600\text{m,}$ 0 m, 0 m, $2.0\mu \text{s})$.

REVIEW Although the experimenters see the explosion at different times, they agree that the explosion actually happened at $t=2.0\mu \text{s}$.