Frame $\text{S}\prime$ has one important distinction. It is the one and only inertial reference frame in which the light clock is at rest. Consequently, it is the one and only inertial reference frame in which the times of both events—the emission of the light and the detection of the light—are measured by the same reference-frame clock. You can see that the light pulse in Figure 36.18a starts and ends at the same position. In Figure 36.20, the emission and detection take place at different positions in frame S and must be measured by different reference-frame clocks, one at each position.

The time interval between two events that occur at the same position is called the proper time $\text{Δ}\tau$. Only one inertial reference frame measures the proper time, and it does so with a single clock that is present at both events. An inertial reference frame moving with velocity $v=\beta c$ relative to the proper-time frame must use two clocks to measure the time interval: one at the position of the first event, the other at the position of the second event. The time interval in the frame where two clocks are required is

The “stretching out” of the time interval implied by Equation 36.9 is called time dilation. Time dilation is sometimes described by saying that “moving clocks run slow,” but this statement has to be interpreted carefully. The whole point of relativity is that all inertial frames are equally valid, so there’s no absolute sense in which a clock is “moving” or “at rest.”

To illustrate, FIGURE 36.21 shows two firecracker explosions—two events—that occur at different positions in the ground reference frame. Experimenters on the ground need two clocks to measure the time interval $\text{Δ}t$. In the train reference frame, however, a single clock is present at both events; hence the time interval measured in the train reference frame is the proper time $\text{Δ}\tau$. You can see that $\text{Δ}\tau <\text{Δ}t$, so less time has elapsed in the train reference frame.

FIGURE 36.21 The time interval between two events is measured in two different reference frames.

In this sense, the “moving clock,” the one that is present at both events, “runs slower” than clocks that are stationary with respect to the events. More generally, the time interval between two events is smallest in the reference frame in which the two events occur at the same position.

EXAMPLE 36.4 From the sun to Saturn

Saturn is $1.43\times {10}^{12}\text{m}$ from the sun. A rocket travels along a line from the sun to Saturn at a constant speed of 0.9c relative to the solar system. How long does the journey take as measured by an experimenter on earth? As measured by an astronaut on the rocket?

MODEL Let the solar system be in reference frame S and the rocket be in reference frame $\text{S}\prime$ that travels with velocity $v=0.9c$ relative to S. Relativity problems must be stated in terms of events. Let event 1 be “the rocket and the sun coincide” (the experimenter on earth says that the rocket passes the sun; the astronaut on the rocket says that the sun passes the rocket) and event 2 be “the rocket and Saturn coincide.”

VISUALIZE FIGURE 36.22 shows the two events as seen from the two reference frames. Notice that the two events occur at the same position in $\text{S}\prime$, the position of the rocket, and consequently can be measured by one clock carried on board the rocket.

FIGURE 36.22 Pictorial representation of the trip as seen in frames S and S$\prime$.

FIGURE 36.22 Full Alternative Text

SOLVE The time interval measured in the solar system reference frame, which includes the earth, is simply

$$\text{Δ}t={\displaystyle \frac{\text{Δ}x}{v}}={\displaystyle \frac{1.43\times {10}^{12}\text{m}}{0.9\times (3.00\times {10}^8\text{m/s})}}=5300\text{s}$$

Relativity hasn’t abandoned the basic definition $v=\text{Δ}x/\text{Δ}t$, although we do have to be sure that $\text{Δ}x$ and $\text{Δ}t$ are measured in just one reference frame and refer to the same two events.

How are things in the rocket’s reference frame? The two events occur at the same position in $\text{S}\prime$ and can be measured by one clock, the clock at the origin. Thus the time measured by the astronauts is the proper time $\text{Δ}\tau$ between the two events. We can use Equation 36.9 with $\beta =0.9$ to find

$$\text{Δ}\tau =\sqrt{1-{\beta }^2}\text{Δ}t=\sqrt{1-0.9^2}(5300\text{s})=2310\text{ s}$$

REVIEW The time interval measured between these two events by the astronauts is less than half the time interval measured by experimenters on earth. The difference has nothing to do with when earthbound astronomers see the rocket pass the sun and Saturn. $\text{Δ}t$ is the time interval from when the rocket actually passes the sun, as measured by a clock at the sun, until it actually passes Saturn, as measured by a synchronized clock at Saturn. The interval between seeing the events from earth, which would have to allow for light travel times, would be something other than 5300 s. $\text{Δ}t$ and $\text{Δ}\tau$ are different because time is different in two reference frames moving relative to each other.