11.2 Simultaneity and time dilation  (Page 3/8)

This time has a separate name to distinguish it from the time measured by the Earth-bound observer.

In the case of the astronaut observe the reflecting light, the astronaut measures proper time. The time measured by the Earth-bound observer is

To find the relationship between $\mathrm{\Delta }{t}_0$ and $\mathrm{\Delta }t$ , consider the triangles formed by $D$ and $s$ . (See [link] (c).) The third side of these similar triangles is $L$ , the distance the astronaut moves as the light goes across her ship. In the frame of the Earth-bound observer,

Using the Pythagorean Theorem, the distance $s$ is found to be

Substituting $s$ into the expression for the time interval $\mathrm{\Delta }t$ gives

We square this equation, which yields

Note that if we square the first expression we had for $\mathrm{\Delta }{t}_0$ , we get $(\mathrm{\Delta }{t}_0{)}^2=\frac{{4D}^2}{c^2}$ . This term appears in the preceding equation, giving us a means to relate the two time intervals. Thus,

Gathering terms, we solve for $\mathrm{\Delta }t$ :

Thus,

Taking the square root yields an important relationship between elapsed times:

where

This equation for $\mathrm{\Delta }t$ is truly remarkable. First, as contended, elapsed time is not the same for different observers moving relative to one another, even though both are in inertial frames. Proper time $\mathrm{\Delta }{t}_0$ measured by an observer, like the astronaut moving with the apparatus, is smaller than time measured by other observers. Since those other observers measure a longer time $\mathrm{\Delta }t$ , the effect is called time dilation. The Earth-bound observer sees time dilate (get longer) for a system moving relative to the Earth. Alternatively, according to the Earth-bound observer, time slows in the moving frame, since less time passes there. All clocks moving relative to an observer, including biological clocks such as aging, are observed to run slow compared with a clock stationary relative to the observer.

Note that if the relative velocity is much less than the speed of light ( ), then $\frac{v^2}{c^2}$ is extremely small, and the elapsed times $\mathrm{\Delta }t$ and $\mathrm{\Delta }{t}_0$ are nearly equal. At low velocities, modern relativity approaches classical physics—our everyday experiences have very small relativistic effects.

The equation $\mathrm{\Delta }t={\gamma \mathrm{\Delta }t}_0$ also implies that relative velocity cannot exceed the speed of light. As $v$ approaches $c$ , $\mathrm{\Delta }t$ approaches infinity. This would imply that time in the astronaut’s frame stops at the speed of light. If $v$ exceeded $c$ , then we would be taking the square root of a negative number, producing an imaginary value for $\mathrm{\Delta }t$ .

There is considerable experimental evidence that the equation $\mathrm{\Delta }t={\gamma \mathrm{\Delta }t}_0$ is correct. One example is found in cosmic ray particles that continuously rain down on the Earth from deep space. Some collisions of these particles with nuclei in the upper atmosphere result in short-lived particles called muons. The half-life (amount of time for half of a material to decay) of a muon is $1\text{.}\text{52}\mu \text{s}$ when it is at rest relative to the observer who measures the half-life. This is the proper time $\mathrm{\Delta }{t}_0$ . Muons produced by cosmic ray particles have a range of velocities, with some moving near the speed of light. It has been found that the muon’s half-life as measured by an Earth-bound observer ( $\mathrm{\Delta }t$ ) varies with velocity exactly as predicted by the equation $\mathrm{\Delta }t={\gamma \mathrm{\Delta }t}_0$ . The faster the muon moves, the longer it lives. We on the Earth see the muon’s half-life time dilated—as viewed from our frame, the muon decays more slowly than it does when at rest relative to us.