Time Dilation

Time Dilation

The consideration of the measurement of elapsed time and simultaneity leads to an important relativistic effect.

Suppose, for example, an astronaut measures the time it takes for light to cross her ship, bounce off a mirror, and return. (See this figure.) How does the elapsed time the astronaut measures compare with the elapsed time measured for the same event by a person on the Earth? Asking this question (another thought experiment) produces a profound result. We find that the elapsed time for a process depends on who is measuring it. In this case, the time measured by the astronaut is smaller than the time measured by the Earth-bound observer. The passage of time is different for the observers because the distance the light travels in the astronaut’s frame is smaller than in the Earth-bound frame. Light travels at the same speed in each frame, and so it will take longer to travel the greater distance in the Earth-bound frame.

To quantitatively verify that time depends on the observer, consider the paths followed by light as seen by each observer. (See this figure (c).) The astronaut sees the light travel straight across and back for a total distance of \(2D\), twice the width of her ship. The Earth-bound observer sees the light travel a total distance \(2s\). Since the ship is moving at speed \(v\) to the right relative to the Earth, light moving to the right hits the mirror in this frame. Light travels at a speed \(c\) in both frames, and because time is the distance divided by speed, the time measured by the astronaut is

\(\Delta {t}_{0}=\cfrac{2D}{c}.\)

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

\(\Delta t=\cfrac{2s}{c}.\)

To find the relationship between \(\Delta {t}_{0}\) and \(\Delta t\), consider the triangles formed by \(D\) and \(s\). (See this figure (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,

\(L=\cfrac{v\Delta t}{2}.\)

Using the Pythagorean Theorem, the distance \(s\) is found to be

\(s=\sqrt{{D}^{2}+{\left(\cfrac{v\Delta t}{2}\right)}^{2}}.\)

Substituting \(s\) into the expression for the time interval \(\Delta t\) gives

\(\Delta t=\cfrac{2s}{c}=\cfrac{2\sqrt{{D}^{2}+{\left(\cfrac{v\Delta t}{2}\right)}^{2}}}{c}.\)

We square this equation, which yields

\((\Delta t{)}^{2}=\cfrac{4({D}^{2}+\cfrac{{v}^{2}(\Delta t{)}^{2}}{4})}{{c}^{2}}=\cfrac{{4D}^{2}}{{c}^{2}}+\cfrac{{v}^{2}}{{c}^{2}}(\Delta t{)}^{2}.\)

Note that if we square the first expression we had for \(\Delta {t}_{0}\), we get \((\Delta {t}_{0}{)}^{2}=\cfrac{{4D}^{2}}{{c}^{2}}\). This term appears in the preceding equation, giving us a means to relate the two time intervals. Thus,

\((\Delta t{)}^{2}=(\Delta {t}_{0}{)}^{2}+\cfrac{{v}^{2}}{{c}^{2}}(\Delta t{)}^{2}.\)

Gathering terms, we solve for \(\Delta t\):

\((\Delta t{)}^{2}(1-\cfrac{{v}^{2}}{{c}^{2}})=(\Delta {t}_{0}{)}^{2}.\)

Thus,

\((\Delta t{)}^{2}=\cfrac{(\Delta {t}_{0}{)}^{2}}{1-\cfrac{{v}^{2}}{{c}^{2}}}.\)

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

\(\Delta t=\cfrac{\Delta {t}_{0}}{\sqrt{1-\cfrac{{v}^{2}}{{c}^{2}}}}={\gamma \Delta t}_{0},\)

where

\(\gamma =\cfrac{1}{\sqrt{1-\cfrac{{v}^{2}}{{c}^{2}}}}.\)

This equation for \(\Delta t\) 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 \(\Delta {t}_{0}\) 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 \(\Delta t\), 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 (\(v\text{<<}c\)), then \(\cfrac{{v}^{2}}{{c}^{2}}\) is extremely small, and the elapsed times \(\Delta t\) and \(\Delta {t}_{0}\) are nearly equal. At low velocities, modern relativity approaches classical physics—our everyday experiences have very small relativistic effects.

The equation \(\Delta t={\gamma \Delta t}_{0}\) also implies that relative velocity cannot exceed the speed of light. As \(v\) approaches \(c\), \(\Delta t\) 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 \(\Delta t\).

There is considerable experimental evidence that the equation \(\Delta t={\gamma \Delta t}_{0}\) 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 \(1\text{.}\text{52}\phantom{\rule{0.25em}{0ex}}\mu \text{s}\) when it is at rest relative to the observer who measures the half-life.

This is the proper time \(\Delta {t}_{0}\). 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 (\(\Delta t\)) varies with velocity exactly as predicted by the equation \(\Delta t={\gamma \Delta t}_{0}\). 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.

Another implication of the preceding example is that everything an astronaut does when moving at \(\text{95}\text{.}0\%\text{}\) of the speed of light relative to the Earth takes 3.20 times longer when observed from the Earth. Does the astronaut sense this? Only if she looks outside her spaceship. All methods of measuring time in her frame will be affected by the same factor of 3.20. This includes her wristwatch, heart rate, cell metabolism rate, nerve impulse rate, and so on. She will have no way of telling, since all of her clocks will agree with one another because their relative velocities are zero. Motion is relative, not absolute. But what if she does look out the window?