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

Time dilation is the phenomenon of time passing slower for an observer who is moving relative to another observer.

Suppose, for example, an astronaut measures the time it takes for light to cross her ship, bounce off a mirror, and return. (See Figure 3.) 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 Figure 3(c).) The astronaut sees the light travel straight across and back for a total distance of 2D2D size 12{2D} {}, twice the width of her ship. The Earth-bound observer sees the light travel a total distance 2s2s size 12{2s} {}. Since the ship is moving at speed vv size 12{v} {} to the right relative to the Earth, light moving to the right hits the mirror in this frame. Light travels at a speed cc size 12{c} {} in both frames, and because time is the distance divided by speed, the time measured by the astronaut is

Δt0=2Dc.Δt0=2Dc. size 12{Δt rSub { size 8{0} } = { {2D} over {c} } } {}

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

Proper time Δt0Δt0 size 12{Δt rSub { size 8{0} } } {} is the time measured by an observer at rest relative to the event being observed.

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

Δt=2sc.Δt=2sc. size 12{Δt= { {2s} over {c} } } {}

(2)To find the relationship between Δt0Δt0 size 12{Δt rSub { size 8{0} } } {} and ΔtΔt size 12{Δt} {}, consider the triangles formed by DD size 12{D} {} and ss size 12{s} {}. (See Figure 3(c).) The third side of these similar triangles is LL size 12{L} {}, the distance the astronaut moves as the light goes across her ship. In the frame of the Earth-bound observer,

L=vΔt2.L=vΔt2. size 12{L= { {vΔt} over {2} } } {}

(3)Using the Pythagorean Theorem, the distance ss size 12{s} {} is found to be

s=D2+vΔt22.s=D2+vΔt22. size 12{s= sqrt {D rSup { size 8{2} } + left ( { {vΔt} over {2} } right ) rSup { size 8{2} } } } {}

(4)Substituting ss size 12{s} {} into the expression for the time interval ΔtΔt size 12{Δt} {} gives

Δt=2sc=2D2+vΔt22c.Δt=2sc=2D2+vΔt22c. size 12{Δt= { {2s} over {c} } = { {2 sqrt {D rSup { size 8{2} } + left ( { {vΔt} over {2} } right ) rSup { size 8{2} } } } over {c} } } {}

(5)We square this equation, which yields

(Δt)2=4D2+v2(Δt)24c2=
4
D
2
c
2
+v2c2(Δt)2.(Δt)2=4D2+v2(Δt)24c2=
4
D
2
c
2
+v2c2(Δt)2. size 12{ \( Δt \) rSup { size 8{2} } = { {4 left [D rSup { size 8{2} } + { {v rSup { size 8{2} } \( Δt \) rSup { size 8{2} } } over {4} } right ]} over {c rSup { size 8{2} } } } = { {4D rSup { size 8{2} } } over {c rSup { size 8{2} } } } + { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } \( Δt \) rSup { size 8{2} } } {}

(6)Note that if we square the first expression we had for Δt0Δt0 size 12{Δt rSub { size 8{0} } } {}, we get (Δt0)2=
4
D
2
c2(Δt0)2=
4
D
2
c2 size 12{ \( Δt rSub { size 8{0} } \) rSup { size 8{2} } = { {4D rSup { size 8{2} } } over {c rSup { size 8{2} } } } } {}. This term appears in the preceding equation, giving us a means to relate the two time intervals. Thus,

(Δt)2=(Δt0)2+v2c2(Δt)2.(Δt)2=(Δt0)2+v2c2(Δt)2. size 12{ \( Δt \) rSup { size 8{2} } = \( Δt rSub { size 8{0} } \) rSup { size 8{2} } + { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } \( Δt \) rSup { size 8{2} } } {}

(7)Gathering terms, we solve for ΔtΔt size 12{Δt} {}:

(Δt)21−v2c2=(Δt0)2.(Δt)21−v2c2=(Δt0)2. size 12{ \( Δt \) rSup { size 8{2} } left (1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } right )= \( Δt rSub { size 8{0} } \) rSup { size 8{2} } } {}

(8)Thus,

(Δt)2=(Δt0)21−v2c2.(Δt)2=(Δt0)21−v2c2. size 12{ \( Δt \) rSup { size 8{2} } = { { \( Δt rSub { size 8{0} } \) rSup { size 8{2} } } over {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } {}

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

Δt=Δt01
−v2c2=γΔt0,Δt=Δt01
−v2c2=γΔt0, size 12{Δt= { {Δt rSub { size 8{0} } } over { sqrt {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } =γΔt rSub { size 8{0} } } {}

(10)where

γ=11−v2c2.γ=11−v2c2. size 12{γ= { {1} over { sqrt {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } } {}

(11)This equation for ΔtΔt size 12{Δt} {} 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 Δt0Δt0 size 12{Δt rSub { size 8{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 ΔtΔt size 12{Δt} {}, 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<<cv<<c size 12{v"<<"c} {}), then v2c2v2c2 is extremely small, and the elapsed times ΔtΔt and Δt0Δt0 size 12{Δt rSub { size 8{0} } } {} are nearly equal. At low velocities, modern relativity approaches classical physics—our everyday experiences have very small relativistic effects.

The equation Δt=
γ
Δ
t0
Δt=
γ
Δ
t0
also implies that relative velocity cannot exceed the speed of light. As
vv size 12{v} {} approaches cc size 12{c} {},
ΔtΔt size 12{Δt} {} approaches infinity. This would imply that time in the astronaut’s frame stops at the speed of light. If vv size 12{v} {} exceeded cc size 12{c} {}, then we would be taking the square root of a negative number, producing an imaginary value for ΔtΔt size 12{Δt} {}.

There is considerable experimental evidence that the equation Δt=γΔt0Δt=γΔt0 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
1.52μs1.52μs size 12{1 "." "52"` "μs"} {} when it is at rest relative to the observer who measures the half-life. This is the proper time
Δt0Δt0 size 12{Δt rSub { size 8{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 (ΔtΔt size 12{Δt} {}) varies with velocity exactly as predicted by the equation Δt=γΔt0Δt=γΔt0 size 12{Δt=γΔt rSub { size 8{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.

Suppose a cosmic ray colliding with a nucleus in the Earth’s upper atmosphere produces a muon that has a velocity v=0.950cv=0.950c size 12{v=0 "." "950"c} {}. The muon then travels at constant velocity and lives 1.52μs1.52μs size 12{1 "." "52"` ital "μs"} {} as measured in the muon’s frame of reference. (You can imagine this as the muon’s internal clock.) How long does the muon live as measured by an Earth-bound observer? (See Figure 4.)

*Strategy*

A clock moving with the system being measured observes the proper time, so the time we are given is Δt0=1.52μsΔt0=1.52μs. The Earth-bound observer measures
ΔtΔt as given by the equation
Δt=
γΔt0
Δt=
γΔt0
. Since we know the velocity, the calculation is straightforward.

*Solution*

1) Identify the knowns. v=0.950cv=0.950c,
Δt0=1.52μsΔt0=1.52μs

2) Identify the unknown. ΔtΔt

3) Choose the appropriate equation.

Use,

where

γ=11−
v2c2
.γ=11−
v2c2
.

(13)4) Plug the knowns into the equation.

First find γγ.

γ
=
1
1
−
v
2
c
2
=
1
1
−
(
0.950
c
)
2
c
2
=
1
1
−
(
0.950
)
2
=
3.20.
γ
=
1
1
−
v
2
c
2
=
1
1
−
(
0.950
c
)
2
c
2
=
1
1
−
(
0.950
)
2
=
3.20.

(14)Use the calculated value of γγ size 12{γ} {} to determine ΔtΔt.

Δ
t
=
γΔt
0
=
(
3.20
)
(
1.52
μ s
)
=
4.87
μ s
Δ
t
=
γΔt
0
=
(
3.20
)
(
1.52
μ s
)
=
4.87
μ s

(15)
*Discussion*

One implication of this example is that since γ=3.20γ=3.20 size 12{γ=3 "." "20"} {} at 95.0%95.0% size 12{"95" "." 0%} {} of the speed of light (v=0.950cv=0.950c size 12{v=0 "." "950"c} {}), the relativistic effects are significant. The two time intervals differ by this factor of 3.20, where classically they would be the same. Something moving at 0.950c0.950c size 12{0 "." "950"c} {} is said to be highly relativistic.

Another implication of the preceding example is that everything an astronaut does when moving at 95.0%95.0% size 12{"95" "." 0%} {} 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?

It may seem that special relativity has little effect on your life, but it is probably more important than you realize. One of the most common effects is through the Global Positioning System (GPS). Emergency vehicles, package delivery services, electronic maps, and communications devices are just a few of the common uses of GPS, and the GPS system could not work without taking into account relativistic effects. GPS satellites rely on precise time measurements to communicate. The signals travel at relativistic speeds. Without corrections for time dilation, the satellites could not communicate, and the GPS system would fail within minutes.

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