134


6.2Effects of relativity
Time dilation
Suppose that an observer in a frame Sâ² that moves relative to a frame S measures the time in between successive ticks of a clock. The clock is at rest in Sâ² and so the measurement of the ticks occurs at the same place, Îxâ² = 0. The result of the measurement is Îtâ². The observer in S will measure a time interval equal to
ÎtÎt=Î³(Îtâ²+vÎxâ²c2)=Î³(Îtâ²+0)=Î³Îtâ²
This shows that the interval between the ticks of a clock that moves relative to the observer in S is greater than the interval measured in Sâ²where the clock is at rest. This is known as time dilation. In this case, the time measured by Sâ² is special because it represents a time interval between two events that happened at the same point in space. The clock is at rest in Sâ² so its ticks happen at the same point in Sâ². Such a time interval is called proper time interval. (This is just a name; it is not implied that this is a more âcorrectâ measurement of time).
135
KEY POINTS
A proper time interval is the time between two events that take place at the same point in space.
The time dilation formula is best remembered as
time intervalÂ =Î³ÃÂ proper time intervalÂ =11âv2c2âââââââÃÂ proper time intervalÂ 
WORKED EXAMPLE 6.5
The time interval between the ticks of a clock carried on a fast rocket is half of what observers on Earth record. Calculate the speed of the rocket relative to earth.
Answer
The rocket observers measure the proper time interval since the events âfirst tickâ and âsecond tickâ occur at the same place in the rocket frame. From the time dilation formula it follows that
2=11âv2c2ââââââââ1âv2c2âââââââ=12â1âv2c2=14âv2c2=34âv=0.866c
WORKED EXAMPLE 6.6
A rocket moves past an observer in a laboratory with speed v = 0.85c. The observer in the laboratory measures that a radioactive isotope (which is at rest in the laboratory) has a half-life of 2.0 min. Calculate the half-life measured by the rocket observers.
Answer
We have two events here. The first is that the laboratory observer sees a container with a given mass of the radioactive isotope. The second event is that the laboratory observer sees a container with half the mass of the radioactive isotope. These events are separated by 2.0 min as far as the laboratory observer is concerned. These two events take place at the same point in space as far as the laboratory observer is concerned, and so the laboratory observer has measured the proper time interval between these two events. Hence the rocket observers will measure a longer half-life of time interval = Î³ Ã proper time interval
=11â0.852ââââââââÃ2.0Â min=3.80Â min
The point of this example is that you must not make the mistake of thinking that proper time intervals are measured by âthe movingâ observer. There is no such thing as âthe movingâ observer: the rocket observer is free to consider themself at rest and the laboratory observer moving with velocity v = â0.85c.
SCIENCE IN CONTEXT
The Global Positioning System (GPS) depends on precisely measuring signals from satellites which move at speeds of about 4 km sâ1. This is a small speed compared to the speed of light. The gamma factor for this speed is greater than 1 by approximately 9 Ã 10â11. In one day there are 86 400 s, so if relativity is not taken into account we would have an error of about 9 Ã 10â11 Ã 86 400 â 8 Î¼s in clocks after one day. For signals travelling at the speed of light this would imply an uncertainty in position of about 8 Ã 10â6 Ã 3 Ã 108 = 2400 m. This would make the GPS worthless. (GPS also uses corrections from general relativity.)
Length contraction
Now consider a rod that is at rest in frame Sâ². An observer in Sâ² measures the position of the ends of the rod, subtracts and finds a length Îxâ² = L0. (Notice that since the rod is at rest, the measurement of the position of the ends can be done at different timesâthe rod is not going anywhere. But for the observer in S the rod is moving. So to measure its length the observer must record the position of the ends of the rod at the same time, i.e. with Ît = 0. Since Îxâ² = Î³(Îx â vÎt) where Îx = L is the length in S, we obtain that
136
EXAM TIP
Note that it is only lengths in the direction of motion that are contracted.
L0L=Î³(Lâ0)=L0Î³
This shows that the rod, which moves relative to the observer in S, has a shorter length in S than the length measured Sâ²where the rod is at rest. This is known as length contraction. A reference frame in which an object is at rest is called that objectâs rest frame.
KEY POINTS
The length Îxâ² = L0 is special because it is measured in a frame of reference where the rod is a rest. Such a length is called a proper length.
The length of an object measured by an inertial observer with respect to whom the object is at rest is called proper length. Observers with respect to whom the rod moves at speed v measure a shorter length: length=proper lengthÎ³
WORKED EXAMPLE 6.7
An unmanned spacecraft leaves earth with speed 0.80c towards a planet that is 40 ly away as measured by earth observers. How long will the trip take according to:
aearth clocks
bspacecraft clocks?
Answer
aAccording to earth the time is simply
40Â ly0.80c=50cÃÂ yearÂ c=50Â yr.
bLeaving earth and arriving at the planet occurs at the same point according to the spacecraft so the spacecraft measures a proper time interval for the trip. Hence it takes a time of 50Î³=5053=30Â yr.
Equivalently we can argue that the spacecraft is at rest and the planet is coming towards the spacecraft at speed 0.80c. The distance between earth and planet is then length contracted to 40Î³=4053=24Â ly and so âthe planet will come to the spacecraftâ in a time of 24Â ly0.80c=30Â yr.
THEORY OF KNOWLEDGE
The time dilation effect described is a ârealâ effect. In the HafeleâKeating experiment, accurate atomic clocks taken for a ride aboard planes moving at ordinary speeds and then compared with similar clocks left behind show readings that are smaller by amounts consistent with the formulae of relativity. Time dilation is also a daily effect in the operation of particle accelerators. In such machines, particles are accelerated to speeds that are very close to the speed of light, and thus relativistic effects must be taken into account when designing these machines. The time dilation formula has also been verified in muon decay experiments.
Addition of velocities
Consider a frame Sâ² that moves at constant speed v in a straight line relative to another frame S. Observers in both frames measure the velocity of an object. Observers in Sâ² measure the velocity to be uâ². Observers in S measure the velocity of the same object to be u. How are u and uâ²related? We saw that the Galilean answer (u = uÊ¹ + v) cannot be correct. Einstein showed that the Lorentz equations lead to the following answer:
u=uâ²+v1+uâ²vc2 or, solving for uÊ¹, uâ²=uâv1âuvc2
Think of frame S as the ground and the frame Sâ² to be a rocket moving at velocity 0.80c relative to S. The rocket launches a missile with speed 0.50c relative to the rocket. This means that v = 0.80c and uâ² = 0.50c. The speed of the missile relative to S is u and is given by
137
u=uâ²+v1+uâ²vc2=0.50c+0.80c1+0.50cÃ0.80cc2=1.30c1.40=0.93c
In Galilean relativity then answer would be u = uâ² + v = 1.30c and so impossible. It can be easily checked that, irrespective of how close uâ² or v are to the speed of light, u is always less than c.
WORKED EXAMPLE 6.8
A rocket moves with speed 0.90c relative to the ground. A beam of light is emitted from the rocket in the forward direction. The rocket measures a speed c for the light beam. Use the velocity addition formula to show that the ground observers also measure a speed c for the light beam.
Answer
Applying the formula with v = 0.90c and uâ² = c we find
u=uâ²+v1+uâ²vc2=c+0.90c1+cÃ0.90cc2=1.90c1.90=c
WORKED EXAMPLE 6.9
Two rockets move away from each other with speeds of 0.80c and 0.90c with respect to the ground, as shown in Figure 6.14. What is the velocity of rocket B as measured by rocket A?

Figure 6.14: For worked example 6.9.

Answer
Let us call the frame of the ground S and that of rocket A, Sâ². Then v = â0.80c. We want to find the velocity of rocket B relative to A, i.e. relative to Sâ², so we are looking for uâ². The velocity of rocket B relative to the ground (S) is u = +0.90c and so
uâ²=uâv1âuvc2=+0.90câ(â0.80c)1â(0.90c)(â0.80c)c2=+1.70c1+0.72=+0.988câ+0.99c
The velocity of rocket A relative to rocket B is expected to be â0.99c. We can calculate this explicitly as follows: again let S be the ground frame and Sâ² the frame of rocket B. Then v = +0.90c and u = â0.80c. We are again looking for uâ² and
uâ²=uâv1âuvc2=â0.80câ0.90c1â(â0.80c)(0.90c)c2ââ0.99c
as expected.
Simultaneity
Another great change introduced into physics as a result of relativity is in the concept of simultaneity. Suppose that as usual we have frames S and Sâ² with frame Sâ² moving to the right with speed v relative to S. Suppose that two events take place at the same time in frame Sâ². The time interval between these two events is thus zero, Îtâ² = 0. What is the time interval between the same two events when measured in S? The Lorentz equations give
Ît=Î³(Îtâ²+vc2Îxâ²)=Î³vc2Îxâ²
If Îxâ² â  0, then the events will not be simultaneous in other frames. On the other hand, if Îxâ² = 0, i.e. if the simultaneous events in Sâ² occur at the same point in space, then they are also simultaneous in all other frames.
138
KEY POINT
Events that are simultaneous for one observer and which take place at different points in space, are not simultaneous for another observer in motion relative to the first.
On the other hand, if two events are simultaneous for one observer and take place at the same point in space, they are simultaneous for all other observers as well.
Imagine an observer T is in the middle of a train carriage that is moving with constant speed to the right with respect to the train station. We are told that two light signals are emitted from the ends of the carriage at the same time as far as the observer T in the train is concerned, Figure 6.15. It is then clear that the two signals arrive at T at the same time as far as T is concerned. What does an observer G at rest on the ground say about this situation?

Figure 6.15: Light leaves A and B at the same time as far as T is concerned.

The emissions from A and B take place at different points in space and are simultaneous for observer T. So they will not be simultaneous for observer G. On the other hand the arrival of the signals at T is simultaneous for T and happen at the same point in space so they will also be simultaneous for G.
Which signal was emitted first according to G? From Gâs point of view, T is moving away from the signal from A. So the signal from A has a larger distance to cover to get to T. If the signals are received at the same time, and moved at the same speed (c), it must be that the one from A was emitted before that from B.
CHECK YOURSELF 5
Two observers, N and M, are at rest on the ground. M is midway between two street lights (Figure 6.16). M measures that the streetlights turn on at the same time.

Figure 6.16: For check yourself question 5.

aDoes M receive light from the two lamps at the same time?
bDo the street lights turn on at the same time according to N?
cDoes N receive light from the two lamps at the same time?
WORKED EXAMPLE 6.10
Let us re-work the situation of Figure 6.15 quantitatively. Take the proper length of the âtrainâ (frame Sâ²) to be 300 m and let it move with speed 0.98c to the right. Determine, according to an observer on the ground (frame S), which light is emitted first and by how much.
Answer
We are told that Îtâ² = 0 and Îxâ² = xâ²B â xâ²A = 300 m. We want to find Ît, the time interval between the events representing the emission of light from A and B. The gamma factor is Î³=11â0.982ââââââââ=5.0. Then
Ît=tBâtA=Î³(Îtâ²+vc2Îxâ²)=5.0Ã(0+0.98cc2Ã300)=+4.9Â Î¼s
The positive sign indicates that tB > tA, i.e. the light from A was emitted 4.9 Î¼s before the light from B.
139
Muon decay
Muons are particles with properties similar to those of the electron except that they are more massive, are unstable and decay into electrons; they have an average lifetime of about 2.2 Ã 10â6 s. (The reaction is Î¼ââeâ+vÂ¯e+vÎ¼.)
This is the lifetime measured when the muon is at restâthis is the proper time interval between the creation of a muon and its subsequent decay.
From the point of view of an observer in the laboratory, however, the muons are moving at high speed; the lifetime is longer because of time dilation. Consider a muon created by a source at the top of a mountain 3.0 km tall, Figure 6.17. The muon travels at 0.99c towards the surface of the earth.

Figure 6.17: a Muons created at the top of the mountain would not have enough time to reach the surface as muons if relativity did not hold. b With relativity the muons last longer and make it to the surface before decaying.

Without the relativistic time dilation effect, the muon would have travelled a distance of only
0.99Ã3Ã108Ã2.2Ã10â6Â m=0.653Â km
before decaying into an electron. Thus a detector at the base of the mountain would record the arrival of electrons, not muons. But experiments show the arrival of muons at the detector. This is because the lifetime of the muon as measured by ground observers is
Â time intervalÂ =Î³ÃÂ proper timeÂ =2.2Ã10â6Â s1â0.992ââââââââ=1.56Ã10â5Â s
In this time the muon travels a distance (as measured by the ground observers) of
0.99Ã3Ã108Ã1.56Ã10â5Â m=4.63Â km
This means that the muon reaches the surface of the earth before decaying.
The fact that muons do make it to the surface of the earth is evidence in support of the time dilation effect.
The muon exists as a muon for only 2.2 Ã 10â6 s in the muonâs rest frame. So how does an observer travelling along with the muon explain the arrival of muons (and not electrons) at the surface of the earth, Figure 6.18?

Figure 6.18: From the point of view of an observer on the muon the mountain is much shorter.

The answer is that the distance of 3.0 km measured by the observers on earth is a proper length for them but not for the observer at rest with respect to the muon. This observer claims that it is the earth that is moving upward and so measures a length-contracted distance of
3.0Ã1â0.992ââââââââÂ km=0.42Â km
to the surface of the earth. The earthâs surface is coming up to this observer with a speed of 0.99c and so the time when they will meet is
0.42Ã1030.99Ã3Ã108Â s=1.4Ã10â6Â s
that is, before the muon decays!
In this sense, muon decay experiments are direct confirmations of time dilation and indirect confirmations of length contraction.
140
WORKED EXAMPLE 6.11
At the Stanford Linear Accelerator (SLAC), electrons of speed v = 0.960c move a distance of 3.00 km.
aCalculate how long this takes according to observers in the laboratory.
bCalculate how long this takes according to an observer travelling along with the electrons.
cFind the speed of the linear accelerator in the rest frame of the electrons.
Answer
aIn the laboratory the electrons take a time of 3.00Ã1030.960Ã3.00Ã108Â s=1.04Ã10â5Â s
bThe arrival of the electrons at the beginning and the end of the accelerator track happens at the same point in space as far as the observer travelling along with the electrons is concerned, and so that is a proper time interval. The gamma factor is Î³=11â0.9602âââââââââ=3.571. Thus
time intervalÂ =Î³ÃÂ proper time interval
proper time interval=1.04Ã10â53.571Â s=2.91Ã10â6Â s
cThe speed of the accelerator is obviously v = 0.960c in the opposite direction. But this can be checked as follows. As far as the electron is concerned, the length of the accelerator track is moving past it and so is length-contracted according to
length=proper lengthÎ³=3.00Â km3.571=0.840Â km
and so has a speed of
Â speedÂ =0.840Ã1032.91Ã10â6Â mÂ sâ1=2.89Ã108Â mÂ sâ1â0.96c
SCIENCE IN CONTEXT
General arguments apply to special cases but not the other way around.
Years before relativity was introduced by Einstein an experiment performed by Michelson and Morley gave a very puzzling result concerning the speed of light. There were frantic attempts to resolve the difficulties posed by this experiment. One attempt was to assume that moving lengths appear shorter. Similarly, Lorentz showed that if one used his Lorentz transformations equations certain difficulties with electromagnetism went away. But it was Einstein who re-derived these transformations equations from far more general principles, the postulates of relativity. By demanding that all inertial observers experience the same laws of physics and measuring the same velocity of light in vacuum the Lorentz equations emerged as the simplest linear equations that could achieve this.
141
TEST YOUR UNDERSTANDING
16An earthling sits on a bench in a park eating a sandwich. It takes him 5 min to finish it according to his watch. He is being monitored by planet Zenga invaders who are passing earth at a speed of 0.90c.
aCalculate how long the aliens reckon it takes an earthling to eat a sandwich.
bThe aliens in the spacecraft get hungry and start eating their sandwiches. It takes a Zengan 5 min to eat her sandwich according to Zengan clocks. They are actually being observed by earthlings as they fly over earth. Calculate how long it takes a Zengan to eat a sandwich according to earth clocks.
17A cube has density Ï when the density is measured at rest. Suggest what happens to the density of the cube when it travels past you at a relativistic speed.
18A pendulum in a fast train is found by observers on the train to have a period of 1.0 s. Calculate the period observers on a station platform would measure as the train moves past them at a speed of 0.95c.
19A spacecraft moves past you at a speed of 0.95c, and you measure its length to be 100 m. Calculate the length you would measure if it were at rest with respect to you.
20Two identical fast trains move parallel to each other. An observer on train A tells an observer on train B that by her measurements (i.e. Aâs) train A is 30 m long and train B is 28 m long.
The observer on train B takes measurements. Calculate what he will find for:
athe speed of train A with respect to train B
bthe length of train A
cthe length of train B.
21An unstable particle has a lifetime of 5.0 Ã 10â8 s as measured in its rest frame. The particle is moving in a laboratory with a speed of 0.95c with respect to the lab.
aCalculate the lifetime of the particle according to an observer at rest in the laboratory.
bCalculate the distance travelled by the particle before it decays, according to the observer in the laboratory.
22The star Vega is about 50 ly away from earth. A spacecraft moving at 0.995c is heading towards Vega.
aCalculate how long it will take the spacecraft to get to Vega according to clocks on earth.
bThe crew of the spacecraft consists of 18-year-old IB graduates. Calculate how old the graduates are (according to their clocks) when they arrive at Vega.
23A rocket approaches a mirror on the ground at a speed of 0.90c as shown in Figure 6.19.
The distance D between the front of the rocket and the mirror is 2.4 Ã 1012 m, as measured by the observers on the ground, when a light signal is sent towards the mirror from the front of the rocket.

Figure 6.19: For question 23.

142
Calculate when the reflected signal is received by the rocket as measured by:
athe observers on the ground
bthe observers on the rocket.
24Two objects move along the same straight line as shown in Figure 6.20. Their speeds are as measured by an observer on the ground.

Figure 6.20: For question 24.

Find:
athe velocity of B as measured by A;
bthe velocity of A as measured by B.
25Repeat question 24 for the arrangement in Figure 6.21.

Figure 6.21: For question 25.

26A particle A moves to the right with a speed of 0.60c relative to the ground. A second particle B moves to the right with a speed of 0.70c relative to A. Calculate the speed of B relative to the ground.
27Particle A moves to the left with a speed of 0.60c relative to the ground. A second particle B moves to the right with a speed of 0.70c relative to A. Find the speed of B relative to the ground.
28A muon travelling at 0.95c covers a distance of 2.00 km (as measured by an earthbound observer) before decaying.
aCalculate the muonâs lifetime as measured by the earthbound observer.
bCalculate the lifetime as measured by an observer travelling along with the muon.
29The lifetime of the unstable pion particle is measured to be 2.6 Ã 10â8 s (when at rest). This particle travels a distance of 20 m in the laboratory just before decaying. Calculate its speed.
30In frame S two light signals are emitted from the same point in space 2.0 Âµs apart. In frame Sâ² the time between the emissions is measured to be 8.0 Âµs.
aWhat is the speed of Sâ² relative to S?
bHow far apart are the emission points according to Sâ²?
31A rocket approaches a space station. The rocket moves with velocity 0.80c relative to the space station. Two lights at the ends of the space station turn on at the same time according to space station observers. Which light turns on first according to rocket observers?
32Two simultaneous events in frame S are separated by a distance Îx = x2 â x1 = 1200 m. Determine the time separating these two events in frame Sâ², stating which one occurs first. Sâ² moves past S with velocity 0.80c.
33A spacecraft leaves earth moving away at speed 0.60c relative to earth. Mandy, an astronaut, has her birthday that day. In a yearâs time (according to spacecraft clocks) she will have her birthday again. Her friends on earth want to celebrate by sending her a light signal. When, by earth clocks, should the light signal be sent so that it arrives at the spacecraft on Mandyâs birthday?
