1. Chapter 20 The special theory of relativity
Albert Einstein ( 1879 ~ 1955 )

2. 20-1 Troubles with classical physics
The kinematics developed by Galileo and the
mechanics developed by Newton, which form the
basis of what we call “classical physics”, had
many triumphs. However, a number of
experimental phenomena can not be understood
with these otherwise successful classical theories.
Troubles with our ideas about time
The pions ( or ) created at rest are observed

3. to decay ( to other particles ) with an average lifetime of only .
In one particular experiment, pions were created in motion at a speed of In this case they were observed to travel in the laboratory an average distance of
before decaying, from which we conclude that they decay in a time given by , much larger than the lifetime measured for pions at rest.
This effect, called “time dilation”, which cannot be explained by Newtonian physics. In Newtonian physics time is a universal coordinate having identical values for all observers.

4. 2. Trouble with our ideas about length
Suppose an observer in the above laboratory placed one marker at the location of the pion’s formation and another at the location of its decay.
The distance between the markers is measured to be 17.4m. Now consider the observer who is traveling along with the pion at a speed of u=0.913c. This observer, to whom the pion appear to be at rest, measures its lifetime to be 26.0ns, and the distance between the markers is
Thus two observers measure different value for the
same length interval.
3. Troubles with our ideas about light

5. 20-2 The postulates of special relativity
1. Einstein offered two postulates that form the
basis of his special theory of relativity.
(I) The principle of relativity: “The laws of physics
are the same in all inertial reference frames.”
(II) The principle of the constancy of the speed of
light : “ The speed of light in free space has the
same value c in all inertial reference frames.”
2. The first postulate declares that the laws of
physics are absolute, universal, and same for all
inertial observers.

6. The Second postulate is much more difficult to
accept, because it violates our “ common sense”,
which is firmly grounded in the Galilean kinematics
that we have learned from everyday experiences.
It implies that “it is impossible to accelerate a
particle to a speed greater than c”.

7. 20-3 Consequences of Einstein’s postulates
1.The relativity of time
We consider two observers: S is at rest on the ground, and S’ is in a train moving on a long straight track at constant speed u relative to S.
The observers carry identical timing devices, illustrated in Fig 20-4, consisting of a flashing light bulb F attached to a detector D and separated by a distance from a mirror M.
The bulb emits a flash of light that travels to the mirror, when the reflected light returns to D, the clock ticks and another flash is triggered.

8. M
The time interval between ticks is:
(20-1)
The interval is observed by
either S or S’ when the
clock is at rest respect to that
observer. F D
Fig 20-4

9. We now consider the situation when one
observer looks at a clock carried by the other. Fig
20-5 shows that S observes on the clock carried
by S’ on the moving train. B C A
Fig 20-5 L L F D F D F D S

10. According to S, the flash is emitted at A, reflected
at B, and detected at C.
This interval is
(20-20)
Substituting for from Eq(20-1) and solving
Eq(20-2) for gives
(20-3)

11. The time interval measured by the observer (S’)
relative to whom the clock is at rest is called the
“proper time(正确时间) ”, and
That is, the observer relative to whom the clock is in motion measures a greater interval between ticks. This effect is called “time dilation”. All observer in motion relative to the clock measure “longer intervals”.
Eq(20-3) is valid for any direction of the relative motion of S and S’.

12. 2. The relativity of length
Fig 20-6 shows the sequence of events as
observed by S for the moving clock which is
on the train sideway, so that the light now travels along the direction of motion of the train.
According to S the length of the clock is L, which
is different from the length measured by S’,
relative to whom the clock is at rest.

13. M
Fig 20-6
(A)
(B) L F D F F S’ S’ D D S F
(C) S’ D

14. In the process from A, B to C, the total time taken is
(20-6)
From Eq(20-3), setting
(20-7)
Setting Eqs(20-6) and (20-7) equal to one another
and solving, we obtain
(20-8)
Eq(20-8) summarizes the effect known as “length
contraction”.

15. (a) The length measured by an observer who is
at rest with respect to the object being measured is
called the “rest length” or “proper length”.
(b) All observers in motion relative to S’ measure a
shorter length, but only for dimensions along the
direction of motion; length measurement transverse to the direction of motion are unaffected.
(c) Under ordinary circumstances, and the
effects of length contraction are too small to be
observed.

16. 3 The relativistic addition of velocities
Let us now modify our timing device, as shown in Fig20-7. The flashing bulb F is moved to the mirror end and is replaced by a device P that emits particles at a speed , as measured by an observer at rest with respect to the device.
particle P
light F D

17. The time interval measured by an observer (such as S’) who is at rest with respect to the device is:
(20-9)
What’s the velocity of the particles, measured by the observer S on the ground?
(20-12)
Eq(20-12) gives one form of the velocity addition law.

18. (a) According to Galileo and Newton, a projectile
fired forward at speed in a train that is moving
at speed u should have a speed relative to an observer on the ground.
This clearly permits speeds in excess of c to be realized.

19. (b) The Eq(20-12) prevents the relative speed from ever exceeding c.If
Thus, Eq(20-12) is consistent with Einstein’s
second postulate

20. Sample Problem 20-2
A spaceship is moving away from the Earth at a speed of 0.80c when it fires a missile parallel to the direction of motion of the ship. The missile moves at a peed of 0.60c relative to the ship. What would be the speed of the missile as measured by an observer on the Earth?
Compare with the predictions of Galilean kinematics.

21. 20-4 The Lorentz transformation
We wish to calculate the coordinates x’ , y’ ,
Z’, t’ of an event as observed by S’ from the
coordinates of x, y, z, t of the same event
according to S.
We simplify this problem somewhat, without
losing generality by always choosing the x
and x’ axes to be along the direction of .
Namely the velocity of O’ is , respective to O.

22. The Lorentz transformation equations are
(20-14)
Where the factor is
(20-15)

23. It is convenient in relativity equation to
introduce the speed parameter , defined as
(20-16)
The inverse Lorentz transformation:
–u u
(20-17)

24. Sample problem 20-3
In inertial frame S, a red light and a blue light are
separated by a distance , with the red
light at the larger value of x. The blue light flashes,
and later the red light flashes. Frame S’ is
moving in the direction of increasing x with a
speed of What is the distance between
the two flashes and the time between them as
measured in S’ ?

25. The velocity in x direction of O’ is u, respective to O.
Table 20-2
The velocity in x direction of O’ is u, respective to O.
Lorentz transformation
Inverse transformation
Interval transformation
Inverse Interval transformation

26. Solution:
The Lorentz parameter is
From table 20-2
and

27. 20-5* Measuring the space-time coordinates of an event

28. 20-6 The transformation of velocities
is the velocity of a particle in S reference
is the velocity of S’ reference relative to S in x direction
is the velocity of the particle in S’ reference
, then

29. (20-18)
In similar fashion,
(20-19)
Eqs(20-18) and (20-19) give the Lorentz velocity
transformation.

30. 1 We now show directly that Lorentz velocity
transformation gives the result demanded by
Einstein’s Second postulate ( the constancy of the
speed of light ).
Suppose that the common event being observed
by S and S’ is the passage of a light beam along
the x direction. Observer S measures and

31. .
Using Eqs (20-18) and (20-19)
Thus the speed of light is indeed the same for all
observers or all frames.

32. 2. When ( or equivalently, when ),
Eqs(20-18) and (20-19) reduce to
and (20-20)
which are the Galilean results.

33. Sample Problem 20-4
A particle is accelerated from rest in the lab until its velocity is 0.60c. As viewed from a frame that is moving with the particle at a speed of 0.60c relative to the laboratory, the particle is then given an additional increment of velocity amounting to 0.60c. Find the final velocity of the particle as measured in the lab frame.

34. 20-7 Consequences of the Lorentz transformation
1.The relativity of time
Fig20-15 shows a different view of the time dilation
effect.
(a) S ……
(b) ……
Fig 20-15

35. (20-21)
, the is a proper time ( )
Note that: the time dilation effect is completely
symmetric. If a clock C at rest in S is observed by
S’, then S’ concludes that clock C is running slow.

36. (a) The relativity of simultaneity
Suppose S’ has two clocks at rest, located at
and A flash of light emitted from a point
midway between the clocks reaches the two clocks
If and , then

37. * * According to S’ ( see Fig 20-16a), .
According to S, the light signal reaches clock 1 before it reaches clock 2, and thus the arrival of the light signals at the locations of the two clocks is not simultaneous to S. 1 2 1 2 c c * *
(a) S
(b) S
Fig 20-16
(b) The twin paradox
2. The relativity of length

38. Sample problem 20-5 An observer S is standing on a platform of length
on a space station. A rocket passes at a
relative speed of 0.8c moving parallel to the edge
of the platform. The observer S notes that the front
and back of the rocket simultaneously line up with
the ends of the platform at a particular instant (Fig
20-29a)
According to S, what is the time necessary for
the rocket to pass a particular point on the platform?

39. (b) What is the rest length of the rocket?
(c) According to S’ on the rocket what is the length D of the platform?
(d) According to S’, how long does it take for observer S to pass the entire length of the rocket?
(e) According to S, the ends of the rocket simultaneously line up with the ends of the platform. Are these events simultaneous to S’?

40. Solution: (a) (a) (b) (c) is rest length. (b) (c) Fig (20-29) 0.8c 65m

41. (d)
(e) According to S’, the rocket has a rest length of
and the platform has a contracted length of From Fig 20-19b and 20-19c, the time interval or

42. 20-8 Relativistic momentum
Here we discuss the relativistic view of linear
momentum. Consider the collision shown in Fig
20-20a, viewed from the S frame. Two particles,
each of mass m, move with equal and opposite
velocity v and –v along the x axis. They collide at
the origin, and the distance between their lines of
approach has been adjusted so that after the
collision the particles move along the y axis with
equal and opposite final velocities (Fig 20-20b).

43. (a) (b) Fig 20-20 (c) (d) y S S Frame S S’ Frame S’ S’ 1 1 2 2
Before collision
After the collision v S’ 1
Frame S’ 1 2 S’ 2 v
Fig 20-20
(c)
(d)

44. The collision to be perfectly elastic, in the S frame
Initial :
Final :
The momentum is conserved in the S frame.
According to S’ which moves relative to the S
frame with speed (Fig 20-20c) , particle 2
is at rest before the collision. Using Eqs (20-18)
and (20-19) we can find the transformed x’ and y’
component of the initial and final velocities.

45. Thus in the S’ frame:
,momentum is not conserved. Therefore,
if we are to retain the conservation of momentum

46. as a general law consistent with Einstein’s first
postulate, we must find a new definition of
momentum, that is
(20-23)
In terms of components,
and (20-24)
There the speed v in the denominator of these
expressions is always the speed of the particle as

47. measured in particular inertial frame. It is not the
speed of an inertial frame.
This new definition restores conservation of
momentum in the collision. In the S frame, the
velocities before and after are equal and opposite,
and thus Eq(20-23) again gives zero for the initial
and final momenta. In the S’ frame,
(20-25)

48. 20-9 Relativistic energy Using the velocity shown in Figs 20-20c and
20-20d, you can show that, with , the total
initial and final kinetic energies are
(20-26)
Thus is not equal to According to S
This situation violates the relativity postulate,
we require a new definition of kinetic energy if we

49. are to preserve the law of conservation of energy
and the relativity postulate.
(b) The classical expression for kinetic energy also
violates the Second relativity postulate by allowing
speed in excess of the speed of light. These is no
limit ( in either classical or relativistic dynamic ) to
the energy we can give to a particle.
(c) Relativistic kinetic energy is
(20-27)

50. Using Eq(20-27), we can show that kinetic energy
is conserved in the S’ frame of the collision of Fig
20-20.
2. Energy and mass in special relativity
We can also express Eq(20-27) as
(20-30)
where the total relativistic energy E is defined as
(20-31)

51. and rest energy
(20-32)
The rest energy can be regarded as the internal
energy of a particle at rest.
According to Eq(20-32), whenever we add
energy to a object that remain at rest, we
increase its mass by an amount
If we compress a spring and increase its potential
energy by an mount , then its mass increases by

52. (b) The total relativistic energy must be conserved in any interaction.
The sun radiates an energy of
every second, and the corresponding
change in the mass is

53. Sample problem 20-8 Two 35g putty balls are thrown toward each other,
each with a speed of 1.7m/s. The balls strike each
other head-on and stick together. By how much
does the mass of the combined ball differ from the
sum of the masses of the two original balls?
Solution:
We treat the two putty balls as an isolated system.
No external work is done on this isolated system.

54. With , where
We have
The corresponding increase in mass is
Such a tiny increase in mass is beyond our ability to measure.

55. 3. Conservation of total relativistic energy
Eq(20-30) can be written as
Manipulation of Eqs(20-23) and (20-31) gives a
useful relationship among the total energy,
momentum, and rest energy

56. Sample problem 20-10 A certain accelerator produces a beam of neutral
Kaons ( ) with kinetic energy 325
Mev. Consider a Kaon that decays in flight two
pion( ) . Find the kinetic energy of
each pion in the special case in which the pions
travel parallel or antiparallel to the direction of the
Kaon beam.
Solution:
From Eq(20-33), the initial total relativistic energy is

57. The initial momentum is
The total energy of the final system consisting of
the two pions is
(20-35)
Thus we have one equation in the two unknowns
and , with conservation of momentum.

58. thus
Using Eqs(20-30) and (20-34) or