Transcript Chapter20 The special theory of relativity.ppt

Chapter 20 The special theory
of relativity
Albert Einstein ( 1879 ~ 1955 )
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.
1. Troubles with our ideas about time
The pions (   or   ) created at rest are observed
to decay ( to other particles ) with an average lifetime of
only 26.0ns .
In one particular experiment, pions were created in
motion at a speed ofv  0.913c. In this case they were
observed to travel in the laboratory an average distance of
D  17.4m before decaying, from which we conclude that
they decay in a time given byD  63.7ns, much larger than
the lifetime measured for pions vat 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.
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
9
(0.913c)( 26.0  10 )  7.1m
Thus two observers measure different value for the
same length interval.
3. Troubles with our ideas about light
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.
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”.
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 L0 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.
M
The time interval t 0 between
ticks is:
t 0 
2 L0
c
(20-1)
The interval t 0 is observed by
either S or S’ when the
clock is at rest respect to that
observer.
L0
F D
Fig 20-4
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
A
C
L
F D
S
'
F D
ut
S
Fig 20-5
L
S'
F D
S'
According to S, the flash is emitted at A, reflected
at B, and detected at C.
This interval t is
t ) 2
2
L

(
u
0
2L
2
t 

c
c
2
(20-20)
Substituting for L0 from Eq(20-1) and solving
Eq(20-2) for t gives
t 
t 0
1  (u ) 2
c
(20-3)
The time interval t 0 measured by the observer (S’)
relative to whom the clock is at rest is called the
“proper time(正确时间) ”, and t 0  t .
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’.
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 L0 measured by S’,
relative to whom the clock is at rest.
M
Fig 20-6
(A)
L0
(B)
F
L
F
S’ D
S

u

F
S’ D
u
ut1

u
F
S’ D
L  ut1  ct1
D
ct 2
ut 2
ct 2  L  ut 2
(C)
In the process from A, B to C, the total time taken is
L
L
2L
t  t1  t 2 


cu cu
c
2 L0
From Eq(20-3), setting t 0 
c
t 0
2 L0
1
t 

c 1  (u ) 2
1  (u ) 2
c
c
1
u 2
1  ( ) (20-6)
c
(20-7)
Setting Eqs(20-6) and (20-7) equal to one another
and solving, we obtain
L  L0 1  (u ) 2
(20-8)
c
Eq(20-8) summarizes the effect known as “length
contraction”.
(a) The length L0 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, u  c and the
effects of length contraction are too small to be
observed.
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 v0 , as
measured by an observer at rest with respect to the device.

P
v0
particle
light
D
L0
F
The time interval t 0 measured by an observer (such as S’)
who is at rest with respect to the device is:
(20-9)
L
L
t 0 
0
v0

0
C
What’s the velocity of the particles, measured by
the observer S on the ground?
v0  u
v 
v0 u
1
2
c
(20-12)
Eq(20-12) gives one form of the velocity addition law.
(a) According to Galileo and Newton, a projectile
fired forward at speed v0 in a train that is moving
at speed u should have a speed v0  u relative to
an observer on the ground.
This clearly permits speeds in excess of c to be
realized.
(b) The Eq(20-12) prevents the relative speed from
ever exceeding c.
(a)If
(b)If
v0  c
v0  c
v0  u
v
 v0  u
v0 u
1 2
c
cu
v
c
cu
1 2
c
Thus, Eq(20-12) is consistent with Einstein’s
second postulate
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.
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 u .

Namely the velocity of O’ is u , respective to O.
The Lorentz transformation equations are
x  ut
x' 
  ( x  ut )
u
1  ( )2
c
(20-14)
y'  y
z '  zt  ux
2
ux
c
t' 
  (t  2 )
c
u 2
1 ( )
c
Where the factor  is

1
u 2
1 ( )
c
1
(20-15)
It is convenient in relativity equation to
introduce the speed parameter  , defined as
  uc
(20-16)
The inverse Lorentz transformation:
x   ( x'ut ' )
y  y'
z  z'
t   (t ' ux'
c
2
)
–u
u
(20-17)
Sample problem 20-3
In inertial frame S, a red light and a blue light are
separated by a distance x  2.45km , with the red
light at the larger value of x. The blue light flashes,
and 5.35s later the red light flashes. Frame S’ is
moving in the direction of increasing x with a
speed of u  0.885c . What is the distance between
the two flashes and the time between them as
measured in S’ ?
Table 20-2
The velocity in x direction of O’ is u, respective to O.
Lorentz
Inverse
Interval
Inverse
transformation transformation transformation Interval
transformation
x'   ( x  ut )
x   ( x'ut ' )
x'   (x  ut ) x   (x'ut ' )
y'  y
y  y'
y '  y
y  y '
z'  z
z  z'
z '  z
z  z '
t '   (t  ux
) t   (t ' ux' 2 ) t '   (t  ux ) t   (t ' ux' )
c
c
c2
c2
2
Solution:
The Lorentz parameter is
1
1


 1.928
1  (0.885) 2
1  (u ) 2
c
From table 20-2
x'   (x  ut )
 1.928[2450m  (0.885)(3.00  10 8 m / s)(5.35  10 6 s)]
 2078m  2.08km
and
t '   (t  ux 2 )  3.15s
c
20-5* Measuring the space-time
coordinates of an event
20-6 The transformation of velocities

v is the velocity of a particle in S reference
u is the velocity of S’ reference relative to S in x
 direction
v ' is the velocity of the particle in S’ reference
x  u
x'
 (x  ut )
t
vx ' 


t '  (t  u x 2 )
u (x )
t
c
1
x
t
 vx , then
c2
vx  u
vx ' 
uv
1 x 2
c
(20-18)
In similar fashion,
vy '
 (1 
vy
uv x
vz ' 
2
)
 (1 
vx
uv x
2
)
(20-19)
c
Eqs(20-18) and (20-19) give the Lorentz velocity
transformation.
c
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 v x  c and
v y  vz  0 .
Using Eqs (20-18) and (20-19)
vx  u
cu
vx ' 

c
uv x
u
1

1
c
c2
v y '  vz '  0
Thus the speed of light is indeed the same for all
observers or all frames.
2. When u  c ( or equivalently, when
Eqs(20-18) and (20-19) reduce to
and
vx '  vx  u
vy ' vy
vz '  vz
(20-20)
which are the Galilean results.
c   ),
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.
20-7 Consequences of the Lorentz
transformation
1.The relativity of time
Fig20-15 shows a different view of the time dilation
effect.


s'
s' u
u
c
'
c
'
(a)
t 2 't1 '
t1 '
S
(b)
x0 '
x0 '
t1
t2
x1
……
Fig 20-15
x2
……
t   (t 'u x'
)
(20-21)
x' 0 , the t ' is a proper time ( t 0 )
t  t '  t 0 
c
2
t 0
u 2
1 ( )
c
 t 0
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.
(a) The relativity of simultaneity
Suppose S’ has two clocks at rest, located at x1 '
and x 2 '. A flash of light emitted from a point
midway between the clocks reaches the two
clocks
t   (t 'u x'
c
2
)
If t ' 0 and x' 0 , then t  0 .
According to S’ ( see Fig 20-16a), t ' 0 .
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
c
(a)

u
*s '
1
c
S
s* '
(b)
Fig 20-16
(b) The twin paradox
2. The relativity of length
2
S

u
Sample problem 20-5
An observer S is standing on a platform of length
D0  65m 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)
(a) According to S, what is the time necessary for
the rocket to pass a particular point on the platform?
(b) What is the rest length L0 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’?
Solution:
(a)
0.8c
L
65m
t0 

 0.27 s
8
0.8c 2.4 10 m / s
S’
(a)
S
108m
(b)
(b)
L0  L 
65m
1  (0.8) 2
 108m
D0

 D0
u 2
1  ( )  39m
c
S’
S
0.8c
39m
(c) D0  65m is rest length.
D
65m
108m
S’
(c)
S
39m
Fig (20-29)
0.8c
(d)
108m
t ' 
 0.45s
0.8c
(e) According to S’, the rocket has a rest length of
L0  108m and the platform has a contracted length
of D  39m . From Fig 20-19b and 20-19c, the
time interval
 (0.8c)( 65m)

x
t '  u
 0.29s
2 
c
c 2 1  0.8 2
108  39
or
t ' 
 0.29s
0.8c
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).

v
y

v
1
S
Frame S

1
S
 v2

v
2
(a)
(b)
After the collision
Before collision


2v
2
1 v
1
c2
u  v

u  v
S’
Frame S’
2
v 1  ( v )2
c
Fig 20-20
(c)

(d)
S’
 v 1  ( v )2
c
v 2  ( v )2
c
1
v
2
v
v 2  ( v )2
c
The collision to be perfectly elastic, in the S frame
Initial : Pxi  mv  m(v)  0 Pyi  0
Final : P  0
Pyf  mv  m(v)  0
xf
The momentum is conserved in the S frame.
According to S’ which moves relative to the S
frame with speed u  v (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.
Thus in the S’ frame:
2v
2mv
Pxi '  m(
)  m(0) 
2
2
v
v
1
1
2
c
c2
Pxf '  mv  mv  2mv
Pyi '  0
Pyf '  mv 1  ( v ) 2  m(v 1  ( v ) 2 )  0
c
c
Pxi '  Pxf ' ,momentum is not conserved. Therefore,
if we are to retain the conservation of momentum
as a general law consistent with Einstein’s first
postulate, we must find a new definition of

mv
momentum, that is 
P
(20-23)
1 ( v )2
c
In terms of components,
mv x
mv y
and Py 
(20-24)
Px 
1  (v )2
c
1  (v )2
c
There the speed v in the denominator of these
expressions is always the speed of the particle as
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,
2mv
Pxi '  Pxf ' 
2
(20-25)
v
1
c2
Pyi '  Pyf '  0
20-9 Relativistic energy
1. Using the velocity shown in Figs 20-20c and
20-20d, you can show that, with k  1 mv 2 , the total
2
initial and final kinetic energies are
2mv 2
2
2
ki ' 
v
2
k f '  mv (2 
(20-26)
2)
2
v
c
(1 
2)
c
Thus k i ' is not equal to k f ' . According to S k i  k f.
(a) This situation violates the relativity postulate,
we require a new definition of kinetic energy if we
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
mc 2
(20-27)
k
 mc 2
2
v
1
c2
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
K  E  E0
(20-30)
where the total relativistic
energy E is defined as
2
E
mc
2
v
1
c
2
(20-31)
and rest energy E0
E0  mc 2
(20-32)
The rest energy E0 can be regarded as the internal
energy of a particle at rest.
(a) According to Eq(20-32), whenever we add
energy E to a object that remain at rest, we
increase its mass by an amount m  E c 2 .
If we compress a spring and increase its potential
energy by an mount U , then its mass increases
by U 2 .
c
(b) The total relativistic energy must be conserved
in any interaction.
26
4

10
J
The sun radiates an energy of
every second, and the corresponding
change in the mass is
26

4

10
J
9

E
m 



4

10
kg
2
8
2
c
(3  10 m / s)
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.
With K  K f  K i , where
We have
K f  0.
1
E 0  K i  2( mv 2 )  (0.035kg)(1.7 m / s ) 2  0.101J
2
The corresponding increase in mass is
m 
E0
0.101J
18


1
.
1

10
kg
2
8
2
c
(3 10 m / s)
Such a tiny increase in mass is beyond our ability
to measure.
3. Conservation of total relativistic energy
Eq(20-30) can be written as
E  K  E0
Manipulation of Eqs(20-23) and (20-31) gives a
useful relationship among the total energy,
momentum, and rest energy
E  ( pc) 2  (mc 2 ) 2
Sample problem 20-10
A certain accelerator produces a beam of neutral
Kaons ( mk c 2  498Mev ) with kinetic energy 325
Mev. Consider a Kaon that decays in flight two
2
m
c
pion(   140 Mev ) . 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
E k  K  mk c 2  325Mev  498Mev  823Mev
The initial momentum is
Pk c  Ek  (mk c 2 ) 2  8232  4982  655Mev
2
The total energy of the final system consisting of
the two pions is
E  E1  E2  ( p1c) 2  (m c 2 ) 2  ( p 2 c) 2  (m c 2 ) 2
 823Mev
(20-35)
Thus we have one equation in the two unknowns
P1 and P2 , with conservation of momentum.
p1c1  p2 c  pk c  655Mev
thus
 13Mev
Using Eqs(20-30) and (20-34)
p1c1  668Mev or
K  ( pc) 2  (m c 2 ) 2  m c 2
K 1  (668Mev ) 2  140 Mev  140Mev  543Mev
K 2  (13Mev )  140Mev  140Mev  0.6Mev
2