Modern Physics
Chapter 18
The Special Theory of relativity
狭义相对论基础
Chapter 19
The Quantization of Light
光的量子性
Chapter 20
Quantum Theory of the Atoms
原子的量子理论
I. Three experimental milestones before the
modern physics
1. The discovery of x-ray by Rontgen in 1895.
x-ray incidents on graphite, the quantization of
light was proved.(Compton scatter)
electron
2. The discovery of radioactivity in
1896 by Becquerel.
With natural radioactive source (particle), Ruthersford proposed the
nuclear model of the atom in 1911.
3. The discovery of electron in 1897 by
J.J.Thomson
II. Three theoretical milestones of the modern
physics
1. The idea that electromagnetic radiation has
particle characteristics as well as wave
properties----Wave-Particle Duality of light.
Plank: Harmonic oscillators radiate energy only
in quanta of energy h. (1900)
Einstein: the electromagnetic field consists of
light quanta ( photon ). (1905)
Bohr: with the nuclear model, the electron and the
particle nature of radiation to provide a
model of the hydrogen atom (1913)
2.The theory of relativity proposed by Einstein
in 1905.
Display a new thinking about space and time
and mass-energy relationship.
3. The microscopic particles have the WaveParticle Duality, too.
De Broglie hypothesis in 1924, Schrodinger
equation in 1926, … Quantum Mechanics were
established.
Chapter 18
The special theory of
relativity
§18-1 The Michelson-Morley experiment
§18-2 The Postulates of Special Relativity
狭义相对论的基本假定
§18-3 The Lorentz transformation
洛仑兹变换
§18-5 The Lorentz Transformation of Velocities
相对论速度变换式
§18-4 Some Consequences of the Lorentz
Transformation
洛仑兹变换的一些结果
§18-6 The Relativistic Dynamic theory
相对论动力学基础
§18-1 The Michelson-Morley experiment
The Michelson interferometer was used to
look for ether.
M1
u--the speed of the earth
relative to the ether.
For the light beam
(2) ：O  M2  O
l2
l2
t2 

c u c u
l1
S
(1)
l2
O
(2)
u
2l 2
1
 (
)
2
2
c 1u / c
P
M2
For the light beam
(1) ：O  M1  O
The actual path of
beam (1) is O  A O
In the OAB,
t1 2
t1 2
2
( c )  ( u )  l1
2
2
2 l1
1
 t1  (
)
c 1  u 2 / c2
M1
A
t1
c
2
l1
O
B
ut1
O
Let l1 = l2 = l and using u<< c
We get
2l
u
t 2  (1  2 )
c
c
2
2l
u2
t1  (1  2 )
c
2c
The time difference of two light beams is
t  t 2  t1  lu / c
2
3
The difference of optical lengths of two light
beams is
  ct  lu / c
2
2
Let the interferometer rotates 900
S
M1
l1
S
(1)
900
l2
O
(2)
u
P
M2
P
u
l1
O
(1)
(2)
l2
M2
t1  t 2
t 2  t1
M1
The time difference of two light beams is
t   t 2  t1 lu / c
2
3
The difference of optical lengths   ct  
The difference of optical lengths changes 2
because of the rotating.
The number of interferometer fringes should
2
shift
2 2lu
N 


c
2
 0.4
Taken c=3108 m/s, l=11m, =5.910-7 m, u=
3104 m/s (the speed of the earth moving around
the sun)
The result of the experiment is
null result
i.e.
The ether does not exists and
the speed of light relative to the earth
is C.
§6-2 The postulates of special relativity
I. Galileo transformation
K
1. The event
u
y
It happens in the
K ' y'
P
space at sometime.
0
x
z
--- recorded by the
0'
coordinates of
ut
z
'
space and time.
When the observer locates in K-system, it is
recorded with P(x, y, z, t)
When the observer locates in K-system, it is
recorded with P(x, y, z, t )
x'
(x, y, z, t)
Relationship?
2. Galileo transformation
Let 0 and 0 coincide
when t=t=0
Then
z 0
x'  x  ut
 y'  y
 z'  z
 t'  t
K
Or
y
(x, y, z, t )
u
K ' y'
P
x 0'
ut
z'
x  x'ut '
 y  y'
 z  z'
 t  t'
x'
Transformations of velocities
and accelerations
v
'

v

u
 x x
v y '  v y
vz '  vz
x'  x  ut
 y'  y
 z'  z
 t'  t
a
'

a
x
 x
a y '  a y
az '  az
 
 a'  a
II. The classical opinions about the time and space
Absolute
space：
K r  ( x2  x1 )  ( y2  y1 )  ( z 2  z1 )
2
2
2
K x2 ' x1 '  ( x2  ut )  ( x1  ut )  x2  x1
  r'  ( x2 ' x1 ' )  ( y2 ' y1 ' )  ( z2 ' z1 ' )
2
2
2
 r
----The measurement of length is constant
in different coordinate system.
Absolute
time:
 t'  t
 t'  t
--The measurement of time interval is
constant in different coordinate system.
Absolute
mass:
m  m


So F '  F
§6-2 The postulates of special relativity
1. The principle of the relativity：the laws of
physics are the same in all inertial reference
frames.
---all inertial reference frames are equivalent.
---there is no preferential frame in the universe.
2. The principle of the constancy of the speed
of light：the speed of light c is the same for
every inertial reference frame and is
independent of any motion of the source.
§18-3 The Lorentz transformation
A event happens in the space at sometime,
It is recorded with P(x, y, z, t) in K-system
and with P(x, y, z, t ) in K-system
K
z
(x, y, z, t)
0
y
u
K ' y'
P
x 0'
ut
z'
Relationship
in relativity?
x'
(x, y, z, t )
A event happens in the space at sometime, As
the space and time have the symmetry and the
homogeneity,
Their transformations should be linear.
Let
x  Ax  Bt
y  y
A, B, E, D are constants.
z  z
t   Ex  Dt
Using the principle of the relativity:
Observing in K-system,
O locates in x=ut at time t.
x =0
x  Ax  Bt
B   Au,
x  A( x  ut )
Inversely, observing in K-system,
O locates in x =-ut at time t.
x =0
x  A( x  ut )
-ut
0
=Ex+Dt
D A
Using the principle of the constancy of the
speed of light:
Assume a light beam is emitted in the origin of
K and K at time t =t  = 0,
The signal of light arrives x =ct in K-system at
time t and x=ct  in K-system at the time t  .
x  A( x  ut )
ct
ct
t   Ex  At
Au
E 2
c
So
x  A( x  ut )
y  y
z  z
u
t   A( t  2 x )
c
Find A=?
Meanwhile, the signal of light arrives a point on
y axis with (y =ct, x=0) in K-system at time t.
And this point is measured with (x, y ) in Ksystem at time t .
i.e., in K-system :
x= 0
y=ct
in K-system :
x   y  c t 
2
2
2
2
x   y  c t 
2
2
=A(0-ut)
2
2
=A(t-0)
=y=ct
We get：
A
x  A( x  ut )
y  y
u
t   A( t  2 x )
c
1
1 u c
2
2
Lorentz
transformation
x' 
its inverse
transformation
x  ut
1 u c
y'  y
z'  z
2
u
t 2 x
c
t' 
2
2
1 u c
x
2
x' ut'
1 u c
y  y'
z  z'
2
2
u
t ' 2 x'
c
t
2
2
1 u c
Notes
 The space and time relate to each other in
L-transformations.
 L-transformations have no scene if v >c
---- c is the ultimate speed of an object.
 When u<<c,
1 u c
2
2
1
x'  x  ut
t'  t
--L-transformation Galileo transformation
If two events happen in the space at sometime,
they are recorded with P1(x1, y1, z1, t1) and
P2(x2, y2, z2, t2) in K-system,
P1(x1, y1, z1, t1 ) and P2(x2, y2, z2, t2 ) in
K-system,
then,
x  x2  x1
t  t 2  t1
x  x2  x1
t   t 2  t1
x  
 x  u t
1 u c
2
2
u
t  2 x
c
t ' 
2
2
1 u c
§18-5 The Lorentz Transformation
of Velocities
 dx' 
dx  udt
1 u c
2
dt ' 
2
dt  udx c
2
1  u2 c 2
dy'  dy
dz'  dz
vx  u
dx  udt
dx'


 vx ' 
2
2
1  uv x c
dt ' dt  udx c
dy' dy 1  u 2 c 2
vy'

2
dt '
dt  udx c
vy 1  u c
2

1  uv x c
2
2
Similarly,
dz' v z 1  u c
vz ' 

2
dt '
1  uv x c
2
2
vx  u
v x 
2
1  uv x c
vy 1 u c
2
v y 
1  uv x c
2
vz 1  u c
v z 
2
1  uv x c
2
2
2
u<<c
v' x  v x  u
vy' vy
vz '  vz
§18-4
Some Consequences of
the Lorentz Transformation
I. The relativity of length
The proper length (固有长度) L0：the length
measured by the observer who is at rest
relative to the object.
y'
K'
A rod is at rest in
K-system,
In K-system：
L0  x2 ' x1 '
K
0
y 0'
u
x2
x1
x'
?
x
the spaceship is moving relative to K-system,
so we must measured x1 and x2 simultaneously
in K-system,
i.e.,
at time
y'
t1  t 2  t
u
y
measure
L  x2  x1
K'
x'
0
'
K
0
x1
x2
x
L0  x2 ' x1 '
x2  ut
x1  ut


2 2
2 2
1 u c
1 u c

x 2  x1
1 u c
2
2

L
1 u c
2
2
 L  L0 1  u c  L0
2
2
--length contraction (长度收缩)
Notes：
The measurement of length is not absolute
and the proper length L0 measured by an
observer who is at rest relative to the object is
the largest.
 When the object is moving relative to the
observer, the coordinates of the two side must
be measured simultaneously.
 Length contraction happens only on the
direction of the object moving.
 All inertial reference frames are equivalent.
2. The relativity of time( the time dilation effect)

Proper time interval (固有时间) t0：the time
interval measured by the observer who is at
rest relative to the two events happening at
the same point of the space.
In K-system:
Event 1
Event 2
x 0 , t1
x0 , t 2
t 0  t 2  t1
K
0
flash
y
x0
x
In K -system:
t   t 2  t1 

t 2  ux0 c 2
1 u c
2
2

t1  ux0 c 2
1 u c
2
2
t 0
1 u
2
 t 0
--time dilation
(时间膨胀)
c
2
K
0
y
K'
y'
u
x'
0'
x0
x
Notes：
 t0 is the smallest interval between two
events that any observer can measure.
A moving clock relative to the observer run
slowly than a static clock relative to the
observer----moving clock run slow (动钟变慢)
K'
y'
the flash travels one
period in K-system,
b
0'
Flash clock
x'
2b
t '  t 0
c
Observing in K-system,
the flash-system is moving
K
y
y'
l
b
l
l
b
K'
l
0
ut
x
u
0'
2l 2 2
ut 2

b (
)
t 
c
2
c
t 0
moving clock run
 t 
1 u c
2
2
slow
x'
3.The relativity of
“simultaneity”
K'
(“同时”的相对性
y 0'
)
Assume
two events
K
happen,
they are measured by
0
an observer locating in
K–system,
their positions are x1, x2 and
y'
u
x'
x1
x2
t  t 2  t1  0
simultaneous
x
In K –system,
K'
t '  t 2  t1

t 2  ux 2 c
1 u c
2


2
K
1 u c
0
2
2
( x1  x2 ) u / c
1 u c
2
u
y 0'
x'
2
t1  ux1 c
2
y'
2
2
0
x1
x2
is not
simultaneous
observing in K
–system
x
Notes：
simultaneity is not an absolute concept but a
relative one, depending on the state of motion
of the observer.
y'
K'
 The sequence of the
u
events happening:
x'
y 0'
t'  t   t 
2

1
( x1  x 2 )u / c
1 u c
2
 t 2  t1
2
2
K
0
0
x1
x2
x
--the event locating on x2 happens
first observing in K –system.
The “start-end” sequence of the event is
absolute in any inertial reference frame.
u
uv
t  2 x t (1  2 )
c
c
 t ' 

2
2
2
2
1 u c
1 u c
as
u c
 t '
and
x
v
t
vc
t have same sign
-- The “start-end” sequence of the event
does not change.
[Example] A observer A saw that two events
locating on x axis happen simultaneously. Their
distance is 4m. Another observer B saw that the
distance between the two events is 5m. Do the
two events happen simultaneously relative to the
observer B? how much is the time interval
measured by B ?  how much is the relative
speed between A and B =?
Solution：let A locates on K-system and B
locates on K-system. B is moving relative
to A with the speed u along x axis.
then
x  4 m
t  0
x' 5 m
x ' 
We get
 x  u t
1 u c
2
2
5 
4
1 u c
2
u  0.6c
u
x
2
0.6  4
c
8
t ' 


10
s
2
2
2 2
1 u c
c 1 u c
2
§18-6
The Relativistic Dynamic theory
1. Relativistic mass
When an object with static mass m0 is moving
with the speed v，its moving mass is
m  m( v ) 
m0
1 v c
2
2
Deduce
Assume there are two same particles A and B，
their static mass is m0 respectively,
and relative speed is v
B
A
If A and B collide in absolute
v
non-elastic,
the momentum of A and B is conservative.
Let B is K-system,
mv  ( m0  m)Vc (1)
Vc--- The speed of A+B’s center of mass
relative to K-system
A is K’-system :
 mv  ( m0  m)Vc( 2)
A
v
B
Vc’--- The speed of A+B’s center of mass
relative to K’-system
Combining (1) and (2), we get
Vc  Vc ( 3)
And using Lorentz transformation of velocity
Vc  v
Vc 
vVc
1 2
c
 ( 4)
Combining (3) and (4) , we get
2
2
c
c
Vc 
 c 2  1  5 
v
v
Substituting (5) into(1), we get
m
m0
2
v
1 2
c
Discussion
 If v<<c ,
 If v
m(v )
c,
m(v )
m0

a
0
---The speed of an object cannot exceed c
2. Relativistic momentum and dynamics equation
Momentum


p  mv 

m0 v
1 v c
2
2
Dynamics equation

m0 v
d
(
)
dt 1  v 2 c 2


 d ( mv )

F

dt
 m dv  v dm
dt
dt
3. Relativistic energy

An object has a displacement dr under the

action of a force F .
The increment of its kinetic energy is

  d ( mv ) 
dE k  F  dr 
 dr
dt
 
 d ( mv )  v
 v dm  mvdv
2
E k   dE k   v dm  mvdv  ?
2
As
m
m0
We get
m c m v m c
2
2 2
v
1 2
c
2 2
2 2
0
Making a differential on the two side of equation
c dm  v dm  mvdv
2
2
E k   dE k 

m
m0
2
c dm
 mc  m0 c
2
Kinetic energy
Total energy
2
Rest energy
Discussion
 Rest energy E0  m0 c 2 contains the total
internal kinetic energies of all particles
moving in the object + the total internal
potential energies
--Internal (intrinsic)energy of the object
--micro-energy
 Kinetic energy E k  E  E0 is the macroenergy when the object has a mechanical
motion as an entirety.
If
,
v  c
E k  mc  m0c  m0 c (
2
2
2
1
1 v c
1 v 2
2
 m0 c [1  ( )    1]
2 c
1
2
2
 m0 v  m0 c
2
2
2
 1)
----Large part of energy stores in the object
 Total energy
E  mc  E0  E k
2
 Mass-energy conservation law
In an isolated system of particles, the total
energy remains constant.
 E   m c =constant
2
i
i
i
i
the total mass of the system is conservative
 m =constant
i
i
The conservation of total energy is
equivalent to the conservation of total mass.
 Rest energy and kinetic energy can convert
each other.
Example: nuclear fission
Before fission, the object has
E  E0  M 0 c
2
( E k 0  0)
After fission, the object break into two particles.
Their total energy is
 m10c  Ek 1  m20c  Ek 2
2
2
Total energy is conservative: E  m1c 2  m2c 2
Total mass is conservative:
M 0  m1  m2
The loss of rest mass :
m0  M 0  m10  m20
The loss of rest energy :
m 0 c
2
Ek 1  Ek 2
The increment of kinetic energy .
4. The relationship between the total energy and
momentum
E
m0 c
2
1 v c
2
2
p
m0 v
1 v c
2
2
Eliminating v , we get
E  m0 c  p c
2
2 4
2 2
 E0  p c
2
pc
E
2 2
E0
动质
能三
角形
E  m0 c  p c
2
5. Photon
m
2 4
m0
1 v
2
 Rest mass m0=0
c
2
E
h
Total (moving) mass m  2  2
c
c
E
h

h
momentum p 


c
c

2 2
[Example] A particle with rest mass mo is moving
at the speed vo=0.4c. If the particle is
accelerated till its final momentum = 10 initial
momentum. What is the ratio of its final speed
and initial speed?
Solution initial momentum is
p0 
m0 v 0
1  v0 c
2
2

0.4m0 c
1  0.4
Final momentum
p  10 p0  4.4m0 c
2
 0.44m0 c
As
p
m0 v
1 v c
2
get
2
v  0.975c
v 0.975
 
 2.87
v0
0.4
[Example] Two particles have rest mass mo
respectively. One particle is at rest and another
is moving with the speed v =0.8c. They collide
and stick together. Find the rest mass of the
compound particle.
解：Assume the mass of the compound
particle is M and speed is V after colliding
The mass of the system is conservative
before and after colliding: m0  m  M
The momentum is conservative:
mv  MV
m0
m0
m
0
m 


2
2
2
1 v c
0.6
1  0.8
8
 M  m0  m  m0
3
mv
V 
 0.5c
M
M0  M 1  V c
2
 2.31m0
2
8
2
 m0 1  0.5
3