§25.1 reference frames and the classical
Galilean relativity
1.what is reference frame
z
A reference frame
consists of (1) a
coordinate system
and (2) a set of
synchronized clocks
distributed
throughout the
coordinate grid and
rest with respect to it.
y
x
1
§25.1 reference frames and the classical
Galilean relativity
§25.1 reference frames and the classical
Galilean relativity
A reference frame has three spatial coordinates
and one time coordinate (x, y, z, t).
Four dimensional space-time
Inertial reference frames
Reference frame
noninertial reference frames
synchronized clocks:
O
A
B
l
l
2
§25.1 reference frames and the classical
Galilean relativity
2. Some fundamental concepts
¾Event is something that happens at a
particular place and instant.
¾Observers belong to particular inertial frames
of reference, they could be people, electronic
instrument, or other suitable recorders.
¾Relativity is concerned with how an event
described in one reference frame is related to its
description in another reference frame. That is
how the coordinates and times of events
measured in one reference frame are related to
the coordinate, time, and corresponding physical
quantities in another reference frame.
§25.1 reference frames and the classical
Galilean relativity
¾The special theory of relativity is concerned
with the relationship between events and
physical quantities specified in different
inertial reference frames.
¾The general theory of relativity is
concerned with the relationship between
events and physical quantities specified in any
reference frames.
¾Transformation equations are
⎛ x⎞
⎛ x′ ⎞
that indicate how the four space
⎜ ⎟
⎜ ⎟
y
⎜ ⎟
⎜ y′ ⎟
and time coordinates specified in
⇔
one reference frame are related to ⎜ z ⎟ ⎜ z ′ ⎟
⎜ ⎟
⎜ ⎟
the corresponding quantities
⎜t⎟
⎜ t′ ⎟
⎝ ⎠
⎝ ⎠
specified in another reference frame.
3
§25.1 reference frames and the classical
Galilean relativity
¾Standard geometry we use for the special
theory of relativity has two inertial reference
frames called S and S’, with their x- and x’coordinate axes collinear. Imagine collections of
clocks distributed at rest throughout each
respective frame; the clocks all are set to 0 s
when the two origins coincide.
Observer is at rest in frame S.
§25.1 reference frames and the classical
Galilean relativity
Observer is at rest in frame S’.
4
§25.1 reference frames and the classical
Galilean relativity
3. Classical Galilean relativity
1the Galilean time transformation equation
In classical physics time is a universal measure
of the chronological ordering of events and the
time interval between them.
Watches in fast sports cars, airplanes,
spacecraft, and oxcarts tick at the same rate as
those at rest on the ground; the time interval
between two events and the rate at which time
passes are independent of the speed of the
moving clock; they are same everywhere.
t = t′
§25.1 reference frames and the classical
Galilean relativity
2the Galilean spatial coordinate transformation
equations
r r r
r ′ = r − rO
r r r
u′ = u − v
x ′ = x − vt
y′ = y
z′ = z
t = t′
5
§25.1 reference frames and the classical
Galilean relativity
3the Galilean velocity component transformation
equations
dx ′
,
u′x =
dt ′
dx d( x ′ + vt )
=
= u′x + v
ux =
dt
dt
u x = u′x + v
u y = u′y
uz = u′z
The velocity components along
a direction perpendicular to
the motion are the same in two
standard inertial reference
frames.
§25.1 reference frames and the classical
Galilean relativity
4the Galilean acceleration component
transformation equations
d 2 x′
a ′x =
,
dt ′ 2
d 2 x d 2 ( x′ + vt )
= a′x
ax = 2 =
2
dt
dt
a ′x = a x
a ′y = a y
a ′z = a z
The acceleration components
are the same in the two inertial
reference frames.
6
§25.1 reference frames and the classical
Galilean relativity
a ′x = a x
a ′y = a y
a ′z = a z
r r
a = a′
m = m′
r
r
r
r
F = m a = m ′a ′ = F ′
Descriptions of
what happens as a
result of the laws
of mechanics may
different from one
inertial reference
frame to another,
but the laws of
mechanics are the
same.
§25.2 the need for change and the
postulates of the special theory
1. Why we need relativity theory?
1troubles with our ideas about time
The pion π+ or π- created in a high-energy
particle accelerator is a very unstable
particle. Its lifetime at rest is 26.0 ns; when it
moves at a speed of v= 0.913c, an average
distance of D=17.4 m are observed before
decaying in the laboratory. We can calculate
the lifetime in this case by D/v=63.7 ns, it is
much larger than the lifetime at rest.
Such an effect cannot be explained by
Newtoneian physics!
7
§25.2 the need for change and the
postulates of the special theory
2troubles with our ideas about length
Suppose an observer in the above laboratory
placed one marker at location of pion’s
formation and another at the location of its
decay. The distance between the two
markers is measured to be 17.4 m.
Another observer who is traveling along with
pion at a speed of u=0.913c . To this observer,
the distance between the two markers
showing the formation and decay of the pion
is (0.913c)(26.0×10-9s)=7.1 m.
Two observers who are in relative motion
measure different values for the same length
interval.
§25.2 the need for change and the
postulates of the special theory
3troubles with our ideas about speed
If the observer A
throws a ball at
superluminal,
speed, the result
observed by
observer O will
be that the
observer B get
the ball before
the observer A
throwing it.
8
§25.2 the need for change and the
postulates of the special theory
4troubles with our ideas
about energy
Electron and positron are
initially toward one
another at a very low
speed.
Electron and positron have
annihilated one another
and give a radiation.
Electron-positron
e+
The walls of the container
absorbed the radiation and
increasing the internal
energy of this system.
e−
radiation
∆ E int
§25.2 the need for change and the
postulates of the special theory
5troubles with our ideas about light
Electromagnetism provides some subtle clues
that all is not right with Glilean relativity.
∂2Ey
∂x 2
= µ 0ε 0
c=
∂2Ey
∂ 2 Bz
∂ 2 Bz
= µ 0ε 0
or
∂x 2
∂t 2
∂t 2
1
= 3.0 × 10 8 m/s
µ 0ε 0
This result is not dependent of the speed v of
the source of the waves in the equation!
9
§25.2 the need for change and the
postulates of the special theory
Experiment is performed in reference frame S:
c
c
Experiment is performed in reference frame S’
which is moving at speed v in the same direction
as light is moving:
r
v
c
c
§25.2 the need for change and the
postulates of the special theory
Therefore the Galilean transformation
equations between two inertial reference
frames need to be modified to accommodate
the observed invariance of the speed of light.
The new transformation equations , called
the Lorentz transformation equations,
together with their implications and
consequences, constitute the special theory of
relativity.
2. The postulates of special theory of relativity
1the speed of light in a vacuum has the same
numerical value c in any inertial reference
frame, independent of the motion of the source
and observer.
10
§25.2 the need for change and the
postulates of the special theory
It means that there is an ultimate speed c,
the same in all directions and in all inertial
reference frames. Light and any massless
particles travel at this speed. No particle that
have mass and carry energy or information
can reach or exceed this limit, no matter how
much or how long it is accelerated.
This postulate is expression of the experimental
result.
(1)in a 1964 experiment, physicist at CERN.
Rapid
moving
source
π 0 →γ +γ
Speed of the light was
same as measured at
rest in the laboratory.
§25.2 the need for change and the
postulates of the special theory
(2)W. Bertozzi experiment(1964)
v = 0.999 999 999 95c
11
§25.2 the need for change and the
postulates of the special theory
2the fundamental laws of physics are the same
for observers in all inertial reference frames.
No frame is preferred.
Descriptions of what happens as a result of the
laws of physics may different from one inertial
reference frame to another, but the underlying
fundamental physical principles and laws are
the same.
Glileo assumed that the laws of mechanics
were the same in all inertial reference frames.
Einstein extended that idea to include all the
laws of physics.
§25.3 consequences of Einstein’s postulates
1. Time dilation
Event 1: emission of the pulse
x1′ = 0, y1′ = 0, z1′ = 0, t1′ = 0
s′
l0
Event 2: reflection of the pulse
x 2′ = 0, y′2 = l 0 , z ′2 = 0, t 2′ = l0 c
Event 3: detection of the pulse
x′3 = 0, y′3 = 0, z ′3 = 0, t 3′ = 2l 0 c
Time interval: ∆t ′ = τ 0 = 2 l 0 c
τ0
Proper time interval
12
§25.3 consequences of Einstein’s postulates
Event 1: emission of
the pulse
s
x1 = 0, y1 = 0, z1 = 0, t1 = 0
l0
Event 2: reflection of
the pulse
vτ
x 2 = vτ 2 , y 2 = l 0 ,
z 2 = 0, t 2 = τ 2
Event 3: detection of
the pulse
x 3 = vτ , y3 = 0,
z 3 = 0, t 3 = τ
τ
l 02 + (vτ 2) 2
=
c
2
2l0 c
τ0
τ=
=
1 − v 2 c2
1 − v2 c2
τ
§25.3 consequences of Einstein’s postulates
Conclusions: the moving clock has a greater
time interval between its ticks than the clock
that is at rest ; a moving clock runs slow.
This is called time dilation.
Time intervals in relativity are not absolute or
universal but depend on whether the clock is
moving or not: we say time is a relative not an
absolute quantity.
The proper time interval is the shortest.
Time dilation has been confirmed by many
experiments:
13
§25.3 consequences of Einstein’s postulates
Microscopic clock: Muons are unstable
elementary particles with a (proper) lifetime of
2.2 µs. they are produced with very high
speeds in the upper atmosphere when cosmic
rays collide with air molecules. Take the
height L0 of the atmosphere to be 90 km in the
reference frame of earth, if the average speed
of the muons is 0.999978c, can the muons
arrive the surface of the earth?
§25.3 consequences of Einstein’s postulates
Solution: According to Newton’s mechanics
L = 0.999978 × 3.0 × 108 × 2.2 × 10 −6 = 660 m
The muons can not arrive the surface of the earth.
According to relativity, the life-time of the muon is
τ=
τ0
1 − (v c )
2
2
=
2.2
1 − (0.999978)
2
= 331.7µs
The distance that muon can travel is
0.999978 × 3 × 108 × 331.7 × 10 −6 = 99507.8m
The muon can arrive the surface of the earth.
14
§25.3 consequences of Einstein’s postulates
Example: your starship passes Earth with a
relative speed of 0.9990c. After traveling 10.0
y(your time), you stop at lookout LP13, turn,
and then travel back to earth with the same
relative speed. The trip back takes another
10.0 y(your time). How long does the round
trip take according to measurements made on
Earth?(neglect any effect due to the
accelerations)
Solution: Event 1: start from Earth
Event 2: stop at LP13
Proper time: τ 0 = ∆t 0 = 10.0 y
§25.3 consequences of Einstein’s postulates
The Earth measurement of the time interval is
τ = ∆t =
τ0
1 − (v c )
2
=
10.0
1 − 0.999
2
= 224 ( y)
On the return trip, we have the same situation
and the same data.thus the round trip requires
∆t total =
2τ 0
1 − (v c )
2
= 448 ( y)
15
§25.3 consequences of Einstein’s postulates
Macroscopic clocks:
In October 1977, Joseph Hafele and Richard
Keating flew four portable atomic clocks
twice around the world on commercial airline,
once in each direction. The prediction of the
theory was verified within 10%.
A few years later, physicist at the university of
Maryland carried out a similar experiment
with improved precision(1%).
§25.3 consequences of Einstein’s postulates
2. Length contraction
1lengths perpendicular to the direction of
motion
y′
y
l0
o′
Measured using
rulers at rest in
S’
x′
l0
o
Measured using
rulers at rest in S
x
We can prove the sticks are the same length.
16
§25.3 consequences of Einstein’s postulates
Prove by contradiction:
Make the hypothesis that a moving stick,
oriented perpendicular to the direction of
motion, is longer than stick at rest.
An observer in reference frame S:
y′
y
l > l0
o′
Measured using
rulers at rest in
S’
x′
l0
Measured using
rulers at rest in S
o
x
§25.3 consequences of Einstein’s postulates
An observer in reference frame S’:
y′
y
l0
o′
l > l0
Measured using
rulers at rest in
S’
x′
Measured using
rulers at rest in S
o
x
The two observer can compare their conclusions
by communication, and find the results are
contradictory, therefore, the hypothesis must be
false.
17
§25.3 consequences of Einstein’s postulates
Conclusion: lengths measured perpendicular
to the direction of motion are unaffected by the
motion.
2length oriented along the direction of motion
y
y′
Measured
by us in lab
τ
O
v=
τ0
l0
τ
=
l0
γτ 0
O′ l
x
l0
v=
l
τ0
l=
l0
γ
Measure by
Fairy godmother
x′
= l 0 1 − (v 2 c 2 )
l0 is the proper length.
§25.3 consequences of Einstein’s postulates
Conclusion:
The length l is shorter than l0when measured
in a frames in which it is moving.
The proper length is longest in all measurement.
Length contraction only occurs for those
lengths (or components of lengths ) oriented
along the direction of motion.
18
§25.3 consequences of Einstein’s postulates
Example 1: Sally (at point A) and Sam’s
spaceship (of proper length L0=230 m) pass
each other with constant relative speed v.
Sally measures a time interval of 3.75 µs for
the ship to pass her (from the passage of
point B to the passage of point C). In terms of
c, what is the relative speed v between Sally
and the ship?
§25.3 consequences of Einstein’s postulates
L L0 γ L0 1 − (v c ) 2
Solution: v =
=
=
∆t
∆t
∆t
Solving this equation for v leads us to
v=
=
L0 c
(c∆t ) 2 + L20
230
( 2.998 × 10 8 )( 3.57 × 10 − 6 ) + ( 230) 2
= 0.210c
19
§25.3 consequences of Einstein’s postulates
Example 2:(a)Can a person, in principle, travel
from Earth to the galactic center (which is
about 23000 ly distant) in a normal
lifetime?Explain, using either time-dilation or
length-contraction arguments. (b)What
constant speed is needed to make the trip in 30
y (proper time)?
Solution:
(a) Do not consider the important problem
such as fuel requirements, stresses to the
human body due to the accelerations, the
answer is yes.
§25.3 consequences of Einstein’s postulates
(b) d=23000 ly=23000c (y)
∆t =
v=
∆t 0
1 − (v c )2
c
=
d
v
=
c
1 + ( ∆t 0 c d ) 2
1 + ( 30 23000) 2
299792458
=
1 + 0.000017013
= 299792203
= 0.99999915c
20
§25.3 consequences of Einstein’s postulates
3. Simultaneity is relative
Station frame: s
Einstein’s train:
Train frame: s ′
I( xA, yA, zA, t)
frameS′ A′
frame S
II( xB , yB , zB , t)
. C′
.
B′
.
.
C
A
u
B
In S frame:
the two light signals arrive C simultaneously.
§25.3 consequences of Einstein’s postulates
I ( xA , yA , z A , t )
frameS′
frame S
II( xB , yB , zB , t )
. C′
A′
.
B′
.
.
C
A
B
I ( x′A , y′A , z′A , t1′ )
frameS′
.
frame S
A
u
II( x′B , y′B , z′B , t2′ )
. C′
A′
.
C
B′
u
.
B
21
§25.3 consequences of Einstein’s postulates
In frame S’：
the two light signals do not arrive C’
simultaneously.
Conclusions:
If two events are simultaneous in one inertial
reference frame, they may not be simultaneous
to any other inertial reference frame moving
with respect to the first.
§25.4 The Lorentz transformation equations
Modify the Galilean relativity:
the new transformation equations must
approach the Galilean equations in the limit
of small speed.
1. The transformation of coordinates
x ′ = Ax + Bt
y
y′
S
For O’ x ′ = 0
x
x ′ = Avt + Bt = 0 O
O′
x = vt
B = − Av
x ′ = Ax − Avt = A( x − vt )
S’
x′
22
§25.4 The Lorentz transformation equations
x ′ = A( x − vt ) (1)
x = A( x ′ + vt ) ( 2)
y
O
y′
S
x
x = vt
S’
O′ x′
1
x′2 x ′
l 0 = x ′2 − x1′ = A( x 2 − x1 ) − Av ( t 2 − t1 )
l 0 = Al
x 2 − x1 = l , t 2 − t1 = 0
l
Q l = 1 − v 2 c 2 l0 = 0
x ′ = γ ( x − vt ) (1)
γ
x = γ ( x′ + vt ) ( 2)
∴A=γ
§25.4 The Lorentz transformation equations
Eliminate the x’ of equations (1) and (2)
t ′ = γ ( t − vx c 2 ) ( 3)
t = γ ( t ′ + vx ′ c 2 ) (4)
Transformation
equations
x ′ = γ ( x − vt )
y′ = y
z′ = z
v
t ′ = γ (t − 2 x )
c
Inverse
equations
x = γ ( x′ + vt ′)
y = y′
z = z′
v
t = γ ( t ′ + 2 x ′)
c
23
§25.4 The Lorentz transformation equations
x ′ = x − vt
y′ = y
z′ = z
t′ = t
Notice that
v << c ,
γ →1
2. Important deductions
1time dilation
then
if
τ = t 2 − t1
t1 = γ ( t1′ + vx0′ c 2 )
t 2 = γ ( t 2′ + vx0′ c 2 )
= γ ( t 2′ − t1′ ) +
v
( x0′ − x0′ )
c2
= γ ( t 2′ − t1′ ) = γτ 0
§25.4 The Lorentz transformation equations
2the relativity of simultaneity
In S reference frame
y
o
S
x
l0
Event 1: Musician 1 Event 2: Musician 2
begins to play
begins to play
x1 = 0 m
t1 = 0 s
x2 = l0 m
t2 = 0 s
∆x = x 2 − x1 = l 0 , ∆t = t 2 − t1 = 0
24
§25.4 The Lorentz transformation equations
Using the Lorentz transformation equations
x ′ = γ ( x − vt )
y′ = y
z′ = z
v
t ′ = γ (t − 2 x )
c
x = γ ( x′ + vt ′)
y = y′
z = z′
v
t = γ ( t ′ + 2 x ′)
c
We can get
∆x ′ = γ ( ∆x − v∆t )
∆t ′ = γ ( ∆t −
∆x = γ ( ∆x ′ + v∆t ′)
v
v
′
∆
x
)
∆
t
=
γ
(
∆
t
+
∆x ′ )
c2
c2
§25.4 The Lorentz transformation equations
In S’ reference frame
Event 1: Musician 1 Event 2: Musician 2
begins to play
begins to play
x1′ = 0 m
t1′ = 0 s
∆x′ = x ′2 = γl 0
∆t ′ = t 2′ = −γvl0 / c 2
Conclusions:
If two events are simultaneous in one inertial
reference frame, they will not be simultaneous
to any other inertial reference frame moving
at non speed v with respect to the first.
Only when ∆t = 0, ∆x = 0 then ∆t ′ = 0
25
§25.4 The Lorentz transformation equations
3. Relativistic velocity addition
1velocity parallel or antiparallel to the direction
of motion of the two inertial reference frames
from
v
x = γ ( x′ + vt ′) and t = γ ( t ′ +
We have
c2
x ′)
dx = γ (dx′ + vdt ′) and dt = γ (dt ′ +
ux =
v
dx ′ )
2
c
dx
γ (dx′ + vdt ′)
=
dt γ (dt ′ + v dx ′)
c2
§25.4 The Lorentz transformation equations
dx ′
+v
u′ + v
′
= x
u x = dt
v dx ′
vu′
1+ 2
1 + 2x
c dt ′
c
2velocity perpendicular to the direction of
motion of the two inertial reference frames
dy ′
dy
u′y =
, uy =
dt ′
dt
v
from
y = y′ and t = γ ( t ′ + 2 x ′)
c
We have dy = dy ′and dt = γ (dt ′ +
v
dx ′ )
c2
26
§25.4 The Lorentz transformation equations
dy
dy ′
dy ′ dt
=
=
dt γ (dt ′ + v dx ′) γ (1 + v dx′ / dt ′)
c2
c2
u′z
u′y
=
u
uy =
like manner z
vu′x
vu′x
+
γ
(
1
)
γ (1 + 2 )
2
c
c
uy =
In special relativity, even though the y- and y’coordinates in the two frames are the same ,
the velocity components are not same because
of the relativity of time( t ≠ t ′ ).
§25.4 The Lorentz transformation equations
Example 1: The distance of two lamps A and
B on a train is 25 m, the train is moving at
the speed of v=20.0 m/s with respect to the
ground. If the observer on the train declare
that the two lamps is turn on simultaneously,
then what is the conclusion for the observer
on the ground?
B
A
v
r
S’
S
27
§25.4 The Lorentz transformation equations
Solution: According to Lorentz’ transformation
v
∆t ′ + 2 ∆x ′
c
= 5.6 × 10 −15 s
∆t = t A − t B =
v2
1− 2
c
The effect of the relativity is not in evidence.
If
v = 2.7 × 10 8 m / s
v
∆t ′ + 2 ∆x ′
c
= 1.7 × 10 − 7 s
then ∆t = t A − t B =
2
v
1− 2
c
The effect of the relativity is in evidence.
§25.4 The Lorentz transformation equations
Example 2: Two events happened
simultaneously at two points of distance 1000 m
on x axis in an inertial reference frame K. The
observer in another inertial reference frame K’
which was moving along the x axis measured
the distance of the two events is 2000 m. Find
the time interval of the two events in K’ frame.
Solution: According to Lorentz’ transformation
v
⎞
⎛
∆ x ′ = γ (∆ x − v∆ t ) ∆t ′ = γ ⎜ ∆ t − 2 ∆ x ⎟
c
⎠
⎝
Analysis: γ → v → ∆t ′
28
§25.4 The Lorentz transformation equations
from ∆x = 1000m ,
One can get
from γ =
∆t = 0, ∆x ′ = 2000 m
′
γ = ∆x = 2
∆x
1
v
⇒ v=
3
c
2
1 − ( )2
c
v
γv∆x
then ∆t ′ = t 2′ − t1′ = γ ( ∆t − 2 ∆x ) = − 2
c
c
3 1000
= −2 ×
×
= −5.77 × 10 − 6 (s )
8
2 3 × 10
What does the “-” means?
§25.4 The Lorentz transformation equations
Example 3: An Earth starship has been sent to
check an Earth outpost on the planet P1407,
whose moon houses a battle group of the often
hostile Reptulians. As the ship follows a straight
line course first past the planet and then past
the moon, it detects a high energy microwave
burst at the Reptulian moon base and then,
1.10s later, an explosion at the Earth outpost,
which is 4.00×108 m from the Reptulian base as
measured from the ship’s reference frame. The
Reptulians have obviously attacked the Earth
outpost, so the starship begins to prepare for a
confrontation with them.
29
§25.4 The Lorentz transformation equations
(a)The speed of the ship relative to the planet
and its moon is 0.980c. What are the distance
and time interval between the burst and the
explosion as measured in the planet –moon
inertial frame?
Solution: Event 1: burst; event 2: explosion
S frame: starship; S’ frame: planet-moon
§25.4 The Lorentz transformation equations
∆x = xe − xb = 4.00 × 108 m
∆t = t e − t b = 1.10 s
According to the transformation equation
∆x ′ = γ ( ∆x − v∆t )
∆t ′ = γ ( ∆t −
γ =
1
1− v c
2
2
=
v
∆x )
2
c
1
1 − (0.980)
2
= 5.0252
∆x ′ = 3.86 × 108 m
∆t ′ = −1.04 s
30
§25.4 The Lorentz transformation equations
(b)What is the meaning of the minus sign for ∆t’?
∆t ′ = t e′ − t b′ = −1.04 s ⇒ t e′ < t b′
It means that the burst occurred 1.04 s after the
explosion in the moon-planet reference frame.
(c)Did the burst cause the explosion, or vice versa?
If there is a causal relationship between the
two events, information must travel from the
location of one event to the location of the
other to cause by it. The required speed of the
information in the ship frame is
v info = ∆x ∆t = 4.00 × 10 8 1.10 = 3.64 × 108 m/s
They are unrelated events!
§25.4 The Lorentz transformation equations
Example 4: An observer S sees a big flash of
light 1200 m from his position and a small flash
of light 720 m closer to him directly in line with
the big flash. He determines that the time
interval between the flashes is 5.00 µs and the
big flash roccurs first. (a) what is the relative
velocity v of a second observer S’ for whom
these flashes occur at the same place in the S’
reference frame? (b) what time interval
between them does S’ measure?(c) From the
point view of S’, which flash occur first?
31
§25.4 The Lorentz transformation equations
Solution:
(a) Event 1: big flash; Event 2: small flash
S
r
v
S′
O ′small
O
big
720 m
x′
x
1200 m
∆x = x2 − x1 = (1200 − 720) − 1200 = −720 m
Q ∆x ′ = 0
∴ ∆x ′ = γ ( ∆x − v∆t ) = 0
v = ∆x / ∆t = −1.44 × 10 8 = −0.48c (m/s )
§25.4 The Lorentz transformation equations
(b) what time interval between them does S’
measure?
v∆ x
∆ t ′ = t 2′ − t1′ = γ ( ∆ t − 2 )
c
− 1.44 × 10 8 × ( − 720 )
−6
5.00 × 10 −
( 2.998 × 10 8 ) 2
=
1 - (0.48) 2
= 4.39 × 10 − 6 s > 0
(c) the time interval in the s’ frame is positive
The order of the flashes is the same in the S’
frame as it is in the S frame.
32
§25.4 The Lorentz transformation equations
Example 5: Two galaxies are speeding away
from the Earth along a line in opposite
directions, each with a speed of 0.75c with
respect to the planet. At what speed are they
moving apart with respect to each other?
Solution:
u′x + v
vu′
S
1 + 2x
c
O
0.75c + 0.75c
= 0.96c
=
1 + (0.75) 2
ux =
Earth
S’
O’
v
•
u′x
§25.4 The Lorentz transformation equations
Example 6: An armada of spaceships that is
1.00 ly long(in its frame) moves with speed
0.800c relative to ground station S. A
messenger travels from the rear of the
armada to the front with a speed of 0.950c
relative to S. How long does the trip take as
measured (a) in the messenger’s rest frame,
(b) in the armada’s rest frame, and (c) by an
observer in frame S?
Solution: (a) in the messenger’s rest frame Sm’
The velocity of the armada is
umx − v
0.80c − 0.95c
′ =
umx
=
= −0.625c
1 − umx v c 2 1 − 0.80 × 0.95
33
§25.4 The Lorentz transformation equations
The length of the armada as measured in S’ is
l
l m = 0 = 1.0(ly ) 1 − (0.625) 2 = 0.781(ly )
γm
The length of the trip is
∆t m′ =
l
0.781(ly )
=
= 1.25( y )
u′mx
0.625c
(b)in armada’s frame Sa’, the velocity of the
messenger is
uax − v a
0.95c − 0.80c
′ =
uax
=
= 0.625c
1 − uax v a c 2 1 − 0.95 × 0.80
§25.4 The Lorentz transformation equations
The length of the trip is
l
1 .0( ly )
∆ t a′ = 0 =
= 1 .6( y )
′
0 .625 c
uax
(c) in system S, the length of the armada is
l
l s = 0 = 1.0(ly ) 1 − (0.80) 2 = 0.60(ly )
γs
The length of the trip is
∆t ′S =
0.60(ly )
ls
=
= 4.0( y )
′
0.95c − 0.80c
u′mx − uax
34
§25.5 The Doppler effect
1. The longitudinal Doppler effect
How the frequency of a light source is
affected by the relative motion of the
source and observer?
Consider a firefly, the proper frequency is
ν0 =
1
τ0
In frame S’(firefly)
Event 0:
Event 1:
x0′ = 0
t 0′ = 0
x1′ = 0
t1′ = τ 0
Event 2:
x ′2 = 0
t 2′ = 2τ 0
§25.5 The Doppler effect
In frame S(ground)
According to the transformation equation
Event 0:
x0 = 0
t0 = 0
x = γ ( x ′ + vt ′)
v
t = γ ( t ′ + 2 x ′)
c
Event 1:
Event 2:
x1 = γvτ 0
t1 = γτ 0
x 2 = 2γvτ 0
t 2 = 2γτ 0
The time interval between the zero flash and
the first flash seen by your eyes at origin O,
when the firefly is receding from you
35
§25.5 The Doppler effect
τ = γτ 0 +
γ vτ 0
c
The time interval between the first flash and
the second flash
∆t = 2τ = 2γτ 0 + 2
γvτ 0
c
The frequency of the flashes seen at the origin
ν=
1
γτ 0 +
γvτ 0
=
c
1
v
c
γτ 0 (1 + )
=
ν0
v
c
γ (1 + )
The receding source is given by
§25.5 The Doppler effect
ν recede
v2
v
ν0 1− 2 ν0 1−
c =
c
=
v
v
(1 + )
1+
c
c
The frequency seen by your eyes at origin O,
when the firefly is approaching from you
ν appro =
ν0 1+
1−
v
c
v
c
36
§25.5 The Doppler effect
2. The transverse Doppler effect
If the firefly is moving transverse to the line
of sight, The proper period of the firefly is
τ0, the proper frequency in its frame S’ is ν0
and is very large.
The frequency of the firefly in frame S is
1
1
v2
ν= =
=ν0 1− 2
τ γτ 0
c
1The transverse Doppler effect is always a red
shift; 2 this effect has no classical analog; 3 it
is always smaller than longitudinal Doppler
effect.
§25.6 Relativistic dynamics
1. Relativistic momentum
Implications of relativity for the dynamics of a
single particle, the effect of forces, the work
done by these forces, and the consequent
changes in momentum and kinetic energy of
the particle.
r
r
Classical momentum pclass = m u
If we accept the postulates of the special
theory of relativity, the classical definition of
the momentum cannot be valid at high speeds.
Because it leads that the law ofr momentum
conservation is violate. Then p = ?
37
§25.6 Relativistic dynamics
y′
m
y
S′
− u0 ˆj ′
S
r
v
O′
u0 ˆj
x′
m
x
O
S frame: according to the definition of the
classical momentum, consider y-axis direction
u
Before collision mu0 ˆj − m 0 ˆj Total
γ
momentum
u
After collision − mu0 ˆj m 0 ˆj is not
conserved!
γ
§25.6 Relativistic dynamics
We define the relativistic momentum of a particle
r
r
p ≡ γm u
rest mass
γ =
1
u2
1− 2
c
m is called rest mass, u is the speed of the
particle.
Now we can verify that the total momentum is
conservative before and after collision of the
two particles. We consider only the direction
of y-axis.
38
§25.6 Relativistic dynamics
S frame: according to new definition of the
momentum
Before collision: For m in S frame
r
m
p1 = γmu0 ˆj =
u0 ˆj
u02
1− 2
c
For m in S’ frame
u0
v2
ux = v ,
u y = − = − u0 1 − 2
c
γ
⎛
1
v 2 ⎞⎟
⎜
p2 = m
− u0 1 − 2
2
2
2 12 ⎜
c ⎟⎠
1 − (ux + u y ) c
⎝
[
]
§25.6 Relativistic dynamics
Substitution of ux and uy,
u0
p2 = − m
= − p1
2
u
1 − 02
c
after collision: following the similar process
r
m
p1 = −
u0 ˆj
For m in S frame
2
u
1 − 02
c
u0
p2 = m
= − p1
For m in S’ frame
2
u
1 − 02
c
39
§25.6 Relativistic dynamics
Conclusion:
1The total y-component of the momentum
of the two particles before and after the
collision is zero.
2when u<<c, γ→1, the relativistic momentum
reduce to the classical expression.
3when u →c, γ→∞, γm →∞,
r
r dpr d(γm ur )
r dγ
du
F=
=
= mu
+ mγ
dt
dt
dt
dt
For constant force, we can not accelerate a
particle infinitely.
§25.6 Relativistic dynamics
2. The CWE theorem revisited
We shall require that the relativistic total
energy E satisfy two conditions:
1the total energy E of any isolated system
is conserved.
2W will approach the classical value when
u/c approaches zero.
The force in relativity
r dpr d(γm vr )
F=
=
dt
dt
40
§25.6 Relativistic dynamics
The kinetic energy KE is the work done by the
net force in accelerating a particle from rest to
some velocity. For one dimension
KE =
∫
v
0
F dx =
∫
v
0
v
d (γ mv )
d x = ∫ vd (γ mv )
0
dt
dp d(γmv )
=
dv
dv
⎡1
v2
v 2 −1 2
v ⎤
m 1 − 2 − mv ⎢ (1 − 2 ) ( −2 2 )⎥
c
c
c ⎦
⎣2
=
v2
1− 2
c
§25.6 Relativistic dynamics
dp d(γmv )
=
=
dv
dv
KE =
∫
v
0
vd (γ mv ) =
= mc 2 (
1
m
v2 3 2
(1 − 2 )
c
∫
v
0
v 2 −3 2
m (1 − 2 ) vd v
c
− 1) = γ mc 2 − mc 2
v2
1− 2
Rest energy E0
c
Expanding γ by the binomial theorem yields
v 2 −1 2
1 v2
γ = (1 − 2 ) ≈ 1 +
+L
c
2 c2
41
§25.6 Relativistic dynamics
Conclusion:
1When v<<c, the expression for the kinetic
energy reduce to the classical expression.
KE = γ mc 2 − mc 2
1 v2
2
≈ mc (1 +
) − mc 2
2
2c
1
= mv 2 = KE class
2
2the relativistic total energy E is then defined
as the sum of the kinetic energy and the rest
energy.
E = γ mc 2 = KE + mc 2 = KE + E 0
§25.6 Relativistic dynamics
3the speed of light is an unreachable upper
bound on the speed of a particle (with
nonzero mass) in special relativity.
If we want v→c, we
need to do an
infinite amount of
work on the particle
to accomplish this
feat!
42
§25.6 Relativistic dynamics
3. The relation of relativistic momentum and
the total relativistic energy
from
p = γmv , and E = γmc 2
v
E
2
c
2
2 2
E − c p = γ 2 m 2 c 4 − c 2γ 2 m 2 v 2
v2
2
2 4
= γ m c (1 − 2 )
c
p=
We find
from
2
We find E − c p = m c = E mc
2
2
2
2 4
2
0
E
pc
§25.6 Relativistic dynamics
Example 1: A particular object is observed to
move through the laboratory at high speed.
Its total energy and the components of its
momentum are measured by lab workers to
be E=4.51017 J, px=3.8×108 kg · m/s, py=3.0 ×
108 kg · m/s, and pz=2.0 × 108 kg · m/s. What
is the object’s rest mass?
Solution: E 2 − c 2 p 2 = m 2 c 4
[
]
m2c4 = (4.5×1017 )2 − (3.8×108 )2 + (3.0×108 )2 + (3.0×108 )2 c2
= 1.74×1035
(1.74×1035)1 2
m=
= 4.6kg
c2
43
§25.6 Relativistic dynamics
Example 2: A proton is moving at speed v=0.900c.
a. find its total energy E;
b. Find its kinetic energy;
c. Determine the magnitude p of its relativistic
momentum.
Solution: a. The total energy
1
1
E = γmc 2 =
mc 2 =
mc 2
v2
1 − 0 .9 2
1− 2
c
= 2.29 × 1.67 × 10 − 27 × ( 3.0 × 10 8 ) 2
= 3.44 × 10 −10 J = 2.15 × 10 3 MeV
§25.6 Relativistic dynamics
b. Find its kinetic energy
KE = (γ − 1)mc 2
= ( 2.29 − 1)(1.67 × 10 − 27 )( 3.0 × 108 ) 2
= 1.94 × 10 −10 J
= 1.21 × 10 3 MeV
c. Determine the magnitude p of its relativistic
momentum
v
p = γmv = 2 E
c
= 2.29(1.67 × 10 − 27 )(0.9 × 3 × 10 8 )
= 1.03 × 10 −18 kg ⋅ m/s
44
§25.6 Relativistic dynamics
Example 3: A proton of mass 938.3 MeV/c2 is
accelerated across a potential difference of
202.2 MeV. Determine its total energy (in MeV)
and momentum (in MeV/c). What is the speed
of the particle?
Solution: E = mc 2 + KE
= 938.3MeV + 202.0MeV
= 1140 MeV
p = E 2 − E 02 / c = 647.5MeV/c
E = γE 0 = E 0 / 1 − v 2 c 2
v = c 1 − E 02 E 2 = c 1 − 983.3 2 1140 2 = 0.5683c
§25.6 Relativistic dynamics
4. Implications of the equivalence between
mass and energy
According to E = γ mc 2 = KE + mc 2
1 A particle can only have kinetic energy KE;
2 A particle can only have rest energy mc2;
3 A particle can have both kinetic energy KE
and rest energy mc2;
KE = γmc 2 − mc 2 = ∆mc 2
It is possible to convert an isolated system of
particles with mass to a system of particle with
less mass, even zero mass, and –remarkably –
vice versa.
45
§25.6 Relativistic dynamics
Example 4: There are several fusion reactions
that convert mass directly into energy and can
power stars and drive hydrogen bombs. One
such process fuses two nuclei of heavy
hydrogen(deuterium) together, resulting in a
still heavier hydrogen(tritium) nucleus, an
ordinary hydrogen nucleus(proton), and the
KE they fly off with. It is customary to write
such a reaction in terms of the neutral atoms
involved(neglecting the tiny amounts of energy
holding the electrons to each atom, ≈10 eV).
2
2
3
1
thus
1 H + 1 H → 1 H + 1 H + KE
Determine the energy liberated per fusion.
§25.6 Relativistic dynamics
The mass of Deuterium is 1876.12 MeV;
The mass of Tritium is 2809.43 MeV ;
The mass of Hydrogen is 938.783 MeV.
Solution:
m D c 2 + m D c 2 = mT c 2 + m H c 2 + KE
187.12MeV + 1876.12MeV
= 2809.41MeV + 938.783MeV + KE
KE = 4.03MeV = 6.45 × 10 −13 J
46
§25.6 Relativistic dynamics
“mass is equivalent to energy”
“mass is converted into energy”
“energy is converted into mass”
47