Transcript The cornerstone of the theory of special relativity

Relativity
Chapter 1
A Brief Overview of Modern Physics
20th Century revolution:
- 1900 Max Planck
Basic ideas leading to Quantum theory
- 1905 Einstein
Special Theory of Relativity
21st Century
Story is still incomplete
Basic Problems
 Newtonian mechanics fails to describe
properly the motion of objects whose speeds
approach that of light
 Newtonian mechanics is a limited theory
– It places no upper limit on speed
– It is contrary to modern experimental results
– Newtonian mechanics becomes a specialized
case of Einstein’s special theory of relativity
when speeds are much less than the speed of
light
Galilean Relativity
 To describe a physical event, a frame of
reference must be established
 There is no absolute inertial frame of
reference
– This means that the results of an experiment
performed in a vehicle moving with uniform
velocity will be identical to the results of the
same experiment performed in a stationary
vehicle
Galilean Relativity
 Reminders about inertial frames
– Objects subjected to no forces will experience no
acceleration
– Any system moving at constant velocity with respect to
an inertial frame must also be in an inertial frame
 According to the principle of Galilean relativity,
the laws of mechanics are the same in all inertial
frames of reference
Galilean Relativity
 The observer in the
truck throws a ball
straight up
– It appears to move in
a vertical path
– The law of gravity
and equations of
motion under uniform
acceleration are
obeyed
Galilean Relativity
 There is a stationary observer on the ground
– Views the path of the ball thrown to be a parabola
– The ball has a velocity to the right equal to the
velocity of the truck
Galilean Relativity – conclusion
 The two observers disagree on the shape of the
ball’s path
 Both agree that the motion obeys the law of
gravity and Newton’s laws of motion
 Both agree on how long the ball was in the air
Conclusion: There is no preferred frame of
reference for describing the laws of mechanics
Frames of Reference and Newton's Laws
The cornerstone of the theory of special
relativity is the Principle of Relativity:
The Laws of Physics are the same in all
inertial frames of reference.
We shall see that many surprising consequences follow
from this innocuous looking statement.
Let us review Newton's mechanics in terms of frames of
reference.
A point in space is specified
by its three coordinates
(x,y,z) and an "event" like,
say, a little explosion by a
place and time – (x,y,z,t).
A "frame of reference" is just a set of coordinates - something
you use to measure the things that matter in Newtonian
mechanical problems - like positions and velocities, so we
also need a clock.
The "laws of physics" we shall consider are those of
Newtonian mechanics, as expressed by Newton's laws
of motion, with gravitational forces and also contact
forces from objects pushing against each other.
_____________________________
For example, knowing the universal gravitational
constant from experiment (and the masses
involved), it is possible from Newton's second law,
force = mass x acceleration,
to predict future planetary motions with great
accuracy.
Suppose we know from experiment that these
laws of mechanics are true in one frame of
reference. How do they look in another frame,
moving with respect to the first frame? To figure
out, we have to find how to get from position,
velocity and acceleration in one frame to the
corresponding quantities in the second frame.
Obviously, the two frames must have a
constant relative velocity, otherwise the law of
inertia won't hold in both of them.
Let's choose the coordinates so that this velocity is along
the x-axis of both of them.
Notice we also throw in a clock with each frame.
Now what are the coordinates of the event (x,y,z,t) in S'?
It's easy to see t' = t - we synchronized the clocks when O‘
passed O. Also, evidently, y' = y and z' = z, from the figure.
We can also see that x = x' +vt. Thus (x,y,z,t) in S
corresponds to (x',y',z', t' ) in S', where
x  x  vt
y  y
z  z
t  t
That's how positions transform - these are known as the
Galilean transformations.
What about velocities ? The velocity in S' in the x'
direction
dx dx d
dx
ux 

 ( x  vt) 
 v  ux  v
dt  dt dt
dt
This is just the addition of velocities formula
u x  ux  v
How does acceleration transform?
du x du x d
du x

 (u x  v) 
dt 
dt
dt
dt
Since v is constant we have
ax  ax
the acceleration is the same in both frames. This
again is obvious - the acceleration is the rate of
change of velocity, and the velocities of the same
particle measured in the two frames differ by a
constant factor - the relative velocity of the two
frames.
If we now look at the motion under gravitational forces, for
example,
 Gm1m2
ˆ
m1a 
r
2
r
we get the same law on going to another inertial frame
because every term in the above equation stays the same.
Note that acceleration is the rate of change of momentum
- this is the same in both frames. So, in a collision, if total
momentum is conserved in one frame (the sum of
individual rates of change of momentum is zero) the
same is true in all inertial frames.
Maxwell’s Equations of Electromagnetism
in Vacuum
Gauss’ Law for Electrostatics

Gauss’ Law for Magnetism

Faraday’s Law of Induction
Ampere’s Law
E  dA 
q
0
B  dA  0

d B
E  d  
dt

d E
B  d   0 0
dt
The Equations of Electromagnetism
Gauss’s Laws
..monopole..
q
1
 E  dA  0
2
 B  dA  0
...there’s no
magnetic monopole....!!
?
The Equations of Electromagnetism
Faraday’s Law
3

d B
E  d  
dt
Ampere’s Law
4

d E
B  d   0 0
dt
.. if you change a
magnetic field you
induce an electric
field.........
.. if you change an
electric field you
induce a magnetic
field.........
Electromagnetic Waves
Faraday’s law:
dB/dt
electric field
Maxwell’s modification of Ampere’s law
dE/dt
magnetic field
d E
 B  dl  00 dt
d B
 E  dl   dt
These two equations can be solved simultaneously.
The result is:
E(x, t) = EP sin (kx-t) ĵ
B(x, t) = BP sin (kx-t) ẑ
Plane Electromagnetic Waves
Ey
Bz
c
E(x, t) = EP sin (kx-t) ĵ
B(x, t) = BP sin (kx-t) ẑ
x
Plane Electromagnetic Waves
Ey
Bz
E(x, t) = EP sin (kx-t) ĵ
B(x, t) = BP sin (kx-t) ẑ
Notes: Waves are in Phase,
but fields oriented at 900.
k=2π/λ.
Speed of wave is c=ω/k (= fλ)
c  1 / 00  3 108 m / s
c
x
At all times
E=cB
It was recognized that the Maxwell equations
did not obey the principles of Newtonian
relativity. i.e. the equations were not invariant
when transformed between the inertial
reference frames using the Galilean
transformation.
Lets consider an example of infinitely long
wire with a uniform negative charge density
λ per unit length and a point charge q
located a distance y1 above the wire.
S’
2k
E
y1
The observer in S and see identical electric field
at distance y1=y1’ from an infinity long wire
carrying uniform charge λ per unit length.
Observers in both S and S’ measure a force
F
on charge q due to the line of charge.
2kq

y1
 0 v 2 q
However, the S’ observer measured and additional force
2y1
due to the magnetic field at y ’ arising from the motion
1
of the wire in the -x’ direction. Thus, the electromagnetic force
does not have the same form in different inertial systems,
implying that Maxwell’s equations are not invariant under a
Galilean transformation.
Speed of the Light
It was postulated in the nineteenth century that
electromagnetic waves, like other waves,
propagated in a suitable material media, called the
ether.
In according with this postulate the ether filed
the entire universe including the interior of the
matter.
It had the inconsistent properties of being
extremely rigid (in order to support the stress of
the high electromagnetic wave speed), while
offering no observable resistance to motion of the
planet, which was fully accounted for by Newton’s
law of gravitation.
Speed of the Light
The implication of this postulate is that a
light wave, moving with velocity c with respect
to the ether, would travel at velocity c’=c +v
with respect to a frame of reference moving
through the ether at v.
This would require that Maxwell’s equations
have a different form in the moving frame so as
to predict the speed of light to be c’, instead of
c
1
0 0
Conflict Between Mechanics and E&M
A. Mechanics
Galilean relativity states that it is impossible
for an observer to experimentally distinguish
between uniform motion in a straight line and
absolute rest. Thus, all states of uniform
motion are equal.
Conflict Between Mechanics and E&M
B. E&M
InitiallyThe initial interpretation of the speed of light
in Maxwell's theory was this c was the speed of
light seen by observers in absolute rest with
respect to the ether.
In other reference frames, the speed of light
would be different from c and could be obtained
by the Galilean transformation.
ProblemIt would now be possible for an
observer to distinguish between different
states of uniform motion by measuring the
speed of light or doing other electricity,
magnetism, and optics experiments.
Possible Solutions
1. Maxwell's theory of electricity and
magnetism was flawed. It was approximately 20
years old while Newton's mechanics was
approximately 200 years old.
2. Galilean relativity was incorrect. You can
detect absolute motion!
3. Something else was wrong with mechanics
(I.e Galilean transformation).
Experimental Results
Most physicists felt that Maxwell's equations
were probably in error.
Numerous experiments were performed to
detect the motion of the earth through the ether
wind.
The most famous of these experiments was
the Michelson-Morley experiment. Because of
the tremendous precision of their interferometer, it
was impossible for Michelson and Morley to miss
detecting the effect of the earth's motion through
the ether unless mechanics was flawed!
The Michelson-Morley experiment is a race between light beams.
The incoming light beam is split into two beams by a half-silvered
mirror. The beams follow perpendicular paths reflecting off full
mirrors before recombining back at the half mirror. Time
differences are seen in the interference pattern on the screen.
Theory
We will simplify the calculations by
assuming that L1 = L2 = L.
The time required to complete path 1
(horizontal path) is given by
L
L
T1 

cu cu
where we have used the Galilean
transformation and velocity = distance/time.
L
2Lc


T1  2
c

u

c

u

2
2
2
c u
c u




2L 
1

T1 
2 

c
u
1    
  c  
Since u<<c, we can use the binomial
approximation:
2L   u 
T1 
1   
c   c 
2



We can determine the time required to complete
path 2 (vertical path) using the distance diagram
below:
u(T2/2)
u(T2/2)
Using the Pythagorean theorem, we have:
2
 T2 
 T2 
2
c
 L  u



2
2




2
2
2
2
c T2
2 u T2
L 
4
4
c
2

2
 u T2  4L
2
2
2
2
2
4L
4L
T2  2 2 
2
c u
 u 
2
c 1    
  c  
2


T2 
2L
u
c 1  
c
2
Again, using the binomial approximation:
2

2L
1u 
T2 
1    
c  2  c  
Thus, the time difference for the two paths is
approximately
2
2


L 1u
Lu
ΔT  T1 - T2  2       
c  2  c   c  c 
We can now calculate the phase shift in
terms of wavelengths as follows:
f λc
λ
c
T
λ  cT
Thus, the phase shift in terms of a fraction of
a wavelength is given by
u
 λ  c T  L  
c
λ Lu
  
λ
λc
2
2
Using a sodium light,  = 590 nm, and a
interferometer with L = 11 m, we have
 4
Δλ 
11 m

 
10
7

λ
 5.90 x 10 m 


2
 0.2
This was a very large shift (20%) and
couldn't have been over looked.
Result - No shift was ever observed
regardless of when the experiment was
performed or how the interferometer was
orientated!
Einstein’s Postulates
• In 1905 Albert Einstein published a paper on the
electrodynamics of moving bodies. In this paper, he
postulated that absolute motion can not be detected
by any experiment. That is, there is no ether. The
reference frame connected with earth is considered
to be at rest and the velocity of the light will be the
same in any direction. His theory of special relativity
can be derived from two postulates:
• Postulate 1: Absolute uniform motion can
not be detected.
• Postulate 2: The speed of light is
independent of the motion of the source.
The Lorentz Transformation
Galilean transformations:
x = x’ + vt’, y = y’, z = z’, t = t’
The inverse transformations are
x’ = x – vt, y’ = y, z’ = z, t’ = t
These equations are consistent with experimental
observations as long as v is much less than c. They lead to
the familiar classical addition law for velocities. If a particle
has velocity ux = dx/dt in frame S, its velocity in frame S’ is
dx dx d
dx
ux 

 ( x  vt)   v  u x  v
dt  dt dt
dt
from here we have
du x dux
ax 

 ax
dt
dt 
The Lorentz Transformation
It should be clear that the Galilean
transformation is not consistent with Einstein’s
postulates of special relativity.
If light moved along the x axis with speed
ux’=c in S’, these equations imply that the speed
in S is ux=c+v rather than ux=c, which is not
consistent with Einstein’s postulates and
experiment.
The classical transformation equations must
therefore be modified .
The Lorentz Transformation
We assume that the relativistic transformation
equation for x is the same as the classical equation
except for a constant multiplier on the right side:
x   ( x' vt' )
where γ is a constant that can depend on v and c
but not on coordinates. The inverse transformation
in this case
x'   ( x  vt)
Lorentz Transformation
The transformation described by these
equations is called the Lorentz transformation.
Lorentz’s equations replace the flawed
Galileo transformation equations in relating
the measurements of two different observers
in uniform motion relative to each other.
Lorentz Transformation
x'  γ x  ut 
y'  y
z'  z
 xu 
t'  γ  t  2 
c 

1
γ
1β2
u
β
c
1
γ
2
u 
1  
 
c
Addition of Velocities
We start by taking the derivative with respect to
t' of the first Lorentz transformation equation:
 dx
dx'
dt 
 γ u 
dt'
dt' 
 dt'
From Calculus, we have that:
 dx dt
dx'
dt 
 γ
u 
dt'
dt' 
 dt dt'
 dx dt
dx'
dt 
 γ
u 
dt'
dt' 
 dt dt'
dx'
 dx
 dt
 γ   u
dt'
 dt
 dt'
Using the definition of velocity, we have:
dt
v '  γ v  u 
dt'
Differentiating the fourth Lorentz equation with
respect to t, we obtain:
dt'  dt u dx 

 γ   2
dt
 dt c dt 
dt' 
u
 γ 1  2
dt
 c
2
 γ
v  
 c2
c2  u v




dt
c
1


2
dt' γ(c  u v) γ1 u v 

2 
c 

We now substitute this result into our velocity
equation to obtain:
 vu 
vu
c
v'

u v  c  βv 
1
2
c
where v is velocity of object seen by unprimed
observer, v ' is velocity of object seen by primed
observer, u is velocity of the primed observer as seen
by unprimed observer.
EXERSICE: Show that both observers agree on
the speed of light.
SOLUTION: We determine the speed v' as seen
by the primed observer for a beam of light v = c
seen by the unprimed observer. Using the velocity
addition formula we have:
cu cu
v'

cu
u
1
1
2
c
c
cu 
cc
 

c

u


Lorentz-Fitgerald (Length) Contraction
Fitzgerald's length contraction is now a direct
consequence of the Lorentz transformation.
Consider two different observers in relative motion
who measure the length of a box as shown below:
S
S’
x1
observer sees stationary
box
observer sees box moving
at u
At a single instant of time t, the unprimed observer
measures the box's length in S as
L0  x2  x1
while the unprimed observer measures the box's length in S’ as
L' 
'
x2
'
 x1
Using the Lorentz transformation equation for x', we have that
 x2
  x1

L'    ut     ut 

 

L' 
x2  x1


ut  ut


x2  x1

Therefore, we have that
L' 
L0

Lorentz-Fitzgerald (Length) Contraction
L  Lo
u
1  
c
2
where L is the improper length
Lo is the proper length
u is the relative speed between the
observers
Length Contraction
Warning: Your everyday use of English terms can
get you in trouble in this material. The term "proper"
is not meant to indicate that its the "correct" or "right"
answer. Two observers might measure the length of
a train as 10 m when they see the train as stationary.
If one observer rides in the train as it moves down
the track, he will measure the train's length as 10 m
while an observer standing by the track sees the
train as shorter than 10 m. Both Observers Are
CORRECT!!! The question "What is the length of
the train?" is meaningless!! You must specify the
frame. There is no such thing as absolute
distance anymore!
Time Dilation
Another interesting aspect of relativity
is that moving clocks always run slower.
This is not an artifact of the clock! It is a
consequence of the nature of time itself
according to Einstein.
All clocks will behave this way include
your biological processes!
Time Dilation
Let us consider a "Light Clock" in which one
unit of time corresponds to the time it takes a
light pulse to travel between two meters a
distance L apart. We will place the clock on a
train traveling at speed u with the unprimed
observer.
Mirror
L
Light Path As Seen By Train Observer
Time Dilation
For the unprimed observer on the train,
the time it takes the light pulse to make a
single tick is given by
L
t
c
Time Dilation
The primed observer standing by the railroad
track see the train and clock pass with a speed
of u. Thus, the path of the light beam appears to
follow the diagonal path shown below:
ct'
L
u t'
u
Time Dilation
From the Pythagorean theorem, t' is given by
c t' 
2
 u t'   L
2
2
c t'  u t'  L
2
c
2
2
2

2
 u t'  L
2
2
2
L
t'  2

2
c u
2

2

2
2
L
2

 u  
2
c 1  


c




Time Dilation
L
t' 
c
u 
1  
c
2
L
γ
c
We now plug in our results for the unprimed
observer and we find that
t'  γ t
Thus, our observers do not agree upon time!
Time Dilation Equation
T 
To
u
1  
c
2
where To is the proper time
T’ is the improper time
Time Dilation as a Function of Speed
An astronomer on Earth observes a meteoroid in the southern
sky approaching the Earth at a speed of 0.800c. At the time of
its discovery the meteoroid is 20.0 ly from the Earth.
Calculate (a) the time interval required for the meteoroid to
reach the Earth as measured by the Earthbound astronomer,
(b) this time interval as measured by a tourist on the
meteoroid, and (c) the distance to the Earth as measured by
the tourist.
An astronomer on Earth observes a meteoroid in the southern
sky approaching the Earth at a speed of 0.800c. At the time of
its discovery the meteoroid is 20.0 ly from the Earth.
Calculate (a) the time interval required for the meteoroid to
reach the Earth as measured by the Earthbound astronomer,
(b) this time interval as measured by a tourist on the
meteoroid, and (c) the distance to the Earth as measured by
the tourist.
(a) The 0.8c and the 20 ly are measured in the Earth frame, so
in this frame,
.
x 20 ly 20 ly 1c
t 

 25.0 yr
v 0.8c
0.8c 1 ly yr
(b) We see a clock on the meteoroid moving, so we do not
measure proper time; that clock measures proper time.
t tp
tp 
t


25.0 yr
1 1-v2 c2
 25.0 yr 1 0.82  25.0 yr 0.6  15.0 yr
An astronomer on Earth observes a meteoroid in the southern
sky approaching the Earth at a speed of 0.800c. At the time of
its discovery the meteoroid is 20.0 ly from the Earth.
Calculate (a) the time interval required for the meteoroid to
reach the Earth as measured by the Earthbound astronomer,
(b) this time interval as measured by a tourist on the
meteoroid, and (c) the distance to the Earth as measured by
the tourist.
(c) Method one: We measure the 20 ly on a stick stationary in
our frame, so it is proper length. The tourist measures it to be
contracted to
Lp
20 ly
20 ly
L


 12.0 ly

1 1 0.82 1.667
Method two: The tourist sees the Earth approaching at
 0.8 ly
yr15 yr  12.0 ly
Classical Doppler Shift
Anyone who has watched auto racing on TV is
aware of the Doppler shift.
As a race car approaches the camera, the
sound of its engine increases in pitch (frequency).
After the car passes the camera, the pitch of
its engine decreases.
We could use this pitch to determine the
relative speed of the car. This technique is used
in many real world applications including
ultrasound imaging, Doppler radar, and to
determine the motion of stars.
I. Moving Observer
Assume that we have a stationary audio
source that produces sound waves of frequency f
and speed v.
y
v
Source

Observer
A stationary observer sees the time between each
wave as
Distance
Time 
Speed
x
I. Moving Observer
1 λ
T 
f v
If the observer is now moving at a velocity u relative to the
source then the speed of the waves as seen by the observer
is
Speed  v  u
where the positive sign is when the observer is moving
toward the source. The time between waves is now
1
λ
T' 

f ' vu
I. Moving Observer
Taking the ratio of our two results we get that
T f ' vu
 
T' f
v
 vu 
f 'f 

 v 
II. Moving Source
We now consider the case in which the source is moving
toward the observer. In this case, the wave's speed is
unchanged but the distance between wave fronts (wavelength)
is reduced. For the source moving away from the observer the
wavelength increased.
y
u
v
uT
l
Source
Observer

λ ' λ  u T
x
II. Moving Source
λ' λuT
T

1  u
λ
λ
λ
λ'
u
1 
λ
λf
λ'
u vu
1  
λ
v
v
c λ' v  u

λ c
v
II. Moving Source
f
vu

f'
v
Thus, the frequency seen by the observer for a moving source
is given by
 v 

f '  f 
 vu 
Moving Observer
 vu 
f 'f 

 v 
Moving Source
 v 

f '  f 
 vu 
Note: The motion of the observer and source create different
effects. For sound, this difference is explained due to motion
relative to the preferred reference frame! This preferred frame
is the reference frame stationary to the medium propagating
the sound (air)!
Relativistic Doppler Effect
For light in vacuum the distinction between
motion the source and detector can not be done.
Therefore the expression we derived for sound can
not be correct for light.
Consider a source of light moving toward the
detector with velocity v, relative to the detector. If
source emits N electromagnetic waves in time ΔtD
(measured in the frame of the detector), the first
wave will travel a distance cΔtD and the source will
travel the distance v ΔtD measures in the frame of
detector.
Relativistic Doppler Effect
The wavelength of the light will be:
ct D  vt D
' 
N
The frequency f ’ observed by the detector
will therefore be:
c
cN
1
N
f '



 ' t D (c  v) 1  v t D
c
Relativistic Doppler Effect
If the frequency of the source is f0 it will emit N=f0ΔtS
waves in the time ΔtS measured by the source. Then
f 0 t S
f0
t S
N
1
f '





v t D
v
v t D

t
D
1
1
1
c
c
c
1
Here ΔtS is the proper time interval ( the first wave and
the Nth wave are emitted at the same place in the source’s
reference frame). Therefore times ΔtS and ΔtD are related as:
t D 
t S 
t S
v2
1
c2
Relativistic Doppler Effect
When the source and detector are moving
towards one another:
f0
1
f '

v 
1
c
f '
f '
v2
1
c2
v
1
c
v
c f
v 0
1
c
1
v
c f
v 0
1
c
1
f0
- approaching
- receding
Relativistic Doppler Shift
Since light has no medium, there should be no
difference between moving the source and the observer.
The problem with our previous derivation when dealing
with light is that we haven't considered that space and time
coordinates are different for the source and observer.
Thus, we must account for the contraction of space and
dilation of time due to motion. After accounting for differences
in time and space, we get the following result for both moving
source and moving observer:
f '
cu
f0
cu
Police radar detects the speed of a car as follows: Microwaves
of a precisely known frequency are broadcast toward the car.
The moving car reflects the microwaves with a Doppler shift.
The reflected waves are received and combined with an
attenuated version of the transmitted wave. Beats occur
between the two microwave signals. The beat frequency is
measured. (a) For an electromagnetic wave reflected back to
its source from a mirror approaching at speed v, show that the
reflected wave has frequency
cv
f  f source
cv
where fsource is the source frequency. (b) When v is much less
than c, the beat frequency is much smaller than the
transmitted frequency. In this case use the approximation
f + fsource ≈ 2 fsource and show that the beat frequency can be
written as fbeat = 2v/λ. (c) What beat frequency is measured for
a car speed of 30.0 m/s if the microwaves have frequency
10.0 GHz? (d) If the beat frequency measurement is accurate
to ±5 Hz, how accurate is the velocity measurement?
(a) Let fc be the frequency as seen by the car: fc  fsource c v
c v
c v
and, if f is the frequency of the reflected wave: f  fc
c v
Combining gives:
f
(b) Using the above result:
 c v
fsource
 c v
f c v  fsource  c v
 f fsource  c  f fsource  v  2 fsourcev
The beat frequency is then:
2 fsourcev
2v
fbeat  f fsource 

c

(c)
fbeat 
 2  30.0 m s 10.0  109 H z  2  30.0 m s

(d)
v 

8
3.00  10 m s
c
fsource

3.00  108 m s
9
10.0  10 H z
 0.030 0 m 
 3.00 cm
fbeat
v
2
fbeat  5 H z  0.030 0 m 
2

2
 2 000 H z= 2.00 kH z
 0.075 0 m s  0.2 m ih
A ball is thrown at 20.0 m/s inside a boxcar moving
along the tracks at 40.0 m/s. What is the speed of
the ball relative to the ground if the ball is thrown
(a) forward (b) backward (c) out the side door?
A ball is thrown at 20.0 m/s inside a boxcar moving
along the tracks at 40.0 m/s. What is the speed of
the ball relative to the ground if the ball is thrown
(a) forward (b) backward (c) out the side door?
(a)
v  vT  vB  60.0 m s
(b)
v  vT  vB  20.0 m s
(c)
v  vT2  vB2  202  402  44.7 m s
A muon formed high in the Earth’s atmosphere
travels at speed v = 0.990c for a distance of 4.60 km
before it decays into an electron, a neutrino, and an
antineutrino    e   v  v . (a) How long does
the muon live, as measured in its reference frame?
(b) How far does the muon travel, as measured in its
frame?


A muon formed high in the Earth’s atmosphere travels at speed
v = 0.990c for a distance of 4.60 km before it decays into an
electron, a neutrino, and an antineutrino    e   v  v .
(a) How long does the muon live, as measured in its reference
frame? (b) How far does the muon travel, as measured in its
frame?

1
For
1


 7.09
v
 0.990
v2
1  .990 2
c
1 2
c
,
(a)
The muon’s lifetime
as measured in the Earth’s rest
.
frame is
4.60 km
t
0.990c
and the lifetime measured in the muon’s rest frame is


1 
4.60  103 m
  2.18 s
tp 


7.09  0.990 3.00  108 m s 


t



A muon formed high in the Earth’s atmosphere
travels at speed v = 0.990c for a distance of 4.60 km
before it decays into an electron, a neutrino, and an
antineutrino    e   v  v . (a) How long does
the muon live, as measured in its reference frame?
(b) How far does the muon travel, as measured in its
frame?

(b)
L  Lp

2
3
L
v
4.
60

10
m
p
 
1   

 649 m
 c

7.09
The identical twins Speedo and Goslo join a migration from
the Earth to Planet X. It is 20.0 ly away in a reference frame
in which both planets are at rest. The twins, of the same age,
depart at the same time on different spacecraft. Speedo’s
craft travels steadily at 0.950c, and Goslo’s at 0.750c.
Calculate the age difference between the twins after Goslo’s
spacecraft lands on Planet X. Which twin is the older?
The identical twins Speedo and Goslo join a migration from
the Earth to Planet X. It is 20.0 ly away in a reference frame
in which both planets are at rest. The twins, of the same age,
depart at the same time on different spacecraft. Speedo’s
craft travels steadily at 0.950c, and Goslo’s at 0.750c.
Calculate the age difference between the twins after Goslo’s
spacecraft lands on Planet X. Which twin is the older?
In the Earth frame, Speedo’s trip lasts for a time:
20.0 ly
x
t

 21.05 yr
v 0.950 ly yr
Speedo’s age advances only by the proper time interval
tp 
t

 21.05 yr 1 0.952  6.574 yr during his trip.
The identical twins Speedo and Goslo join a migration from the Earth to Planet X.
It is 20.0 ly away in a reference frame in which both planets are at rest. The
twins, of the same age, depart at the same time on different spacecraft. Speedo’s
craft travels steadily at 0.950c, and Goslo’s at 0.750c. Calculate the age
difference between the twins after Goslo’s spacecraft lands on Planet X. Which
twin is the older?
Similarly for Goslo,
20.0 ly
x
v2
tp 
1 2 
1 0.752  17.64 yr
v
0.750 ly yr
c
.
While Speedo has landed
on Planet X and is waiting for his
brother, he ages by:
20.0 ly
20.0 ly

 5.614 yr
0.750 ly yr 0.950 ly yr
Then
G oslo
ends up older by
17.64 yr  6.574 yr 5.614 yr  5.45 yr
An atomic clock moves at 1 000 km/h for 1.00 h as
measured by an identical clock on the Earth. How
many nanoseconds slow will the moving clock be
compared with the Earth clock, at the end of the
1.00-h interval?
An atomic clock moves at 1 000 km/h for 1.00 h as
measured by an identical clock on the Earth. How many
nanoseconds slow will the moving clock be compared with
the Earth clock, at the end of the 1.00-h interval?
Solution:
t
t tp 
p
1 v2 c2


v2 
v2 
tp   1 2  t  1 2  t

c 
2c 

2 
2 
2


v
v
v
t  t p  t  t 1  2   t 1  1  2   t 2
 2c 


2
c
2c




An atomic clock moves at 1 000 km/h for 1.00 h as
measured by an identical clock on the Earth. How many
nanoseconds slow will the moving clock be compared with
the Earth clock, at the end of the 1.00-h interval?
Solution:
1.00  106 m
v  1000 km h 
 277.8 m s
3 600 s
v 277.8m / s
7


9
.
27

10
c 3  108 m / s
2
t  v 
3600s
t  t p    
(9.27  107 )2  1.54  109 s  1.54ns
2 c
2
A spacecraft is launched from the surface of the Earth with a
velocity of 0.600c at an angle of 50.0° above the horizontal
positive x axis. Another spacecraft is moving past, with a
velocity of 0.700c in the negative x direction. Determine the
magnitude and direction of the velocity of the first spacecraft
as measured by the pilot of the second spacecraft.
A spacecraft is launched from the surface of the Earth with a
velocity of 0.600c at an angle of 50.0° above the horizontal
positive x axis. Another spacecraft is moving past, with a
velocity of 0.700c in the negative x direction. Determine the
magnitude and direction of the velocity of the first spacecraft
as measured by the pilot of the second spacecraft.
Solution:
Let frame S be the Earth frame of reference. Then v 
S
0.7c
ux   0.6c cos50  0.386c
uy   0.6c sin 50  0.459c
As measured from the S’ frame of the second spacecraft:
A spacecraft is launched from the surface of the Earth with a
velocity of 0.600c at an angle of 50.0° above the horizontal
positive x axis. Another spacecraft is moving past, with a
velocity of 0.700c in the negative x direction. Determine the
magnitude and direction of the velocity of the first spacecraft
as measured by the pilot of the second spacecraft.
Solution:
As measured from the S’ frame of the second spacecraft:
0.386c  0.7c
ux  v
1.086c
ux 


 0.855c
2
2
1.27
1 uxv c
1  0.386c 0.7c c 
S
0.459c 1  0.7
0.459c 0.714
uy 


 0.258c
2
1  0.386 0.7
1.27
 1 uxv c

Iu’I =
tan1
uy

 0.855c
2
2
  0.285c  0.893c
2
0.258c
 16.8 above the x -axis
0.855c
A moving rod is observed to have a length of 2.00 m and to
be oriented at an angle of 30.0° with respect to the direction
of motion, as shown in Figure. The rod has a speed of
0.995c. (a) What is the proper length of the rod? (b) What is
the orientation angle in the proper frame?
A moving rod is observed to have a length of 2.00 m and to
be oriented at an angle of 30.0° with respect to the direction
of motion, as shown in Figure. The rod has a speed of
0.995c. (a) What is the proper length of the rod? (b) What is
the orientation angle in the proper frame?
 
1
1 v2 c2

1
1 0.9952
 10.0
L1x  L1 cos1   2.00 m
 0.867  1.73 m
L1y  L1 sin 1   2.00 m  0.500  1.00 m
.
L1x
is a proper length, related to
L2x  10.0L1x  17.3 m
L2x
by
L1x 
L2x

L2y  L1y  1.00 m
We have seen that Einstein’s postulates
require important modifications in our ideas of
simultaneity and our measurements of time
and length. Einstein’s postulates also require
modification of our concepts of mass,
momentum, and energy.
Relativistic Momentum
In classical mechanics, the momentum of a
particle is defined as a product of its mass and its
velocity, mu.
In an isolated system of particles, with no net
force acting on the system, the total momentum of
the system remains the same. However , we can see
from a simple though experiment that the quantity
Σmivi is not conserved in isolated system.
We consider two observers: observer A in reference frame S
and observer B in frame S’, which is moving to the right in the
x direction with speed v with respect to the frame S. Each
has a ball of the mass m.
One observer throws his ball up with a speed u0 relative to
him and the other throws his ball down with a speed u0
relative to him, so that each ball travel a distance L, makes an
elastic collision with the other ball, and returns.
Classically, each ball has vertical momentum of magnitude
mu0. Since the vertical components of the momenta are
equal and opposite, the total vertical momentum are zero
before the collision. The collision merely reverses the
momentum of each ball, so the total vertical momentum is
zero after the collision.
Relativistically, however, the vertical components of the
velocities of the two balls as seen by the observers are not
equal and opposite. Thus, when they are reversed by the
collision, classical momentum is not conserved. As seen by A in
frame S, the velocity of his ball is uAy=+u0. Since the velocity of
ball B in frame S’ is u’Bx=0, uBy’=-u0, the y component of the
velocity of ball B in frame S is uBy=-u0/γ. So, the Σ miui is
conserved only in the approximation that u<<c.
We will define the relativistic momentum p of a particle to have
the following properties:
1. In collisions , p is conserved.
2.As u/c approaches 0, p approaches mu.
We will show that quantity
mu
p
u2
1 2
c
is conserved in the elastic collision and we take this equation
for the definition of the relativistic momentum of a particle.
Conservation of the Relativistic Momentum
p
mu
u2
1 2
c
One interpretation of this equation is that the mass of
an object increases with the speed. Then the
quantity
m
mrel 
u2
1 2
c
is called the relativistic mass. The mass of a particle
when it is at rest in some reference frame is then
called its rest mass .
mrel= γmrest
Conservation of the Relativistic
Momentum
The speed of ball A in S is u0, so the y component of its
relativistic momentum is:
p Ay 
mu0
u02
1 2
c
The speed of B in S is more complicated. Its x component
is v and its y component is –u0/γ.
2 

v 
2
2
2
2

u B  u Bx  u By  v   u0 1  2


c


2 2
u0 v
2
2
2
u B  v  u0  2
c
2
Using this result to compute √1-(u2B/c2), we obtain
2
B
2
2
2
0
2
2
u
v u
2 v
1
 1  2   u0 4 
c
c
c
c
 v 2  u02 
 1  2 1  2 
 c  c 
and
2
B
2
2
2
0
2
u
u
u
v
1
1
 1 2  1

1
c
c
c

c
2
0
2
Conservation of the Relativistic
Momentum
The y component of the relativistic momentum of
ball B as seen in S is therefore
 mu0 
 

muBy
 

pBy 

2
2
uB
u0
1
1 2
1 2
c

c
p By 
 mu0
u02
1 2
c
Conservation of the Relativistic
Momentum
Since pBy=-pAy’ the y component of the total
momentum of the two balls is zero. If the speed
of each ball is reversed by the collision, the total
momentum will remain zero and momentum will
be conserved.
p  mu
with

1
2
u
1 2
c
Relativistic momentum as given by equation
p
mu
u2
1 2
c
versus u/c, where u is speed of the object relative to
an observer. The magnitude of momentum p is plotted
in units of mc. The fainter dashed line shows the
classical momentum mu.
Relativistic Energy
In classic mechanic: The work done by the net
force acting on a particle equals the change in the
kinetic energy of the particle.
In relativistic mechanic: The net force acting on a
particle to accelerate it from rest to some final
velocity is equal to the rate of change of the
relativistic momentum. The work done by the net
force can then be calculated and set equal to the
change of kinetic energy.
Relativistic Energy
As in relativistic mechanic, we will define the kinetic
energy as the work done by the net force in
accelerating a particle from rest to some final
velocity uf.




dp
 mu 
K   Fnet ds   ds   udp   ud 
2 
dt
 1  u 
2
c


where we have used u=ds/dt.
 
 

 m d 

 

 


mu
1
1
d
du  

u2
u2
u2
1 2
1 2
1 2

c
c
c














 
1   2u 
1
1




m u  

 du 

 1
2
3 
2 
2
c

2


u 
 
u 

1






 c2 
 
c 2 




 

















2
u2 


1
u
1
  1  2 1 
  du 
 m  2 1 


3
3
c 

c 
 u2  
 u 2  






 c2 
 
 c2 
 





 

 








 m 1 




1
 u2

 c2





3



 du







u 



Relativistic Energy
Substituting the last result in the equation for kinetic
energy we obtain:
K
mc2
u2
1 2
c
 mc2
The second part of this expression, mc2, is independent of
the speed and called the rest energy Eo of the particle.
E0=mc2
The total relativistic energy:
E  K  mc2 
mc2
u2
1 2
c
Relativistic Energy
The work done by unbalanced force increases the
energy from the rest energy to the final energy
mc2
u2
1 2
c
where
mrel 
is the relativistic mass.
 mrel c 2
m
2
u
1 2
c
Relativistic Energy
We can obtain a useful expression for the
velocity of a particle by multiplying equation for
relativistic momentum by c2:
p
pc 
2
mu
2
u
1 2
c
2
mc u
2
u
1 2
c
 Eu
u pc

c E
Energies in atomic and nuclear physics are
usually expressed in units of electron volts
(eV) or mega-electron volts (MeV):
1eV = 1.602 x 10-19J
A convenient unit for the masses of atomic
particles is eV/c2 or MeV/c2, which is the rest
energy of the particle divided by c2.
The expression for kinetic energy
K
mc2
u2
1 2
c
 mc2
does not look much like the classical expression
½(mu2). However when v<<c, we can approximate
1/√1-(u2 /c2) using the binomial expansion:
(1+x)n =1 + nx + n(n-1)x2/2 + …….≈ 1 + nx
Then
 u2 
 1  2 
2
u
 c 
1 2
c
1

1
2
1 u2
 1
2 c2
and when v<<c




2

 1
1
1
u


2
2
2

K  mc 
 1  mc 1 

1

mu
2
 2
2
2
c


 1  u

c2


The equation for relativistic momentum:
p
mu
u2
1 2
c
and equation for the total energy:
E  K  mc2 
mc2
u2
1 2
c
can be combined to eliminate the speed u:
E2 = p2c2 +(mc2)2
E2 = p2c2 +(mc2)2
We received the relation for total energy,
momentum, and rest energy.
If the energy of a particle is much grater than its
rest energy mc2, the second term on the right
side of the last equation can be neglected,
giving the useful approximation:
E ≈ pc (for E>>mc2)
An unstable particle at rest breaks into two
fragments of unequal mass. The mass of the first
fragment is 2.50 × 10–28 kg, and that of the other is
1.67 × 10–27 kg. If the lighter fragment has a speed
of 0.893c after the breakup, what is the speed of
the heavier fragment?
An unstable particle at rest breaks into two fragments of
unequal mass. The mass of the first fragment is 2.50 × 10–28
kg, and that of the other is 1.67 × 10–27 kg. If the lighter
fragment has a speed of 0.893c after the breakup, what is the
speed of the heavier fragment?
Relativistic momentum of the system of fragments must be conserved. For
total momentum to be zero after as it was before, we must have, with
subscript 2 referring to the heavier fragment, and subscript 1 to the lighter,
p2  p1
 2m 2u2   1m 1u1 


1.67  1027 kg u2
1  u2 c
2
2.50  1028 kg
1  0.893

2

  0.893c
 4.960  1028 kg c
An unstable particle at rest breaks into two fragments of
unequal mass. The mass of the first fragment is 2.50 × 10–28
kg, and that of the other is 1.67 × 10–27 kg. If the lighter
fragment has a speed of 0.893c after the breakup, what is the
speed of the heavier fragment?
27
2
 1.67  10
u2 
u22
 1 2


28
c
c
 4.960  10
12.3
u22
2
c
1
u2  0.285c
An unstable particle with a mass of 3.34 × 10–27 kg is
initially at rest. The particle decays into two fragments
that fly off along the x axis with velocity components
0.987c and –0.868c. Find the masses of the
fragments. (Suggestion: Conserve both energy and
momentum.)
An unstable particle with a mass of 3.34 × 10–27 kg is
initially at rest. The particle decays into two fragments
that fly off along the x axis with velocity components
0.987c and –0.868c. Find the masses of the
fragments. (Suggestion: Conserve both energy and
1
momentum.)
1 
 2.01
1  0.868
2 
2
1
1  0.987
2
 6.22
E1  E2  Etotal
 1m 1c2   2m 2c2  m totalc2
27
2.01m 1  6.22m 2  3.34  10
kg
An unstable particle with a mass of 3.34 × 10–27 kg is
initially at rest. The particle decays into two fragments
that fly off along the x axis with velocity components
0.987c and –0.868c. Find the masses of the
fragments. (Suggestion: Conserve both energy and
momentum.)
27
2.01m 1  6.22m 2  3.34  10 kg
This reduces to:
m 1  3.09m 2  1.66  1027 kg
p1  p2
 1m 1u1   2m 2u2
 2.01 0.868c m 1   6.22 0.987c m 2
m 1  3.52m 2
m 1  8.84  1028 kg
m 2  2.51 1028 kg
A pion at rest (mπ = 273me) decays to a muon (mμ =

207me) and an antineutrino ( m v  0 ). The reaction is
written       v . Find the kinetic energy of the
muon and the energy of the antineutrino in electron volts.
(Suggestion: Conserve both energy and momentum.)
A pion at rest (mπ = 273me) decays to a muon (mμ =

207me) and an antineutrino ( m v  0 ). The reaction is
written       v . Find the kinetic energy of the
muon and the energy of the antineutrino in electron volts.
(Suggestion: Conserve both energy and momentum.)
pbefore decay  pafterdecay  0
E  Ev  E
pv  p   m  u    207m e u
 m  c  pvc  m  c
2
2
pv
  207m e 
 273m e
c
u
  207m e    207m e  273m e
c
1 u c
 u  273
u
 1.74
  1  
 1.32
 0.270

1 u c
c 207
c
A pion at rest (mπ = 273me) decays to a muon (mμ =

m
207me) and an antineutrino ( v  0 ). The reaction is





 v . Find the kinetic energy of the
written
muon and the energy of the antineutrino in electron volts.
(Suggestion: Conserve both energy and momentum.)

K     1 m  c2    1 207 m ec2



1

K 
 1 207 0.511 M eV 
 1  0.270 2



Ev  E  E
K   4.08 M eV
Ev  m  c2   m  c2   273  207  m ec2


207
  0.511 M eV 
Ev   273 
2

1

0.
270




Ev  29.6 M eV
SUMMARY
1. Einstein’s Postulates:
• Postulate 1: Absolute uniform motion
can not be detected.
• Postulate 2: The speed of light is
independent of the motion of the
source.
SUMMARY
• 2. The Lorentz Transformation:
x'  γ x  ut 
y'  y
z'  z
 xu 
t'  γ  t  2 
c 

SUMMARY
6. The Velocity Transformation:
'
v u
vx  x
'
uv
1 x
c2
v

u
v'x  x
uv
1 x
2
c
SUMMARY
• 3. Time Dilation:
• 4. Length Contraction:
• 5. The Relativistic
Doppler Effect:
T 
To
u
1  
c
L' 
2
L0

cu
f '
f0
cu
approaching
cu
f '
f0
cu
receding
SUMMARY
7. Relativistic Momentum:
8. Relativistic Energy:
p
u2
1 2
c
E  mc2  K  mc2 
mc2
u2
1 2
c
E0  mc 2
9. Rest Energy:
10. Kinetic Energy
mu
K  mc2 
mc2
u2
1 2
c
 mc 2 (1   )
SUMMARY
10. Useful Formulas for Speed, Energy, and Momentum:
 
E  p c  mc
2
E  pc
2 2
for
2 2
E  mc
u pc

c E
2