Transcript CHAPTER 2: Special Theory of Relativity
Slide 1
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
Slide 2
CHAPTER 2
Special Theory of Relativity II
Relationship Between Energy and
Momentum
Compton Scattering
Discovery of the Positron
Albert Einstein (1879-1955)
If you are out to describe
the truth, leave elegance
to the tailor.
The most incomprehensible thing about the
world is that it is at all
comprehensible.
- Albert Einstein
Slide 3
Relativistic Momentum
Because physicists believe
that the conservation of
momentum is fundamental,
we begin by considering
collisions without external
forces:
S
u
dP/dt = Fext = 0
v
y
z
x
v
S’
S’
Frank is at rest in S and
throws a ball of mass m
in the -y-direction. Mary
(in the moving system)
similarly throws a ball in
system S’ that’s
moving in the
x direction with
velocity v with
respect to
system
S.
Slide 4
Relativistic Momentum
What does Frank measure for the
change in momentum of his own
ball?
What does Frank measure for the
change in momentum of Mary’s ball?
Slide 5
Relativistic Momentum
The conservation of linear momentum
requires the total change in momentum
of the collision, ΔpF + ΔpM, to be zero.
The addition of these y-momenta
is clearly not zero.
S
v
S’
Linear momentum is not conserved if we use the conventions for
momentum from classical physics—even if we use the velocity
transformation equations from special relativity.
There is no problem with the x direction, but there is a problem with
the y direction the ball is thrown in each system.
Slide 6
Relativistic Momentum
S
The failure of the conservation of momentum in
the collision cannot be due to the velocities,
because we used the Lorentz transformation to
find the y components. It must have something
to do with mass!
v
S’
We can reconcile this discrepancy by using
pg mv
where:
g
1
1 v /c
2
2
Important: note that we’re using g in this formula, but the v in g is
really the velocity of the object, not necessarily that of its frame.
Exercise 4-15: Using the above modification show that the initial y
component of Frank ball cancels out with that of Mary’s
Slide 7
Relativistic momentum
Slide 8
At high velocity, does the mass
increase or just the momentum?
Some physicists like to refer to the mass as the rest mass m0 and
call the term m = gm0 the relativistic mass. In this manner the
classical form of momentum, m, is retained. The mass is then
imagined to increase at high speeds.
Most physicists prefer to keep the concept of mass as an
invariant, intrinsic property of an object. We adopt this latter
approach and will use the term mass exclusively to mean rest
mass. Although we may use the terms mass and rest mass
synonymously, we will not use the term relativistic mass.
Slide 9
Relativistic Energy
We must now redefine the concepts
of work and energy.
So we modify Newton’s second law to
include our new definition of linear
momentum, and force becomes:
d
F
(g m u )
dt
dt
dt
dp
d
mu
1 u
2
2
c
where, again, we’re using g in this formula, but it’s really the
velocity of the object, not necessarily that of its frame.
Slide 10
Relativistic Energy
Exercise 4-16: Show that the kinetic energy Ek and hence the work
done by a net force in accelerating a particle from rest to some
velocity u is;
E k m c ( g 1)
2
Exercise 3: Show that for u=v<
Slide 11
Relativistic Energy
1. Even an infinite amount of energy is not enough to achieve c.
2. For u<
Electrons accelerated to high energies in an electric field
Slide 12
Total Energy and Rest Energy
Ek g mc mc
2
Manipulate the energy equation:
2
The term mc2 is called the Rest Energy
The sum of the kinetic and rest energies is the total energy of
the particle E and is given by:
E Ek mc g mc
2
2
mc
1
2
u
2
c
2
and that for a particle at rest u= 0
E
1
2
mu mc
2
2
Becomes the famous
E mc
2
Slide 13
Invariant Mass
Square the momentum equation, p = g m u, and multiply by c2:
Substituting for u2
using b 2 = u2 / c2 :
But b 2 1
And:
1
g
2
p c g m c m c
2
2
2
2
4
2
4
Slide 14
Invariant Mass
p c g m c m c
2
2
2
2
4
2
The first term on the right-hand side is just E2, and the second is E02:
Rearranging, we obtain a relation between energy and momentum.
or:
This equation relates the total energy of a particle with its momentum.
The quantities (E2 – p2c2) and m are invariant quantities.
Note that when a particle’s velocity is zero and it has no momentum,
this equation correctly gives E0 as the particle’s total energy.
4
Slide 15
Massless particles
Exercise 4-17 : Starting from;
E p c m c
2
2
2
2
4
Show that any massless particle must travel at the speed of light.
Slide 16
Compton Effect
When a photon enters matter, it can interact
with one of the electrons. The laws of
conservation of energy and momentum
apply, as in any elastic collision between
two particles. The momentum of a particle
moving at the speed of light is:
p
E
h
c
c
The electron energy is:
This yields the change in
wavelength of the scattered
photon, known as the
Compton effect:
h
hc /
Ee
hc /
Slide 17
Pair Production and Annihilation
If a photon can create an electron, it
must also create a positive charge to
balance charge conservation.
In 1932, C. D. Anderson observed a
positively charged electron (e+) in
cosmic radiation. This particle, called a
positron, had been predicted to exist
several years earlier by P. A. M. Dirac.
A photon’s energy can
be converted entirely
into an electron and
a positron in a process
called pair production:
Paul Dirac
(1902 - 1984)
Slide 18
Pair Production
in Empty Space
h
Conservation of energy for pair
production in empty space is:
E
E+
h E E
The total energy for a particle is:
So:
E p c
This yields a lower limit on the photon energy:
Momentum conservation yields:
h p c p c
h p c cos( ) p c cos( )
This yields an upper limit on the photon energy:
h p c p c
A contradiction! And hence the conversion of energy and momentum
for pair production in empty space is impossible!
Slide 19
Pair Production
in Matter
In the presence of matter, the
nucleus absorbs some energy
and momentum.
The photon energy required for
pair production in the presence
of matter is:
h E E K . E .( nucleus )
h 2 m e c 1.022 M eV
2
Slide 20
Pair Annihilation
A positron passing through matter
will likely annihilate with an
electron. The electron and positron
can form an atom-like configuration
first, called positronium.
Pair annihilation in empty space
produces two photons to conserve
momentum. Annihilation near a
nucleus can result in a single
photon.
Slide 21
Pair Annihilation
Conservation of energy:
2 m e c hv1 hv 2
2
Conservation of momentum:
hv1
hv 2
c
0
c
So the two photons will have the
same frequency:
v1 v 2 v
The two photons from positronium
annihilation will move in opposite
directions with an energy:
hv m e c 0.511 M eV
2
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
Slide 2
CHAPTER 2
Special Theory of Relativity II
Relationship Between Energy and
Momentum
Compton Scattering
Discovery of the Positron
Albert Einstein (1879-1955)
If you are out to describe
the truth, leave elegance
to the tailor.
The most incomprehensible thing about the
world is that it is at all
comprehensible.
- Albert Einstein
Slide 3
Relativistic Momentum
Because physicists believe
that the conservation of
momentum is fundamental,
we begin by considering
collisions without external
forces:
S
u
dP/dt = Fext = 0
v
y
z
x
v
S’
S’
Frank is at rest in S and
throws a ball of mass m
in the -y-direction. Mary
(in the moving system)
similarly throws a ball in
system S’ that’s
moving in the
x direction with
velocity v with
respect to
system
S.
Slide 4
Relativistic Momentum
What does Frank measure for the
change in momentum of his own
ball?
What does Frank measure for the
change in momentum of Mary’s ball?
Slide 5
Relativistic Momentum
The conservation of linear momentum
requires the total change in momentum
of the collision, ΔpF + ΔpM, to be zero.
The addition of these y-momenta
is clearly not zero.
S
v
S’
Linear momentum is not conserved if we use the conventions for
momentum from classical physics—even if we use the velocity
transformation equations from special relativity.
There is no problem with the x direction, but there is a problem with
the y direction the ball is thrown in each system.
Slide 6
Relativistic Momentum
S
The failure of the conservation of momentum in
the collision cannot be due to the velocities,
because we used the Lorentz transformation to
find the y components. It must have something
to do with mass!
v
S’
We can reconcile this discrepancy by using
pg mv
where:
g
1
1 v /c
2
2
Important: note that we’re using g in this formula, but the v in g is
really the velocity of the object, not necessarily that of its frame.
Exercise 4-15: Using the above modification show that the initial y
component of Frank ball cancels out with that of Mary’s
Slide 7
Relativistic momentum
Slide 8
At high velocity, does the mass
increase or just the momentum?
Some physicists like to refer to the mass as the rest mass m0 and
call the term m = gm0 the relativistic mass. In this manner the
classical form of momentum, m, is retained. The mass is then
imagined to increase at high speeds.
Most physicists prefer to keep the concept of mass as an
invariant, intrinsic property of an object. We adopt this latter
approach and will use the term mass exclusively to mean rest
mass. Although we may use the terms mass and rest mass
synonymously, we will not use the term relativistic mass.
Slide 9
Relativistic Energy
We must now redefine the concepts
of work and energy.
So we modify Newton’s second law to
include our new definition of linear
momentum, and force becomes:
d
F
(g m u )
dt
dt
dt
dp
d
mu
1 u
2
2
c
where, again, we’re using g in this formula, but it’s really the
velocity of the object, not necessarily that of its frame.
Slide 10
Relativistic Energy
Exercise 4-16: Show that the kinetic energy Ek and hence the work
done by a net force in accelerating a particle from rest to some
velocity u is;
E k m c ( g 1)
2
Exercise 3: Show that for u=v<
Slide 11
Relativistic Energy
1. Even an infinite amount of energy is not enough to achieve c.
2. For u<
Electrons accelerated to high energies in an electric field
Slide 12
Total Energy and Rest Energy
Ek g mc mc
2
Manipulate the energy equation:
2
The term mc2 is called the Rest Energy
The sum of the kinetic and rest energies is the total energy of
the particle E and is given by:
E Ek mc g mc
2
2
mc
1
2
u
2
c
2
and that for a particle at rest u= 0
E
1
2
mu mc
2
2
Becomes the famous
E mc
2
Slide 13
Invariant Mass
Square the momentum equation, p = g m u, and multiply by c2:
Substituting for u2
using b 2 = u2 / c2 :
But b 2 1
And:
1
g
2
p c g m c m c
2
2
2
2
4
2
4
Slide 14
Invariant Mass
p c g m c m c
2
2
2
2
4
2
The first term on the right-hand side is just E2, and the second is E02:
Rearranging, we obtain a relation between energy and momentum.
or:
This equation relates the total energy of a particle with its momentum.
The quantities (E2 – p2c2) and m are invariant quantities.
Note that when a particle’s velocity is zero and it has no momentum,
this equation correctly gives E0 as the particle’s total energy.
4
Slide 15
Massless particles
Exercise 4-17 : Starting from;
E p c m c
2
2
2
2
4
Show that any massless particle must travel at the speed of light.
Slide 16
Compton Effect
When a photon enters matter, it can interact
with one of the electrons. The laws of
conservation of energy and momentum
apply, as in any elastic collision between
two particles. The momentum of a particle
moving at the speed of light is:
p
E
h
c
c
The electron energy is:
This yields the change in
wavelength of the scattered
photon, known as the
Compton effect:
h
hc /
Ee
hc /
Slide 17
Pair Production and Annihilation
If a photon can create an electron, it
must also create a positive charge to
balance charge conservation.
In 1932, C. D. Anderson observed a
positively charged electron (e+) in
cosmic radiation. This particle, called a
positron, had been predicted to exist
several years earlier by P. A. M. Dirac.
A photon’s energy can
be converted entirely
into an electron and
a positron in a process
called pair production:
Paul Dirac
(1902 - 1984)
Slide 18
Pair Production
in Empty Space
h
Conservation of energy for pair
production in empty space is:
E
E+
h E E
The total energy for a particle is:
So:
E p c
This yields a lower limit on the photon energy:
Momentum conservation yields:
h p c p c
h p c cos( ) p c cos( )
This yields an upper limit on the photon energy:
h p c p c
A contradiction! And hence the conversion of energy and momentum
for pair production in empty space is impossible!
Slide 19
Pair Production
in Matter
In the presence of matter, the
nucleus absorbs some energy
and momentum.
The photon energy required for
pair production in the presence
of matter is:
h E E K . E .( nucleus )
h 2 m e c 1.022 M eV
2
Slide 20
Pair Annihilation
A positron passing through matter
will likely annihilate with an
electron. The electron and positron
can form an atom-like configuration
first, called positronium.
Pair annihilation in empty space
produces two photons to conserve
momentum. Annihilation near a
nucleus can result in a single
photon.
Slide 21
Pair Annihilation
Conservation of energy:
2 m e c hv1 hv 2
2
Conservation of momentum:
hv1
hv 2
c
0
c
So the two photons will have the
same frequency:
v1 v 2 v
The two photons from positronium
annihilation will move in opposite
directions with an energy:
hv m e c 0.511 M eV
2