Chapter 2
Relativity II. Selected Problems
2.1
Problem 2.5 (In the text book)
Recall that the magnetic force on a charge q moving with velocity v in a magnetic field B
is equal to qv × B. If a charged particle moves in a circular orbit with a fixed speed v in
the presence of a constant magnetic field, use the relativistic form of Newton’s second law
to show that the frequency of its orbital motion is
r
v 2
qB
1−
f=
2πm
c
Solution
Newton’s second law is:
dp
dt
where F is the force and p is the momentum. Using the relativistic form of momentum gives
the relativistic form of Newton’s second law.
!
mv
d
p
F=
dt
1 − v 2 /c2
F=
In the case of a charged particle moving in a magnetic field, the force F = qv × B. If |v| = v
is constant and B and v are perpendicular, then the force F will perpendicular to both B
and v. In this case the force will not do any work on the charged particle, thus the speed
2
CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
stays constant in magnitude but will change direction under the effect of the force, we then
have:
d
qvB =
dt
!
m
p
1 − v 2 /c2
m
dv
= p
2
2
1 − v /c dt
(2.1)
where dv/dt is the centripetal acceleration produce by th force and given by v 2 /r, Equation
(2.1) becomes:
m
qvB = p
1 − v 2 /c2
v2
qvBr
= p
m
1 − v 2 /c2
r
v2
qBr
1− 2
v =
m
c
v2
r
(2.2)
Under the effect of the centripetal force the particle moves in a circular path and covers a
distance of one full circumference in a time T (the period), where f = 1/T , is the frequency
of the circular motion or the number of full rotations per second. So the speed v is then:
2πr
T
2πr
T =
v
v
f =
2πr
v =
(2.3)
Using Equation (2.2) in Equation (2.3) we get:
r
qB
v2
f=
1− 2
2πm
c
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.2. PROBLEM 2.16 (IN THE TEXT BOOK)
2.2
3
Problem 2.16 (In the text book)
As noted in Section 2.2, the quantity E 2 − p2 c2 is an invariant in relativity theory. This
means that the quantity E 2 − p2 c2 has the same value in all inertial frames even though
E and p have different values in different frames. Show this explicitly by considering the
following case. A particle of mass m is moving in the +x direction with speed u and has
momentum p and energy E in the frame S.
(a) If S 0 is moving at speed v in the standard way, find the momentum p0 and energy E 0
observed in S 0 . (Hint: Use the Lorentz velocity transformation to find p0 and E 0 . Does
E = E 0 and p = p0 ?
(b) Show that E 2 − p2 c2 is equal to E 02 − p02 c2 .
Solution
(a) In the the S frame, the speed of the particle is u, momentum is p and energy is E such
that:
p = p
E = p
mu
1 − u2 /c2
mc2
1 − u2 /c2
(2.4)
(2.5)
Using Lorentz speed transformation we get in the S 0 frame u0 , p0 , and E 0 :
u0 =
p0 =
=
E0 =
=
Physics 205:Modern Physics I, Chapter 2
u−v
1 − uv/c2
mu0
p
1 − u02 /c2
m[(u − v)/(1 − uv/c2 )]
p
1 − [(u − v)/(1 − uv/c2 )]2 (1/c2 )
mc2
p
1 − u02 /c2
mc2
p
1 − [(u − v)/(1 − uv/c2 )]2 (1/c2 )
Fall 2004
(2.6)
(2.7)
(2.8)
(2.9)
(2.10)
Ahmed H. Hussein
4
CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
Comparing Equations (2.4) and (2.8) we see that p 6= p0 and comparing Equations (2.5)
and (2.10) we find E 6= E 0 .
(b) Using Equations (2.5) and (2.6) we get in the S frame:
m2 c4
m 2 u2 c2
−
1 − u2 /c2 1 − u2 /c2
u2
m2 c4
1− 2
=
1 − u2 /c2
c
2 4
= mc
E 2 − p2 c2 =
(2.11)
Using Equations (2.7) and (2.9) we construct E 02 − p02 c2 as:
m2 c4
m2 u02 c2
−
1 − u02 /c2 1 − u02 /c2
m2 c4
u02
=
1− 2
1 − u02 /c2
c
= m2 c4
E 02 − p02 c2 =
(2.12)
It follows from Equations (2.11) and (2.12) that:
E 2 − p2 c2 = E 02 − p02 c2
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.3. PROBLEM 2.19 (IN THE TEXT BOOK)
2.3
5
Problem 2.19 (In the text book)
Calculate the binding energy in M eV per nucleon in the isotope 126 C. Note that the mass of
this isotope is exactly 12 u, and the masses of the proton and neutron are 1.007276 u and
1.008665 u, respectively.
Solution
Atomic mass unit (u) is defined as exactly the mass of
1u =
=
=
=
=
=
=
The nucleus of
12
6C
12
6C
divided by 12, i.e.
1.99264648 × 10−26
kg
12
1.66053873 × 10−27 kg
1.66053873 × 10−27 × c2 J/c2
1.66053873 × 10−27 × (2.99792458 × 108 )2 J/c2
1.49241778 × 10−10 J/c2
1.49249287 × 10−10
M eV /c2
1.602176362 × 10−13
931.494 M eV /c2
has six protons and 6 neutrons so its binding energy BE is:
BE =
=
=
=
∆mc2 M eV
m126 C − 6mp − 6mn
(6 × 1.007276 + 6 × 1.008665 − 12) × 931.494 M eV
89.09 M eV
A nucleon is a generic name for a neutron or a proton, so 126 C has 12 nucleons, so the binding
energy per nucleon is:
BE/nucleon =
Physics 205:Modern Physics I, Chapter 2
89.11
= 7.42 M eV
12
Fall 2004
Ahmed H. Hussein
6
2.4
CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
Problem 2.21 (In the text book)
An electron having kinetic energy K = 1.000M eV makes a head-on collision with a positron
at rest. (A positron is an antimatter particle that has the same mass as the electron but
opposite charge.) In the collision the two particles annihilate each other and are replaced
by two γ rays of equal energy, each traveling at equal angles θ with the electron’s direction
of motion. (Gamma rays are massless particles of electromagnetic radiation having energy
E = pc.) Find the energy E, momentum p, and angle of emission θ of the γ rays.
Solution
The annihilation process is shown in Figure (2.1).
Figure 2.1: Annihilation process of an electron and a positron.
Conservation of mass-energy requires:
K + 2me c2 = 2E
1
E =
K + m e c2
2
= 0.500 + 0.511
= 1.011M eV
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.4. PROBLEM 2.21 (IN THE TEXT BOOK)
7
Where K is the kinetic energy of the electron, me is the rest mass of the electron and the
positron, and E is the energy of each gamma ray.
Conservation of momentum requires:
pe = 2p cos θ
where pe is the momentum of the incident electron and p is the momentum of each gamma ray.
Since the particles of the gamma rays (photons) are massless, then p = E/c = 1.011 M eV /c,
using:
Ee2 = p2e c2 + (me c2 )2
where Ee is the total energy of the electron, Ee = K + me c2 , then
pe c =
=
=
pe =
=
=
p
E 2 − (me c2 )2
p e
(K + me c2 )2 − (me c2 )2
p
K 2 + (me c2 )2 + 2Kme c2 − (me c2 )2
1p 2
K + 2Kme c2
c
1p
(1.00)2 + 2 × 1.00 × 0.511
c
1.422 M eV /c
Finally the angle of gamma ray emission is:
cos θ =
=
=
=
θ =
Physics 205:Modern Physics I, Chapter 2
pe
2p
pe
2E/c
1.422
2 × 1.011
0.703
45.3◦
Fall 2004
Ahmed H. Hussein
8
2.5
CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
Problem 2.30 (In the text book)
The creation and study of new elementary particles is an important part of contemporary
physics. Especially interesting is the discovery of a very massive particle. To create a particle
of mass M requires an energy M c2 . With enough energy, an exotic particle can be created
by allowing a fast-moving particle of ordinary matter, such as a proton, to collide with a
similar target particle. Let us consider a perfectly inelastic collision between two protons: An
incident proton with mass m, kinetic energy K, and momentum magnitude p joins with an
originally stationary target proton to form a single product particle of mass M . You might
think that the creation of a new product particle, 9 times more massive than in a previous
experiment, would require just 9 times more energy for the incident proton. Unfortunately,
not all of the kinetic energy of the incoming proton is available to create the product particle,
since conservation of momentum requires that after the collision the system as a whole still
must have some kinetic energy. Only a fraction of the energy of the incident particle is thus
available to create a new particle. You will determine how the energy available for particle
creation depends on the energy of the moving proton. Show that the energy available to
create a product particle is given by
r
K
2
2
M c = 2mc 1 +
2mc2
From this result, when the kinetic energy K √
of the incident proton is large compared to its
rest energy mc2 , we see that M approaches 2mK/c. Thus if the energy of the incoming
proton is increased by a factor of 9, the mass you can create increases only by a factor of 3.
This disappointing result is the main reason that most modern accelerators, such as those at
CERN (in Europe), at Fermilab (near Chicago), at SLAC (at Stanford), and at DESY (in
Germany), use colliding beams. Here the total momentum of a pair of interacting particles
can be zero. The center of mass can be at rest after the collision, so in principle all of the
initial kinetic energy can be used for particle creation, according to
K
2
2
2
M c = 2mc + K = 2mc 1 +
2mc2
where K is the total kinetic energy of two identical colliding particles. Here, if K mc2 ,
we have M directly proportional to K, as we would desire. These machines are difficult to
build and to operate, but they open new vistas in physics.
Solution
The collision two-protons collision processes are shown in Figure (2.2). The top left is the
situation before collision in the fixed target experiment and the top right is the situation
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.5. PROBLEM 2.30 (IN THE TEXT BOOK)
9
after collision in the fixed target case. The bottom left is the situation before collision in
the colliding beams experiments and the bottom left is the situation after collision in the
colliding beams case.
Figure 2.2: Collision of two protons. Top: fixed target collision. Bottom: colliding beams
collision. Left: before collision. Right: after collision.
In the fixed target experiments the target proton is at rest while the projectile proton is
moving toward the target, as a rest to conserve momentum, the final particle should have
equal momentum and in the same direction as the projectile momentum.
Initially, we then have:
E1
E2
E12
p2
Physics 205:Modern Physics I, Chapter 2
= K1 + mc2
= mc2
= p21 c2 + m2 c4
=0
Fall 2004





(2.13)




Ahmed H. Hussein
10
CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
and in the final state we have:
Ef = Kf + M c2
Ef2 = p2f c2 + M 2 c4
)
(2.14)
Mass-energy conservation requires:
Ef = E1 + E2
Ef2 = E12 + 2E1 E2 + E22
p2f c2 + M 2 c4 = p21 c2 + m2 c4 + 2(K1 + mc2 )mc2 + m2 c4
(2.15)
and momentum conservation requires:
pf = p1
(2.16)
Using Equation (2.16) in Equation (2.15) we get:
M 2 c4 = m2 c4 + 2K1 mc2 + 2m2 c4 + m2 c4
= 4m2 c4 + 2K1 mc2
K
2 4
= 4m c 1 +
2mc2
r
K
M c2 = 2mc2 1 +
2mc2
(2.17)
In colliding beams experiments, two beams of protons (actually, in most cases a proton and
anti-proton) with equal energies are directed toward each other to collide head-on. Since
the two particles have equal and opposite momenta, the initial momentum is zero, and as a
result the final moment should be zero as well, i.e. K1 = K2 = 12 K, where K is the same as
the incident proton energy in the fixed target experiments, and p1 = −p2 , Kf = 0, pf = 0.
Initially, we have:

1
2 
E1 = K + mc 
2
(2.18)
1
2 

E2 = K + mc
2
and in the final state we have:
Ef = M c2
(2.19)
Conservation of mass-energy requires:
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.5. PROBLEM 2.30 (IN THE TEXT BOOK)
Ef = Ei + E2
1
1
K + mc2 + K + mc2
M c2 =
2
2
= 2mc2 +K
K
2
= 2mc 1 +
2mc2
11
(2.20)
Note that we chose the total kinetic energy in the initial state to be the same in both cases.
In the fixed target case, one proton carries a total kinetic energy of K and in the colliding
beams case, the two protons carry a total kinetic energy of K, i.e. each carry 12 K.
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein