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 q v × 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 qB f =
2 πm r
1 − v c
2
Solution
Newton’s second law is:
F = d p 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.
d
F = dt mv p
1 − v 2 /c 2
!
In the case of a charged particle moving in a magnetic field, the force F = q v × 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: qvB =
= d dt m p
1 − v 2 /c 2
!
m dv p
1 − v 2 /c 2 dt
(2.1) where dv/dt is the centripetal acceleration produce by th force and given by v
2
/r , Equation
(2.1) becomes: qvB = qvBr m
= v = m p
1 − v 2 /c 2 v 2 p
1 − v 2 /c 2 qBr r
1 − m v 2 c 2 v
2 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: v =
T = f =
2 πr
T
2 πr v v
2 πr
Using Equation (2.2) in Equation (2.3) we get: f = qB
2 πm r
1 − v 2 c 2
(2.3)
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein

2.2. PROBLEM 2.16
(IN THE TEXT BOOK)
2.2
Problem 2.16
(In the text book)
As noted in Section 2.2, the quantity E 2 means that the quantity E 2 − p 2 c 2
− p 2 c 2 is an invariant in relativity theory. This 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 p
0 observed in S
0
. (Hint: Use the Lorentz velocity transformation to find p
0
E = E
0 and p = p
0
?
and energy E
0 and E
0
. Does
(b) Show that E
2 − p
2 c
2 is equal to E
0 2 − p
0 2 c
2
.
3
Solution
(a) In the the S frame, the speed of the particle is u , momentum is p and energy is E such that: p =
E = mu p
1 − u 2 /c 2 mc
2 p
1 − u 2 /c 2
Using Lorentz speed transformation we get in the S
0 frame u
0
, p
0
, and E
0
:
(2.4)
(2.5) u
0 p
E
0
0
=
=
=
=
= u − v
1 − uv/c 2 mu
0 p
1 − u 0 2 /c 2 m [( u − v ) / (1 − uv/c 2 )] p
1 − [( u − v ) / (1 − uv/c 2 )] 2 (1 /c 2 ) mc
2 p
1 − u 0 2 /c 2 mc 2 p
1 − [( u − v ) / (1 − uv/c 2 )] 2 (1 /c 2 )
(2.6)
(2.7)
(2.8)
(2.9)
(2.10)
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein

4 CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
Comparing Equations (2.4) and (2.8) we see that p = p
0 and (2.10) we find E = E
0
.
and comparing Equations (2.5)
(b) Using Equations (2.5) and (2.6) we get in the S frame:
E
2 − p
2 c
2 m
2 c
4
=
1 − u 2 /c 2 m
2 c
4
=
1 − u 2 /c 2
= m
2 c
4 m
2 u
2 c
2
−
1 − u 2 /c 2
1 − u
2 c 2
Using Equations (2.7) and (2.9) we construct E
0 2 − p
0 2 c 2 as:
E
0 2 − p
0 2 c
2
=
= m 2 c 4
1 − u 0 2 /c 2
1 m
−
2 u 0 c
2
4
/c 2 m 2 u
0 2 c 2
−
1 − u 0 2 /c 2
1 − u
0 2 c 2
= m
2 c
4
It follows from Equations (2.11) and (2.12) that:
E
2 − p
2 c
2
= E
0 2 − p
0 2 c
2
(2.11)
(2.12)
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein

2.3. PROBLEM 2.19
(IN THE TEXT BOOK)
2.3
Problem 2.19
(In the text book)
Calculate the binding energy in M eV per nucleon in the isotope 12
6
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.
5
Solution
Atomic mass unit ( u ) is defined as exactly the mass of
12
6
C divided by 12, i.e.
1 .
99264648 × 10
− 26
1 u =
12
= 1 .
66053873 × 10
− 27 kg kg
= 1 .
66053873 × 10
− 27
× c
2
J/c
2
= 1 .
66053873 × 10
− 27
× (2 .
99792458 × 10
8
)
2
J/c
2
= 1 .
49241778 × 10
− 10
J/c
2
1 .
49249287 × 10
− 10
=
1 .
602176362 × 10 − 13
= 931 .
494 M eV /c
2
M eV /c
2
The nucleus of 12
6
C has six protons and 6 neutrons so its binding energy BE is:
BE = ∆ mc
2
M eV
= m
12
6
C
− 6 m p
− 6 m n
= (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 energy per nucleon is:
12
6
C has 12 nucleons, so the binding
BE/ nucleon =
89 .
11
12
= 7 .
42 M eV
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein

6 CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
2.4
Problem 2.21
(In the text book)
An electron having kinetic energy K = 1 .
000 M 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 + 2 m e c
2
= 2 E
E =
1
K + m e c
2
2
= 0 .
500 + 0 .
511
= 1 .
011 M 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, m e is the rest mass of the electron and the positron, and E is the energy of each gamma ray.
Conservation of momentum requires: p e
= 2 p cos θ where p e 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:
E e
2
= p
2 e c
2
+ ( m e c
2
)
2 where E e is the total energy of the electron, E e
= K + m e c 2 , then p e p c e
=
= p
E 2 e
− ( m e c 2 ) 2 p
( K + m e c 2 ) 2 − ( m e c 2 ) 2
=
=
= p
K 2 + ( m e c 2 ) 2 + 2 Km e c 2 − ( m e c 2 ) 2
1 p
K 2 + 2 Km e c 2 c
1 p
(1 .
00) 2 + 2 × 1 .
00 × 0 .
511 c
= 1 .
422 M eV /c
Finally the angle of gamma ray emission is: p e cos θ =
=
2 p p e
2 E/c
1 .
422
=
2 × 1 .
011
= 0 .
703
θ = 45 .
3
◦
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein

8 CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS
2.5
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 c
2
. 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
M c
2
= 2 mc
2 r
K
1 +
2 mc 2
From this result, when the kinetic energy K of the incident proton is large compared to its rest energy mc 2 , we see that M approaches 2 mK/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
M c
2
= 2 mc
2
+ K = 2 mc
2
K
1 +
2 mc 2 where K is the total kinetic energy of two identical colliding particles. Here, if K mc 2 , 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:
E
1
E
E
1
2 p
2
2
= K
1
+ mc
2
= mc
2
= p
2
1
= 0 c
2
+ m
2 c
4









(2.13)
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein

10 CHAPTER 2. RELATIVITY II. SELECTED PROBLEMS and in the final state we have:
Mass-energy conservation requires:
E f
= K f
+ M c
2
E
2 f
= p
2 f c
2
+ M
2 c
4
)
(2.14)
E f
E f
2 p
2 f c
2
+ M
2 c
4
= E
1
+ E
2
= E
1
2
+ 2 E
1
E
2
= p
2
1 c
2
+ m
2 c
4
+ E
2
2
+ 2( K
1
+ mc
2
) mc
2
+ m
2 c
4 and momentum conservation requires: p f
= p
1
Using Equation (2.16) in Equation (2.15) we get:
(2.15)
(2.16)
M
2 c
4
M c
2
= m
2 c
4
= 4 m
2 c
4
= 4 m
2 c
4
+ 2 K
1 mc
2
+ 2 K
1 +
1 mc
K
2
+ 2
2 mc 2 m
2 c
4
+ m
2 c
4
= 2 mc
2 r
K
1 +
2 mc 2
(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.
K
1
= K
2
=
1
2 the incident proton energy in the fixed target experiments, and p
K , where K is the same as
1
= − p
2
, K f
= 0 , p f
= 0.
Initially, we have:
E
1
E
2
=
=
1
K + mc
2
2
1
K + mc
2
2





(2.18) and in the final state we have:
E f
= M c
2
(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) 11
E f
M c
2
= E i
+ E
2
=
1
2
K + mc
2
= 2 mc
2
+ K
+
1
2
K + mc
2
= 2 mc
2
1 +
K
2 mc 2
(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
1
2
K .
Physics 205:Modern Physics I, Chapter 2 Fall 2004 Ahmed H. Hussein