532
9 Particle Physics – I
9.105 (a) E
∗2
= (E 1 + E 2 )
2
− ( p 1 − p 2 )
2
= E 1
2
+ 2 E 1 E 2 + E 2
2
− p 1
2
+ 2 p 1 p 2 − p 2
2
=
E 1
2
− p 1
2
+
E 2
2
− p 2
2
+ 2(E 1 E 2 + p 1 p 2 )
= m 1
2
+ m 2
2
+ 2(E 1 E 2 + p 1 p 2 )
If E 1 >> m 1 and E 2 >> m 2 , m 1
2 and m 2
2 can be neglected and p 1 ≈
E 1 , p 2 ≈ E 2
Therefore E
∗2
≈ 4E 1 E 2
If the beams cross at an angle θ then
E
∗2
= (E 1 + E 2 )
2
−
p 1
2
+ p 2
2
− 2 p 1 p 2 cos θ
≈ 2E 1 E 2 (1 + cos θ) = 4 E 1 E 2 (1 + cos θ )/2
Thus the available energy in the CMS is reduced by a factor of (1 +
cos θ)/2 compared to head-on-collision.
(b) In the CMS E
∗
= 25 + 25 = 50 GeV
With the fixed proton target
E
∗
= (m
2
+ m
2
+ 2 T 1 m)
1/2
= 50 GeV
Substituting m = 0.94 GeV, we find T 1 = 1329 GeV
9.106 When the protons travel toward each other with equal energy, and therefore
with the same speed, their net momentum is zero. In that case the Lab system
is reduced to the C-M system, and the observer is watching the events sitting
in the C.M. system. The total energy is then,
E
∗
= 10
10 eV + 10
10 eV + 10
9 eV + 10
9 eV = 22 × 10
9 or 22 GeV
Let E 1 be the energy of a proton in the lab system moving toward the target
proton originally at rest, then if the total energy available in the CMS has to
be the same as E
∗
= 22 GeV,
m 1
2
+ m 2
2
+ 2 E 1 M 1
1/2 = E
∗
= 22 GeV
(1
2
+ 1
2
+ 2 E 1 × 1)
1/2
= 22
Or E 1 = 241 GeV
Therefore required kinetic energy = 241 − 1 = 240 GeV
9.107 Total energy available in the CMS is
E
∗
=
m 1
2
+ m 2
2
+ 2 E 1 m 2
1/2
where E 1 is the total energy of projectile of mass m 1 and m 2 is the mass of
the target.
E
∗
= 4M 0 + 2 M 0 = 6M 0
m 1 = m 2 = M 0
6M 0 =
M 0
2
+ M 0
2
+ 2 E 1 M 0
1/2
whence E 1 = 17 M 0
Or T 1 = E 1 − M 0 = 16 M 0
The significance of this result is that in the colliders a lot more energy is
available than in fixed target experiments.
9 Particle Physics – I
9.105 (a) E
∗2
= (E 1 + E 2 )
2
− ( p 1 − p 2 )
2
= E 1
2
+ 2 E 1 E 2 + E 2
2
− p 1
2
+ 2 p 1 p 2 − p 2
2
=
E 1
2
− p 1
2
+
E 2
2
− p 2
2
+ 2(E 1 E 2 + p 1 p 2 )
= m 1
2
+ m 2
2
+ 2(E 1 E 2 + p 1 p 2 )
If E 1 >> m 1 and E 2 >> m 2 , m 1
2 and m 2
2 can be neglected and p 1 ≈
E 1 , p 2 ≈ E 2
Therefore E
∗2
≈ 4E 1 E 2
If the beams cross at an angle θ then
E
∗2
= (E 1 + E 2 )
2
−
p 1
2
+ p 2
2
− 2 p 1 p 2 cos θ
≈ 2E 1 E 2 (1 + cos θ) = 4 E 1 E 2 (1 + cos θ )/2
Thus the available energy in the CMS is reduced by a factor of (1 +
cos θ)/2 compared to head-on-collision.
(b) In the CMS E
∗
= 25 + 25 = 50 GeV
With the fixed proton target
E
∗
= (m
2
+ m
2
+ 2 T 1 m)
1/2
= 50 GeV
Substituting m = 0.94 GeV, we find T 1 = 1329 GeV
9.106 When the protons travel toward each other with equal energy, and therefore
with the same speed, their net momentum is zero. In that case the Lab system
is reduced to the C-M system, and the observer is watching the events sitting
in the C.M. system. The total energy is then,
E
∗
= 10
10 eV + 10
10 eV + 10
9 eV + 10
9 eV = 22 × 10
9 or 22 GeV
Let E 1 be the energy of a proton in the lab system moving toward the target
proton originally at rest, then if the total energy available in the CMS has to
be the same as E
∗
= 22 GeV,
m 1
2
+ m 2
2
+ 2 E 1 M 1
1/2 = E
∗
= 22 GeV
(1
2
+ 1
2
+ 2 E 1 × 1)
1/2
= 22
Or E 1 = 241 GeV
Therefore required kinetic energy = 241 − 1 = 240 GeV
9.107 Total energy available in the CMS is
E
∗
=
m 1
2
+ m 2
2
+ 2 E 1 m 2
1/2
where E 1 is the total energy of projectile of mass m 1 and m 2 is the mass of
the target.
E
∗
= 4M 0 + 2 M 0 = 6M 0
m 1 = m 2 = M 0
6M 0 =
M 0
2
+ M 0
2
+ 2 E 1 M 0
1/2
whence E 1 = 17 M 0
Or T 1 = E 1 − M 0 = 16 M 0
The significance of this result is that in the colliders a lot more energy is
available than in fixed target experiments.