1000 Solved Problems in Modern Physics

532 9 Particle Physics – I

9.105 (a)E∗^2 =(E 1 +E 2 )^2 −(p 1 −p 2 )^2

=E 12 + 2 E 1 E 2 +E 22 −p 12 + 2 p 1 p 2 −p 22

=

(

E 12 −p 12

)

+

(

E 22 −p 22

)

+2(E 1 E 2 +p 1 p 2 )

=m 12 +m 22 +2(E 1 E 2 +p 1 p 2 )

IfE 1 >>m 1 andE 2 >>m 2 ,m 12 andm 22 can be neglected andp 1 ≈

E 1 ,p 2 ≈E 2

ThereforeE∗^2 ≈ 4 E 1 E 2

If the beams cross at an angleθthen

E∗^2 =(E 1 +E 2 )^2 −

(

p 12 +p 22 − 2 p 1 p 2 cosθ

)

≈ 2 E 1 E 2 (1+cosθ)= 4 E 1 E 2 (1+cosθ)/ 2

Thus the available energy in the CMS is reduced by a factor of (1+

cosθ)/2 compared to head-on-collision.

(b) In the CMSE∗= 25 + 25 =50 GeV

With the fixed proton target

E∗=(m^2 +m^2 + 2 T 1 m)^1 /^2 =50 GeV

Substitutingm= 0 .94 GeV, we findT 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∗= 1010 eV+ 1010 eV+ 109 eV+ 109 eV= 22 × 109 or 22 GeV
LetE 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 asE∗=22 GeV,
(
m 12 +m 22 + 2 E 1 M 1

) 1 / 2

=E∗=22 GeV

(1^2 + 12 + 2 E 1 ×1)^1 /^2 = 22

OrE 1 =241 GeV

Therefore required kinetic energy= 241 − 1 =240 GeV

9.107 Total energy available in the CMS is
E∗=

(

m 12 +m 22 + 2 E 1 m 2

) 1 / 2

whereE 1 is the total energy of projectile of massm 1 andm 2 is the mass of

the target.

E∗= 4 M 0 + 2 M 0 = 6 M 0

m 1 =m 2 =M 0

6 M 0 =

(

M 02 +M 02 + 2 E 1 M 0

) 1 / 2

whenceE 1 = 17 M 0

OrT 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.