9.3 Solutions
533
9.108 The CMS energy is calculated from the invariance of E
2
− p
2
E
∗2
= E
2
− p
2
=
E p + E e
2 −
p p − p e
2
= E
2
p − p
2
p + E
2
e − p
2
e + 2
E p E e + p p p e
≈ m p
2
+ m e
2
+ 4E p E e
Since m p << E p and m e << E e
E
∗
≈
4E p E e =
√
4 × 820 × 30 = 314 GeV
Note that the HERA accelerator is different from other colliders in that the
energy of the colliding particles (protons and electrons) is quite asymmetrical. It has been possible to achieve high momentum transfer in the CMS
(20,000 GeV
2 ), necessary for the studies of proton structure.
9.109 Total CMS energy, E
∗
≈ [2 E 1 E 2 (1 + cos θ )/2]
1/2
Substituting E 1 = 20 GeV, E 2 = 300 GeV, and θ = 10
0
we find E
∗
= 109 GeV
If the same energy (E
∗
= 109 GeV) is to be achieved in a fixed target
experiment, then the electron energy in the lab-system would be
(2 E M + M
2
+ m
2 )
1/2
= E
∗
Neglecting M
2 and m
2
E = (E
∗ )
2
/2M = (109)
2
/(2 × 0.94) ≈ 6, 300 GeV
533
9.108 The CMS energy is calculated from the invariance of E
2
− p
2
E
∗2
= E
2
− p
2
=
E p + E e
2 −
p p − p e
2
= E
2
p − p
2
p + E
2
e − p
2
e + 2
E p E e + p p p e
≈ m p
2
+ m e
2
+ 4E p E e
Since m p << E p and m e << E e
E
∗
≈
4E p E e =
√
4 × 820 × 30 = 314 GeV
Note that the HERA accelerator is different from other colliders in that the
energy of the colliding particles (protons and electrons) is quite asymmetrical. It has been possible to achieve high momentum transfer in the CMS
(20,000 GeV
2 ), necessary for the studies of proton structure.
9.109 Total CMS energy, E
∗
≈ [2 E 1 E 2 (1 + cos θ )/2]
1/2
Substituting E 1 = 20 GeV, E 2 = 300 GeV, and θ = 10
0
we find E
∗
= 109 GeV
If the same energy (E
∗
= 109 GeV) is to be achieved in a fixed target
experiment, then the electron energy in the lab-system would be
(2 E M + M
2
+ m
2 )
1/2
= E
∗
Neglecting M
2 and m
2
E = (E
∗ )
2
/2M = (109)
2
/(2 × 0.94) ≈ 6, 300 GeV