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9 Particle Physics – I
Energy in the center of mass system (CMS)
In the collision of particle of mass m 1 , of total energy E 1 in the lab system with m 2
initially at rest, total energy E
∗ available in the CMS is given by
E
∗
=
m
2
1 + m
2
2 + 2m 2 E 1
1 /2
(9.5)
Division of energy in the decay A → B + C, at rest.
Total energy carried by B,
E B =
m
2
A + m
2
B − m
2
C
2m A
(9.6)
Circular accelerators
In static magnetic field, a charged particle is not accelerated but is only bent into
a circular path if the field is perpendicular to the plane of the path. Otherwise the
particle goes into a helical path.
The radius of curvature r is related to the momentum p by
P = 0.3 Br
(9.7)
where p is in Gev/c, r in meters, and the field B in Tesla. (1 Tesla = 10
4 gauss).
If n = γ − 1 then R =
mc
qB
(n
2
+ 2n)
1 /2
(9.8)
Betatron
Energy gained
ΔT = e
ΔΦ
Δt
(9.9)
where Δϕ/Δt is the rate of change of flux.
Cyclotron
ω 0 =
qB
m
(Resonance condition)
where ω 0 = 2π f 0
Energy at the extraction point
T =
(qBR)
2
2m
(9.10)
Synchrocyclotron
ω 0 =
qB
m 0
(9.11)
ω =
qB
m
(9.12)