486 9 Particle Physics – I
Energy in the center of mass system (CMS)
In the collision of particle of massm 1 , of total energyE 1 in the lab system withm 2
initially at rest, total energyE∗available in the CMS is given by
E∗=(
m^21 +m^22 + 2 m 2 E 1)^1 / 2
(9.5)
Division of energy in the decayA→B+C,atrest.
Total energy carried byB,
EB=m^2 A+m^2 B−m^2 C
2 mA(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 curvatureris related to the momentumpby
P= 0. 3 Br (9.7)
wherepis in Gev/c, r in meters, and the fieldBin Tesla. (1 Tesla= 104 gauss).
Ifn=γ−1 thenR=mc
qB(n^2 + 2 n)(^1) / 2
(9.8)
Betatron
Energy gained
ΔT=e
ΔΦ
Δt(9.9)
whereΔφ/Δtis the rate of change of flux.
Cyclotron
ω 0 =qBm (Resonance condition)
whereω 0 = 2 πf 0
Energy at the extraction point
T=
(qBR)^2
2 m(9.10)
Synchrocyclotron
ω 0 =qB
m 0(9.11)
ω=qB
m