Lorentz Transformations
(px,py,px,E) are components of a 4-vector which has
same Lorentz transformation
px’ = g (px + uE/c2)
u = velocity of transform
py’ = py
between frames is in
pz’ = pz
x-direction. If do px’ -> px
E’ = g (E + upx) + -> - Use common sense
also let c = 1
Frame 1
Frame 2 (cm)
Before and after scatter
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center-of-momentum frame
Sp = 0. Some quantities are invariant when going
from one frame to another:
py and pz are “transverse” momentum
Mtotal = Invariant mass of system dervived from
Etotal and Ptotal as if just one particle
How to get to CM system? Think as if 1 particle
Etotal = E1 + E2 Pxtotal = px1 + px2 (etc)
Mtotal2 = Etotal2 - Ptotal2
gcm = Etotal/Ptotal and bcm = Ptotal/Etotal
1
2 at rest
p1 = p 2
1
2
Lab
CM
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Particle production
convert kinetic energy into mass - new particles
assume 2 particles 1 and 2 both mass = m
Lab or fixed target Etotal = E1 + E2 = E1 + m2
Ptotal = p1 --> Mtotal2 = Etotal2 - Ptotal2
Mtotal = (E12+2E1*m + m*m - p12).5
Mtotal= (2m*m + 2E1*m)0.5 ~ (2E1*m)0.5
CM: E1 = E2 Etotal = E1+E2 and Ptotal = 0
Mtotal = 2E1
1
2 at rest
p1 = p 2
1
2
Lab
CM
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p + p --> p + p + p + pbar
what is the minimum energy to make a protonantiproton pair?
• In all frames Mtotal (invariant mass) at threshold is
equal to 4*mp (think of cm frame, all at rest)
Lab Mtotal = (E12+2E1*m + m*m - p12).5
Mtotal= (2m*m + 2E1*m)0.5 = 4m
E1 = (16*mp*mp - 2mp*mp)/2mp = 7mp
CM: Mtotal = 2E1 = 4mp or E1 and E2 each =2mp
1
2 at rest
p1 = p 2
1
2
Lab
CM
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Transform examples
• Trivial: at rest E = m p=0. “boost”
velocity = v
E’ = g(E + bp) = gm
p’ = g(p + bE) = gbm
• moving with velocity v = p/E and then
boost velocity = u (letting c=1)
E’ = g(E + pu)
p’ = g(p + Eu)
calculate v’ = p’/E’ = (p +Eu)/(E+pu)
= (p/E + u)/(1+up/E) = (v+u)/(1+vu)
• “prove” velocity addition formula
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• A p=1 GeV proton hits an electron at rest.
What is the maximum pt and E of the
electron after the reaction?
• Elastic collision. In cm frame, the energy
and momentum before/after collision are the
same. Direction changes. 90 deg = max pt
180 deg = max energy
 bcm = Ptot/Etot = Pp/(Ep + me)
· Pcm = gcmbcm*me
(transform electron to cm)
· Ecme = gcm*me
(“easy” as at rest in lab)
·
• pt max = Pcm as elastic scatter same pt in lab
•
Emax = gcm(Ecm + bcmPcm)
= gcm(gcm*me + gbb*me)
= g*g*me(1 + b*b)
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• p=1 GeV proton (or electron) hits a
stationary electron (or proton)
mp = .94 GeV me = .5 MeV
incoming target bcm gcm
Ptmax
Emax
p
e
.7
1.5 .4 MeV 1.7 MeV
p
p
.4
1.2 .4 GeV 1.4 GeV
e
p
.5
1.2 .5 GeV 1.7 GeV
e
e
.9995 30 15 MeV 1 GeV
 Emax is maximum energy transferred to stationary
particle. Ptmax is maximum momentum of (either)
outgoing particle transverse to beam. Ptmax gives
you the maximum scattering angle
•
a proton can’t transfer much energy to the electron
as need to conserve E and P. An electron scattering
off another electron can’t have much Pt as need to
conserve E and P.
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