Transcript Chapter 1, Lecture 3 - University of Hawaii Physics and Astronomy

Alessandro Bettini
Introduction to
Elementary Particle Physics
SECOND EDITION
Cambridge University Press
2014
CHAPTER 1: Preliminaries
Today’s plan
• Review calculations of s, the energy available
in the CM frame to produce new particles.
• Reminder about q2 (momentum transfer for a
virtual photon)
• Introduce hadrons, leptons, quarks and
fundamental interactions.
• Introduce cross-section, σ, used in particle and
nuclear physics calculations of interaction
probabilities (next time)
Diphoton mass
Find the mass of this system when
the two photon energies are equal
What is the first step ?
What relativistic invariant should we use ?
Do you see why ?
s = 2E - 2 p cosJ = 2E (1- cosq )
2
2
2
✔
CM and Fixed Target Frames
In the CM frame, the momenta sum
to zero. What is s ?
s = (Ea + Eb )
2
In the special case, the
beam energies are equal
s = (2E) = 4E
2
2
s = 2E ,
this is the available energy for making new particles
CM and Fixed Target Frames
In the fixed target frame, the target
particle is at rest. What is s ?
s = (Ea + mb ) - pa
2
2
Do you see
why ?
s = (Ea + mb )2 - pa 2 = Ea 2 + 2mb Ea + mb 2 - pa 2
s = 2Ea mb ; assuming E a >> ma , mb
this is the available energy for making new particles
Fixed target vs CM conceptual question
For making new particles, which kind of
experiment/accelerator is most effective ?
Is there any tradeoff ?
Ans: s (available energy for making new particles) is
much larger with head-on collisions but
luminosity/intensity can be orders of magnitude larger
with fixed target geometry.
Fixed target vs CM example
Fermilab, located in Batavia, Il (suburbs of
Chicago) used to collide 1 TeV protons on
1 TeV anti-protons. What is the available
energy to produce new particles ?
s = 2TeV = 2000GeV
What energy proton beam would be
required to achieve the same available
energy in fixed target ?
Fixed Target Example
p+ p® p+ p+ p+ p
What is the threshold (minimum) energy for making
an anti-proton in this reaction ?
Question:
BTW why do we
need 4 particles ?
Let’s calculate s for the initial and final states.
si = (Ea + mb )2 - pa 2
si = (Ea + mb )2 - pa 2 = 2mb Ea + mb 2 + ma 2
What is the invariant s for the final state ?
s f = (4m p )
2
si = 2m p Ea + m p 2 + m p 2 = 16m p 2 = s f
N.B. The 4
nucleons are at
rest in the CM
frame not the lab
frame
2m p Ea = 14m p 2 Þ Ea = 7m p = 7(938MeV ) = 6.56GeV
Discovery of the anti-proton
Berkeley, California 1955
Emilio Segre
Owen Chamberlain
Problem 1.5: How to calculate q2
Figure from
Bettini
Use this geometry and the fact that p’=p
to obtain,
q
q = 2 psin( )
2
x vs t representation of
particle interaction
We are probing the proton with
a “virtual photon”. The
wavelength of the photon
(deBroglie wavelength
associated with the momentum
transfer) determines the length
scale that is probed.
Problem 1.7: GZK energy cutoff
This reaction should
limit the maximum
energy of protons in
cosmic rays
1966: Greisen,
Katsepin and Kuzmin
Need to take into account
interaction with the cosmic
microwave background
Early claims of post GZK
cosmic rays by AGASA in
Japan not confirmed by
Auger
N.B. these cosmic ray plots are on a log-log scale
Mandelstam Invariants
Scattering
processes in
particle physics
s: Available energy in CM for making new
particles
t: 4-momentum transfer squared
Stanley Mandelstam,
UC Berkeley (19282016)
Note
s≥0, t≤0,
u≤0
s channel
t-channel
u-channel
Classification of “Elementary” Particles
Section 1.8 of Bettini
In quantum mechanics, angular momentum is quantized. In particular,
there is intrinsic spin associated with particles.
Question: Are leptons composite or point-like ?
Question: Are baryons bosons or fermions ?
Question: Are mesons bosons or fermions ?
Now pay close
attention to the
propagator
particles