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![Page 1: Announcements Homework returned now 9/19 Switching to more lecture-style class starting today Good luck on me getting powerpoint lectures ready every day.](https://reader035.fdocuments.net/reader035/viewer/2022070410/56649ea95503460f94bad05f/html5/thumbnails/1.jpg)
Announcements
Homework returned now
9/19
• Switching to more lecture-style class starting today• Good luck on me getting powerpoint lectures ready every day
• Schedule slowing down a bit• Monday – reading assignment only• Wednesday – homework only
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Questions from the Reading Quiz“Can you explain how to obtain and the meaning of the Feynman invariant amplitude? ”
• How to obtain it – depends on the theory• For * theory – read chapter 5
• What it means• I don’t know• It’s an amplitude – the amplitude that you go into a particular state• But with some factors taken out to make it simpler
iM
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Questions from the Reading Quiz“I'm also kind of confused on the concept of D, mainly whether or not is has any physical significance”
• Definition:
324 4
out in3 out
22 2
i
i i
dD p p i
E
pM
What is its physical significance?•It is most of the factors in the probability for a process to occur•It allows for any number of particles in the final state• Works for 2 or 3 (or more) particle final states
•It ignores details of the incoming state• Can work for decay rates or cross-sections
•Mostly, it’s just a calculational tool
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Questions from the Reading Quiz“How is it that we just assume using first order perturbation theory will work?”
• It is an approximation• In general, for perturbation theory to work, we need a small parameter• The small parameter in this theory is g• If g is not small, then this approximation is poor
• In QED, for example, the parameter is e• e = 0.3, which is kind of small• As you go to higher order, you get factors of about e2/162
• In QCD, the coupling is significantly larger than one• Perturbation theory fails
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How to calculate everythingFermi’s Golden Rule
• The general formula for probability is:
24 4out in
in out
1 12
2 2j ij i
P VT p p iE V EV
M
• For 1-2 particles in the initial state, and 2-3 particles in the final state, these become:
2 1 1 2
2
4
D
MD
E E
p p
2
2
2
1 2 1 125
two ,16
1three .
8 2
cm
pD d i
E
D dE dE d d i
M
M
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Comments on D, and :
• The quantity D is Lorentz invariant• Usually calculated in the c.m. frame
• The quantity p is the momentum of either particle in this frame• The quantity Ecm in any frame is just s
2
2two
16 cm
pD d i
E M
• Decay rate is not Lorentz invariant• Listed value is in comoving (c.m.) frame• In arbitrary frame, formula becomes D/2E
• Cross section is Lorentz invariant in any frame where particles are colinear• Head-on collisions or stationary target• Then magnitude symbols not really necessary
2
D
M
2 1 1 24
D
E E
p p
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Differential cross-sections and decays:
• Combining these:
2
D
M 2
2two
16 cm
pD d i
E M
2
2 2two
32
pd i
M M
• Sometimes you are asked for the “differential decay rate”• That means: don’t do the integral 2
2 232
d pi
d M
M
• Similarly for cross-sections
2 1 1 24
D
E E
p p
2
22 1 1 264 cm
pd i
E E E
p pM
2
22 1 1 264 cm
d pi
d E E E
p p
M
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The general procedure• Find the Feynman invariant amplitude
• Multiply it by its complex conjugate• Later, this will be harder than it looks now
• Rewrite this quantity in terms of the given or final energies or angles
• Calculate D using one of the two formulas we have
• Find the decay rate or cross-section
iM
2 *i i iM M M
2 2
1 2 1 1252
1two , three .
16 8 2cm
pD d i D dE dE d d i
E
M M
2 1 1 2
, .2 4
D D
M E E
p p
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An easy problem:
Calculate the rate for decay in the * theory. *k p p
• This is a decay, so we use the decay formula• Two particles in final state, so we need that formula too
2
D
M 2
2.
16 cm
pD d i
E M
• Center of mass energy is the mass
i igM 2 2i gM
2
28
pg
M
• Final particles have equal and opposite 3-momentum• Final particles have identical energies
12E M 2 21
2p M m
22 2
24
16
gM m
M
2
232 cm
pgd
ME
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A hard problem:
Calculate the rate for muon decay. Treat all final state particles as massless.
1 2 3ep e p p p 2 22 1 364 Fi G p p p p M
• This is a decay, so we use the decay formula• Three particles in final state, so we need that formula too
2
D
m
2
1 2 1 125
1.
8 2D dE dE d d i
M
We need to:•Write all quantities in terms of the final energies or angles•Determine limits of integration•Perform all integrals
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Rewriting the Amplitude
1 2 3ep e p p p 2 22 1 364 Fi G p p p p M
• Conservation of momentum
1 2 3p p p p 2 1 3p p p p
• Square this quantity, remembering everything except the muon is massless
2 2
2 1 3p p p p
2 2 2p p E E p p
• Working in the rest frame of the muon, so
22 1 32 2m p p p p
211 3 22p p m m E
2 2 2 22 232 2Fi G m m E E M
2m E
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Limits on integration
2 2 2 22 232 2Fi G m m E E M
2
D
m
2
1 2 1 125
1.
8 2D dE dE d d i
M
• Particle 1 can go any direction we want 1 4d • The azimuthal angle runs from 0 to 2
12 2d • The energy integrals are tricky:• The total momentum is zero: they form a triangle• The total energy is m: this is the perimeter• No side can have more than half the total perimeter
2 2 21 2 1 12 2 25
12
16 FG dE dE d d m E m E
p1
p2
p3
1 1 11 2 32 2 2, ,E m E m E m
11 2 2E E m
3 1 2E m E E
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Announcements
Homework returned in boxes
9/21
• Small error in problem 4.9• Note my solutions are online for past homework problems
(a formula for momentum of the final particles was found in problem 2.8b)
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Finishing the Problem
2 2 21 2 2 23
12
2 FG dE dE m E m E
11 2
12 2
11 2 2
,
,
.
E m
E m
E E m
1 12 2
122
2 2 22 2 2 13 0
12
2
m m
F m EG m E m E dE dE
122 2 2 3
2 2 23 0
12
2
m
FG m E m E dE
122 2 3 41 2
2 23 43 0
1
2
m
FG m E m E
2 5
3
1.
192 FG m
2 2 21 2 1 12 2 25
12
16 FG dE dE d d m E m E
1 4d 12 2d
Calculate the rate for muon decay. Treat all final state particles as massless.
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Turning it Into Numbers
2 53
1
192 FG m
Calculate the rate for muon decay. Treat all final state particles as massless.
5 21.16637 10 GeV ,
105.658 MeV 0.105658 GeVFG
m
2 553
11.16637 10 GeV 0.105658 GeV
192
19 103.00091 10 GeV 3.0091 10 eV
161
10
6.5821 10 eV s
3.0091 10 eV
62.1874 10 s 2.1874 s exp 2.1970 s
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The general procedure1. Find the Feynman invariant amplitude2. Multiply it by its complex conjugate• Later, this will be harder than it looks now
3. Rewrite this quantity in terms of the given or final energies or angles4. Calculate D using one of the two formulas we have
5. Find the decay rate or cross-section
iM
2 *i i iM M M
2 2
1 2 1 1252
1two , three .
16 8 2cm
pD d i D dE dE d d i
E
M M
2 1 1 2
, .2 4
D D
M E E
p p
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Sample Calculation – Step 3The amplitude for the scattering process e-(p1) + -(p2) e-(p3) + -(p4)is given at right, in the limit where all masses are negligible. Find the differential cross-section if they are colliding head-on, each with energy E.
42 1 2 3 4
21 4 2 31 3
2 p p p pei
p p p pp p
M
3. Rewrite this quantity in terms of the given or final energies or angles
p1 p2
p4
p3
1
2
,0,0,
,0,0,
p E E
p E E
3
4
, sin cos , sin sin , cos
, sin cos , sin sin , cos
p E E E E
p E E E E
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The dot products
42
1 2 3 4 1 4 2 32
1 3
2ei p p p p p p p p
p p
M
1
2
,0,0,
,0,0,
p E E
p E E
3
4
, sin cos , sin sin , cos
, sin cos , sin sin , cos
p E E E E
p E E E E
2 21 2 2p p E E E E
2 2 2 2 2 2 2 2 23 4 sin cos sin sin cosp p E E E E 22E
2 21 4 cosp p E E
2 22 3 cosp p E E 2 2
1 3 cosp p E E
42 2
2
24 1 cos
1 cos
ei
M
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Sample Calculation – Steps 4 and 5…Find the differential cross section…
4. Calculate D using one of the two formulas we have
42 2
2
24 1 cos
1 cos
ei
M
2
2two
16 cm
pD d i
E M
2
216 2
Ed i
E M
24
22
4 1 cos
16 1 cos
ed
5. Find the decay rate or cross-section
2 1 1 24
D
E E
p p 28
D
E
24
22 2
4 1 cos
128 1 cos
ed
E
24
22 2
4 1 cos
128 1 cos
d e
d E
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Comments on this problem
• Total cross section is infinity• Caused by = 0• Classically, because particles always scatter a little• Experimentally, small angles cannot be detected
• This expression assumes you are in the c.m. frame• Since cross-section is invariant, good idea to write in terms of s
24
22 2
4 1 cos
128 1 cos
d e
d E
e-(p1) + -(p2) e-(p3) + -(p4)
2cms E 2 22 4E E
24
22
4 1 cos
32 1 cos
d e
d s
• Unfortunately, is not Lorentz invariant• If we had already found total cross-section,
this problem would not occur
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Sample Calculation – Step 3The amplitude for H e+e- is given by the formula at right, where p and p’ are the momenta of the final state particles, m is the mass of the electron, and v is a constant. What is the rate for this decay? Let M be the mass of the Higgs.
2
2 22
mi p p m
v M
H k e p e p
k p p
2 2 2 2k p p p p
2 22 2M m p p
2 2
2 2 2 2 2 2122 2
42
m mi M m m M m
v v M
3. Rewrite this quantity in terms of the given or final energies or angles
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Sample Calculation – steps 4 and 5
2 2
2 2 2 2 2 2122 2
42
m mi M m m M m
v v M
4. Calculate D using one of the two formulas we have
H k e p e p
2
2two
16 cm
pD d i
E M
22 2
2 24
16 2
p md M m
M v
2
2 22
48
pmD M m
Mv
2
D
M
5. Find the decay rate or cross-section
2
2 22 2
416
pmM m
M v
2 214p M m
2
3/22 22 2
432
mM m
M v
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Announcements
Homework not yet graded
9/24
• Wednesday: Problems 4.9 & 4.10• Friday: Reading quiz I and problems 5.1, 5.2, 5.3, 5.5, and 5.6
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Feynman DiagramsWhere they fit:
1. Find the Feynman invariant amplitude2. Multiply it by its complex conjugate• Later, this will be harder than it looks now
3. Rewrite this quantity in terms of the given or final energies or angles4. Calculate D using one of the two formulas we have5. Find the decay rate or cross-section
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How to draw Feynman Diagrams• Make a list of initial particles on the left and final particles on the
right• Label them by their four-momenta (and spin)
• Start drawing the right type of line for each initial and final particle• Line for particle• Arrow right for particle• Arrow left for * particle
• Find every possible way to connect everythingtogether using the allowed couplings• To tree level only
• You now have a series of Feynman diagrams• You need to calculate each one• You need to add their contribution
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How to calculate Feynman DiagramsFOR EACH DIAGRAM•Conserve four-momentum at each vertexMultiply the following factors:•Include one factor of – ig for each vertex•Include one factor of i/(k2-M2) for each propagator•Include one factor of i/(p2-m2) for each propagatorAdd all the diagrams together