Physics 535 lecture notes: - 24 Nov 27th, 2007
Homework due Next Thursday:
1) Derive the cross sections for  e d  e u and e u  e  d
Note that since we do not have quarks interacting with quarks this interaction is more like
a typical weak scattering process rather than like pion decay.
What is the ratio of these cross sections. Do they obey a crossing relationship?

2) What quarks or anti-quarks
does 
e not interact with.
3) 10.12
1) The weak force.

Charged W interactions
igw 
 1  5 
2 2
ig  q q / MW2 
Weak W vertex factor:
Weak W propagator:

q 2  MW2
ig
For small q this propagator reduces to: 2
MW

2gW2
and the Fermi constant can be defined: GF 
8M W2

2) Consider two scattering interactions for electrons:

t channel neutrino electron scattering: e e  e e ,12  34
This is essentially the same as the neutrino muon scattering we studied before, since there
are still no alternative diagrams.

Write currents between the two electrons and neutrinos. They have the same index since
there are related by the W

ig 

ig
ig
iM  u (3) w   1  5 u(1) 2 u (4) w   1  5 u(2)

 M 

2 2
2 2
2
g
M  W 2 u (3)  1  5 u(1) u (4)  1  5 u(2)
8MW
2gW2
The terms in front can be identified with the Fermi coupling constant GF 
8M W2
G
M  F u (3)  1  5 u(1) u (4)  1  5 u(2)
2
The conjugate terms have an independent index















*
GF2
u (3)  1  5 u(1) u (3)  1  5 u(1) u (4)  1  5 u(2) u (4)  1  5 u(2)
2
GF2
2
M 
u (3)  1  5 u(1)u (1)  1  5 u(3) u (4)  1  5 u(2)u (2)  1  5 u (4)
2
M 
2





Averaging over the spins of the initial electron, adds a factor of ½, and evaluating the
traces

 

GF2
Tr   1  5 p1  me   1  5  p3  me  Tr   1  5 p2  1  5 p4
4
Terms to one order in the electron mass are zero because they have an odd number of
gamma matrices. Terms to two orders cancel out. Evaluated the traces gives.
M 
2

M  64GF2 p1  p2  p3  p4 
2
In CM frame




1/ 2 
 
2 
m
e
p1  p2   p4  p3   E e Ee  (E  m ) Ee  Ee E e 1 1 2   2E 2
  E e  
1/ 2 
 
2 
m
2
2 1/ 2
e
p1  p2   E e Ee  (E e  me ) Ee  Ee E e 1 1 2   2E 2
  E e  
2
M  256GF2 E 4
2
e
2 1/ 2
e
At high energy in CM frame we could also phrase this in terms of s, t and u
s  p1  p2   2 p1  p2   4 E 2
2
M  16GF2 s 2
2

s channel anti-neutrino electron scattering: ee  ee ,12  34


Write currents between the two electrons
and neutrinos. They have the same index since
there are related by the W

ig 

ig
ig
iM  (2) w   1  5 u(1) 2 u (3) w   1  5  (4)

 M 

2 2
2 2
2
g
M  W 2 (2)  1  5 u(1) u (3)  1  5  (4)
8MW
G
M  F (2)  1  5 u(1) u (3)  1  5  (4)
2
The conjugate terms have an independent index









*





*
GF2
(2)  1  5 u(1) (2)  1  5 u(1) u (3)  1  5  (4) u (3)  1  5  (4)
2
GF2
2
M 
(2)  1  5 u(1)u (1)  1  5  (2) u (4)  1  5  (3)(3)  1  5 u (4)
2
M 
2





Averaging over the spins of the initial electron, adds a factor of ½, and evaluating the
traces. Also ignoring the mass of the electron.
M 
2

 
GF2
Tr   1  5 p1  1  5 p2 Tr   1  5 p3  1  5 p4
4

M  64GF2 p1  p3  p2  p4 
2


In CM frame
p1  p3   p2  p4   E 2 1 cos
M  64 E 4 1 cos  
2




2
t  p1  p3   2p1  p3 
2
M  16GF2 t 2
2
In electron neutrino scattering s channel has the angular dependence and t is isotropic.
Also the s channel diagram can be obtained from the t channel diagram by an s to t
crossing.
3) Scattering cross sections
M
d
1 M pi
1


2
2
2
d 64  4 E p f 256 E 2
2
2
t channel neutrino electron scattering: e e  e e ,12  34



M  256GF2 E 4
d
1
 2 GF2 E 2
d 
2

4


GF2 E 2
s channel anti-neutrino electron scattering: ee  ee ,12  34


*
d
1
2

GF2 E 2 1 cos 
2
d 4


1
2
GF2 E 2  1 cos  sin 
2
2
  4GF2 E 2  sin   2cos sin   cos2  sin d



1 2 2 
1
GF E cos  cos2   cos3  


2
3

4 2 2
GF E
3


There is also a difference in the total cross section!


 ee  ee  1

 e e  e e  3
Why is anti-neutrino scattering one third as likely? We can make a helicity agreement.
The backward scattering interaction for the anti-neutrino basically violates helicity
conservation. This relationship would have been different if the W+ didn’t interact with
neutrinos in a pure V-A way.
Helicity conservation is an important factor in how neutrinos interact that has to be
considered when detecting neutrinos and anti-neutrinos.
3) Pion decay    ee
Since the anti u and d quarks in the pion interact via the strong force it is incorrect to
consider this as an annihilation diagram. However, it is instructive.

du  ee , 12  34


ig 

ig
ig
iM  (2) w   1  5 u(1) 2 u (3) w   1  5  (4)
 M 

2 2
2 2
 
2
G
2
M  F (2)  1  5 u(1)u (1)   (2) u (3)  1  5  (4)(4)1  5 u(3)
2


M 
2






GF2
Tr   1  5 p1  md   1  5  p2  mu    1  5 p4   1  5 p3
2

Event though the mass terms will dominate they cancel along with any terms odd in a
gamma matrix and we get.
M  64GF2 p1  p4  p2  p3 
2

In CM the quark momentums will be approximately 0 and we will gets terms with the
quark mass times the energy of the outgoing particles.

However, this is not quite a correct treatment. The distribution of momentums of the
quarks, and thus the possible angular distributions of terms like p1  p4 , will be
constrained by the strong force and we should treat them in a more generalized way.
G
2
M  F F  u (3)  1  5  (4)
2

Where F will be a function of the pion and quark momentums p   p1  p2 times a
scalar function of p 2 , f  p2  f  m .
G
2
M  F F  u (3)  1  5  (4)

2
2 2
G
f
2
M  F  p  p u (3)  1  5  (4)( 4)  1  5 u(3)
2
2 2
G f
2
M  F  p  p Tr   1  5 p4   1  5 p3
2
2 2
G f
2
M  F  p  p 8p4 p3  p3 p4  g  p3  p4  8ip1 p3  
2
2
M  4GF2 f 2 2p  p3 p  p4   p2 p3  p4 

using

p  p1  p2  p3  p4
p  p3   p3  p4  p3   me2  p3  p4 
p  p4   p3  p4  p4   p3  p4 




p 2  p3  p4   p32  p42  2p3  p4   me2  2p3  p4   m2
2p3  p4   m2  me2















2
 
1
1
1 1
1 
1 
2
M  4GF2 f 2 2me2  m2  me2  m2  me2  m2  m2  me2 
2
2
2 2
2 
2 
 


M  2GF2 f 2 m2  me2 m2  me2  m2 m2  me2 
2
M  2GF2 f 2 me2 m2  me2 
2


The decay width is


pf
pf 
1
m2  me2 

2m


M
2
8m 
Where using homework 3.16
(mA,0,0,0) = (EB + EC,0,0,0)
(mA,0,0,0) = (EB + (pB2 + mC2)1/2,0,0,0)
(mA,0,0,0) = (EB + (EB2 - mB2 + mC2)1/2,0,0,0)
m A = EB + (EB2 - mB2 + mC2)1/2
(m A - EB)2 = EB2 - mB2 + mC2
m A2 -2 m A EB + EB2 = EB2 - mB2 + mC2
EB = (m A2 + mB2 - mC2)/2 m A
2
1
GF2 f 2 me2 2
2
2
2 2 2
2
2
m

m
2G
f
m
m

m

m  me2 

e
F 
e 
e
3  
3
16m
8m
If we compare the decay width to muons and electrons

   ee 
me2 m2  me2 
2

2
     m 2 m2  m2 


Again we can make a helicity argument. The electron or muon is forced to be in the
wrong helicity state which can happen more often the muon.