Lecture 25: April 24th 2012
Reading for next time: Griffiths 9.1,9.2
Homework due Thursday: 9.3, 9.10
1) Charged W weak decays and scattering at low or medium energy.
For decay
m(1) ® n m (3)e(4)n e (2)
M=
GF é
u(3)g m (1- g 5 ) u(1)ùûéëu(4)g m (1- g 5 ) n (2)ùû
ë
2
For the scattering:
m(1)n e (2) ® n m (3)e(4)
M=
[
where GF =
M=
][
]
][
]
GF
u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
2
[
2gw2
8M W2
GF
u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
2
[
][
] [u(4)g (1- g )u(2)][u(4)g (1- g )u(2)]
M =
GF2
u (3)g m (1- g 5 ) u(1) u (3)g n (1- g 5 ) u(1)
2
M =
GF2
u (3)g m (1- g 5 ) u(1)u (1)g n (1- g 5 ) u(3) u (4)g m (1- g 5 ) u(2)u (2)g n (1- g 5 ) u (4)
2
2
2
[
*
m
5
n
][
5
]
Perform the sums and averages over spins. Introduce a factor of ½ for the average over
muon spins. Note that when we average over the initial muon states there are two
possibilities while when we sum over the final electron states there are two possibilities
for the electron but only one possible spin state for the neutrinos.
GF2 é m
M =
Tr ëg (1- g 5 ) ( p/ 1 + mm ) g n (1- g 5 ) p/ 3 ùûTr éëg m (1- g 5 ) p/ 2g n (1- g 5 ) ( p/ 4 + me )ùû
4
2
*
Similar traces to those we found for the EM force except that we have the gamma 5
terms. Mass terms drop out since they have an odd number of gamma matrices. g 5 is
four gamma matrices. Evaluating this trace:
M = 64GF2 ( p1 × p2 )( p3 × p4 )
2
Note that the convention of s, t and u factors is not as useful here since the mass of the W
dominates in the propagator, where for the electromagnetic force the mass of the photon
was zero and the electron mass was typically negligible and q was the dominant term in
the denominator.
The matrix element for decay is the same. There was no crossing type exchange of two
particles. You have an electron antineutrino in the final state instead of a neutrino in the
initial state. This would introduce a negative sign in front of the mass term from the sum
over electron antineutrino states. However, the mass terms are certainly negligible for
neutrinos.
2) Weak decay and lifetimes.
m(1) ® n m (3)e(4)n e (2)
M=
GF é
u(3)g m (1- g 5 ) u(1)ùûéëu(4)g m (1- g 5 ) n (2)ùû
ë
2
M = 64GF2 ( p1 × p2 )( p3 × p4 )
2
In the case of the lifetime of the muon we have three final state particles and therefore
have to perform three sets of integrals over the phase space of the final state particles.
There are three interesting steps along the way.
First the matrix element
( p1 × p2 ) = mm E 2
for ( p3 × p4 ): ( p3 + p4 ) = me2 + 2( p3 × p4 )
( p3 + p4 ) = ( p1 - p2 ) = mm2 - 2( p1 × p2 ) = mm2 - 2mm E 2 = me2 + 2( p3 × p4 )
( p3 × p4 ) =
1 2 1 2
mm - me - mm E 2
2
2
æ
æ1
ö
1
m2 ö
2
M = 64GF2 ( p1 × p2 ) ( p3 × p4 ) = 64GF2 mm E2 ç mm2 - me2 - mm E2 ÷ = 32GF2 mm2 E2 çç mm - 2E2 - e ÷÷
è2
ø
2
mm ø
è
where typically the mass of the electron is taken as small.
Second the differential width, before you integrate over the electron energies.
The width will involve 3 integrals over the final state particles. We would like to
integrate over the momentums of the two neutrinos, which will not be observed, in order
to find the differential width as a function of the observed electron energy.
The delta function conserving overall energy and momentum will allow us to eliminate
one neutrino momentum and energy integral. The integral over the second neutrino runs
from ½ the muon mass minus the first neutrino momentum(the electron goes one way
and both neutrinos go the other) to ½ the muon mass (the second neutrino goes one way
and both the electron and the first neutrino go the other way). Performing the second
integral and the angular integrals leaves.
dG
GF2 2 2 æ 4 E ö
=
mm E çç1÷÷
dE 12p 3
è 3mm ø
Where E is the energy of the electron and runs from 0(electron sits still and the neutrinos
go in opposite directions) to mm/2(electron goes one way and the neutrinos go the other).
If you didn’t know the mass of the muon you could measure it from the “slope” of the
energy distribution of the electron.
If you didn’t know the mass of the electron, but did know the mass of the muon, this
expression would be modified by the me2 / mm term above. You would be most sensitive
when the electron energy was small and me2 / mm the term contributed a relatively large
amount modifying the slope of the electron energy distribution near E=0. Similarly with
a known electron mass you could try to extract the neutrino mass.
These limiting or “endpoint” cases are very interesting in 3 body decays involving new
physics particles decaying to visible standard model particles and invisible new physics
and standard model particles. In the three body case these relationships will be modified
by the mass of the third particle.
An interesting case: c 20 ® l +l - c10
Consider the limit c 10 where is produced at rest and the lepton pair gets all the resulting
energy. Calculate the total energy of the lepton pairs which will be equal to the mass
difference between c 20 and c 10 .
Many of these endpoint expressions can be derived even in complex decays.
Third the final width and lifetime:
G=
1
t
=
GF2 mm5
192p 3
Note that before weak theory was formulated we could only measure GF from the
experiments to measure the lifetime of the muon and didn’t know that we could calculate
GF given the mass of the W boson and a W . Alternatively we can calculate a W given the
mass of the W particle and GF . We can only predict the mass of the W boson once we
determine how a W runs so we can predict it’s value at low energy given a W and a at
high energy.
Using =2.2x10-6 sec and m=105.7 MeV and
GF =
= 6.582x10-22 MeVs
192p 3
=1.16 ´10-11 MeV -2 =1.16 ´10-5 GeV -2
5
mmt
and using GF =
2gW2
8MW2
2gW2
2 4paW
1
if aW =
=
= 81.4GeV ,
8GF
8GF
29.5
3) Consider two scattering interactions for electrons:
MW =
t channel neutrino electron scattering: e-n e ® n ee-,12 ® 34
This is essentially the same as the neutrino muon scattering we studied before, since there
are still no alternative diagrams via the W boson. We will study the Z boson contribution
later.
Write currents between the two electrons and neutrinos. They have the same index since
there are related by the W
é
ù ig é
ù
-ig
-ig
-iM = êu (3) w g m 1- g 5 u(1)ú mn2 êu (4) w g m 1- g 5 u(2)ú
ë
ûM ë
û
2 2
2 2
2
g
M = W 2 u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
8MW
2gW2
The terms in front can be identified with the Fermi coupling constant GF =
8MW2
G
M = F u (3)g m 1- g 5 u(1) u (4)g m 1- g 5 u(2)
2
(
)
[
[
(
][
(
) ][
]
(
)
]
The conjugate terms have an independent index
)
[
][
] [u(4)g (1- g )u(2)][u(4)g (1- g )u(2)]
GF2
M =
u (3)g m (1- g 5 ) u(1) u (3)g n (1- g 5 ) u(1)
2
2
M =
2
[
*
m
5
5
n
][
]
GF2
u (3)g m (1- g 5 ) u(1)u (1)g n (1- g 5 ) u(3) u (4)g m (1- g 5 ) u(2)u (2)g n (1- g 5 ) u (4)
2
Averaging over the spins of the initial electron, adds a factor of ½, and evaluating the
traces
M =
2
[
] [
GF2
Tr g m (1- g 5 )( p/ 1 + me )g n (1- g 5 )( p/ 3 + me ) Tr g m (1- g 5 ) p/ 2g n (1- g 5 ) p/ 4
4
]
Consider an intermediate relativistic range with large enough momentums that the terms
involving the electron mass are small but not large enough that the momentum dependent
part of the numerator of the propagator is important. Note that the effect propagator is
significantly dependent on the momentum at energies high enough for the W to be on
mass shell. In CM this would be on order 81 GeV and in lab frame the neutrino would
have to be on order 3200GeV! This leaves a large range of momentums where the terms
involving the electron mass are negligible but the propagator is still dominated by the W
mass squared in the denominator.
Neglecting the terms to second order in the electron mass. Evaluated the traces gives.
M = 64GF2 ( p1 × p2 )( p3 × p4 )
2
In CM frame
æ æ
2 ö1/ 2 ö
m
e
( p1 × p2 ) = ( p4 × p3 ) = E e Ene + (E - m ) Ene = Ene E e çç1+ ç1- 2 ÷ ÷÷ = 2E 2
è è Ee ø ø
2
M = 256GF2 E 4
2
e
2 1/ 2
e
At medium energy in CM frame we could also phrase this in terms of s, t and u
s = ( p1 + p2 ) = 2( p1 × p2 ) = 4E 2
2
M =16GF2 s2
2
4) s channel anti-neutrino electron scattering via annihilation: e-n e ® n ee-,12 ® 34
Write currents between the two electrons and neutrinos. They have the same index since
there are related by the W
é
ù ig é
ù
-ig
-ig
-iM = ên (2) w g m 1- g 5 u(1)ú mn2 êu(4) w g m 1- g 5 n (3)ú
ë
ûM ë
û
2 2
2 2
(
)
(
)
*
gW2 é
M=
n (2)g m (1- g 5 ) u(1)ùûéëu(4)g m (1- g 5 ) n (3)ùû
2 ë
8MW
G
M = F éën (2)g m (1- g 5 ) u(1)ùûéëu(4)g m (1- g 5 ) n (3)ùû
2
The conjugate terms have an independent index
M =
*
*
GF2 é
m
5
ùén (2)g n (1- g 5 ) u(1)ù éu(4)g (1- g 5 ) n (3)ùéu(4)g (1- g 5 ) n (3)ù
n
(2)
g
1g
u(1)
(
)
m
n
ûë
ûë
ûë
û
2 ë
M =
GF2
n (2)g m (1- g 5 ) u(1)u (1)g n (1- g 5 )n (2) u (4)g m (1- g 5 )n (3)n (3)g n (1- g 5 ) u (4)
2
2
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.
[
] [
GF2
M =
Tr g m (1- g 5 ) p/ 1g n (1- g 5 ) p/ 2 Tr g m (1- g 5 ) p/ 3g n (1- g 5 ) p/ 4
4
2
]
M = 64GF2 ( p1 × p3 )( p2 × p4 )
2
In CM frame
( p1 × p3 ) = ( p2 × p4 ) = E 2 (1- cosq)
M = 64 E 4 (1- cosq)
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.