Lagrangian with ElectroWeak Unification The Standard Model assumes that the mass of the neutrino is zero and that it is left handed travelling with its spin pointing opposite to its direction ID: 461228
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Slide1
Standard Model Lagrangianwith Electro-Weak Unification
The Standard Model assumes that the mass of the neutrino is zero and that
it is “left handed” -- travelling with its spin pointing opposite to its direction
of motion.
Since in this case there would be no “right handed” neutrino, the “flavor”
partner of the neutrino must be a “left handed” electron. This changes
the structure of the Standard Model
Lagrangian
– which is assumed to treat
only left handed flavor doublets.Slide2
The spinor for the
neutrino
is the same as for the
electron,if one substitutes m = 0 :
neutrino
spinorSlide3
The helicity operator,
p
, satisfies the following condition
So when acting on the
spinor
It’s a projection!
2
= u
u
1Slide4
left hand side gives
(p) (p)/
pp
= 1 (
2x2 unit matrix
)
1
From this we can determine conditions on a and bSlide5
So, we have the following condition on a and b:Slide6
These are the only
spinors
allowed for a zero mass neutrino!
positive
helicity
negative
helicity
The neutrino, if it has a zero mass can only have its spin
pointing along (or opposite to) it’s momentum.Slide7
Non-conservation of parity: Wu 1957
J
J-1Slide8
number of
electrons
e
JSlide9
The neutrino could have both values of helicity, and Wu’s experiment, while confirming non conservation of parity (left-right symmetry broken), did not conclusively determine the neutrino’s helicity
.
In 1958
Goldhaber determined the helicity of the neutrino in the K capture of an electron:
63Eu152 (J=0) 62Sm*
152 (J=1) + . The Sm* decays giving off a photon with the same
helicity
as the neutrino.
Goldhaber
measured the
helicity
of the photon by passing it through magnetized iron. If the photon has the same direction of spin as the magnetized iron, it would pass through, otherwise it would produce a spin flip. He reported a
helicity
of -1.
Helicity of the neutrino: Goldhaber
1958Since 1998 it has been accepted that the neutrino has a small mass. This produces some corrections in the Standard Model. For this
description of the Standard Model, it is assumed that the neutrino has no mass. Slide10
Each term in the SM
Lagrangian
density containing quarks and the leptons can be rewritten using the following expression. For
the neutrinos, however, only the left handed term exists.
In the following slide we use the notation d R
= dRSlide11
The Lagrangian density withthe U(1), SU(2) and SU(3)gauge particle interactions
neutral vector boson
heavy vectors bosons (W
, W
3
)
8 gluons
YSlide12
The following is the interaction Lagrangian density for the first generation of particles with the left and right handed parts shown explicitly.
B
B
B
B
B
B
W
1
W
1
W
2
+
W
2
+
W
3
W
3
a
G
a
sum over a = 1,2,…8
a
G
a
B
Sum over all leptons and quarks interacting
with
B
Sum over all left handed leptons and quarks interacting
with W
Sum over all quarks interacting
with
G
a
Slide13
Weinberg’s decomposition of the B and W
:
sin
2
W 0.23
-- to be determined experimentally!
W
= Weinberg angleSlide14
Next steps: rewriting interaction
Lagrangian
density so
that interactions with the photon are identified.
The neutrino has zero charge and can’t interact with the photon.
1.
2
.
3
.Slide15
After substituting the expressions for B and W0 (which takes some work),one can identify factors which equal e, the electronic charge, or the up quark charge, etc. This permits one to find relationships between
sin
W
, cos
W , e, g
2
and
g
1
.
One finds that:
g
2 = e /
sinW
g 1 =
e / cosW
Y
L
= -1
Y
R
= 2 Y
L
Also one defines:
T
3
f
= + 1/2
for the
u
L
= -
1/2
for the
d
L
=
0 for
u
R
=
0 for
d
R
Slide16
(E & M) QED interactions
weak neutral current interactions
weak flavor changing interactions
QCD color interactions
+
+
The Standard Model Interaction
Lagrangian
for the 1
st
generationSlide17
The U(1) and SU(2) interaction terms
A
Z
+
Z
W
+
W
-
weak
neutral current interactions
(E & M)
QED
interactions
weak
flavor changing interactions
g
2
g
2
e
g
2
=
e
/
sin
W