Lecture 8: Feb 16th 2012
Reading: Griffiths chapter 3,4
Homework due Thursday Feb 23rd: Griffiths: 4.8, 4.10, 4.11, 4.29
1) Ice cube
The typical neutron decay diagram can be rotated to give us the interaction of neutrinos
with matter.
n->p+e-
d -> uW- -> u e- anti-
becomes
d -> dW+e- -> u e- or d -> uW- -> ueActually both diagrams at once in a non-time ordered way since the W is in the
intermediate state.
The electron is then detected through Cherenkov radiation. The neutrino energy will
also be related to the electron velocity and thus the Cherenkov angle. You typically want
a detector that can detect the Cherenkov ring of light. However it takes a very large
amount of matter to get a significant total probability of these interactions taking place.
And of course anit-neutrinos can be detected as well: Anti-u -> d e+
It would also be useful to have a detector that can distinguish electrons from positrons but
this is difficult in a large area detector like icecube, but easy in a smaller detector where
you look for photons from the positrons annihilating on electrons or use a tracking
detector with a magnetic field.
Note: the weak interaction is not C conserving and these processes have different cross
sections by a factor of 3 and different angular distributions!
For the heavier generations
d -> u - Where absorbing material and tracking detectors can be used to identify the
muons and their charge.
d -> u - where the hadronic tau decays can be identified using a calorimeter.
Note for these interactions to take place the neutrino must have enough momentum or
equivalent energy to produce the mass of the muons to tau.
These three are typically referred to as the charged current interactions.
And the neutral current scattering interactions via a Z boson . This interaction is identical
for three generations of neutrinos and can measure the total neutrino flux. Technically it
is difficult to detect since you have to observe the recoil of the nucleus.
q -> q
2) Typical high energy physics units take hbar =1 c=1
c = d/t, for a particle traveling at c if c=1 then if distance is in cm then time is understood
in units of the unit of time it takes the particle to travel 1cm. You will sometimes see
lifetime specified in units of distance.
The use of hbar removes a large factor of order 10-22.
The considering energy and wavelength E=hbar*c 2/, if hbar and c=1 a photon with
wavelength 2 has an energy of 1. Often we talk about the Compton wavelength bar
such that E=hbar*c/bar,
The real convenience of this notation is that when dealing with mass, momentum and
energy they all have the same values since c=1. Then it’s easy to see that given 2Tev of
momentum you can convert that into 2TeV on energy or particle with 2TeV of mass.
Typically you have to understand things in units of distance (or distance squared for a
cross section) and time(for lifetimes).
For conversion back to standard units useful numbers are
c = 197 MeVfm
hc = 1240 eVnm
= 6.582x10-22 MeVs
Consider a particle of rest mass m = 197 MeV
To get to standard units of energy divide by c2
m = 197 MeV/c2, so E = mc2 = (197 MeV/c2)*(c2) = 197MeV
To find the Compton wavelength bar = hbarc/E = 197Mevfm/197MeV = 1fm
Or for a photon:
for a photon with a wavelength of 1nm E = hc/ =1240eVnm/1nm = 1240eV
In particle physics we will typically leave the hbar and c out of our calculation
completely. The solution to a give calculation will involve dimensionless constants such
as  and factors of momentum/energy/mass which we will quote in units of
energy(MeV).
Example calculation and interpretation in terms of lifetimes in seconds or cross sections
in fm2.
To interpret you will have to determine what factors of
197 MeVfm you need to multiply by.
= 6.582x10-22 MeVs and
For instance the muon lifetime calculation results in an answer of:
muon decay:  ~ 192/W2m5 ~ 192mW4/2m5 =
192*(81000MeV)4*1372/(105.7MeV)5 = 3.7x1015(1/MeV) 
c=
3.7x1015(1/MeV) * 6.582x10-22 MeVs = 2.0x10-6 sec
3) Relativity
Typically in particle physics we don’t use Lorentz transformations and concentrate on
momentum and energy conservation and invariant quantities. Though sometimes we
have to deal with time dialation effects for particle velocities (for TOF detectors) and
lifetimes (for b hadron Id).
Time dilatation and lifetimes. Particles decay according to an exponential lifetime
distributions, which can be characterized by a lifetime parameter. For a relativistic
particle this lifetime, proper lifetime, is in it’s rest frame and the lifetime can be dilated in
other frames.
t = , where  = 1/(1-v2)1/2
Relativistic energy and momentum are going to be
p = mv
E = m
note v = p/E
The invariant quantity associated with momentum and energy is mass
E2 – p2 = m2
The mass of a particle is the same in all frames.
If we don’t take c=1
E2 – p2c2= m2c4
Note that if energy is measured in GeV then to make the units work out
Mass is GeV/c2
Momentum in GeV/c
To solve problems.
Choose a frame. CM is often best unless you are asked for a result in the rest frame. You
prefer not to do a Lorentz transform at the end because the quantities you measure are
momentum, energy and mass, not velocity. The velocity may not be known so avoid
solving the problem in the wrong frame and trying to convert at the end.
Write down the energy and momentum four vectors before and after. Simplify to a two
vector if possible if the momentum can all be in the x direction. Use these to write down
the conservation of momentum and energy equations.
Write down the invariant quantities. Here it may be useful to consider other frames.
Example relativity problem
A pion decays into an electron and an antineutrino with masses m, me and m. Calculate
the velocity of the antineutrino in the rest frame of the electron purely in terms of the
masses of the three particles. Interpret you answer in the case where m is 0.
Let the 4 momentum of the particles be p, pe and p
The from energy momentum and conservation invariant quantities we have:
(pe + p)2 = p2 = m2
where these are the four vectors. i.e. with four momentum in terms of energy and 3
momentum p42 = E2 - p32 = m2
In the rest from of the electron, reducing to a two vector by setting the momentum to be
in the x direction, the “four” momentum are:
pe = (me,0) and p = (E,px)
and
(pe + p)2 = me2 + m2 + 2meE = m2
again four vectors
E = (m2 - me2 - m2)/2 me
and with v = px/ E and px = (E2 - m2)1/2
v = (E2 - m2)1/2/ E
v = (1 - m2/ E2)1/2
v = (1 - 4 me2 m2/(m2 - me2 - m2)2) ½
4) Symmetries
Many symmetries we observe in nature are associated with conservation laws.
Some Examples:
Time invariance: Conservation of energy
Translation invariance: Momentum
Rotational Invariance: Angular Momentum
Gauge Transformation: Charge
Time invariance and energy conservation is interwoven in the way we treat quantum
mechanics using a Hamiltonian or more typically a Lagrangian. The Schrödinger
equation contains this explicitly (with a time independent potential).
-
d 2 Y(x,t)
dY(x,t)
+ U(x)Y(x,t) = i
2
2m dx
dt
2
HY(x,t) = i
dY(x,t)
= EY(x,t)
dt
Solutions of the Schrödinger equation
Y(x,t) = y(x)e iwt
Has a constant as a function of time, or conserved property  that we identified with the
energy: E = w
This example also illustrates another point. For each conserved property you should be
able to write down an operator, in this case H, the Hamiltonian. The wave equation
solutions will be eigenstates of this operator with an eigenvalue representing the value of
the conserved property for that eigenstate.
Next consider a gauge transformation invariance.
Gauge transformation:   ei and anti-particle y , e-i y
If gauge invariance is a symmetry that leaves interactions in nature unchanged then it
should leave both probabilities and the Lagrangian unchanged. Probabilities are
calculated using the wave functions as y y , or yy for an antiparticle (think bra-ket like
notation), are clearly unchanged.
Consider a relativistic Lagrangian chosen so that the Euler Lagrange equations reproduce
the Klein Gordon equation, which is the simplest quantinization of the relativistic energy
and momentum relationship.
( )
L= ¶my (¶ my ) + m 2yy
This is clearly left unchanged by the gauge transformation. We will find that the
conserved property associated with the gauge transformation is the charge. Each of the
forces has it’s own gauge group of transformation and charges.
We will demonstrate conservation of charge by considering the effect of the gauge
transformation on the Euler Lagrange equations.
5) Symmetries, group theory
Often these kinds of symmetries are categorized by group theory. The gauge
transformation is a unitary transformation in one dimension, U(1). The transformation
can also be represented as a matrix, just a constant in one dimension. There are many
group transformations we encounter in particle physics
U(n) Unitary transformation in n dimensions, nxn matrix, UU+=I, U+ ajoint or transpose
conjugate.
SU(n) Special unitary transformation in n dimensions, nxn matrix with determinant 1.
O(n) Orthogonal transformation in n dimensions, nxn matrix. Os transpose and inverse
are the same.
SO(n) Special orthogonal transformation in n dimensions, nxn real matrix with
determinant 1.
Examples of these in nature
U(1) Gauge symmetry and conservation of charge
SU(2) spin and isospin symmetry and conservation
SO(3) rotational symmetry and conservation of angular momentum.
The operations involving SO(3) and SU(2) matrices are the same except for a minus sign.
Thus the same in probabilities, which is why we combine them later.