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Final Exam (129A), Dec 13, 5–8 pm
- Explain the branching fractions of the W -boson. [15]
The W -boson couples to anything with weak isospin, namely all left-handed fermions in the standard model. However, the top quark is too heavy to be pro- duced in the decay of the W -boson. Therefore, the possibilities are:
W −^ → e−^ ν¯e, μ−^ ν¯μ, τ −^ ¯ντ , d′^ ¯u, s′¯c. (1)
The quarks come in three colors. Of course, d′^ and s′^ contain all three genera- tions of down quarks d, s, b. But their masses are negligible compared to the W m^2 b /m^2 W 1, and hence we can ignore difference among them. Therefore we can treat d′^ and s′^ “mass eigenstates” of zero mass. Ignoring all masses of fermions relative to the W boson, there are nine final states, while the quark final states are enhanced due to the additional emission of a gluon by a factor of (1 + αs/π). Therefore,
BR(W −^ → e−^ ν¯e) = BR(W −^ → μ−^ ν¯μ) = BR(W −^ → τ −^ ¯ντ )
=
3 + 6(1 + αs/π)
BR(W −^ → hadrons) =
6(1 + αs/π) 3 + 6(1 + αs/π)
I used αs(mZ ) = 0.117. They agree with the data within the error bars.
- A very high-energy cosmic ray proton would absorb the cosmic- microwave background photons to become a Delta resonance, which quickly decays into a nucleon and a pion, thereby losing its energy. Assuming all CMBR photons have the energy E = kT , what is the maximum proton energy for this not to happen? (There is a puzzling report that we see cosmic rays above this so-called GZK cutoff.) [10]
The temperature of CMBR photons is T 0 = 2.725K. This translates to the energy of the photon E = kT 0 = (8. 617 × 10 −^5 )eV/K × 2 .725K = 2. 348 × 10 −^4 eV. The center-of-momentum energy of the photon-proton collision is the greatest when the collision is head-on. It can be calculated as
E CM^2 = (pp + pγ )^2 = (Ep + Eγ )^2 − (Epβp − Eγ )^2 = m^2 p + Ep(1 + βp)Eγ. (4)
Because mp Ep and βp ≈ 1 in this extreme situation, we find
ECM =
√ 2 EpEγ < m∆ = 1232MeV. (5)
Figure 1: Data from AGASA and Hi-Res experiments on ultra-high-energy cosmic rays. Taken from http://arXiv.org/abs/hep-ex/0208024.
Therefore, Ep < m^2 ∆/(2Eγ ) = 3. 2 × 1021 eV. Right now, two experiments have probed cosmic rays in this energy range, AGASA in Japan and Hi-Res in Utah. They seem to diagree with each other. Note that we are talking about the event rate of one per square kilometer per century per steradian.
- KamLAND experiment reported 54 reactor anti-electron-neutrino event for 86. 8 ± 5 .6 events (5.6 is the systematic error) expected without neutrino oscillation. Assume all reactors are at the distance of 180 km and the neutrino energy of Eν ∼ 3 MeV. (1) What is the reaction used to detect reactor anti-electro-neutrino? [5] (2) If ∆m^2 is relatively high and the oscillation is averaged out, what is the preferred value of sin^2 2 θ with statistical and systematic errors? [5] (3) In order for a sizable oscillation effect to be present as observed, estimate what the minimum value of ∆m^2 is in eV^2. [5]
The key equation is the neutrino survival probability
Psurv = 1 − sin^2 2 θ sin^2
∆m^2 4 E
L = 1 − sin^2 2 θ sin^2
(
- 27
∆m^2 eV^2
GeV E
L
km
)
. (6)
(1) The reaction used for the detection is ¯νep → e+n, where the neutron is later captured np → dγ.
(2) If ∆m^2 is high and the oscillation is averaged out, we have Psurv = 1 − 1 2 sin
(^2 2) θ. The statistical error of the observed event is √54 = 7.3, and hence
may decay. Grand unified theories indeed predict it. List five possible decay modes of the proton consistent with all other conservation laws. [15]
For example, p → e+γ, e+π^0 , π+ν, K+ν, e+η.
- Using K resonances, identify states on the leading Regge trajectory [5], obtain the Regge slope [5], and the force between the quark and the anti-quark in Newton [5].
The Particle Data Group lists many mesons with S = ±1 and I = 1/2. As clear from the list, the subscript refers to the spin, while the asterisk disinguishes parity. Recall P = (−1)L+1, and hence higher L excitations would give you al- ternating parity. Because we are interested in the leading Regge trajectory, we look for the highest spin state at the given mass. The isospin splitting is ingored. They are: K(496)(0−), K∗(892)(1−), K 2 ∗ (1430)(2+), K 3 ∗ (1780)(3−), K 4 ∗ (2045)(4+), K∗ 5 (2380)(5−). (The last one is only in the big book, not in the booklet.) It is clear that the first one does not belong to this list because the parities do not alternate. Hence, we keep the last five. Indeed, they more-or-less fall on a straight line.
0 1 2 3 4 5 6 J
0
2
4
6
8
(^2) m (GeV)
Figure 3: The leading Regge trajectory for kaons.
The slope is approximately α′^ = (5 − 1)/(2. 3802 − 0. 8922 ) = 0.82 GeV−^2. The force between the quark and the anti-quark then is the tension T = 1/(4α′) = 0 .31 GeV^2 using Eq. (4) in “Strong Interactions II.” To convert it to the unit of Newton, we divide T by ¯hc = 0.197 GeV fm and find T = 1. 57 × 1015 GeV/m=
- 5 × 105 J/m= 2. 5 × 105 N. It is a very strong force acting on an elementary particle, as strong as the gravity on a mass of ten ton.
- Experiments at LEP-II e+e−^ collider searched for the Higgs boson, found a hint, but finished with a lower bound of 114.4 GeV (95% CL). Draw the Feynman diagram which could have been relevant for the Higgs boson production at LEP-II. [5]
Z
Z
e +
e –
h
Figure 4: The Feynman diagram of e+e−^ → Zh.
- Using the Friedmann equation
( (^) ˙ R R
) 2 = 83 π GN ρ, the first law of thermodynamics d(ρR^3 ) = −pd(R^3 ), and the equation of state p = wρ, determine the evolution of the scale parameter R(t) for Universe dominated by relativistic matter w = 1/3, non-relativistic matter w = 0, and the cosmological constant w = −1. Show that the Universe accelerates R >¨ 0 only for the last case. [15]
Using the first law of thermodynamics and the equation of state, we find
d(ρR^3 ) = −wρd(R^3 ), (7)
and hence dρ ρ
= −(1 + w)
d(R^3 ) R^3
and ρ ∝ R−3(1+w). (9)
This establishes why the “cosmological constant” corresponds to w = −1 which makes the energy density constant as the Universe expands. We now use the Friedmann equation, ( (^) ˙ R R
) 2
8 π 3
GN ρ ∝ R−3(1+w). (10)
Therefore, R˙ ∝ RR−3(1+w)/^2 = R−(1+3w)/^2. (11)
