PHY 557 Midterm Eexamination Fall 1999

  1. Short Answer Problems:   (5 points each)
    1. List all known quarks (not antiquarks), their approximate masses, charges (in terms of e); and their isospin, strangeness, etc. quantum numbers.

      See Particle Data Booklet, page 24.

    2. Calculate the lifetime (in seconds) of the r+ vector meson, which decays into p+p0, and has a width G=150 MeV. Is this decay mediated by the EM, Weak, or Strong force?

      t = 1/G = 6.67×10-3 MeV-1 = (multiply by h=6.58×10-22 MeV·s) = 4.39×10-24 s

    3. What, specifically, is responsible for the short range and weakness of the "weak force"? Compare the propagators descibing the exchange of the quanta of the EM and the Weak forces and explain.

      The range of a force and the propagator are linked to one another by a simple Fourier Transformation. The W boson propagator is (in the limit for mass MW»q²) proportional to 1/(q²+M²) » 1/M²; i.e. the propagator's q-dependence is masked by the large W mass. While the coupling constant gW = e/sinqW » 2e is not very different from e, it is the mass MW which is responsible for the weakness of the weak force.
      The Fourier transform of the propagator leads to a potential that has a form proportional to e-MWr/r, i.e. has a range given by the inverse W mass: ravg » 1/80 GeV = 2.5×10-3 fm.

    4. Give the value of the isospin I, and its third component I3, and the quark composition of the K+ meson. Show that the formula I3 = Q - ½(B+S+C+T) works in this case.

      The K+ and its partner the K0 form an Isospin doublet. This is because, while the s-bar quark has I=0 and S=+1, the u quark has I,I3=½,+½ and S=0. For the K+ meson the above formula for the third component of Isospin gives:

      I3 = Q - ½(B+S+C+T) = +1 - ½(0+1+0+0) = +½,
      which proves the case.

    5. Explain why quarks never appear as free particles; what particular aspect of the color interaction is responsible for this?

      The inter-quark color force grows (about linearly) with distance; thus, when trying to pry two quarks apart, a growing energy density develops between the quarks, which, at some point, is energetic enough to spontaneously produce a quark-antiquark pair. The new quarks pair-off with the original quarks. In the end, a string of quark-antiquark pairs and (anti)quark triplets develops along the pulling direction, and expends any energy that is put into trying to separate two quarks into creating new quarks (bound, color-neutral pairs and triplets).

    6. The experiment by Wu et al. showed that electrons in the decay of polarized 60CoÝ ® 60Ni + e- + anti-ne (where the arrow Ý indicates the J=5 polarization direction of 60Co) are emitted preferentially anti-parallel to the polarization direction of 60Co. Show that this result implies the violation of Parity invariance, earlier postulated by Yang and Lee.

      The Parity operation on the experiment of Wu et al. would result in an experiment that gives off electrons that are preferentially emitted parallel to the magnetic field (direction of 60Co spin), and Parity invariance therefore requires a symmetric emission pattern with respect to the plane perpendicular to the magnetic field. This is not observed, and thus parity invariance is not a symmetry of the weak Lagrangian!

    7. Indicate which interactions, - EM, Weak, and Strong - satisfy the experimentally observed conservation laws listed below. Also, indicate which conservation laws invariances are known to be following from local or global symmetries of the Lagrangian describing the interaction, and the conservation laws that do not seem to be related to any known symmetry.
      Q (charge cons.)S, EM, Weak                            
      Le(epton number) S, EM, Weak no known corresponding symmetry
      B(aryon number) S, EM, Weak no known corresponding symmetry
      C(harge cojugation inv.) S, EM  
      P(arity conservation) S, EM  
      T(ime invariance) S, EM  
      I(sospin inv.) S (EM preserves I3 only)  
      S(trangeness cons.) S, EM no known corresponding symmetry
    8. Show that the phase-space element dp/E is a Lorentz scalar; i.e. invariant under Lorentz transformations. Note:
      æ
      è      		 E'
      p'      		 ö
      ø      		 = æ   g     -bg ö
      è -bg     g   ø      		 × æ
      è      		 E
      p      		 ö
      ø

      see HW; use dE = dÖ(p²+m²) = pdp/E:
      dp' = d(gp - bgE) = gdp - bgdE = gdp - bgpdp/E = dp(gE - bgp)/E = dp(E')/E; q.e.d.

    9. The Dirac equation leads to four independent solutions (which are representation-dependent). What is their physical interpretation?

      They represent spin-½ particles (spin up and down), and spin-½ anti-particles (spin down and up).

    10. Give the helicity operator in the Dirac representation.

      See HW:

      æ½s·p/|p|     0     ö
      è     0     ½s·p/|p|ø

     

     

     

     

     

     

  2. Consider the annihilation of electron and positron as a pure first-order QED process:
    e- + e+ ® m- + m+
    In the following, you may safely ignore all masses in view of the large energies of the particles.
    1. Draw the Feynman diagram(s), and label/number the in- and outgoing particle lines. (7 points)

    2. In the center of mass of the e-e+ system, show that the magnitudes of the initial and final momenta pi* and pf* simply equal ½Ös. (7 points)

    3. The "Lepton Tensor" for the electron (1®3) can be calculated, using trace properties of the Dirac gamma matrices, as:
      L(e)mn = 2p1mp3n + 2p3mp1n + gmnq²,
      and similarly for the muon tensor (2®4). Calculate the matrix element squared and averaged, as given in d) below, from:     (10 points)
      |Mfi|² (e-e+® m-m+) = (e/q)4 L(e)mn Lmn(m)

    4. Given the matrix element squared (and averaged) for e-m-®e-m-:
      |Mfi|² (e-m-®e-m-) = (e/q)4 [8(p1·p2)² + 8(p1·p4)²] ,
      show that the matrix element squared and averaged for e-e+®m-m+ equals:     (8 points)
      |Mfi|² (e-e+® m-m+) = (e²/s)2 [8(p1·p3)² + 8(p1·p4)²] .

    5. Given that ds/dW* = (1/64p²s)|Mfi|² (ignoring masses), calculate the value of stotal(e-e+®m-m+) at center-of-mass energy Ös = 20 GeV.     (10 points)
      Note: a º e²/4p = 1/137.

    6. How would the result above be changed for the annihilation of a red up-type quark (uR) with a anti-red up-bar quark (uR), which carry momentum fractions x1 and x2 respectively, e.g. from a proton-antiproton interaction?     (8 points)
      Note: if the total center-of-mass energy squared of the proton-antiproton system (with p1 and p2) is s, then the center-of-mass energy squared of the quark-antiquark system is:
      (x1p1 + x2p2)² = x1x2s .

h = 6.582×10-22 MeV·s,     hc = 197.3 MeV·fm,     (hc)² = (197.3 MeV·fm)² = 0.389 GeV²·mb.
For a particle of mass m and energy-momentum fourvector (E,p), the Lorentz transformation to its rest-system contains the parameters b=p/E and g=E/m .