IntroductiontoElementaryParticles Instructor’s Solution Manual

1 Introduction to Elementary ParticlesInstructor s Solution Manual29th August 2008 VAcknowledgments:I thank Robin Bjorkquist, who wrote and typeset many ofthe solutions in the first four chapters; Neelaksh Sadhoo, who typeset solu-tions from the first edition; and all those who sent me solutions or have tried to make every entry clear and accurate, but please: if you find anerror, let me know I will post errata on my web Set of Solutions for Introduction to Elementary Particles. David GriffithsCopyrightc 2005 WILEY-VCH Verlag GmbH & Co.

2 KGaA, WeinheimISBN: 3-527-XXXXX-XVIC ontents1 Historical Introduction to the Elementary Particles12 Elementary Particle Dynamics93 Relativistic Kinematics174 Symmetries375 Bound States576 The Feynman Calculus797 Quantum Electrodynamics978 Electrodynamics and Chromodynamics of Quarks1479 Weak Interactions17110 Gauge Theories209 ContentsVII11 Neutrino Oscillations23312 What s Next237 AThe Dirac Delta Function247 Partial Set of Solutions for Introduction to Elementary Particles. David GriffithsCopyrightc 2005 WILEY-VCH Verlag GmbH & Co.

3 KGaA, WeinheimISBN: 3-527-XXXXX-X11 Historical Introduction to the Elementary ParticlesProblem an undeflected charged particle,qE=qvB= v= just a magnetic field,qvB=mv2R= qm=vBR= 15m; h= 10 22 MeV s;c= 108m/s;som= h2r0c=( hc2r0)1c2= =138 by a factor of 15m; h= 10 22 MeV s;c= 108m/s;me= MeV/c2. x p h2sopmin= h2r0=( hc2r0)1c= p2minc2+m2ec4= energy of an electron emitted in the beta decay of tritium is < 17 Historical Introduction to the Elementary ParticlesProblem =13[2(mN+m ) m ].

4 MN= ;m = ;m = =13[2( ) ]= = by =13[2(m2K+m2K) m2 ]=13(4m2K m2 ).mK= ;m = =13[ 105]= 105 m = = by M =1232 1385= 153M M =1385 1533= : 151. M =M +151=1533+151=1684 =1672 by (a) n+ or +K0 + p+ K0; ++ 0; ++ ; 0+ +; + +; 0+K+ 0+K ; + K0; +K ; 0+ ; + 0; + 3(b)Kinematically allowed: n+ + ++ 0; 0+ +; + + 0+ ; + 0 Problem (a)With a strangeness of 3, the would have to go to( 0+K )or( + K0)to conserveSandQ. But the Kcombination is too heavy(at least 1808 MeV/c2, whereas the is predicted see Problem to have a mass of only 1684).

5 (b) cm;t=d/v=(5 10 3m)/(3 107m/s) =2 10 10s.(Actually,t= 10 10s.)Problem + = n+ 0 = + }3% lists a total of 30 meson types; in the first column is the particle name atthe time, in the second column the quoted mass (in MeV/c2), and in the thirdits current Historical Introduction to the Elementary Particlesmesonmassstatusexotic 138 K496KK31630deadyes 21340f0(1370)? 31275f1(1285)?K 1260K1(1270)?f1253f2(1270)?K 51150deadyes 11045a0(980)? 21040dead 11020 (1020) 5990deadyesK 1888K (892) 3885 (985)?

6 781 (782)mesonmassstatusexotic 755 (770) 2780f0(600)? 1720f0(600)? 4760deadyesK 1730dead 645dead 625dead 3597deadyes 556dead 549 2520dead 2440deadyes 1395dead 1330deadyes ABC317deadProblem the last column in Problem I count 7 exotic species, all of them nowdead. Of the surviving particles (of course) none is quark (u):one(u u); 2 quarks (u,d):four(u u,u d,d u,d d);3 quarks (u,d,s):nine; 4 quarks (u,d,s,c):sixteen;5 quarks (u,d,s,c,b):twenty-five; 6 quarks (u,d,s,c,b,t) general formula fornflavors quark (u)= 1baryon (uuu);2 quarks (u,d)= 4baryons (uuu,uud,udd,ddd);3 quarks (u,d,s)= 10baryons (baryon decuplet).

7 5 Fornquarks, we can haveall three quarks the same :nwaystwo the same, one different :n(n 1)waysall three different :n(n 1)(n 2)/6 ways.[For the third type of combination, divide by six to cover the equivalent per-mutations (uds=usd=dus=dsu=sud=sdu).]So the total isn+n(n 1)+n(n 1)(n 2)/6=n+n2 n+n(n 1)(n 2)/6=n6[6n+(n 1)(n 2)]=n6(6n+n2 3n+2)=n6(n2+3n+2)=n(n+1)(n+2) for four quarks we have20baryon types, for five quarks,35, and for sixquarks, hasC=3uddddcsccdddusc3 haveC=2uusdscudssscdds6 haveC=1ussdsssss10 haveC=061 Historical Introduction to the Elementary ParticlesProblem uc uu cc cu dc sd cd uc ds cd d3 haveC=1 3 haveC= 1u ss ud ss ds s10 haveC=0(includingc c)Problem qmesonmassyearu u 0(*) d + d 0(*) sK+ s (*)

8 DD+ qmesonmassyearc sD+ c c(1S) bB+ bB+c62861998b b (1S) masses are in MeV/c2; (*) indicates that the particle is a combination ofdifferent quark ++ 12321951uus + ++ + + ++ccdcc + +ccccc ++ccc7qqqbaryonmassyearuub + 0b5624?ddb 0b57921995dsb bucb +cbdcb 0cbscb 0cbccb +ccbqqqbaryonmassyearubb 0bbdbb bbsbb bbcbb 0ccbbbb bbbBlank spaces indicate that the particle has not yet been found (2008).Problem : 3 336 62=946; actually 939; : 2 336+540 62=1150; actually 1193; error: : 2 336+540 62=1150; actually 1116; : 336+2 540 62=1354; actually 1318; and leptons:ccc(q= 1):e c c c(q=1):e+ccn(q= 23): u c c n(q=23):ucnn(q= 13):d c n n(q=13): dnnn(q=0): e n n n(q=0): e(The neutrinos could be switched, but this seems the most natural as-signment.)

9 Mediators:ccc n n n(q= 1):W nnn c c c(q=+1):W+ccc c c c(q=0):Z0nnn n n n(q=0): }or vice versaGluons:We need matching triples of particles and antiparticles,81 Historical Introduction to the Elementary Particlesccn c c n(3 different orderings for each triple, so 9 possibilities);cnn c n n(3 different orderings for each triple, so 9 possibilities);leading to a total of 18 are at least four distinct answers, depending on the particle in question: Antiparticles (such as the positron)annihilatewith the correspondingparticle (the electron, in this case), and since there are lots of electronsin the lab, positrons don t stick around long enough to have any role inordinary chemical processes.

10 But if you could work in a total vacuumyou could make atoms and molecules of antimatter, and all of chemistrywould proceed just the same as with ordinary matter. Most elementary particles (such as muons, pions, and intermediate vec-tor bosons) are intrinsicallyunstable; they disintegrate spontaneouslyin a tiny fraction of a second not long enough to do any serious chem-istry. You can make short-lived exotic atoms , with (say) muons in orbitaround the nucleus instead of electrons. Some of these systems last longenough to do spectroscopy.