Part I: Short Answer [50 points] For each of the following, give a short answer (2-3 sentences, or a formula). [5 points each]

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Soluions o Midrm Exam Nam: Paricl Physics Fall 0 Ocobr 6 0 Par I: Shor Answr [50 poins] For ach of h following giv a shor answr (- snncs or a formula) [5 poins ach] Explain qualiaivly (a) how w acclra paricls o high nrgy and (b) how w g hm o go in a circl Equaions ar nic bu no ncssary Elcromagnic forcs ar usd o acclra and guid paricls using h Lornz forc quaion Fq EvB Elcric fild acclra hm incrasing hir nrgy whil magnic filds guid hm in a circl Suppos I wr pracicing archry and I was rying o hi sphrical war balloons of radius R wih small arrows Wha would b h cross-scion for hs args? Th cross-scion is h ara as viwd from on dircion Sinc a sphr has h silhou of a circl i will hav cross-scion ara of R Which of h following quaions ar manifsly Lornz covarian? (a) p p pp - Ys (b) F F - No indics can b rpad down (c) pq pq - Ys (d) p m - Unmachd indx () - Can mach up wih down 4 Explain physically or using quaions wha i mans if your hory rspcs pariy If pariy is rspcd h laws of physics will look h sam if you s i in a mirror Mahmaically h asis way o impos pariy is o l xx 5 Explain brifly according o h Dirac hory why i is ha posiiv nrgy lcrons do no normally dcay and bcom ngaiv nrgy lcrons According o h Dirac hory all hs ngaiv nrgy sas ar alrady filld and hrfor h lcron can fall down o such a sa by h Pauli xclusion principl

6 Th unsabl Z-boson is lisd as having a mass of 99 GV This implis ha for a Z a rs H Z p0 E Z p 0 wih E 99 GV Explain why w know for sur ha in fac h Z-boson is no an ignsa of h full Hamilonian An xac ignsa volvs according o ie 0 If h Z-boson wr an xac ignsa of h full Hamilonian hn i would nvr dcay bu h problm says ha i is unsabl 7 In wak inracions you can g non-zro marix lmns of h form 0 W u q whr W - is h W-boson wih charg u is h up quark wih charg +/ and q is an ani-quark of som sor Tll m h charg of q and h charg of h corrsponding quark q Charg mus b consrvd so h sum of h chargs on h righ mus add o zro If w mak x h charg of h unknown ani-quark hn 0 x so x Th corrsponding quark will hav charg 8 Suppos you ar masuring h cross-scion for som procss and you g vry clos o h mass of an inrmdia paricl (rsonanc) Wha happns qualiaivly o h cross-scion? Whn you g clos o rsonanc hr will b a vry larg incras in h cross-scion 9 Whn you hav a rsonanc how can you ll from a graph of cross-scion vs nrgy wha h dcay ra or widh is for ha inrmdia paricl? Th widh is approximaly h full widh a half maximum of h rsonanc Tha is find h highs poin in h cross scion find h placs whr h curv is half h pak and ak h diffrnc in nrgis of hos wo placs Tha is (approximaly) 0 In h Fynman diagrams a righ h arrow is a frmion and h solid lin is a boson Would you add or subrac h conribuions o h Fynman ampliud and why? Bcaus h wo diagrams diffr by h xchang of an xrnal frmion lin you would subrac hm

Par II: Calculaion [50 poins] Each problm has is corrsponding poin valu markd Solv h quaions on spara papr [5] According o h paricl daa book h K + mson has a mass of 497 MV and a 8 man lifim of 8 0 s (a) If a K + mson had an nrgy of E = 50 MV how long would i las and how far would i go if i lass on avrag lifim? W sar wih h quaion E m v W firs s ha E 50 MV 476 m 497 MV W hn g h amoun of im i acually lass as 8 8 476 8 0 s 589 0 s W will nd h vlociy which w can find pry radily from v v v 476 09777 Disanc ravld is hn vlociy ims im which works ou o 8 8 d v 09777 5890 s 9980 m/s 7 m (b) Wha is h widh of a K + in V? Th widh is givn by 6 6580 V s 8 80 s 8 57 0 V (c) Th branching raio for K 0 0 is 07% Wha is K in V? Th parial widh can hn b compud from K BRK K 0 0 007 570 8 V 00 0 8 V

[5] Th diffrnial cross-scion for a high nrgis is givn by (a) Calcula h oal cross-scion d cos d 6E W simply ingra his ovr angls o yild d d d cos dcos cos cos d 6E 6E 8 8E 8E E cos 0 cos (b) Convr o barns if E = 00 GV Us 7 W hav 7 00 GV 097 GV fm b 00 fm 0 65 0 b 065 nb N (c) If an collidr is opraing wih ach bam a E = 00 GV a a luminosiy of L 567 b s how many pairs will i mak in on day? Th numbr producd is 0 6 Ld 650 b 567 b s 0 b/b 4 h 60 m/h 60 s/m 06 [0] A collision procss aks h form A p B p C p D p 4 whr h masss of h four paricls ar m m m and m 4 rspcivly Show ha p p4 can b wrin in rms of p p By consrvaion of four-momnum w know ha p p p p 4 Rarranging his slighly w hav p p p p 4 Squaring his xprssion w hav p p p p 4 4 4 4 4 pp4 m m pp m m4 pp4 pp m m4 m m p p p p p p p p m m p p m m p p

4 [0] Considr h procss p p k k in h cnr of mass fram L h mass of h s b m and h mass of h s b M (a) Assum h iniial paricls hav nrgy E in h cnr of mass fram Tll m h nrgy of h final paricls and h magniud of h momnum of h iniial and final paricls You mus giv argumns for your answrs no jus h answrs Sinc w ar in h cnr of mass fram h momna will b qual and opposi Sinc h iniial paricls hav qual momna and qual masss hy will hav qual nrgis E Th oal nrgy E will b dividd bwn h wo final paricls Sinc h final paricls hav maching momna and maching masss hy will also hav qual nrgis Hnc ach of hm gs nrgy E Th momna can b found by h fac ha E p m so w hav p p p E m k k k E M (b) Wri ou xplicily all four componns of all four momna You may assum h iniial paricls ar coming in along h x axs Th final paricls will go in an arbirary dircion W simply includ facors corrsponding o h various dircions and wri 00 sincos sinsin cos p E p k E k k k p E00 p k E ksin cos ksin sin kcos (c) Calcula all six do-producs of iniial and final momna xplicily i ll m p p k k p k p k p k p k Ths ar sraighforward: p p E p p k E kp p k E kp p k E kp p k E kp k k E k sin cos k sin sin k cos E k sin k cos E k cos cos cos cos

5 [5] A high nrgy ani-nurino m 0 wih nrgy E collids wih an lcron a rs ( m 05 MV ) (a) Find a simpl formula for h cnr of mass nrgy squard s in rms of E and m If w l p and p b h wo momna hn w hav s p p p p p p 0m m E W found h do produc by using h fac ha p E00 E p m 000 p p m E (b) How big mus E b o produc a W boson ( m 8040 GV ) via h procss W? W For h procss w nd s p M Solving for h nrgy w hav W W me sm 6 65 0 GV 65 PV 8040 GV 00005 GV sm MW m E m m 00005 GV I s no ha ofn w g o us PV 6 [5] Considr a rnormalizabl hory wih wo chargd spin 0 paricls wih charg + and wih charg + Thy ar no quivaln o hir ani-paricls and (a) Wri down all possibl rnormalizabl marix lmns of h form 0 X whr X has mor han wo paricls and figur ou which ons mus b ral To b rnormalizabl hr mus b no mor han four paricls Th oal charg mus b zro On way o do his is o hav xacly qual numbrs of wih and wih which yilds h following hr marix lmns: 0 0 0 W can us h Hrmiian and ani-paricl propry o show hs ar ral:

0 0 0 0 0 0 0 0 0 W can also ry o balanc charg by having say on xra and canclling i wih a bunch of s or w can do h sam hing wih an xra and canclling i wih s Th rsuling marix lmns w will nam as 0 g 0 h Ths marix lmns ar complx conjugas of ach ohr: g 0 0 0 h ` (b) Mak up a diagrammaic noaion for h paricls and Draw all possibl vrics for his hory and giv m h corrsponding facor ha should b includd for his hory A normal noaion would b a singl arrow for and a ripl arrow for bu ha s hard o draw so I ll us an opn arrow for and a closd arrow for Thr ar fiv marix lmns and hnc fiv possibl vrics in his hory as skchd a righ i i ig ig (c) Considr h scaring Draw h rlvan i Fynman diagram and giv m h rlvan Fynman ampliud Thn find h diffrnial and oal cross-scion raing all paricls as masslss if h nrgy of ach of h iniial paricls is E Thr is only on r-lvl diagram skchd a righ Th Fynman ampliud is i ig W would hn procd o h cross scion in h usual way ig p E ig g cm D i d d d 4 E k E p 8E 6 E 8E 6 E 56 E d g d 56 E p k W now ingra ovr angls bu noing ha hr ar idnical paricls in h final sa w nd o hrow in a facor of ½ o avoid doubl couning so w hav g 8 E

7 [5] Working in h full hory wih psudoscalar couplings skch all svn r-lvl diagrams for h procss p p k k k g Thn carfully wri h Fynman invarian ampliud for wo of 5 i hm: on which dos involv h coupling and on which dos no involv i You don hav o do h ohr diagrams nor do you hav o do anyhing wih h rsuling ampliuds Th Fynman ruls for h only wo allowd vrics ar givn abov Th svn rlvan diagrams ar skchd blow Only h las on involvs h coupling For h ohr diagrams h uppr frmion propagaor has momnum p ki whr k i is h momnum of h aachd o h uppr vrx and h lowr frmion propagaor has momnum kj p Th las diagram has a scalar propagaor wih momnum p p Puing i all oghr w find h Fynman ampliud is i i g j i pki m kj p m v k p m p k m u ig i v u 5 5 5 i j p p M Th sum ovr i and j is ovr all possibl pairs chosn from bu picking i j so six rms in all 8 [5] I is possibl (hough no likly) ha on of h dcays of h op quark will b p s b p s h p whr and b ar h op and boom quark (boh frmions) and h + is a chargd scalar Assum h ampliud for his procss aks h form i u i u 5 whr and ar ral consans Calcula h dcay ra bh Assum h op quark has mass m h boom quark has mass 0 and h scalar fild has mass m h Th ampliud can b simplifid so w sar by finding So h squar of h ampliud is i u i u 5

Tr i u i u u i u uu i u u i 5 5 5 5 Now h iniial op quark has random spin so w avrag ovr his spin For h boom quark w would sum ovr spins So w ar acually inrsd in s s 5 5 i Tr p m i p i Kping in mind ha p ani-commus wih 5 his can b rwrin as 5 5 Tr 5 i Tr p m p i i p m p s s Tr p p m p p p 0 p p Now from consrvaion of four-momnum w know ha p p p Rarranging his slighly w hav p p p Squaring his w hav p p p p p p p mh m 0 pp h p p m m Subsiuing his ino our prvious xprssion w hav i m m s s h W now procd o h dcay ra which is givn by D p p i d m mh d m m 6 E m cm s s Thr ar no complicaions involving h final sa paricls so w g a simpl facor of 4 Thr is howvr h complicaion ha w nd o figur ou wha h momnum of h final sa paricls is Bcaus w ar raing h boom as masslss his is h sam as h nrgy of h boom Sinc h op quark is a rs p m 000 so ha p p me b From our prvious argumns w hrfor hav me b m mh m mh Eb m As mniond his is h sam as h momnum so puing i all oghr w hav p m mh m mh 8m 6m