Andre Schneider P621

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Priscilla Atkinson
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1 ndr Schnidr P61 Probl St #03 Novbr 6, Srdnicki 10.3 Vrtx for L 1 = gχϕ ϕ. Th vrtx factor is ig. ϕ ig χ ϕ igur 1: ynan diagra for L 1 = gχϕ ϕ. Srdnicki 11.1 a) Dcay rat for th raction ig igur : ynan diagra for th raction. Sytry factor is S =. Th total dcay rat is Γ= 1 S dγ (.1) whr dγ = 1 E 1 T dlips (k 1 ). (.) In th cntr of ass fra E 1 =, s =, which iply that k 1 = = 1 4 (.3) 1

2 and LIPS (k 1 )= k 1 16π s dω M = 1 3π 1 4 dω M. (.4) ro th ynan diagra ruls it = ig and th total dcay rat bcos ( ) 1/ Γ= g 1 4 8π. (.5) b) Dcay rat for th raction ϕ χ χ ϕ ig igur 3: ynan diagra for th raction ϕ χ. Sytry factor is S = 1. χ c) Now, ( ) 1/ Γ= g 1 4 χ 16π. (.6) ϕ 3 Srdnicki 11. a) In th T fra k =(, k ) and k =(, k ) (3.1) k γ =(, k γ ) and k γ =(, k γ) (3.) whr k = 0, k γ = and k γ =. Thus, s = (k γ + k ) =( + ) = + (3.3) u = (k k γ) =( ) = + (3.4) b) Sinc s + t + u = w hav t = s u =( ) (3.5) which cobind with t = + cos θ T cos θ T = 1 + ( 1 1 ) (3.6) c) Th fra indpndnt cross sction is dσ dt = 1 64πs k 1 T (3.7) M

3 whr th atrix lnt T, using th rsults for u and s fro (a), is T =3π α 4 + ( ) ( ( ) 4 ) () (3 6 + ) ( ( ) 4 ) ( ) + ( ) ( )( ) = 3π α ( + + )+ ( + ) ( + ) ( 1 = 3π α 1 ) ( ) + = 3π α (cos θ T 1) + (cos θ T 1) + = 3π α sin θ T +. (3.8) Using part (b) w hav t = (1 cos θ T ) or t =( ) (3.9) dt = d (1 cos θ T ) + d(cos θ T ) or dt =d. (3.10) and using 1 cos θ T = (1/ 1/ ) in th last lin lads to d = d(cos θ T) dt = d(cos θ T )= π dω T. (3.11) Now, w plug in th rsults (3.8), (3.11) toghtr with k 1 T = s k 1 M s k 1 M = (3.1) into (3.7) to obtain as xpctd. dσ = 3π α dω T 64π = α sin θ T + sin θ T + (3.13) 4 Srdnicki 11.3 a) Using th dfinition for dlips n (k) w hav that dlips 3 (k 1 ) = (π) 4 δ 4 (k 1 k 1 k k 3) dk 1 dk dk 3 = (π) 4 δ 4 (k 1 k 3 k 1 k ) dk 1 dk dk 3 = dlips (k 1 k 3) dk 3 (4.1) 3

4 which iplis that Γ= 3G = 3G = 3G (k 1 k )(k 1 k 3)dLIPS 3 (k 1 ) k 1µ k µ k 1 ν k 3ν dlips (k 1 k 3 ) dk 3 dk 3 k 1µk 3ν k µ k 1 ν dlips (k 1 k 3 ) (4.) b) With a dinsional analysis w can s that th units on th right hand sid hav to b k. lso, th rsult of th intgral has to b sytric undr intrchang of µ and ν indics. Thus, th solution can b writtn as a su of a tr that dpnds on µ and ν and a tr that dos not ix of diagonal trs, that is, proportional to g µν. c) Th intgral dlips (k) is Lorntz invariant and thus its valu is fra indpndnt. or siplicity w pick th M fra whr th rsult can b usd. Thus, 1 dlips (k) = 4(π) E 1 δ(e E 1 + E s)δ 3 (k 1 k )d 3 k 1d 3 k (4.3) dlips (k) = 1 16π δ(e 1 + E s)δ 3 (k 1 + k ) d 3 k 1d 3 k E 1 E = 1 δ(e1 k 0 ) 16π E 1 k 1 d k 1 dω M ( = 1 ) 1 d(e 1 k 0 ) = 1 4π de 1 8π E1=k 0 / whr w usd th fact that E 1 = E = k 1 sinc 1 = = 0. d) ontracting g µν and k µ k ν with th rsult fro part (b) w hav that g µν k 1 µ k ν dlips (k) =g µν g µν k + g µν k µ k ν k 1 µ k µ dlips (k) =δ µ µ k + k µ k µ (k 1 k )dlips (k) = (4 + )k (4.5) (4.4) and k µ k ν k 1 µ k ν dlips (k) =k µ k ν g µν k + k µ k ν k µ k ν k µ k 1 µ k ν k ν dlips (k) =k µ k µ k + k µ k µ k ν k ν (k 1 k)(k k)dlips (k) = ( + )k k. (4.6) Now subtracting rsult (4.6) fro k (4.5) w obtain k 3k k = (k 1 k ) (k 1 k)(k k) dlips (k). (4.7) Subtracting rsult (4.6) fro 4k (4.5) w obtain k 3k k = (k 1 k ) 4(k 1 k)(k k) dlips (k). (4.8) 4

5 Sinc dlips (k) δ 4 (k k 1 k ) w hav that k = k 1 + k { k = k 1 + k +k 1 k =k 1 k k = k 1 k + k k, (4.9) which in th M fra, that is, E 1 = E and k 1 = k = 0, bcos Now w can s that k 1 k = k / and k 1 k = k k = k /. (4.10) = 1 96π and = 1 48π. (4.11) ) Now that w hav th valus for and toghtr with th rsult fro (b) w hav that th dcay rat of th uon is Γ= 3G dk 3k 1µ k 3ν g µν (k 1 k 3) + (k 1 k 3) µ (k 1 k 3) ν. (4.1) In th approxiation whr 0, k 3 = k =(E, k ) k = 0, and considring th rst fra of th uon, k 1 =(, 0), th dcay rat bcos Γ = 3G d 3 k (π) 3 k 1µ k ν g µν (k 1 k ) + (k 1 k ) µ (k 1 k ) ν E µ=0 = 3G d 3 k E (π) 3 ( ( E ) + k )+( E )( E ( E ) k ) E = 3G d 3 k E (π) 3 ( ( E ) + E ) ( E )(E ( E )+E ) E = 3G d 3 k (π) 3 E (E ) E ( E ) E = 3G (π) 3 d 3 4E 3 k 96π = G 6π(π) 3 (3 4E ) k d k dω M = G 1π 3 (3 4E )E de. (4.13) Thus, ( dγ = G de 1π 3 E 3 4 E ). (4.14) whr th axiu lctron nrgy E happns whn in th M fra th lctron has ontu opposit to that of th two nutrinos, that is, whn k = E =3/4. f) Intgrating (4.13) fro zro to E ax Γ= G 1π 3 = G 5 1π 3 = / th total dcay rat is / 0 E 3 E4 3E 4 E3 / E =0 de = G 5 19π 3 (4.15) 5

6 g) If th lifti of th uon is τ 1/ = s and its ass is = GV thn Γ= τ 1 1/ GV and 19π 3 G = 5 Γ = GV. (4.16) h) Th nrgy spctru of th lctron is Plot of P (E ) Μ P E 4 1 dγ P (E ) Γ = 16 de E ( 3 4 E ) (4.17) E Μ igur 4: Plot of µ P (E E ) vrsus E /. Not hat vnthough w can plot th curv byond E /µ = 1/ th rsults in that rgion ar non-physical sinc th axiu nrgy th itd lctron can hav is E = /. 5 Srdnicki 11.4 Sinc th intraction tr is of th for g, only vrtics containing, and contribut to th scattring aplitud. Th diagras ar of th for a) igur 5: ynan diagras for L 1 = g. sinc no vrtics can happn for this procss. T =0 (5.1) 6

7 b) T =0. (5.) sinc only on vrtic can happn in this procss (th othr vrtx will always hav two s). c) T = g 1 t + 1 u sinc th only uniqu contributing diagras ar (5.3) igur 6: Possibl ynan diagras for th procss. d) T =0 (5.4) sinc on of th vrtics will always hav two lins or two lins or two lins. ) T = g 1 s + 1 t sinc th only uniqu contributing diagras ar (5.5) igur 7: Possibl ynan diagras for th procss. f) T =0 (5.6) sinc at last on of th vrtics will always hav two lins or two lins or two lins. 7

Calculus II (MAC )

Calculus II (MAC ) Calculus II (MAC232-2) Tst 2 (25/6/25) Nam (PRINT): Plas show your work. An answr with no work rcivs no crdit. You may us th back of a pag if you nd mor spac for a problm. You may not us any calculators.