Tests of b-decay theoryTests of l(pe) † lpe( )≡dlpe( )dpeªg22h7p3c3Mfi2F Z', pe( )S pe, pn( )Ef- Ee( )2pe2lpe( )F Z', pe( )pe2= K Ef- Ee( )lpe( )F Z', pe( )S pe, pn( )pe2= K Ef- Ee( )Kurie plot for “allowed”Kurie plot for “forbidden” -- Find endpoint energy/momentumPlot --Tests of l(pe)• If a (allowed) Kurie plot is linear then --– The transition (decay) is “allowed” which impliesis a constant; does not depend on kb and/or kn.– The extrapolated x-axis crossing (y=0) is at Ee = Efknown as the endpoint energy.– The slope of the line is --† Mfi2 † g22h7p3c3Mfi2† Mfi2Weak interactioncoupling constantNuclear matrixelementTwo unknowns ---c.f., Fig. 9.4Tests of l(pe)• If a (allowed) Kurie plot is not linear then --– The transition (decay) is not “allowed” which implies is a not constant; does depend on kb and/or kn. The decay must be “first forbidden” or“second forbidden” … (see next slide)– A functional correction can be applied to straighten theline and thereby represent the momentum dependence,called the “shape correction” -- (see slide 2)† Mfi2† Mfi2† S pe, pn( )c.f. Fig. 9.5Tests of l(pe)Lepton wavefunctions -- † jbªeir k e⋅r r V1 / 2 ª1V1 / 21+ ir k e⋅r r +r k e⋅r r ( )22+ ⋅ ⋅⋅È Î Í Í Í ˘ ˚ ˙ ˙ ˙ ª1V1 / 2 “Allowed term”“First forbidden term”“Second forbidden term”etc….Tests of l † l=lpe( )0pmaxÚdpe=dlpe( )dpe0pmaxÚdpel=g22h7p3c3Mfi2F Z', pe( )Ef- Ee( )2pe20pmaxÚdpef Z', po( )≡1me5c7F Z', pe( )Ef- Ee( )2pe20pmaxÚdpel=g2me5c72h7p3c3Mfi2f Z', po( )“Fermi integral”Can be calculated!c.f., Fig. 9.8Total decay rateTests of l † l=g2me5c42h7p3Mfi2f Z', po( )l=Ln2t1/ 2 ; Ln2t1/ 2=g2me5c42h7p3Mfi2f Z', po( )ft1/ 2= Ln2( )2h7p3g2me5c4Mfi2Comparative half-life-- or --ft-value- or (log10 ft) value† Mfi2Differences in ft1/2must be due todifferences inTests of lFor what nuclear decays might be ~the same?How would you know they were the same?† Mfi2Log ft would be ~same ÆÆ If can be estimated then g can be determined!!† Mfi20+ Æ 0+ decaysc.f., Table 9.2Neutrino mass (Why?)† Q = MPN- MDN- me- mn( )c2mn> 0 Æ Qwill decrease!† Q = TD+ Te+ TnTe= Q - TD- TnTemax= Q - TD ; Tn= 0mn> 0 Æ Temaxwill decrease!Limits on mn are ~few eVNeed to know everything to aprecision/accuracy £ ~few eVpDpe † lpe( )≡dlpe( )dpeªg22h7p3c3Mfi2F Z', pe( )S pe, pn( )Ef- Ee( )2pe2Need to know nuclear wavefunctions!!† 3H Æ 3He +b-+n