HST rate CC up to z=0.7 and SNe Ia up to z=1.6 42 SNe (!) - 17 CC 0.31 (consistent with Type Ia progenitor model) past measurements data from GOODS survey (ACS) 2 fields, 150 arcmn2/field, up to m(F850LP) 25.9, 5 epochs, 45 days redshifts: 29/42 have spectro redshifts photo-z for 12 1 no host (visible) types: use all information (dont say how/what) 25 "consistent" with Ia, 17 CC 20 Ia confident 7 confident CC (include misclassification in systematics) simulation: generate SN K-cor following KGP CC (follow Richardson 2002) subtypes : Ibc, Ibc-bright, IIL, IIL-Bright, IIP, IIn all gaussian distrib with peak mag Fraction of subtypes from Dahlen et al. (1999) SN spectra= black body at 25000 K near explosion to 5000K late LC from Fillipenko 1997 Extinction model from Hatano et al 1998 Type Ia subtypes: 20% bright (-19.6 s=0.3) 64% normal (-19.3 s=0.45) 16% faint (-17.8, s=0.5) Kcor from 1994D (how does that do at z=1.6?) LC mean B-band LC Extinction model from Hatano et al. 1998 Results CC rates in 2 z bins 0.1-0.5 (6SNe) and 0.5-0.9 (10SNNe) 2+/-0.8 (stat)+07-1.8(syst) 4+/-1(stat)+2-2.6(syst) higher by factors 4 and 7 compared to local rate of Capellaro et al. (converted in volume) Ia rates in 4 z bins 0.21 consistent with declining factor 5 increase at z=1 consistent with SFR Relation SFR-SNR various models Systematics: they conclude that nothing matters Discussion they find rate consistent with SFR + increase then decrease z>1 referee D. Maoz --------paper II Contains more info bins 0.3 0.5 0.7 0.9 1.1 1.3 1.5 nsn 1 2 6 8 1 6 1 chosen 0.2-0.6 3 0.6-1.0 14 1.0-1.4 7 1.4-1.6 1 why 2 papers ? much better written - although verbose (main conclusion 4Gy for Ia) --------------------------------------------------- Rate recalculation (after the meeting) Exercise: Change binning from 0.2-0.6, 0.6-1.0, 1.0-1.4, 1.4-1.8 to 0.4-0.8, 0.8-1.2, 1.2-1.6 N(z)= SNR(z)*dt(z)*dv(z)/(1+z) 1) calculate dt(z) from their data dt(0.4)=3*1.4/0.69/4.32 =1.41 dt(0.8)=14x1.8/1.57/10.74 = 1.49 dt(1.2)=6*2.2/1.15/15.56 = 0.74 dt(1.6)=2*2.6/0.44/18.36 = 0.64 2) interpolate dt(z) at 0.6, 1.0, 1.4 dt(0.6)=1.45 dt(1.0)=1.12 dt(1.4)=0.69 3) compute dv at 0.6, 1.0, 1.4 and calculate SNR(z) SNR(0.6)=8x1.6/1.45/7.6 = 1.16 SNR(1.0)=9x2/1.12/13.43 = 1.20 SNR(1.4)=7x2.4/0.69/17.19 = 1.41 each number has about 35% statistical uncertainty Conclusions: compare to our rate of 0.52 at =0.55 = 0.58 +/-0.15 extrapolated at z=0.6 their rate of 1.16+/-0.41(stat) is about 1.5 sigma higher no sharp increase below z=1 no drop above z=1 continue increase if anything (1.16 at z=0.6, 1.20 at z= 1.0, 1.41 at z=1.4) Note however that this in not the correct way to analyse this data. R.Pain June 29, 04