Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 1 Tests of Fast Timing Detectors in the Meson Test Beam (T979) MTest : May 27 th – June 2 nd Mike Albrow, Sasha Pronko, Erik Ramberg, Anatoly Ronzhin, Andriy Zatserklyaniy + detector simulations by Hans Wenzel & Earle Wilson (student) Motivations for ~ ps / 10 ps timing detectors Set-up and triggers etc. Detector configurations: A)A – B – C in line B)A+B transverse bars – C C) Q-bar1(A) – Q-bar2(B) – C D) Aerogel(C) – B E) Si-PMTs F) Photonis1 – Photonis2 – C in line G) Q-bar1(A) – Q-bar2(B) – 8.7 m flight path --- C in line Thoughts about next steps To be explained! Only if you ask!

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 2 Motivations: Timing on single particles σ(t) typically >~ 100 ps A factor 10 – 100 improvement likely to have unforeseen benefits. We know of some (foreseen), e.g.: Particle ID in beams E.g. at 25 GeV/c, over 15m: Δt(π-K) = 20 ps Δt(K-p) = 50 ps or at 10 GeV/c, over 30m: Δt(π-e) = 10 ps ================== Areas ~few cm2, want thin. Particle ID in large detectors (~CDF-like or ILC) E.g. at 6 GeV/c, over 1.5m: Δt(π-K) = 17 ps Δt(K-p) = 43 ps ================== Areas ~several m2, want thin. Pile-up reduction e.g. in FP420: Extensions to CMS & ATLAS in prepn. p + p  p + H + p + nothing else Measure p’s  M(H), J, C, P, Γ CMS H pp 240m … 420m Argonne-Chicago- (Henry Frisch et al.) Fermilab group PET-TOF  β+e  Δt = 10ps : Δ z = 3mm 4

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 3 to CMS Exec Board Summer 2008 (ATLAS is also reviewing) ? z(vertex) from pp == z(vertex) central Pile-up reduction in FP420 Want L ~ 10^34, ~ 25/x cf σ(z)vtx ~ 60 mm

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 4 At MTest, 120 GeV/c p, ~40,000/spill Simple trigger (schematic): 2mm x 2mm scint. VETO w/hole 2 PMTs in AND 2 PMTs in OR Calibrate electronics resolution with same pulse  start & stop: σ ~ 3 ps Cerenkov light in Quartz window. HV ~ 4.5 kV, G ~ 5.10^5 Dark & shielded box PHOTEK MCP, 10mm Φ PHOTEK MCP, 40mm Φ 210 First A-B-C in line MCP- PMT-A ORTEC 566, 567 TAC/SCA ORTEC AD114 ADC ATTENUATOR ADC ATTENUATOR MCP- PMT-B DAQ ADC C BA Schematic DAQ : T1

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 5 A-B-C in-line results: Cerenkov light in PMT windows All numbers “preliminary”, to be double-checked ADC distributions: cut out tails and stragglers (~ 10%) T1 = tA – tB T2 = tA – tC T3 = tB – tC ======= Check Ti(PH A,B) Make slewing corrections Unfold: A B C etc. PMT-1 (Photek-210, 4.7 kV)=12.0 ps PMT-2 (Photek-210, 4.6 kV)=12.0 ps PMT-3 (Photek-240, 4.2 kV)=7.7 ps Cerenkov light in PMT windows

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 6 Double Q-bar Quartz (fused silica) bars 6mm x 6mm x 90mm  PHOTEK 210 Mounted at Cherenkov angle θc ~ 48deg. on opposite sides. dz = 6mm/sin(48) = 8.1mm. Some light direct to PMT, ~1/2 TIR to PMT Black “sock” over bars just to avoid light sharing C B A Unfold: σ(A) = 22.3 ps σ(B) = 30.5 ps Includes electronics (~3 ps) and 2 mm beam width smear (A,B) Δt = 2 mm x (10 ps/2 mm) Combining [AB] removes beam spread (later, tracking)

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 7 σ = 6.04 ch = 18.7 ps Unfold C = 7.7 ps, σ(AB) = 17.0 ps Resolution of double-Qbar 2 mm x-spread not to be subtracted (only 3 ps electronics) Resolution of Double-Qbar as one device * * Derivation in back-up

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 8 Switched on, saw signals! A = Aerogel B Corrected T2 = A-B = 10.8 ch = 33.5 ps (before unfolding)

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 9 Aerogel results: Unfolding indirect because only 2 PMTs in run. A (Aerogel on 240) and B(210 in beam) T1 = t(A) – t(B) corrected for smearing: 10 mm aero  σ(T1) = 43.7 ps 20 mm aero  σ(T1) = 45.3 ps 30 mm aero  σ(T1) = 33.5 ps Unfold with σ(1) = 12 ps from in-line σ (Aerogel 30 mm) ~ 31 ps = 46 ch. (10mm)  72 ch. (30 mm) Aerogel + mirror ~ massless & short (~ 5 cm), simple. Can have several in line, independent  √N BUT: have large 240 tube close to beam Possibilities to focus light : smaller tube farther away, to be simulated AEROGEL MCP-PMT

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 10 Tests of SiPMs = silicon photomultipliers Eight Hamamatsu SiPMs, 3mm x 3mm In beam with quartz Cherenkov radiators several thicknesses (4 – 12mm), mirrored and not mirrored. Best conditions σ(t) ~ 33 – 37 ps photoelectrons Channels Between SiPMs and C. Slewing correction applied

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 11 World’s Best Beamline Time-of-Flight System? 24 psec resolution positron peak, Using average of A & B times Can measure momentum of a proton with 2 MCP-PMTs! (if you know it’s a proton!) Start = Double-Q-bar Stop = Photek 240 Start-stop dist. = 8.7 m Predictions of proton positions

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 12 Possible Next Steps For FP420 a σ(t) = 10 ps edgeless detector  we learnt a way including CMS-compatible electronics/DAQ with reference time signals (jitter <~ 5 ps) θc = 48deg Q-bars onto PHOTEK 240 MCP-PMTs 40 mm diam. MCP 6mm x 6mm bars TIR: isolated p 20mm I “beam” only 6mm vert., 20 mm horiz. MCP Should get < 8 ps + More aerogel? To test in Fall?

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 13 We thank: PHOTEK Ltd (UK) for loan of MCP-PMTs Accelerator Division Ops for beam, and patience over many accesses Jim Pinfold and Don Summers for gift of aerogel Carl Lindenmeyer and John Korienek for Q-bar support Rick Coleman for low energy beams Hogan Ngyuen & SiDet dept. for many things From FP420 R&D doc. LHC beam

Mike AlbrowAll Experimenters, June 8 th 2009T979: Tests of Precision Timing MTest 14 T1 = A - B T2 = A - C T3 = B - C T1 + T2 = A - B + A - C 3 x T3 = 3B - 3C so T1 + T2 + 3x T3 = A - B + A - C + 3B - 3C = 2A + 2B - 4C and 1/4 ( T1 + T2 + 3x T3 ) = (A+B)/2 - C Back Up Why: