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Review: Prospects of detection of relic antineutrinos by resonant absorption in electron capturing nuclei. J D Vergados & Yu N Novikov, J. Phys. G: Nucl.

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Presentation on theme: "Review: Prospects of detection of relic antineutrinos by resonant absorption in electron capturing nuclei. J D Vergados & Yu N Novikov, J. Phys. G: Nucl."β€” Presentation transcript:

1 Review: Prospects of detection of relic antineutrinos by resonant absorption in electron capturing nuclei. J D Vergados & Yu N Novikov, J. Phys. G: Nucl. Part. Phys. 41 (2014) Kim, Hanbeom

2 Introduction The relic neutrino The cosmic neutrino background
Analogous to the cosmic background radiation Neutrino decoupling about 1 second after the Big Bang, 1010 K Very low average energy 𝐸 𝜈 β‰ˆ eV (corresponding to T = 1.95 K) (during the calculation, 𝐸 𝜈 β‰ˆ eV ) KIMS

3 Introduction Antineutrino absorption Ordinary electron capture
Exothermic Week interaction 𝜈 𝑒 + 𝑒 βˆ’ + 𝑝 + →𝑛 𝜈 𝑒 +(𝐴,𝑍, 𝑍 𝑒 )β†’ (𝐴,π‘βˆ’1, 𝑍 𝑒 βˆ’1) βˆ— Ordinary electron capture 𝑝 + + 𝑒 βˆ’ →𝑛+ 𝜈 𝑒 KIMS

4 The Formalism Cross section for a neutrino of given velocity Ο… 𝜎 𝐸 𝜈 =2πœ‹ 1 𝑣 𝑀𝐸 𝐸 π‘₯ nuc 2 πœ™ 𝑒 2 𝐺 𝐹 2 2 𝛿 𝐸 𝜈 + π‘š 𝜈 +Ξ”βˆ’ 𝐸 π‘₯ βˆ’π‘ 𝐸 𝜈 +𝑏β‰₯Ξ”+ π‘š 𝜈 π‘š 𝜈 β‰ˆ1 eV πœ™π‘’: electron wavefunction Ξ”: mass difference of the two neutral atoms 𝑏: electron binding energy 𝐸π‘₯: final state energy 𝐺𝐹: Fermi’s constant KIMS

5 The Formalism 𝑁 𝜈 𝐸 𝜈 = Ξ¦ 𝜈 𝜎 𝐸 𝜈 = 𝑛 𝜈 ( π‘Ÿ 0 )2πœ‹ 𝑀𝐸 𝐸 π‘₯ nuc 2 πœ™ 𝑒 2 𝐺 𝐹 2 2 𝛿 𝐸 𝜈 + π‘š 𝜈 +Ξ”βˆ’ 𝐸 π‘₯ βˆ’π‘ 𝑁= 𝑁 𝜈 𝐸 𝜈 𝑓 𝐸 𝜈 𝑑 𝐸 𝜈 =πœ‰ 𝑛 𝜈 2πœ‹ 𝑀𝐸 𝐸 π‘₯ nuc 2 πœ™ 𝑒 2 𝐺 𝐹 2 2 𝑓( 𝐸 π‘₯ +π‘βˆ’ π‘š 𝜈 βˆ’Ξ”) πœ‰= 𝑛 𝜈 ( π‘Ÿ 0 )/ 𝑛 𝜈 𝑓 𝐸 = 1 π‘˜ 𝑇 0 𝑒 βˆ’ 𝐸 π‘˜ 𝑇 0 , π‘˜ 𝑇 0 =1.5Γ— 10 βˆ’4 eV 𝑛 𝜈 ( π‘Ÿ 0 ): the density of neutrinos in our vicinity 𝑛 𝜈 : the relic neutrino density 𝑓(𝐸): neutrino energy distribution (gravitationally bound) KIMS

6 The Formalism Neutrino capture
𝑁= 𝑁 𝜈 𝐸 𝜈 𝑓 𝐸 𝜈 𝑑 𝐸 𝜈 =2πœ‹πœ‰ 𝑀𝐸 𝐸 π‘₯ nuc 𝐺 𝐹 πœ™ 𝑒 π‘˜ 𝑇 0 𝑒 βˆ’ 𝐸 π‘₯ +π‘βˆ’ π‘š 𝜈 βˆ’Ξ” π‘˜ 𝑇 𝑛 𝜈 Electron capture with the final state energy 𝐸 π‘₯ β€² 𝑁 π‘’βˆ’capture = 1 2πœ‹ 𝑀𝐸 𝐸 π‘₯ β€² nuc 2 πœ™ 𝑒 𝐺 𝐹 (Ξ”βˆ’ 𝐸 π‘₯ β€² βˆ’π‘) 2 πœ† π‘Ž πœ† 𝑐 = 𝑁 𝑁 π‘’βˆ’capture = 2πœ‹ 𝑀𝐸 𝐸 π‘₯ nuc 2 𝑀𝐸 𝐸 π‘₯ β€² nuc 2 πœ‰ 𝑛 𝜈 (Ξ”βˆ’ 𝐸 π‘₯ β€² βˆ’π‘) 2 π‘˜ 𝑇 0 𝑒 βˆ’ 𝐸 π‘₯ +π‘βˆ’ π‘š 𝜈 βˆ’Ξ” π‘˜ 𝑇 0 KIMS

7 The Formalism Let πœ–= 𝐸 π‘₯ +π‘βˆ’ π‘š 𝜈 βˆ’Ξ”
𝑁 𝑁 π‘’βˆ’capture = 2πœ‹ 𝑀𝐸 𝐸 π‘₯ nuc 2 𝑀𝐸 𝐸 π‘₯ β€² nuc 2 πœ‰ 𝑛 𝜈 (Ξ”βˆ’ 𝐸 π‘₯ β€² βˆ’π‘) 2 π‘˜ 𝑇 0 𝑒 βˆ’ πœ– π‘˜ 𝑇 0 Choose 𝐸 π‘₯ β€²β‰  𝐸 π‘₯ since 𝐸 π‘₯ >βˆ†βˆ’π‘+ π‘š 𝜈 , 𝐸 π‘₯ <βˆ†βˆ’π‘ Given a very fine setting: πœ–β‰ˆπ‘˜ 𝑇 0 β‰ˆ 10 βˆ’3 eV,Ξ”βˆ’ 𝐸 π‘₯ β€² βˆ’π‘=100 keV and known parameters: 𝑛 𝜈 β‰ˆ56 cm βˆ’3 & assuming nuclear matrix elements of the same order 𝑁 𝑁 π‘’βˆ’capture =0.4Γ— 10 βˆ’11 πœ‰ KIMS

8 The Formalism The Uncertainty Principle
Suppose that there is a resonance in the final nucleus at an energy πœ– above the value Ξ”βˆ’ 𝑏+ π‘š 𝜈 with a width Ξ“= πœ– 1+𝛿 , 𝛿β‰ͺ1 𝛿 𝐸 𝜈 + π‘š 𝜈 +Ξ”βˆ’ 𝐸 π‘₯ βˆ’π‘ β†’ 2 πœ‹ Ξ“ ( 𝐸 π‘₯ βˆ’(Ξ”βˆ’π‘+πœ–+ π‘š 𝜈 )) 2 + ( Ξ“ 2 ) 2 Integrate from 𝐸 π‘₯ =Ξ”βˆ’π‘+ π‘š 𝜈 to 𝐸 π‘₯ =Ξ”βˆ’π‘+ π‘š 𝜈 +Ξ“ KIMS

9 The Formalism Antineutrino capture
𝑁=2πœ‹πœ‰ 𝑀𝐸 𝐸 π‘₯ nuc 𝐺 𝐹 πœ™ 𝑒 π‘˜ 𝑇 0 𝑒 βˆ’ 𝐸 π‘₯ +π‘βˆ’ π‘š 𝜈 βˆ’Ξ” π‘˜ 𝑇 𝑛 𝜈 β†’2πœ‹πœ‰ 𝑀𝐸 𝐸 π‘₯ nuc 𝐺 𝐹 πœ™ 𝑒 𝑛 𝜈 1 π‘˜ 𝑇 0 𝐾 𝛽,𝛿 𝐾 𝛽,𝛿 =βˆ’ 1 πœ‹ 𝑖 𝑒 βˆ’ 1 2 𝑖 𝛿+1 π›½βˆ’π›½ βˆ’ 𝐸 1 1βˆ’ 𝑖 2 𝛿+1 𝛽 + 𝑒 𝑖 𝛿+1 𝛽 𝐸 1 1βˆ’ 𝑖 2 𝛿+1 𝛽 βˆ’ 𝐸 𝑖 𝛿+ 1+2𝑖 𝛽 + 𝐸 𝛿+1 βˆ’π‘–βˆ’ 2 𝛿+1 𝛽 + 𝑒 𝑖 𝛿+1 𝛽 𝐸 1 1βˆ’ 𝑖 2 𝛿+1 𝛽 βˆ’ 𝐸 𝑖 𝛿+ 1+2𝑖 𝛽 βˆ’ 𝐸 1 1βˆ’ 𝑖 2 𝛿+1 𝛽 + 𝑒 𝑖 𝛿+1 𝛽 𝐸 1 1βˆ’ 𝑖 2 𝛿+1 𝛽 βˆ’ 𝐸 𝑖 𝛿+ 1+2𝑖 𝛽 + 𝐸 𝛿+1 βˆ’π‘–βˆ’ 2 𝛿+1 𝛽 𝐸 γ…œ = 1 ∞ 𝑒 βˆ’π‘§π‘‘ 𝑑 𝑛 𝑑𝑑 ,𝛽= πœ– π‘˜ 𝑇 0 KIMS

10 The Formalism Electron capture
𝑁 π‘’βˆ’capture = 1 2πœ‹ 𝑀𝐸 𝐸 π‘₯ β€² nuc 2 πœ™ 𝑒 𝐺 𝐹 (Ξ”βˆ’ 𝐸 π‘₯ β€² βˆ’π‘) 2 β†’ 1 2πœ‹ 𝑀𝐸 𝐸 π‘₯ β€² nuc 2 πœ™ 𝑒 𝐺 𝐹 πœ– 0 2 Ξ›( πœ– πœ– 0 ,𝛿) Ξ› πœ– πœ– 0 ,𝛿 = 4 πœ‹ πœ– πœ– 𝛿 𝛿+1 + tan βˆ’ 𝛿+1 βˆ’ tan βˆ’ β‰ˆ 4 πœ‹ πœ– πœ– 𝛿(1βˆ’ 𝛿) KIMS

11 The Formalism KIMS

12 The Formalism 𝑁 𝑁 π‘’βˆ’capture = (2πœ‹) 3 πœ‰ 𝑛 𝜈 πœ– π‘˜ 𝑇 0 𝐾 𝛽,𝛿 Ξ› πœ– πœ– 0 ,𝛿 Average energy available for de- excitation after ordinary π‘’βˆ’capture= Ξ”βˆ’π‘+ πœ– , πœ– =πœ– 4 πœ‹ 3 5 𝛿(1βˆ’ 𝛿) KIMS

13 Some Results π‘˜ 𝑇 0 β‰ˆ 10 βˆ’3 eV, 𝑛 𝜈 β‰ˆ56 cm βˆ’3 β†’ 𝑛 𝜈 πœ– π‘˜ 𝑇 0 =1.6Γ— 10 βˆ’13 Ξ΅/keV Ξ΄ 0.02 0.04 0.06 0.08 0.10 0.4 N/N(e-capture)/10-17ΞΎ 10.9 5.68 3.92 3.04 2.56 0.1 6.96 3.60 2.52 1.96 1.64 0.05 5.56 2.88 2.00 1.56 1.32 KIMS

14 Some Results (7.7Γ— 10 βˆ’22 , 5.8Γ— 10 βˆ’23 , 1.4Γ— 10 βˆ’23 ) obtained for Q=2.3, 2.5, 2.8 keV respectively for the target 163Ho L. Lusignoli and M. Vignati, Phys. Lett. B 697, 11 (2011), arXiv:1012/0760 (hep-ph) 6.6Γ— 10 βˆ’24 for tritium A. Cocco, G. Magnamo, and M. Messina, JCAP 0706, 015 (2007), ; J. Phys. Conf. Ser (2008) 08214, arXiv:hep/ph/ However…… KIMS

15 Specific Example – 157TB 157Tb (71 y)β†’ 157Gd (g.s.) & 157Gd (54 keV)
𝑄 𝐸𝐢 =60.1 βˆ’62.9 keV (Ξ”), 𝑏 𝐾 =50.24 keV, 𝑏 𝐿1 =8.38 keV , 𝑏 𝐿2 =7.93 keV , 𝑏 𝐾 = 7.24 keV Assume 𝑏 𝐿 =8 keV Only L-capture to the excited state is allowed The ratios of K- & L- capture = 7.36 KIMS

16 Specific Example – 157TB Branching Ratio
𝑁 𝑁 π‘’βˆ’capture = (2πœ‹) 3 πœ‰ 𝑛 𝜈 πœ– π‘˜ 𝑇 0 𝐾 𝛽,𝛿 Ξ› πœ– πœ– 0 ,𝛿 Adopt the view that the branching ratio for L-capture toe the 54 keV state < that dictated by the phase-space vector by a factor of 10 Ex) πœ–=50 eV, 𝛿=0.02, π‘š 𝜈 =1 eVβ†’ πœ– =6 eV β†’ 𝑇 𝜈 = πœ– βˆ’ π‘š 𝜈 =5 eV (1/10)(5/104)2=2.5β¨―10-8 (104 eV: K-capture energy to the g.s. of 157Gd KIMS

17 Specific Example – 157TB Branching Ratio (πœ‰= 10 6 ) KIMS

18 Discussion If exists a resonance around (Ξ”βˆ’π‘+πœ–+ π‘š 𝜈 ) with a width πœ– 1+𝛿 , there can be a relatively large rate for 𝜈 absorption. However, EC capture just below (Ξ”βˆ’π‘+ π‘š 𝜈 ) cannot be suppressed completely for 𝛿≠ 0. If the final state is populated by EC capture, the average energy available for de- excitation is Ξ”βˆ’π‘βˆ’ πœ– ( πœ– =πœ– 4 πœ‹ 3 5 𝛿 1βˆ’ 𝛿 ), smaller than (Ξ”βˆ’π‘+πœ–+ π‘š 𝜈 ) If 𝛿≀ 10 βˆ’10 & πœ‰ is large enough, πœ† π‘Ž πœ† 𝑐 can be larger than 1. KIMS

19 Conclusion The possibility of observing relic neutrino with neutrino absorption in a nucleus strongly depends on the properties of the target nuclide. Trituim, 187Re in beta-decay sector and 163Ho – considered as possible candidates But the rate is too small If some resonance conditions are met, a considerable enhancement of the associated rates can be obtained. Mass difference = b πœ–=relic neutrino total energy 157Tbβ†’157Gd meets the conditions, while the nuclide above don’t. KIMS


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