[Another look at] decay in astrophysical neutrinos (work in - PowerPoint PPT Presentation

[Another look at] decay in astrophysical neutrinos (work in progress) Mauricio Bustamante Lehrstuhl für Theoretische Physik II Institut für Theoretische Physik und Astrophysik Universität Würzburg Würzburg, June 12, 2012 M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 1 / 21

Contents 1 Introduction 2 The naive solution 3 The proper solution 4 Flavour-transition probability 5 GRB neutrino fluxes 6 To do M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 1 / 21

Introduction Contents 1 Introduction 2 The naive solution 3 The proper solution 4 Flavour-transition probability 5 GRB neutrino fluxes 6 To do M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 1 / 21

Introduction Introduction ◮ The very long baselines of astrophysical ν ’s ( � 50 Mpc) make them ideal for studying the possibility that they are unstable and decay ◮ We will focus on the redshift dependence of the decay ◮ For a population of mass eigenstates, dN τ = m λ = 1 τ 0 E ≡ κ dt = − λ N with E where τ 0 : lifetime in the rest frame τ : lifetime in the lab frame t : time ellapsed since creation (in lab frame) κ ≡ m /τ 0 ◮ Lower bounds on κ − 1 ≡ τ 0 / m : � 10 5 s eV − 1 (from SN1987A) ◮ For ν 1 : κ − 1 1 � 10 − 4 s eV − 1 (from solar ν ’s) ◮ For ν 2 : κ − 1 2 � 10 − 10 s eV − 1 (from atm. and long-baseline ν ’s) ◮ For ν 3 : κ − 1 3 M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 2 / 21

Introduction Two ways of solving the decay equation dN dt = − λ N for astrophysical ν ’s: 1 Naive solution: introducing the redshift dependence after solving the equation (usual way) 2 Proper (we think!) solution: introducing the redshift dependence before solving the equation These lead to very different behaviours for the surviving population N ( z ) of neutrinos created at redshift z . Also, it is commonly assumed that the decay of cosmological ν ’s is complete when they reach Earth – we have found some caveats. M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 3 / 21

The naive solution Contents 1 Introduction 2 The naive solution 3 The proper solution 4 Flavour-transition probability 5 GRB neutrino fluxes 6 To do M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 3 / 21

The naive solution Deduction The naive solution Step 1: Forget about the z -dependence and solve for N as usual: dN L = ct dt = − λ N ⇒ N ( t ) = N 0 e − λ t → N ( L ) = N 0 e − λ L − − − Step 2: Now, introduce the z -dependence through: ◮ L = light-travel distance = L ( z ) e − λ L − → e − λ L ( z ) ◮ relation between energy at production ( E ) and at detection ( E 0 ): E = ( 1 + z ) E 0 λ = κ κ → λ ( z ) = E − E 0 ( 1 + z ) So, the naive solution is L ( z ) � � − κ N z , 1 ( E 0 , z ) = N 0 exp E 0 ( 1 + z ) M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 4 / 21

The naive solution Two observations Two observations 1 The light-travel distance is bounded by the Hubble horizon: 5 10 light-travel distance luminosity distance 4 10 3 10 2 10 L ( z ) [Gpc] 1 10 L H = 3.89 Gpc 0 10 -1 10 -2 10 -3 10 -3 -2 -1 0 1 2 3 10 10 10 10 10 10 10 z So, e − L ( z ) does not reach zero, even when z → ∞ . κ 2 For a fixed E 0 , the decay constant λ ( z ) = E 0 ( 1 + z ) will be lower for more distant sources. This helps ν ’s from farther away to have a higher chance of surviving. M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 5 / 21

The naive solution Behaviour Behaviour of the naive solution Rewrite it as N z , 1 ( E 0 , z ) = N 0 [ Z 1 ( z )] − γ/ E 0 with γ ≡ κ L H and � I 1 ( z ) I 1 ( z ) = L ( z ) � Z 1 = exp , 1 + z L H Z 1 carries all of the redshift dependence. � a � � b � 1.4 14 I 1 � z � Z 1 � exp 1.2 1 � z 12 1.0 10 1 � z , Z 1 � z � � 1 � z � , I 1 � z � 0.8 8 0.6 I 1 � z � 6 0.4 4 I 1 � z � 1 � z I 1 � z � 2 0.2 1 � z 0 0.0 10 � 3 10 � 2 10 � 1 10 0 10 1 10 2 10 3 10 � 3 10 � 2 10 � 1 10 0 10 1 10 2 10 3 z z M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 6 / 21

The naive solution Problems Problems with the naive solution 1 Z 1 ( z � 10 ) = 1 implies that N z , 1 ( z � 10 ) ≃ N 0 , regardless of E 0 , while one expects that the more energetic neutrinos should live longer than the less energetic ones. 2 The redshift suppression Z 1 was expected to be a monotonically growing function of z , so that N z , 1 ∝ 1 / Z 1 falls with z , since neutrinos from more distant sources should have more time to decay. M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 7 / 21

The proper solution Contents 1 Introduction 2 The naive solution 3 The proper solution 4 Flavour-transition probability 5 GRB neutrino fluxes 6 To do M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 7 / 21

The proper solution Deduction The proper solution Step 1: replace E = ( 1 + z ) E 0 in the differential equation: dN κ dL = − λ ( z ) N = − E 0 ( 1 + z ) N Step 2: replace dN / dL = ( dN / dz ) ( dz / dL ) , so that dN dL N dz = − κ E 0 dz 1 + z The proper solution is N z , 2 ( E 0 , z ) = N 0 [ Z 2 ( z )] − γ/ E 0 , with � z dL I 2 ( z ) = 1 1 Z 2 ( z ) = e I 2 ( z ) z ′ � � , L H ( 1 + z ′ ) dz 0 M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 8 / 21

The proper solution Behaviour Behaviour of the proper solution 2.0 � 1.7106 a 1.5 I 2 � z � , Z 2 � z � 1.0 Z 2 � exp � I 2 � z �� 0.5 I 2 � z � 0.0 10 � 3 10 � 2 10 � 1 10 0 10 1 10 2 10 3 z a = 1 . 71   Z 2 ( z ) ≃ a + be − cz b = 1 − a = − 0 . 71 with in Λ CDM c = 1 . 27  a + be − cz � − γ/ E 0 ⇒ N z , 2 ( z ) ≃ � Surprisingly, at z ≫ 1, N z , 2 ( z ≫ 1 ) ≃ N 0 a − γ/ E 0 � = 0 , i.e., the decay is not automatically complete at large z , because L ( z ) ≤ L H . M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 9 / 21

The proper solution Behaviour Different redshift dependence than in the naive approach: 1.8 Z 2 1.6 Z 1 , Z 2 1.4 1.2 Z 1 1.0 10 � 3 10 � 2 10 � 1 10 0 10 1 10 2 10 3 z a + be − cz � − γ/ E 0 behaves as expected: N z , 2 ≃ � ◮ It falls with z (down to N z , 2 ≃ N 0 a − γ/ E 0 ), since for higher z there is more time for the decay to occur ◮ It grows with E 0 , since the decay constant λ ∝ E − 1 0 N z , 2 actually behaves like a (physically) proper solution! M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 10 / 21

The proper solution Why is the decay not necessarily complete at high z ? Why is the decay not necessarily complete at high z ? ◮ According to the usual, non-redshift dependent, solution, N ( E , L ) = N 0 e − κ L / E , the population (for fixed E ) disappears if L ≫ 1, i.e., N ( E , L ≫ 1 ) → 0. ◮ In contrast, N z , 2 ( z ≫ 1 ) ≃ N 0 a − γ/ E 0 ◮ Reason: light-travel distance L ( z ) is bounded by L H , which makes Z 2 bounded (since N z , 2 ( z ) ∝ Z − 1 ( z ) ) 2 ◮ If we instead use L ( z ) = luminosity distance (unbounded), we obtain an 2 and N ∗ z , 2 ( z ≫ 1 ) → 0: unbounded Z ∗ 10 8 Z 2 � � � z � 6 Z 2 � z � , Z 2 4 Z 2 2 10 � 3 10 � 2 10 � 1 10 0 10 1 10 2 10 3 z M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 11 / 21

The proper solution Proper condition for complete decay Proper condition for complete decay From N z , 2 ( E 0 , z ) = N 0 [ Z 2 ( z )] − γ/ E 0 since Z 2 ( z ≫ 1 ) � ∞ , the condition for complete decay is γ/ E 0 ≫ 1, or, m ≪ L H ≈ 4 × 10 8 κ − 1 ≡ τ 0 E 0 [ GeV ] s eV − 1 . E 0 � 10 5 , 10 12 � In the relevant range, E 0 ∈ GeV, this can be satisfied together with the current bounds on τ i / m i . ∴ complete decay is still allowed, but it will not be due to very long baselines M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 12 / 21

Flavour-transition probability Contents 1 Introduction 2 The naive solution 3 The proper solution 4 Flavour-transition probability 5 GRB neutrino fluxes 6 To do M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 12 / 21

Flavour-transition probability Flavour-transition probability Flavour-transition probability Average probability for astrophysical neutrinos: 3 � 2 [ Z 2 ( z )] − γ/ E 0 , | U α i | 2 � P αβ ( E 0 , z ) = � � U β i � i = 1 with U the PMNS matrix. Note that � eV s − 1 � ≃ 4 × 10 8 κ γ . E 0 E 0 [ GeV ] We have fixed sin 2 θ 12 = 0 . 31 , sin 2 θ 23 = 0 . 51 , sin 2 θ 13 = 0 . 01 , δ CP = 0 Implicit assumption: all mass eigenstates decay with κ i ≡ κ = m /τ 0 M. Bustamante (Uni. Würzburg) Decay in astrophysical neutrinos 13 / 21