1 The pion bump in the gamma reay flux

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1 1 The pion bump in the gamma reay flux Calculation of the gamma ray spectrum generated by an hadronic mechanism (that is by π decay). A pion of energy E π generated a flat spectrum between kinematical limits E min,max γ = E π (1 β π) = E π ( 1 1 m π E π ) (1) The emission spectrum is: dn π γ de γ (E γ, E π ) = 1 E π β π () The rate of emission of gamma rays of energy E γ can be obtained integrating the rate of pion decay, times the spectrum of gamma rays generated in the decay: Ṅ γ (E γ ) = E min π (E γ) de π Ṅ π 1 E π β π (3) One has to compute kinematical limits. It is simple to see that Eπ max. The minimum pion energy that can create a gamma ray of energy E γ can be obtained solving a simple equation. For gamma rays with energy (E γ < m π /): E γ = Eγ min (E π ) = E π ( 1 1 m π E π ) (4) For gamma rays with energy (E γ > m π /) one has to solve the equation E γ = Eγ max (E π ) = E π ( m π E π ) (5) In both cases the solution is Eπ min (E γ ) = E γ + m π (6) 4 E γ Note that for E γ = m π / one has Eπ min (m π /) = m π (that is all pions can create a photon of energy E γ = m π /) Eπ min (E γ ) = m π [ Eγ m π + m ] π E γ The energy E min π is equal for two energies that are symmetric around E γ = m π / (7) 1

2 Hadronic Interactions.1 Threshold of inelastic interactions The threshold for inelastic pp interactions corresponds to the opening of the channels: p p p p π p n π + Assuming that one proton is at rest, this implies that minimum total energy for the projectile proton is: E p = m p + m π + m π m p m p MeV (8) This equation can obtained requiring the the Lorentz invariant s = (p i + p ) (with p 1 and p the 4 momenta of the two protons), that is equal to the square of the total energy in the center of mass frame, must be: s (m p + m π ), and expressing s in the rest frame of the target in the form s = m p + m p E p. The inelastic cross section for pp interactions grows rapidly with E (the kinetic energy of the projectile), reaching quickly a value of approximately mbarn (one mbarn is equal to 10 7 cm ), it stays approximately constant growing only logarithmically (go fits to the high energy cross section have a functional form (log s) The pp inelastic cross section has been measured is at LHC at s = 8 TeV (that corresponds to E p ev ( GeV), by TOTEM σ inel (8 TeV) = 74.7 ± 1.7 mbarn ATLAS has recently measured the cross section at s = 13 TeV σinel (13 TeV) = 74.7±1.7 mbarn. [σ tot (13 TeV) = 96.07±0.18±0.85±0.31 mbarn. Auger estimate at s = 57 TeV of the pp cross section: σ inel (57 TeV) = 9 ± 7 (stat) (sys) ± 7 (Glauber) mbarn.. Final state particles 3 Spin zero Mesons Nonet (octet + singlet) Spin zero mesons: π + = [u d] π = [d u] π 0 = 1 [ u u d d ] η = 1 6 [ u u + d d ss ] η = 1 3 [ u u + d d + ss ] Strange mesons

3 K + = [u s] K = [d s] K = [s u] K = [s d] K s = 1 [K 0 + K 0 ] = 1 [d s + s d] K L = 1 [K 0 K 0 ] = 1 [d s s d] m π MeV m π ± MeV Very large difference in lifetime (electromagnetic decay versus Weak decay) τ π ± sec Decay length of a particle τ π sec τ π ±/τ π l dec (E) = c τ β γ c τ γ == c τ E m ( E l π ±(E) 55.9 ) km 1 TeV ) cm ( E l π 0(E) TeV 3.1 High energy Inelastic interaction of a high energy, relativistic proton with a nucleus. Generate a large number of particles (hadrons). Rapid decay of resonances, one remains with the ground state particles SU(3) Nonet of scalar mesons: π +, π, π K +, K K, K η, η Let us consider a pp interactions in the c.m. frame. The c.m. energy of the reaction is E cm = s. s = E cm = (p 1 + p ) (9) 3

4 (where p 1 and p are the 4 momenta of the two initial state protons). In the c.m. frame the two protons are antiparallel, and the direction of the two momenta p 1 = p define an axis (the axis z) The final state particles are produced with a a small transverse momentum p = p x + p y: p GeV (10) while the longitudinal momentum p z has a very broad distribution that extends in the entire allowed kinematical range. Note that one has: s s p z (11) We are interested in the inclusive single particle distribution, that is the distribution of all particles of a certain type (π ±, π, K ±,...) produced in inelastic interactions. dn d 3 p ( p, s) = pp a 1 σ inel (s) dσ d 3 p ( p, s). (1) pp a Note that integrating over all momenta one finds the average multiplicity for particle of type a. d 3 p dn d 3 p ( p, s) = N a ( s) (13) pp a It is important to consider quantities that are Lorentz invariant. For this one can use the fact that: d 3 p = Relativistic invariant (14) E The transverse momentum p is manifestly Lorentz invariant for boosts along the z direction, and therefore one also has that: dp z E = dy = Relativistic invariant (15) (for boosts along z). It is straightforward to check explicitely the fact that dp z /E is a Lorentz invariant with a direct calculation: Under a Lorentz transformation one has: E = γ(e + β p z ) p z = γ(p z + β E) Let us consider two longitudinal momenta p z and p z1 = (p z + dp z ). After transformation 4

5 one has: p z1 = γ(p z + β E) p z1 = γ[p z1 + β E 1 ) = γ [ (p z + dp z ) + β E ( 1 + de )] dp z dp z ( = γ(p z + β E) + dp z γ + β de dp z ( = p z + dp z γ + β p ) z E = p z + dp z E E In the last steps we have used: de/dp z = p z /E and E = γ(e + β p z ). The bottom line is: dp z = dp z E /E, or dp z /E = dp z/e. The relativistic invariant dp z /E suggests to introduce a new kinematical variable that is the rapidity y. The rapidity is defined by the equation: dy = dp z E Integrating this equation, with the boundary condition that p z = 0 corresponds to y = 0 one obtains: y = 1 [ ] E + ln pz. (17) E p z Note: It is easy to verify that this is right performing the derivative with respect to p z (and using the fact that E = p z + p + m ), to verify that One obtains: ) (16) dy dp z = 1 E. (18) The fact that dy is a relatistic invariant implies that under a Lorentz transformation corresponds to add a constant value to the rapidity: y = y + y(β) (19) 5

6 This can be easily verified performing the transformation explicitely: y = 1 [ E ln + p ] z E p z = 1 [ ] γ(e + β ln pz ) + γ(p z + βe) γ(e + β p z ) γ(p z + βe) = 1 [ ] γ(1 + β)(e + ln pz ) γ(1 β)(e p z ) = 1 [ ] E + ln pz + 1 [ ] 1 + β E p z ln 1 β = y + 1 ln [ 1 + β 1 β where we have obtained the value of the shift for a Lorentz transformation. Note that one can rewrite the shift in rapidity simply as: y = 1 [ ] 1 + β ln 1 β = 1 [ ln (1 + β) ] (1 β) (1 + β) = 1 [ ] (1 + β) ln (1 β ) [ ] 1 + β = ln ln [γ] 1 β ] As a illustration the rapidity in the c.m. frame, and in the laboratory frame (that are related by a Lorentz transformation with γ = s/( m p ) differ by a shift: [ ] s y lab = y cm + ln. (0) It is also useful to note that one can also write the rapidity in the form: y = 1 [ ] E + ln pz = 1 [ E p ln (E + p z ) ] z (E p z )(E + p z ) [ ] E + p z = ln = ln E + p z E p z m + p Pseudorapidity Experimentally it is often difficult to measure the rapidity, because it implies the measurement of the momentum and the energy of a particle. Thre is however a quantity that is much easier to measure, and has approximately the same physical meaning, and is called pseudorapidity. m p 6

7 The pseudorapidity η corresponds to estimate the rapidity calculated assuming that the particle has mass m = 0, so that E p. It immediately follows that the pseudorapidity only depends on the particle direction, or more exactly on the angle θ with respect to the z direction. In fact: 3. Feynman scaling η = 1 [ ] p + ln pz p p z = 1 [ ] p(1 + cos θ) ln = 1 [ cos ] p(1 cos θ) ln (θ)/ sin (θ/) [ ] [ cos(θ/) = ln = ln tan θ ] sin(θ/) The distribution. of transverse and longitudinal momenta are approximately (but not exactly) factorized, so one has: E dn d 3 p f(p z, s) g(p ) (1) The function g(p ) (that without loss of generality can be normalized to give unity after integration over all p ) has approximately gaussian form: g(p ) 1 [ ] π σ exp p σ () [The details of this question are not discussed here. The distribution has a power law tail at large p. The average p (one has approximately p = π/ σ) grows logarithmically with s]. In general one can write, the inclusive distribution of rapidity of final state particle in the interactions at c.m. energy s will depend on s: dn/dy = f(y, s). In 1969, Richard Feynman proposed that the inclusive rapidity distribution satisfies (aymptotically, that is for high energy interactions) the scaling law: dn dy (y, s) F (x F ) (3) The key point of this equation is that the distribution now depends only on one variable (and not two). In this equation the quantity x F (or x Feynman) is defined x F = p z (4) s (where p z is the longitudinal momentum in the c.m. frame), Since p z takes values in the interval: s/ p z + s/ the variable x F has range: 1 x F +1 (5) A first important consequence is that the rapidity spectrum has a flat plateau. That is there is a region of y (around y 0 in the c.m. frame) where dn/dy is approximately constant. If 7

8 dndy 1.0 dndy Figure 1: Example of Feynman scaling the scaling function F (x F ) has a finite limit for x F 0, then in all the region where E s/, one has x F 0, and dn F (0) constant (6) dy This also implies that in the c.m. frame, the spectrum in the energy region where E s/ and E m + p has approximately the form 1/E. In this region one has: E p p z, and: 3.3 String Fragmentation dn dp z = 1 E Modeling of e + e qq (Production of many hadrons). dn dy F (0) E. (7) Iterative string fragmentation generates a Feynman scaling distribution. 8

9 Figure : Iterative string fragmentation 1.5 s 0.1, 1, 10 TeV dnchargeddy Figure 3: Example of the rapidity distributions obtained from the fragmentation of a uu string, calculated with Montecarlo methods using the LUND algorithm [Sjostrand:001yu]. 9

10 3.4 Scaling Violations 6 5 dnchargeddη Figure 4: Pseudorapidity distributions measured in pp interactions by the UA5 experiment at the CERN SpS collider at s = 53, 00, 546 and 900 GeV, and in pp interactions by ALICE at LHC at s = 0.9,.36 and 7 TeV.... Some data on the inclusive production of secondaries in pp interaction from an experiment at Fermilab [Brenner et al. Phys. Rev. D 6, 1497 (198), Experimental Study of Single Particle Inclusive Hadron Scattering and Associated Multiplicities, ] are shown in fig. 5. One can see that effectively one has that the cross section is scaling. The inclusive spectra have been fitted with the functional form F (x) = C (1 x) n /x. The inclusive spectra are fitted with different exponents n. For example at E 0 = 175 GeV n π + = 3.39 ± 0.05 n π = 4.39 ± 0.10 n K + =.77 ± 0.10 n K = 5.48 ± 0.4 Qualitative understanding of these differences on the basis of the structure of the different hadrons. For example the K + is a state [us] and is formed by a quark that is a proton constituent (the proton is the state [uud], while K = [su] and both constituents must be created from the vacuum. 10

11 100 pp Π E GeV 10 E GeV dσdxf F Figure 5: Inclusive spectra of π + produced in pp interactions. For laboratory energies E 0 = 100 GeV and 175 GeV. 100 Π Π pp a E GeV 10 dσdxf 1 K 0.1 K F Figure 6: Fits to the inclusive spectra of π + π K + K in pp interactions for a beam energy E 0 = 175 GeV. 11

12 4 Astrophysical source Let us consider an object that contains a population of relativistic particles N p (E) confined by magnetic fields, and there is a density n of gas. What is the emission of neutrinos and photons from this source. We want to compute the quantities Ṅ να (E ν ) (8) Ṅ γ (E γ ) (9) Number of particles interacting per unit time (p π + ν µ ) (30) (p π µ ν µ ) (31) (p K + ν µ ) (3) Ṅ int (E p ) = N p (E p ) σ pp (E p ) β c n (33) Let us calculate Ṅπ +(E π). This can be performed with the integration Ṅ π +(E π ) = de p N p (E p ) σ pp (E) β c n dn pp π + (E π, E p ) (34) de π Calculation of Ṅ π +(E π ): Ṅ π + ν µ (E ν ) = de π Ṅ π +(E π ) dn π + ν µ de ν (E ν, E π ) (35) dn pp π+ Ṅ π +(E π )(E π ) = de p N p (E p ) σ pp (E p ) β c n (E π, E p ) E π de π [ ] dnpp π + = σ pp n c de p N p (E p ) (E π, E p ) E π de π [ ( )] ( ) = σ pp n c de p K E α 1 Eπ p F E p π + p E p = K E α π E π σ pp n c Where we have used the following assumptions: 1 0 dz z α 1 F p π +(z) = K σ pp n c Z p π +(α) Eπ α = K Eπ α (36) N p (E) K Ep α (37) (E π, E p ) 1 ( ) Eπ F de π E p π + p (38) dn pp π + E p 1

13 In the last steps we have transformed variable introducing z = E π E p (39) used the transformation de p = dz z E π (40) The result is that the pion flux is simply proportional to the parent proton flux with a proportionality factor (the Z factor ) Z p π (α). In general we denote as: Z a b (α) = 1 0 dz F a b (z) z α 1 (41) the momentum of order (α 1) of the inclusive scaling distribution for the production of particle b from particle a. Note that Z a b (1) gives the average multiplicity of particle b, while Z a b () gives E b /E a, that is the fraction of the parent particle energy carried in the final state by particle of type b. Continuing the process one finds: Ṅ p π + ν µ (E ν ) = K σ pp c n Z pp π + Z pi + ν µ (4) 13