PHY-494: Applied Relativity Lecture 5 Relativistic Particle Kinematics

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1 PHY-494: Applied Relativity ecture 5 Relativistic Particle Kinematics Richard J. Jacob February, 003. Relativistic Two-body Decay.. π 0 Decay ets return to the decay of an object into two daughter objects. We will consider the kinematically simple case of π 0 γ + γ. (.) This is an electromagnetic interaction. The neutral pion, π 0, has a mean life of only (8.4 ± 0.6) 0 7 s, some 0 9 times shorter than that of its charged siblings, π + and π, which decay through weak interactions. The decay of charged pions will be the subject of a homework problem. The neutral pion mass is m π 0 ( ± ) GeV/c. The decay products are photons with zero inertial mass. Photons do carry energy, ω, and momentum, k, however, related by the m 0version of the mass-shell condition: ω kc. We will take c from here on out. Masses and momenta will have units of energy, taken as electron-volts most of the time. In the pion rest frame, the initial momentum is zero and so then is the Þnal. Neutral pion decay in the center-of-momentum frame

2 Thus, k C ωc kc ωc. Conservation of energy tells us then that k C ωc kc ωc m π 0 ( ± ) GeV GeV (.) and that the photons leave back-to-back as shown in the Þgure. But since the pion has zero spin, there is no vector to deþneapreferreddirectionofdecay,andthe photons can go in any direction; pion decay in the CM frame is isotropic in a quantum mechanical sense. The probability of photon emission within a given solid angle is independent of the direction. The unique energy of a daughter photon from this decay is a clear signal for the presence of a neutral pion. Uniqueness of daughter particle energies in the CM frame is a characteristic of two-body decay. et us now determine the energy and angle distributions of the photons in a frame (called the ab frame) in which the pion momentum is p ẑp. Neutral pion decay in the lab frame We could use conservation and momentum, along with orentz transformation elements, to compute the ab frame quantities straightforwardly. But we will instead simplify the problem considerably by using invariants. One invariant available to us is ν (p k ). (.3) The four-vector momenta in the ab frame are p ε, 0, 0, p, (.4) We have (recall that k k 0) k k, k. (.5) ν p p p k + k k

3 m π 0 ε k p k +0 m π 0 k ε p cos θ. (.6) In the pion s rest frame, on the other hand, and p C (m π 0, 0, 0, 0), (.7) k C k C, kc, (.8) ν p C p C p C k C + k C k C m π m 0 π 0kC 0 (.9) since k m π 0/. Thus, from Eqn..6, we have k ε p cos θ m π 0 or k γ β cos θ mπ 0, (.0) where γ and β are the usual parameters describing the state of the pion s ab frame motion. They are the same γ and β that appear in the orentz transformation elements representing the boost from the pion s rest frame to the ab frame. This equation relates the energy (as represented by γ) and the emission angle of the Þrst photon. We need another expression that will relate this angle to the emission angle in the CM frame. Since there is nothing in p that allows us to compare any other direction with ˆp ẑ, we invent a four-vector that will allow us to do this, to wit, our basis vector ẑ (0, 0, 0, ) eˆ3, ẑ ẑ. (.) Boosting back to the pion s rest frame, this four-vector becomes We then form the invariant ẑ k : ẑ C (βγ, 0, 0, γ). (.) ab : ẑ k k cos θ, (.3) CM : ẑ C k C βγk γk cos θ C. (.4) 3

4 Neutral pion decay in the CM frame Thus, k cos θ γ m π 0 (cos θ β), (.5) which gives the desired relationship between θ and θ. Now, for given ε, k and cos θ are dependent, according to Eqn Photon distributions. et Γ (ε, k, θ ) be the probability density of having a photon of lab energy k andemittedatlabangleθ. Then, we have as angular and energy distribution functions, and. That is to say, d cos θ dk d cos θ d cos θ is the probability of emission into θ d within range cos θ sin θ dθ. Similarly, dk dk is the probability of photon emission with lab energy k within a bin of width dk. is a (normalized) number of photons, and therefore an invariant. We can thus write d cos θ d cos θ C d cos θ d cos θ. 4

5 From (.0) and (.5), we have after eliminating k, cos θ β cos θ γ cos θ C β. (.6) Taking the implicit derivative with respect to cos θ yields β cos θ +cosθ β β cos θ γ d cos θc d cos θ or d cos θ C d cos θ β cos θ. (.7) Since d cos θ C C for isotropic decay, (the determination of C is left to the homework) we have as the angular distribution in the ab frame, d cos θ C γ β cos θ. (.8) The distribution of decay photons is peaked in the forward direction. At high energy (large γ) the ratio of forward to backward photons is forward backward Ã! +β 6γ. (.9) β We can determine the energy distribution in much the same way. dk d cos θ d cos θ. (.0) dk From Eqn..0, γ β cos θ k γβ d cos θ dk 0, 5

6 or so that d cos θ dk β cos θ βk γ β cos θ βm π 0, (.) C dk γ β cos θ βm π 0 C γβm π 0 C p. (.) This is a ßat distribution between the minimum and maximum values k, max m π 0 γ ( β), cos θ +, (.3) k,min m π 0 γ ( + β), cos θ. (.4) In the high energy limit, k, max m π 0 4γ ε 4γ, (.5). Two-body Scattering à k, min ε 4γ!. (.6) The scattering of particles in the ab frame is illustrated in the Þgure. The three momentum vectors deþne a plane according to p A p C + p D. (.) We take this plane to be the z-x plane and set the azimuthal angle ϕ 0. We have as the appropriate four-vector momenta: p A ε A, 0, 0, p A p B (m B, 0, 0, 0) (.) p C ε C,p C sin θ C, 0, p C cos θ C p D ε D, p D sin θ D, 0, p D cos θ D 6

7 Two-body scattering in the ab frame The four-momenta obey the mass-shell conditions p i p i m i,i A, B, C, D. (.3) And, of course, we have conservation of four-momentum: of which Eqn.. is the spatial part... Mandelstam variables. p A + p B p C + p D, (.4) Ordinarily, there are 6 variables among the 4 four-momenta. But this number reduces considerably through the above conditions as well as various spatial symmetries. The mass-shell conditions and the conservation of four-momenta give us 8 constraints. We can always choose one of the particles to be at rest, eliminating 3 more momentum components. Finally, there are two arbitrary angles for orienting the system and one more for orienting the scattering plane. That is to say, we can boost and rotate ourselves to an arbitrary inertial frame. Thus, the original 6 variables becomes only independent variables, upon which real physics can depend. These two variables can be any of the original 6 or some combination of them. In an experiment, one often chooses the projectile particle s kinetic energy or momentum and the scattering angle of one of the Þnal state particles. In theoretical work, it is often more useful to select two of the available invariants that can be formed from the four-momenta. Such a set are the Mandelstam variables, used throughout nuclear and particle physics. s (p A + p B ) (p C + p D ), (.5) 7

8 t (p A p C ) (p B p D ). (.6) The second form in each case is obtained from the Þrst by using Eqn..4. No special frame labeling is necessary in these deþnitions. However, in the ab frame, s ε A + m B p A ε A +ε A m B + m B ε A + m A ε A m B + m B + m A, or Also in the ab frame, ε A s m A m B m B. (.7) t p A + p C p A p C m A + m C ε A ε C +p A p C cos θ C. (.8) We see that s is related to the lab energy and t is related to the scattering angle. t is called the four-momentum transfer. There is a third invariant that, while not independent of s and t, isusefulin many calculations: u (p A p D ) (p B p C ). (.9) One can easily show that.. Center-of-momentum frame. s + t + u m A + m B + m C + m D. (.0) The Mandelstam variables have an even more transparent interpretation in the CM frame. Here, p C A ε C A, 0, 0, pc A p C B ε C B, 0, 0, p C A (.) p C C ε C C,pC C sin θc C, 0, pc C cos θc C p C D ε C D, pc C sin θc C, 0, pc C cos θc C 8

9 Two-body scattering in the CM frame Thus, as in the two-body decay that we examined in ecture 4, s ε C A + εc B W, (.) the square of the total CM energy. Since s is an invariant, we can go to Eqn..7 and write ε A W m A m B. (.3) m B.3. Example: Compton Scattering Compton scattering from an electron is the reaction γ + e γ 0 + e (.4) where be γ 0 we mean a photon of different energy than the initial one. We use four-momentum notation as indicated in the conservation equation: k + p k 0 + p 0 (.5) The DeBroglie relationship between a photon s wavelength and its momentum is λ h k, (.6) where h is Planck s constant. We do the calculation in the ab frame, but will not use a lab-frame superscript. We have p 0 (k k 0 )+p. 9

10 p 0 p 0 m e k + k 0 + p +(k k 0 ) p k k m e +(k k 0 ) p k k 0, or Now, and Thus or (k k 0 ) p k k 0. (.7) (k k 0 ) p m e (k k 0 ) k k 0 kk 0 kk 0 cos θ. kk 0 ( cos θ) m e (k k 0 ) sin θ m e k k0 kk 0 from which we get the Compton relationship 3. Threshold Calculations k 0 k h (λ0 λ), λ 0 λ m e h sin θ. (.8) We often need to compute the threshold energy, that is the minimum necessary amount, for reactions of the type A + B X where X stands for any number of particles. The threshold condition is that in the CM frame, all particles in X are produced but there is no energy left over to give any of them any motion. Thus, the condition on s is s threshold X m i, f 0

11 where the sum is over all particles represented in X. According to Eqn..7, then, the threshold lab energy is ε A threshold P f m i m A m B m B. (3.) Example: π 0 production in proton-proton collisions, stationary target. In the reaction p + p p + p + π 0, the threshold energy for the projectile proton is ε p (m π +m p ) m p threshold m p m π +m p (m π + m p ). m p Given that roughly, m p 7m π, this becomes ε p threshold 3 4 m π. This requires then a kinetic energy of κ p ε threshold p m p threshold 5 4 m π Example: Antiproton production in proton-proton collisions, stationary target. In order to conserve charge and baryon number, the minimum reaction for producing antiprotons in a collision of two protons is p + p p + p + p + p. The threshold energy in the ab frame is thus ε p threshold (4m p) m p m p 7m p.

12 This then requires a kinetic energy of κ p ε threshold p m p threshold 6m p.