8.701 Introduction to Nuclear and Particle Physics, Recitation 2 Solutions

Massachusetts Institute of Technology Department of Physics Discussion Problems Problem 1: Cosmic rays Greisen, Zatespin, and Kuzmin (GZK) predicted that there is a maximum energy for which any cosmic ray (which are predominantly protons) will be observed. Their argument was that if energetic enough, cosmic rays will lose energy via the process pγCM B → pπ 0 , where γCM B is a cosmic microwave background photon, whose energies are ECM B ≈ 3 × 10−4 eV. Therefore, the threshold energy for which this process can occur sets the maximum expected energy a cosmic ray should be observed to have (called the GZK cutoff). Calculate the GZK cutoff energy in eV, and compare your result to the observed cosmic ray energy spectrum (see Fig. 29.8 of the PDG review on Cosmic Rays or below). [Hint: You can take the energy to be much larger than the proton mass and assume that the frame in which we want to know this energy is the one where the proton and CMB photon collide head on.] 1 © Regents of the University of California. All rights reserved. • 2 Problem 2: Mandelstam variables Consider the reaction between two particles with 4-momenta pµ1 and p2µ . The outgoing particles have 4-momenta pµ3 and pµ4 . Discuss the variables s, t, and u for the centerof-mass frame and a fixed-target frame where the second particle is at rest. Assume that the masses involved are much smaller than the energies (mi  Ei ). • Kinematic in center-of-mass frame: p~∗1 + p~2∗ = p~3∗ + p~4∗ = 0 E1∗ + E2∗ = E3∗ + E4∗ (1) Kinematic in fixed-target frame: p~2 = 0 p~1 = p~3 + p~4 E1 + m2 = E3 + E4 (2) Assuming that m1 , m2  E1 , i.e. small masses relative to the energies of the scattering process: E ≈ |p~| E1∗ ≈ E2∗ ≈ E3∗ ≈ E4∗ For s and t: 3 s= (pµ1 + pµ2 )2 = 2 pµ1 p2 µ + m12 + | m22  ≈2 E1 p~1  m2 ~0 {z Fixed-target frame  = 2 m2 E1 } 2 E1∗ + E1∗ = = 4 E1∗ 2 p~1∗ − p~1∗ {z } | Center-of-mass frame  2 t = (pµ1 ∗ − pµ3 ∗ ) = −2 pµ1 ∗ p∗3 µ + m21 + m32  ∗  ∗  E1 E3 ≈ −2 ∗ p~1 p~3∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ = −2 E1 E3 − p~1 p~3 = −2 E1 E3 − |p~1 | |p~3 | cos θ   s  cos θ∗ − 1 = 2 E12 ∗ cos θ∗ − 1 = 2 t= (p1µ − p3µ )2  = E1 − E 3 p~1 − p~3 2  = E4 − m 2 p~4 2 = E42 − p~24 + m22 − 2 m2 E4 ≈ −2 m2 E4   = −2 m2 E1 + m2 − E3   = 2 m 2 E3 − E1 Note the range of values for t: −∞ < t ≤ 0. For u and s + t + u: 4 2 u = (pµ1 ∗ − pµ4 ∗ ) = −2 pµ1 ∗ p∗4 µ + m21 + m42  ∗  ∗  E4 E1 ≈ −2 p~∗1 p~4∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ = −2 E1 E4 − p~1 p~4 = −2 E1 E4 − |p~1 | |p~4 | cos (π + θ )    s = −2 E12 ∗ cos θ∗ + 1 = − cos θ∗ + 1 2 u= (p1µ − p4µ )2  = E1 − E 4 p~1 − p~4 2  = E3 − m2 p~3 2 = E32 − p~23 + m22 − 2 m2 E3 ≈ −2 m2 E3  2  2  2 s + t + u = pµ1 + pµ2 + pµ1 − pµ3 + pµ1 − pµ4 = 4 X i=1 4    X m2i + 2 p1 µ pµ1 + pµ2 − p3µ + pµ4 = m2i | {z } i=1 ≡0 5