Discussion 11: Particle physics reactions with 4-vector kinematics Names: Section: This week we will practice using 4-vectors, energy-momentum conservation, and the invariance of the relativistic dot product to solve problems in relativistic kinematics. Far from being artificial examples, these are the most common use of relativity in particle physics, and also have applications in nuclear physics as well. We will use natural units (c = 1) throughout. To save some space, here is another convenient notation for the dot product of a 4-vector with itself: P = P . P = Pup (1) You've seen the latter two expressions before; I've avoided the first one in class because it can possibly be confused with the y-component of the 4-vector itself, but we will never be dealing with individual spatial components in this discussion. So you can always read p' as "p-squared." Part 1: Particle decays revisited In lecture we studied the example of pion decay, a - u + v. Let's generalize this example to the following situation: a particle of mass M decays to two particles with masses mi and mz. 1.1. Using energy-momentum conservation and the invariant dot product, find the energy of m1 in the rest frame of M. Hint: this derivation proceeds exactly like the one in lecture, except one of the dot products will be m; instead of 0. 1.2. Now do the same to find the energy of my. Verify that the sum of the final-state energies equals M; this is a good check that you are applying energy conservation correctly! 1.3. Set my = 0 and verify that you recover the result for the energy of the massless neutrino we studied in lecture. Then set my = my = 0 and re-derive your result for 7 decaying to massless photons from problem 3.4 of Discussion 8. 1.4. Prove that M can only decay into my and my if M 2 mi + m2. Decays into heavier final-state particles would violate conservation of energy! 1.5. An important application of these calculations is in nuclear decay. A plutonium nucleus can undergo alpha decay: 235 U +? He. 91 Pu - 32 The mass of the Pu nucleus is M = 222.676 GeV (where 1 GeV = 103 eV), and the masses of the daughter particles are my = 218.942 GeV for ,U and my = 3.727 GeV for $He. (For historical reasons, the helium nucleus ,He is also known as an alpha particle.) Compute the kinetic energy of each of the decay products, T = E - m. What fraction of the rest energy of the plutonium nucleus is converted to kinetic energy in this decay? This is E = me in action!Part 2: Invariant niass Most elementary particles [besides the familiar proton. neutron. and electron} are extremely short- lived. Therefore. it is often not possible to observe them directly. and instead we look for their decay products and try to reconstruct the presence of the original particle. A very common technique to do so is by considering the invariant mass of a set of decay products. In the case where the parent particle decays to two daughter particles [as in Part 1 above}. with 4-momenta it] and km the invariant mass is dened as mic I \"51 + 1'32}?- [2} 2.1. If in and k2 came from the decay of a particle of mass M. explain why mfg = M2. and justify the use of the term \"invariant.\" 2.2. Since the invariant mass is invariant. we can compute its value in any frame. Suppose that 3:1 and it; have the same mass m. and in the laboratory frame. in and k2 have energies E1 and E2. respectively. Compute mfg as a function of E1. E2. or. and cos 312. the angle between the 3-moinenta of it] and kg. 2.3. 1|lei-"hen the decay products are photons. we can set in = 3. Repeat problem 2.2 in the case where m. = (l. 2.4. The Higgs boson was rst discovered in 2312 via its decay to photons. H } TF- Suppose the photon energies are 53 GeV and son GeV. Given that the mass of the Higgs is mg, 2 125 GeV. what is the angle between the two photons in radians? It turns out this angle is quite small. so feel free to use a Taylor expansion to make the math easier. Part 3: Relativistic kinematics at colliders A tried-and~true technique for making new particles is by smashing other particles together. For example. positive and negative muons can be created by colliding electrons and positrons. e++c' } p+ + p'. The electron c and positron e+ have the same mass. m... and the positive muon p+ and negative muon n. have the same mass in\". 1eth assign :1-vectors to each particle in this collision and write the energy-momentum conservation equation as P1+Pz=kl+i2 [3} 3.1. In a typical collider. the electron and positron have equal and opposite momenta. so that the lab frame and the CM frame coincide. Letis write 191 = [Ef and p2 = [E .-" 2. f:]. Find |f:r'| in terms of E and m... 3.2. Typically. the colliding particles are highly relativistic. E E: m... Perform a Taylor expansion of |pi from 3.1 above. If m... 2 3.5 l'vieV and E = lflil Geif. what is the size of the leading-order term in mEfE'.' This justies treating highly relativistic electrons and positrons as if they were inassless. in... so fl. 3.3. The :1-vector corresponding to the initial state energy-momentum is P = p] + p: 2 {Evil} Energy-momentum conservation thus reads P m it] + k2. which looks just like the decays you studied in Part 1. Use this fact to nd the energies of the outgoing Jnuons. Hint: you should find that the ;1+ and n have the some energy