Problem Set # 2 SOLUTIONS

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Camron Wright
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1 Wissink P640 Subatomic Physics I Fall 007 Problem Set # SOLUTIONS 1. Easy as π! (a) Consider the decay of a charged pion, the π +, that is at rest in the laboratory frame. Most charged pions decay according to the scheme π + µ + + ν In this expression, µ represents a muon, a lepton similar to the electron, but with a rest energy m µ c MeV. The symbol ν is for a neutrino, which we will assume (for now) to have zero rest mass. When the above process occurs, the charged muons can be detected, and are found to have a kinetic energy of 4.1 MeV. Based on just the information given here, what is the mass of the π +? (b) Now consider the decay of a neutral pion which is moving in the laboratory frame. About 98.8% of all neutral pions decay into a pair of photons, π 0 γ + γ The rest energy of the π 0 is approximately 135 MeV. If the pion has a kinetic energy of 0.6 GeV as measured in the lab, then what are the measured photon energies if the two photons are emitted back-to-back along the pion s line of motion, that is, one going forward and one going straight back? (c) At what angle is each photon emitted if the two photons are emitted at equal angles left and right of the original pion momentum? You can solve for parts b and c (and get the same answer!) either working directly in the lab frame, or work in the pion s rest frame, and then boost to the lab frame. SOLUTION (a) In this problem, we examine the weak decay of a charged pion into a muon and a neutrino. In the pion rest frame, the µ and the ν must be emitted with equal momentum, but in opposite directions. We also are told that the neutrino is massless, while the outgoing 1

2 muon has a rest mass energy of MeV and a kinetic energy of 4.1 MeV. It is useful to summarize what we know about E and p for each particle; with c 1, we have: π + : p π 0, and so E π m π µ + : E µ m µ + T µ MeV p µ [ E µ m µ] 1/ [ (109.8) (105.7) ] 1/ 9.7 MeV ν : m ν 0, and so E ν p ν We now apply conservation of linear momentum and total energy to the decay process. For momentum, we see that p ν p µ 9.7 MeV, in the direction opposite that of p µ. To conserve total energy, we have: E π E µ + E ν But E µ MeV, and E ν p ν 9.7 MeV, so with p π 0, we find: m π E π MeV which is very close to the correct (PDG) value. (b) We next examine the electromagnetic decay of a neutral pion into two photons, as observed in a frame where the pion has a kinetic energy of 0.6 GeV. Like most problems of this sort, they can be approached in multiple ways. I ll present two here: using brute force E and p conservation in the lab frame, or solving the problem in the easier pion rest frame, and then boosting to the frame of interest. Method I lab frame In the lab, we are given that T π 600 MeV. Knowing m π 135 MeV, we solve for the pion s total energy as E π m π + T π 735 MeV, and so p π [ E π m π] 1/ [ (735) (135) ] 1/ 7.5 MeV To determine the kinematics of the decay products, we use conservation of both linear momentum and total energy. For convenience, I ll use subscripts 1 and to label the outgoing photons. If we know the two photons are emitted along the pion s momentum direction, with one going forward and one going back, then p π p 1 + p p π p 1 p Using E p for each photon, we conclude p π 7.5 MeV E 1 E

3 Conservation of energy gives us directly: E π 735 MeV E 1 + E Combining these last two equations yields our final results E 1 79 MeV E 6 MeV Method II pion rest frame A more elegant way to approach this problem (though not necessarily any easier algebraically) is to work in the rest frame of the pion, where p π 0, and then transform the energy and momentum values we obtain into the lab frame. In general, this is a much more powerful and intuitive way to solve these kinds of problems. We first note that in the pion rest frame (which I will denote as the primed frame), momentum conservation requires that the two photons be emitted with equal and opposite momenta. Because E π m π 135 MeV in this frame, and E p for photons, we see immediately E 1 p 1 E p E π / m π/ 67.5 MeV To transform these quantities to the lab frame, we note that the relative velocity v f between the two frames is just the velocity of the pion as measured in the lab frame, so β f β π p π /E π 7.5 / and : γ f γ π E π /m π 735 / You can verify that β and γ indeed obey their usual relationship. This gives us everything we need to now transform to the lab frame. Because the two 67.5 MeV photons are moving in the +x and x directions, we see E 1 E γ (E + βp x ) 5.44( ) 79 MeV γ (E βp x) 5.44( ) 6 MeV (c) Finally, we consider the case where the two photons are emitted symmetrically left and right of the pion s momentum axis at some angle θ. Because of this symmetry, we know that p 1 p (same magnitudes) and the transverse components (p sin θ) must be equal and opposite. Thus, we only need to concern ourselves with the momentum components along the original pion direction. These statements all hold in both frames. Working in the lab frame first: p π p 1 cos θ + p cos θ Combining this with our results from part b, we see 3

4 p π 7.5 MeV E γ cos θ For energy conservation, we again use symmetry to obtain E π 735 MeV E 1 + E E γ Combining these last two results, we get our final answer: cos θ 7.5 / θ 10.6 Or, we can start in the pion rest frame. To be left/right symmetric and back-to-back means θ 90, and so p x 0 in this case. We then have: and p x γ (p x + βe ) 5.44( ) 361 MeV p y p y 67.5 MeV/c And so we see: tan θ p y /p x 67.5 / θ 10.6, as before. 4

5 . I Need Some First η! Consider the decay of the eta meson, which only comes in a neutral variety, and decays according to η 0 γ γ. Let E 1 and E be the energies of the daughter photons, and φ o be the opening angle between them, all measured in the lab. Using invariant masses, show sin(φ o /) m η E η 1 z where z is the so-called energy-sharing parameter, defined as z E 1 E E 1 + E Note that this means that for a highly relativistic π 0 or η 0, there is a minimum γ opening angle given approximately by φ min o m/e. SOLUTION Working in the lab frame, the invariant mass of the system is given by M inv m η m 1 + m + E 1E (1 β 1 β cos φ o ) E 1 E (1 cos φ o ) where we used the fact that m 0 and β 1 for each photon to obtain the second line. If we use the trig identity 1 cos θ sin (θ/), we obtain immediately m η sin(φ o /) E 1 E sin(φ o /) m η / E 1 E With no loss of generality, we can let E 1 be the larger, and E the smaller, of the two photon energies. Then using conservation of total energy, and our definition of z, we see E 1 + E E η and E 1 E z E η so E 1 (E η /) (1 + z) and E (E η /) (1 z) Using these in our previous equation, we obtain the desired result sin(φ o /) m η (E η /) (1 z ) m η E η 1 z 5

6 3. Mary Found a Little Λ (a) Prove that if m 3 + m 4 > m 1 + m for the reaction (so the final rest mass energy exceeds that in the initial state), then if m (the target ) is initially the rest, the reaction can only proceed if m 1 (the beam ) has a lab kinetic energy greater than T min 1 (m 3 + m 4 ) (m 1 + m ) m (b) Now consider a Λ particle, with mass m Λ 1116 MeV, produced at threshold via the reaction γ p K + Λ, e.g., by having a beam of real photons impinge on a fixed hydrogen target. If you know that the Λ has a proper mean lifetime τ Λ s, what is the probability that the produced Λ gets at least 5 cm (lab frame) away from its production point before decaying? (c) A common decay mode for the Lambda is Λ p π. Using your results from part b, what is the maximum possible angle between the momenta of the parent Λ and the daughter proton, as measured in the lab? SOLUTION (a) Like many problems in kinematics, there are several ways to solve this, all of which are correct, but some of which will be much simpler algebraically than others. So, rather than just jumping in and using brute force E and p conservation in the lab (which will work), we start by realizing that bringing in just enough kinetic energy to allow the reaction to proceed means that we ll have no KE available in the final state. Thus, when T 1 is at T1 min, particles 3 and 4 will created at rest in the c.m. frame. Thus, E cm s m 3 + m 4 Here we see the power of working with Lorentz invariants: this is not only the total finalstate energy in all frames, but is also the total initial-state energy in all frames. Working out the latter in the lab (unprimed) frame, where p 0, gives us E cm [ (E 1 + E ) ( p 1 + p ) ] 1/ [ m 1 + m + m E 1 ] 1/ Equating these two expressions for E cm, squaring each side, and then using E 1 m 1 + T 1, gives us the nice result (m 3 + m 4 ) m 1 + m + m 1m + m T 1 Rearranging this, and recognizing T 1 as our threshold energy, we obtain the final result: T min 1 (m 3 + m 4 ) (m 1 + m ) m 6

7 (b) We now apply this result to Λ photo-production at threshold, bombarding a fixed proton (hydrogen) target with photons. To address the question of interest, we will obviously need to know the velocity of the produced Λ in the lab frame, which will depend on how much total energy is available. For the reaction γ p K + Λ, we see m 1 0, m MeV, m MeV, and m MeV. Sticking all of these into our result from part a tells us that E γ T min 1 (1609.4) (938.3) which is also p γ (as usual, I am taking the lab frame to be the unprimed frame). When created right at threshold, the outgoing Λ will be at rest in the c.m. frame, and so will be moving in the lab frame at velocity β Λ β cm p 1 E 1 + m where I have used an expression given in lecture for the c.m.-to-lab relative velocity. Now, an observer in the lab frame will see the Λ decay with a dilated mean lifetime τ Λ γ cm τ (0.493) s We can easily convert this to a distance, the lab-frame mean decay length, via l Λ β Λ c τ Λ (0.493) ( ) ( ) 4.47 cm Putting this all together, the probability for a produced Λ to travel at least 5 cm before decaying or equivalently, the fraction of all Λ s that survive this distance is given by P survive e d/ l e 5.0/ that is, about 1/3 of all Λ s will travel at least this far from the collision point where they are created before they undergo weak decay. This sets the scale for the accuracy required in the vertex reconstruction (primary and secondary) for such an experiment. (c) Fuhgeddaboudit! 7

Lecture 9 - Applications of 4 vectors, and some examples

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