PHYSICS 474: Introduction to Particle Physics Homework 1 Solution
(a) [10 points] How many different meson (qq¯) and baryon (qqq) combinations can be made with n quark flavors?
For n quark flavors, there are n choices for q and n forq ¯, so total n2 meson combinations . For example, 1 quark (u): 1 (uu¯) 2 quarks (u, d): 4 (uu,¯ ud,¯ du,¯ dd¯) 3 quarks (u, d, s): 9 (uu,¯ ud,¯ us,¯ du,¯ dd,¯ ds,¯ su,¯ sd,¯ ss¯), and so on.
As for the baryons, we can have three types of combinations: (i) All three quarks the same (e.g. uuu, ddd, sss, ··· ): n ways. (ii) Two same, one different (e.g. uud, uus, dds, ··· ...): n(n − 1) ways. (iii) All three different (e.g. uds, udb, cds, cdb, ··· ): n(n − 1)(n − 2)/6 ways. Here the factor 6 is to cover the equivalent permutations which give the same baryon (e.g. uds = usd = dus = dsu = sud = sdu). So the total is n+n(n−1)+n(n−1)(n−2)/6 = n(n + 1)(n + 2)/6 baryon combinations .
(b) [5 points] The Standard Model has 6 flavors of quarks (u, d, c, s, t, b). From your answer to part (a), how many varieties of mesons and baryons would you expect in Nature?
For 6 quarks, we expect 62 = 36 mesons and 6 · 7 · 8/6 = 56 baryons.
(c) Optional: Why don’t we see that many? (You can find the list of all mesons and baryons seen so far in the Particle Data Book.)
According to the Particle Data Book, all hadrons seen so far are various excitations of only 25 mesons and 23 baryons (plus anti-baryons) . No bound states involving the top quark have been found so far, mainly because top quark is too short-lived to form
1 bound states that can be observed with current experimental techniques.1 So from the above calculation, with 5 quark flavors, we should have seen 52 = 25 mesons and 5·6·7/6 = 35 baryons. So (only) 12 baryons are missing at the moment. If you want to know which ones, here they are: ucc, scc, ccc, ucb, dcb, scb, ccb, ubb, dbb, sbb, cbb, bbb. Hopefully they will be found soon!2
2. Nuclear β-decay:
(a) [10 points] In the early 1900s, the β-decay of an unstable nucleus was thought A A − to be a two-body decay: Z X →Z+1 Y + e . Find the energy and momentum of the outgoing electron in this case.
µ µ µ µ µ µ From four-momentum conservation, pX = pY + pe , or pY = pX − pe . Taking the scalar product of each side with itself, we get
2 2 2 pY = pX + pe − 2pX · pe . (1)
2 2 2 2 2 2 2 2 2 2 2 2 Since p = m c , we have pe = mec , pX = mX c and pY = mY c . Also in the rest 2 mX c Ee frame of X, pX = c , 0 and pe = c , pe . So m c2 E p · p = X e − 0 · p = m E . (2) X e c c e X e Substituting these into Eq. (1), we get
2 2 2 2 2 2 mY c = mX c + mec − 2mX Ee, 2 2 2 mX − mY + me 2 or, Ee = c . (3) 2mX
2 We also get the energy of Y using the energy conservation, i.e. mX c = EY + Ee, or 2 2 2 2 mX + mY − me 2 EY = mX c − Ee = c . (4) 2mX
As for the three-momentum, |pY | = |pe| due to momentum conservation. So we can use the energy-momentum relation on either Y or e (should expect the result to be
1 The lifetime of top quark is 4.7 × 10−25 sec, too short even for a strong interaction, which has typical
lifetime of 10−23 sec. 2 The counting for baryons is not completely correct, because they are fermions and we need to anti-
symmetrize the total wave function. This actually leads to more possibilities than the ones listed here.
We will discuss more about this at a later point in the course.
2 symmetric in mY and me):
2 2 2 2 4 Ee = |pe| c + mec , 2 2 2 2 2 2 2 Ee 2 2 (mX − mY + me) c 2 2 or, |pe| = 2 − mec = 2 − mec , c 4mX q c 4 4 4 2 2 2 2 2 2 or, |pe| = mX + mY + me − 2(mX mY + mX me + mY me) 2mX c 1/2 2 2 2 or, |pe| ≡ λ (mX , mY , me) , (5) 2mX where λ(x, y, z) = x2 + y2 + z2 − 2(xy + yz + zx) is the so-called triangle function.
(b) [5 points] Soon after Pauli’s proposal in 1930 that a (anti)neutrino must accompany the electron to conserve energy in the β-decay process, there was some confusion whether the neutrino could be coming from inside the nucleus. Use Heisenberg’s uncertainty relation to estimate the minimum momentum and energy of a neutrino confined to the nucleus (typical size 1 fm). Compare this with the typical neutrino energy in β-decay (a few keV) to show that the neutrino could not have come from inside the nucleus, but must have been produced in the decay process itself.
From Heisenberg’s uncertainty relation,
∆x · ∆p ≥ ~ , or ∆p ≥ ~ . (6) 2 2∆x
Taking the position uncertainty to be the size of the nucleus, we have ∆x = r0 = 1 fm = 10−15 m. So the minimum momentum of the neutrino coming from inside the nucleus should be
~ |p| & = 98.7 MeV/c . (7) 2r0
This means the minimum energy of such a neutrino should be
p 2 2 2 4 E = |p| c + mνc & 98.7 MeV . (8)
Compare this with the observed neutrino energy in β-decay which is less than a few keV, i.e. almost 10,000 times smaller! This suggests that the neutrino could not have been rattling around inside the nucleus, but must be produced in the decay process itself.
3 3. Pion decay:
A charged pion traveling at speed 0.9c decays into a muon and an antineutrino: π− → − µ +ν ¯µ.
(a) [10 points] If the neutrino emerges at 90◦ to the original pion direction, at what angle does the muon come off?
Four-momentum conservation requires pπ = pµ + pν, or pµ = pπ − pν (the trick is to choose the right combination that eliminates some unknown quantities). Taking the scalar product of each side with itself, we get
2 2 2 pµ = pπ + pν − 2pπ · pν . (9)
2 2 2 2 2 2 2 Since neutrinos are (almost) massless, pν = 0. Similarly, pπ = mπc and pµ = mµc . Also
E E p · p = π ν − p · p . (10) π ν c c π ν
◦ Since the neutrino emerges at 90 to the pion direction, pπ ⊥ pν, so pπ · pν = 0. Hence, Eq. (9) becomes
E E m2 c2 = m2 c2 − 2 π ν . (11) µ π c c
2 p 2 2 But Eπ = γmπc (where γ = 1/ 1 − v /c , v being the speed of the pion) and
Eν = |pν|c. So we get
2 2 2 2γmπ|pν|c = (mπ − mµ)c , 2 2 (mπ − mµ)c or, |pν| = . (12) 2γmπ
Also, |pπ| = γmπc. So if the muon comes off at an angle θ to the original pion direction, then from momentum conservation, we must have (see Figure 1
|pµ| cos θ = |pπ| , and |pµ| sin θ = |pν| . (13)
Thus,
2 2 2 2 |pν| (mπ − mµ)c 1 − mµ/mπ tan θ = = 2 2 = 2 , (14) |pπ| 2γ mπv 2βγ
4 ν
v 90○ π θ
μ
FIG. 1. Pion decay to muon and neutrino. The angle θ indicates the direction of muon.
2 2 where β = v/c. Substituting β = 0.9, mµ = 105.66 MeV/c and mπ = 139.57 MeV/c , we obtain θ = 2.6◦.
(b) [10 points] Find the energy and momentum of the muon.
From energy conservation, we have
2 2 2 2 2 (mπ − mµ)c Eµ = Eπ − Eν = γmπc − |pν|c = γmπc − = 307.2 MeV . 2γmπ (15)
From 3-momentum conservation [cf. Eq. (13)], we get s (m2 − m2 )c2 p 2 2 2 π µ |pµ| = |pπ| + |pν| = (γmπv) + = 288.5 MeV/c . (16) 2γmπ
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