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~ [u(kh"(l- y5)u(p)][u(p'h,,(1- y5)v(k')], (12.31)
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Weak Interactions
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see (12.11), where the spinors are labeled by the particle momenta. Recall that the outgoing Pe is described by v (k '). The muon decay rate can now be obtained using (4.36), (12.32) where the invariant phase space is
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d p' d k d k' (2'17')4 15 (4)( P _ p' _ k - k') (2'17' )3 2E , (2'17' )3 2 ", (2'17' )3 2 ",'
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d 1p' d 3k' 2 --8(E - E' - "")c5(p - p' - k') ), (2'17')5 2E' 2",'
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(12.33)
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with pO == E, kO == "', and so on, and where in reaching the last line we have performed the d 3k integration using
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= fd4k8("')c5(k2).
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(12.34)
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EXERCISE 12.8 Derive (12.34) by performing the dw integration on the righthand side (see Exercise 6.7).
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Using (12.31) and (12.29), we find the spin-averaged probability is
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= 64G 2(k p')(k', p),
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(12.35)
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where p = p' + k + k' on account of the d 4k integration performed in (12.33). Since m,. > 200m e , we can safely neglect the mass of the electron.
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EXERCISE 12.9 Verify (12.35). Neglect the mass of the electron, but not that of the muon. EXERCISE n.1O
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Show that
2(k p')(k', p) = (p - k,)2(k', p) = (m 2 - 2m",')m",'
(12.36) in the muon rest frame, where p
(m, 0, 0, 0).
Gathering these results together, the decay rate in the muon rest frame is
G 2 d 3p' = -- --
d 3k' --m",'(m 2 - 2m",') 2m '17' 5 2E' 2",'
xc5(m 2
2mE' - 2m",'
2E'",'(1 - cos8)),
(12.37)
Muon Decay
and, as for f3-decay, we can replace d 3p' d 3k' by
4'17E,2 dE'2'17w,2 dw'dcos 8.
We now use the fact that
S(
2E'w'cos8)
_l_ S ( ... - cos8) 2E'w'
to perform the integration over the opening angle 8 between the emitted e - and and obtain
--3dE'dw'mw'(m - 2w'). 2'17
(12.38)
The S-function integration introduces the following restrictions on the energies E " w', stemming from the fact that - 1 :$ cos 8 :$ 1:
!m -
!m, 0:$ E' :$ !m.
(12.39) (12.40)
These limits are easily understood in terms of the various limits in which the three-body decay 1L ~ efieP,. becomes effectively a two-body decay. For example, when the electron energy E' vanishes, (12.39) yields w' = m12, which is expected because then the two neutrinos share equally the muon's rest energy. To obtain the energy spectrum of the emitted electron, we perform the w' integration of (12.38):
2 df - mG - - _ - - f.!m d' ,( m- 2 w ') ww 3 !m-E' dE' 2'17
~m2E'2(3 3
12'17
_ 4E'). m
(12.41)
This prediction is in excellent agreement with the observed electron spectrum. Finally, we calculate the muon decay rate (12.42) Inserting the measured muon lifetime Fermi coupling G. We find
'T =
2.2 X 10 - 6 sec, we can calculate the
(12.43)
G - 10- 5 Im~.
Comparison of the values of G obtained in (12.24) and (12.43) supports the assertion that the weak coupling constant is the same for leptons and nucleons, and hence universal. It means that nuclear f3-decay and the decay of the muon have the same physical origin. Indeed, when all corrections are taken into
Weak Interactions
account, Gp and G,. are found to be equal to within a few percent: G,. = (1.16632 0.00002) X 10- 5 Gey- 2,
Gp = (1.136 0.003) X 10- 5 Gey-2.
(12.44)
The reason for the small difference is important and is discussed in Section 12.11 [see (12.107)].
EXERCISE 12.ll Draw a diagram showing the particle helicities in the p, - rest frame in the case where the emitted electron has its maximum permissible energy. In this limit, explain why the electron angular distribution has the form 1 .- P cos a, where P is the polarization of the muon and a is the angle between the polarization direction and the direction of the emitted electron:
where N are the numbers of spin-up, spin-down muons.
EXERCISE 12.12 "Predict" the rate for the decay -r- ~ e- PeP where the -r-Iepton has mass 1.8 GeY. The observed branching ratio of this decay mode is approximately 20%. Calculate the lifetime of the -r-Iepton. Can you explain this branching ratio