173 - pr 05 - physics of a massive vector boson: Add to QED a massive massive photon field B B µ of mass M, which couples to electrons via ,
∫
µ
3
∆ H = gψγ ψ B µ ⋅ d x; g ≡ [massive-boson/massless-fermion coupling constant]
(1.1)
A massive photon photon in the initial initial or final state has three three possible possible physical polarizations corresponding corresponding to the three spacelike unit vectors in the boson’s rest frame. These can be characterized invariantly in terms of the boson’s 4-momentum k µ as the three vectors ε µ ( i ) , satisfying,
(ε (i ) ) µ • ε ( j ) = −δ ij → (ε µ ) • ε ( j ) = −δ ij ; k • e(i ) = 0;
(1.2)
1 The four vectors ( M complete orthonormal orthonormal basis. Because Because B µ couples to the conserved current ψγ µ ψ , k µ form a complete
the Ward identity implies that k µ dotted into the amplitude for B-production gives zero thus we can replace:
∑ε i
(i )
µ
ε ν( j )* → − g µν . This gives a generalization to massive bosons of the “Feynman trick” for photon-
polarization-vector polarization-vectorss and simplifies the calculation calculation of B production production cross cross sections. sections. (Warning: This This trick does does not work so simply for “non-Abelian gauge fields”…). Let’s do a few of these computations, using always the approximation of ignoring the mass of the electron.
+ −
a) Compute the cross section for the process: e e → B . Diagram for this process and its time-reversed counterpart (for consideration of decay, later) appears as,
(1.3) Cross section necessitates a computation of amplitude, and we use QFT 02 – 106 – [4.79], which reads, [4.79] → dσ =
1 (∏ f 2 E f
d 3 p f
(2 π )
3
)
M ( p A ,
pB → { p f }
2
2 E A ⋅ 2 EB ⋅ v A − vB
(2π ) 4 ⋅ δ ( 4 ) ( p A + pB −
∑
f
(1.4)
p f )
Corresponding amplitude is obviously iM = v ( p2 )[−igγ µ ]u ( p1 )ε µ* (k ) ≡ −igv2γ µ u1ε µ * , where this time we have a ′
′
polarization-vector; its square summed over polarizations, using vv =ɺ p , ε µ* ε ν = − g µν , γ µ γ µ γ µ = −2γ µ 1 , and Tr[1 p2 µ p1 µ ] = p1 • p2 , then appears as,
∑
M
2
g 2 Tr[v2v2γ µ u1u1γ ν ]ε µ*ε ν = 14 g 2 Tr[ p2γ µ p1γ ν ]( − g µν ) =
=
1 4
=
− p2 p1
µ′
µ ′′
− p2 p1
4
g 2 Tr[γ µ ′γ µ γ µ ′′γ µ ]
(1.5)
spins µ ′
4
µ′′
2
g Tr[γ µ ′ ( −2γ µ ′′ )] =
µ′
µ ′′
− p2 p1
4
2
µ
2
2
( −2) g Tr[4 g µ′µ ′′ ] = 2 g Tr[1 p2 p1µ ] = g ⋅ 2 p1 • p2
Meanwhile: the cross section (1.4) is computed in the center of mass frame, which is characterized by, E f = M ; EA = EB = 12 M ; vA − vB = 2; pi = p f = 0; p1 • p2 = ( M2 )2 − − ( M )2 = 12 M 2 ; 2
(1.6)
Thus, the cross section (1.4) is,
σ = ∫ dσ =
1
1
2 ⋅ 12 M ⋅ 2 ⋅ 12 M ⋅ 2 2M
∑ M spins
2
(2π )
4 −3
⋅δ
( 4 − 3)
( Ei − M ) =
π 2M 3
2
2
g ⋅ 2 ⋅ M δ ( Ei − M ) = 1 2
π g 2 2M
δ ( Ei − M ) (1.7)
Compute the lifetime of this B vector-particle assuming that it decays only to electrons . The lifetime is the inverse of the decay rate. The decay rate is given by QFT 02 – 107 – [4.86], appearing as,
1
4 (4) ∏ M ( m A → { p f } (2π ) δ ( pB − ∑ p f ) 3 2m A f (2π ) 2 E f
[4.86] → d Γ =
3
=
1
3
d p f
3
1 d p1d p2 1 2 M (2π )
3+ 3
2
1
M 2
2
2
∑ T [M ]
M 2 spins
2
(2π ) 4 δ (4) ( p B − E1 − E2 )
(1.8)
Here, T [M ] is the time reversed amplitude of (1.3), and it is computed using QFT 02 – 67 – [3.133]. The sizes of the spin-spaces: Particle B is a spin-1 particle and the products are spin-1/2 particles. The vectorparticle B can have 0 ⊕ 0 ⊕ 0 = 1 polarizations, and e+e- can have (0 ⊕ 0) ⊗ (0 ⊕ 0) = ( 12 ) ⊗ ( 12 ) = 4
polarizations, and the “polarization degeneracy” has the effect, 2 1 1 4 2 2 2 2 * 2 M M = 4 × × g ⋅ 2 p1 • p2 = T [M ] = g M ; = 4deg × × 3deg spins 3 3 spins spins
∑
∑
∑
(1.9)
Putting (1.9) into (1.8) with d 3 p2δ (3) (p B − p1 − p 2 ) = 1 for decay in the CM-frame, we get a decay rate,
∫
Γ = dΓ =
2 p1 d Ωdp1 4
1
∫ 2 M
3
(2π )
3+ 3− 3
2
2
1
g M (2π ) δ ( M − 2 p1 ) =
3
( 12 M ) 2 4π 4
1 3
2 M
(2π )
3
3
2
2
Mg 2
1
g M (2π ) =
12π
≡ Tlifetime
−1
(1.10)
Verify the relation discussed in Section 5.3, QFT 02 – 151 – [5.58] . this relation specialized (well, not really…) to our case appears as, + − + − + − 2 3Γ ( B → e e ) 2 2 + − 2 Γ( B → e e ) [5.58] → σ ( e e → B) = 4π ⋅ ⋅ δ ( ECM − M ) → σ ( e e → B ) = 12π δ ( s − M 2 ) (1.11) M M
Computing
12π 2 M
2
+
−
2
Γδ = 12M π Γ ( B → e e )δ ( s − M ) and using δ ( f ( x)) ≡
12π 2 M
Γδ =
12π 2 Mg 2 M
12π
2
2
δ ( Ei − M ) = π g
2
1 ∇ f
δ ( x − xi ) , we get,
δ (E − M ) + δ (E + M ) 2 Ei − 0
=
π g2 2 M
δ ( Ei − M ) = σ
(1.12)
(b) Compute the differential cross section in the center-of-mass system for the process: e+ e− → γ + B .
The differential cross section is given by, [4.85] →
dσ CM dΩ
=
M 2
2 2 CM
64π E
→
d σ d cos θ
=
1 128π ( p1 + k1)
2
∑ helicities
M
2
=
1 128π ( p1 + k1)
2
∑ s1 s2 r1 r2
M
2
;
(1.13)
By crossing-symmetry, you have contribution from the t and u Mandelstam-channels (c.f., QFT 02 -157). The consequence of this: for any pair-annihilation process you have contributions from the diagrams given by QFT 02 – 168, as,
(1.14)
Correspondingly: for [ε µ , γ ν ] = 0 , k 12 = 0 = [photon-mass] , the massive k2 2 = m B 2 , and p12 = m 2 , the Mandelstam variables 2 p1 µ k1 µ = − t and 2 p1 µ k2 µ = − u , and finally considering s, p12 >> m (i.e., we neglect the fermion masses) the amplitudes are, p1 − k1 + m ν µ * * M1 = v2 (−igγ ) (−ieγ )u1ε 1 µε 2ν 2 2 ( p1 − k1 ) − m = − gev2γ
p1 − k1 + m
ν µ
2
−2 p1 k1 µ + m + 0 − m
;
γ µ u1ε 1* µε 2*ν
2
ν
→ M1 + M 2 = gev2γ (
p1 − k2 + m
ν
= v2 (− ieγ )
M 2
= − gev2γ
−( k1 − p1 ) + 0 −t
2
( p1 − k 2 ) − m
*
ν µ
;
2
2
−2 p1 k 2 µ + m + m B − m −u
*
(−igγ )u1ε 1µε 2ν
p1 − k2 + m
−( k2 − p1 ) + 0
+
µ
2
2
γ µ u1ε 1*µε 2*ν (1.15)
)γ µ u1ε 1* µε 2*ν
2
On neglecting the fermion mass: We notice that m cancels in the denominator of the two amplitudes, so ignorance of the fermion-mass loses first-order physics, O1 (m) .
The amplitude in the differential cross section, then, is, seen in (1.4), which necessitates we compute, M
2
2
2
* * * = (M1 + M2 ) (M1 + M2 ) = M1 + M2 + M1 M2 + M1M2 ; M
2
→
∑
helicities
M
2
;
(1.16)
Interlude of trace-taking: Preparing ourselves for an obnoxious interlude of working out these terms, we first cite Casimir’s trace-trick,
∑
QFT02, [5.4] →
M
2
e4
=
q
helicities
4
Tr[( p ′ − me+ )γ µ ( p + me− )γ ν ]Tr[( k + me− )γ µ ( k ′ − me+ )γ ν ];
∑ [u ( p )Γ u ( p )][u ( p )Γ u( p )]
QFT00, [7.125] →
a
b
1
a
(1.17)
*
b
2
= Tr Γ1 ( pb + mb ) Γ 2 ( pa + ma ) ; Γ ≡ γ Γ 2 γ ; 0
†
0
all spins
Then, computing the amplitudes of (1.16) using (1.17) upon (1.15) and ΣM =
= = ΣM ≡
1 4e2 g 2 1
∑
M
2
= Tr[
spins
ν
g µµ ′ gνν ′ Tr[(γ (
4 g µµ′ gνν ′ 4 g µµ′ gνν ′ 4
4e 2 g 2
k1 − p1 t
ν
Tr[
1
µ
γ ( k1 − p1 )γ k1γ
ν′
+
∑
M
1
]=
4
spins
k2 − p1 u
( k1 − p1 )γ
µ′
k2
t 2
′ ′
2
+
′ ′
Tr[
∑
k1 − p1 t
+
k2 − p1 u
(i )
µ
i
εν( j )* = ∑ spins ε µ( i )εν( j )* → − g µν , 2
)γ µ u1 ⋅ ε 1* µε 2*ν ]
spins ′
)γ µ k1 )(γ ν ( ν
v2γ ν (
∑ε
k1 − p1 t
ν ′
µ
+
k 2 − p1 u
µ′
ν
′
)γ µ k2 )]
ν′
µ′
(k1 − p1 ) γ k2
ut
′ ′
1 u2
′ ′
T3µνµ ν ]
(1.18) µ
γ ( k1 − p1 ) γ k1γ ( k 2 − p1 ) γ k2 + γ ( k2 − p1 )γ k1γ
1 Tr[ t12 T1µνµ ν + ut (T2 µνµ ν + T2′µνµ ν ) +
+
ν
µ
ν′
µ′
γ ( k2 − p1 ) γ k1 γ (k 2 − p1 ) γ k2 u2
]
In which, ′ ′ ′ ′ ′′ ′′ ′′′ ′′′ ′′ ′′′ ′ ′ Tr T1 µνµ ν = Tr[γ ν ( k1 − p1 )γ µ k1γ ν ( k1 − p1 )γ µ k2 ] = (k1ν − p1ν )(k1ν − p1ν )k1µ k 2 µ Tr[ γ ν γ ν ′′γ µ γ µ′′γ ν γ ν ′′′γ µ γ µ′′′ ]; ′ ′
′
′
′′
′′
′′′
′′′
′′
′′′
′
′
γ ν γ ν ′′γ µ γ µ′′γ ν γ ν ′′′γ µ γ µ′′′ ]; Tr T2 µνµ ν = Tr[γ ν ( k1 − p1 )γ µ k1γ ν ( k 2 − p1 )γ µ k 2 ] = (k 1ν − p1ν )(k 2ν − p1ν )k1µ k 2 µ T r[ (1.19) ′ ′ ′′′ ′′′ ′′ ′′ ′′ ′′′ ′ ′ Tr T2′ µνµ ν = ( k2ν − p1ν )(k1ν − p1ν )k1µ k 2 µ Tr[ γ ν γ ν ′′γ µ γ µ′′γ ν γ ν ′′′γ µ γ µ′′′ ]; µνµ ′ν ′
T3
ν ′′′
= (k2
ν ′′′
ν ′′
ν ′′
µ′′
− p1 )(k 2 − p1 )k1 k 2
µ′′′
ν
µ
ν ′
µ ′
γ γ ν ′′γ γ µ ′′γ γ ν ′′′ γ γ µ ′′′ ]; Tr[
We already worked out what the trace of an 8-fold product of independent-gammas is (and we wish to not do it 1 again); please consult QFT 02 - 170 - pr 02 – Bhabha scattering – (1.16) to (1.19) and (1.26) ). The results are, ′ ′ ′ ′ g µµ′ gνν ′ Tr T1 µνµ ν = 32( k1 • p1 )(k 2 • p1 ) = ( −322) 2 ut = g µµ ′ gνν ′ Tr T2 µνµ ν = 32(k1 • p1 )(k 2 • p1 ); (1.20) ′ ′ ′ ′ g µµ′ gνν ′ (Tr T2 µνµ ν + Tr T2′ µνµ ν ) = 64( k1 • k 2 ) ( k1 • k 2 − k1 • k1 − k 2 • p1 ) = ( −642) 2 s ( s + t + u ) ; Putting (1.20) into (1.19), and subsequently into (1.18), we get an amplitude, in which we can use s + t + u = (M B + Eγ ) 2 , where we consider Eγ << M B so s + t + u = M B 2 , and k1 • p1 ⋅ k 2 • p1 = ut , yielding, 1
∑ M 4
2
2
=
e g
spins
4
2
[(
1 t
2
+
1 u
2
)32( k1 • p1 )( k2 • p1 ) +
16 s( s + t + u ) ut
u t M s ] = 4e 2 g 2[2( + ) + B ] t u ut
(1.21)
Subsequently: use t = ( M B 2 − s) sin 2 θ 2 and u = ( M B 2 − s ) cos 2 θ2 = t cot 2 θ 2 , we get the cross section by putting (1.21) into (1.13), and, non-dimensionalizing using 1 ≤ x = M B 2 / s ≤ 0 , we get, dσ d cos θ
2
=
e g
2
u
2
t
M B s
[2( + ) + ]= 2π s t u ut
2
α g (1 − M B
2
M 2 / s
2 θ 2 B / s) 1 + (2 cos 2 − 1) + 4 (1− M 2 / s ) 2 B
2sin θ2 cos θ 2
2s
2
log @ D_XS *sê g 2 D
D_XS *sê g2 x=0.01
2 x α g 2 (1 − x) 1 + cos θ + 4 (1− x ) (1.22) = 2s sin 2 θ
50
x=0.01
x=0.3 x=0.3
40
x=0.9
10
x=0.9
x=0.999
x=0.999
30
5
20 q -3
-2
-1
1
2
3
10
-5
q -3
-2
-1
1
2
3
;
(1.23)
+ − 0 Afterword: The good news: this calculation goes over almost unchanged to the real process: e e → γ + Z . 0 This allows one to measure the number of decays of the Z into unobserved final states which is, in turn, proportional to the number of neutrino species .
(c) Notice that the cross section of part b (1.22) diverges as θ → 0, π . Let us analyze the region near θ = 0 , where the dominant contribution comes from the t-channel diagram and corresponds intuitively to the emission of a photon from the electron line before e+e- annihilation into a B-particle (see (1.14), righthand-diagram). Let us rearrange the formula (1.22) in such a way as to support this interpretation. First: note that the divergence as θ → 0 is cut off by the electron mass. Let the electron momentum be k1 µ = ( E , 0, 0, k ) and let the photon momentum be p1 µ = ( xE , xE sin θ , 0, xE cos θ ) with k = E 2 − me 2 . Under these conditions:
conservation says p2 µ = ((1 − x ) E , − xE sin θ , 0, − xE cos θ ) we have, and the Mandelstam variables appear as,
1
Please note that square brackets indicate Equation-numbers in the text , vs. the parentheses I use here which indicate equation numbers in my homework. In sh ort: look at Problem 5.2 worked in this assignment-submission.
CM
s = (k1
CM
CM
2
+ k2 ) = ( p1
2
+ p2 ) = 4 E ; t = (k1 − p1 ) = ( E (1 − x), − xE sin θ , 0, k − xE cos θ ) ; CM
2
2
2
2
(1.24)
2
u = (k1 − p2 ) = ( (1 − 1 − − x) E , 0 − − xE sin θ , 0, k − − xE cos θ ) = ( xE , xE sin θ , 0, k + xE cos θ ) ; 2
Show that the denominator of the photon (book typo) propagator then never becomes smaller than
O ( me
2
/ s)
for all x. Correspondingly: the photon-propagator from QFT 00 – 244-246, in which θ = O ( m / s ) is the
tensor-element G = Gt = −ig µν q −2 = g µν / (it ) , in which t is given by (1.24) and so we have, 2
t = E 2 (1 − x) 2 − ( k − xE cos θ ) 2 − ( xE sin θ ) 2 = E 2 (1 − x) 2 − ( k − xE ) 2 − x 2 E 2 2
2
2
2
2
2
2
2
2
2
2
≈ E + E x − 2 xE − E + m − x E + 2 E − m xE − 14 x m
m s
2
+ O (θ )
2
(1.25)
t / m 2 = 2 x( Em ) ( mE ) 2 − 1 − 2 x( mE ) 2 + (1 − 14 x 2 ) = 2 x χ χ 2 − 1 − 2 x χ 2 + (1 − 14 x 2 ) >
1 4
χ −2 =
m2 2
4 E
= m 2 / s
Now: integrate (1.22) over forward-angles, cutting off the θ -integral at θ 2 ~ me 2 / s , keeping only the leading logarithmic term proportional to log( me 2 / s ) . Effecting this cut-off-integration upon (1.22) for ξ = cosθ and
the χ 0 = s / m 2 , we get, σ =∫
π
0
=
dσ d cos θ
sin θ dθ =
α g 2 (1 − x ) 2s
∫
+1− χ 0
−1+ χ0
2
(
x +1
( x − 1)
2
2 ln
dσ dξ
2 − χ 0
χ 0
dξ =
α g 2 (1 − x ) 2s
∫
+1− χ 0
−1+ χ 0
− (2 − 2 χ 0 )) =
1 + ξ 2 + 4 (1− x x )2 1−ξ
2
α g 2 (1 − x) x 2 + 1 s
[
( x − 1)
2
ln
d ξ =
α g 2 (1 − x ) x 2 + 1
2 − χ 0
χ 0
2s
(
( x − 1)
− (1 − χ 0 )] =
2
ln
1+ ξ 1−ξ
α g 2 x 2 + 1 x −1
s
ln
+1− χ
− ξ ) −1+ χ 00
(1.26)
χ 0 + O( χ 0 ) 2 − χ 0
Show that in this approximation the cross section for forward photon emission can be written as, +
−
Σ ≡ σ (e e → γ + B ) ≈
∫
ECM = (1− x ) s
s α 1 + (1 − x) 2 log 2 ; (1.27) f ( x )σ (e e → B )dx = lim ∫ f ( x )σ ⋅ dx; f ( x ) = x → 0 x me 2π +
−
where the annihilation cross section is evaluated for the collision of a positron of energy E and an electron of energy (1 − x) E and the function f ( x) being the Weiszacker-Williams distribution function, as given. Given the
four-momenta introduced in (1.24) and before, we have that σ (e + e− → B) is just the x = 0 case of the full cross section σ (e+ e − → γ + B) . Computing the antiderivative of the RHS of (1.27) as instructed by the integral, and putting (1.26) for x = 0 into (1.27) we get, χ 0 χ 0 1 + (1 − x )2 α 1 + (1 − x ) 2 α g 2 x2 + 1 α 2 g 2 ln χ 0 ln ln χ 0 ln lim Σ = lim ⋅ dx = ⋅ dx x →0 2π 2 − χ0 2 − χ 0 x→0 x s x −1 −2π s x (1.28) ???? χ 0 χ 0 α 2g2 α g 2 x 2 + 1 + − ln χ 0 ln ln x = ... = ln = = σ ( e e → γ + B) 2 − χ0 2 − χ 0 s x −1 −π s
∫
∫
Afterword: This function arises universally in processes in which a photon is emitted collinearly from an electron line, independent of the subsequent dynamics. We will meet it again in another context in Problem 6.2.
