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Subatomic Physics

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11.14. The Weak Current of Hadrons at High Energies 371<br />

In the c.m. all momenta have the same magnitude so that the square of the<br />

momentum transfer becomes<br />

−q 2 =(p ν − p µ) 2 =2p 2 ν(1 − cos ϑ), (11.87)<br />

where ϑ is the c.m. scattering angle. With Eq. (11.87), the solid-angle element<br />

dΩ =2π sin ϑdϑ can be written as<br />

so that<br />

dΩ =− π<br />

p2 dq<br />

ν<br />

2 ,<br />

dσ = −1<br />

4π�4c2 E<br />

W |〈µ−N ′ |Hw|νN〉| 2 dq 2 . (11.88)<br />

The central problem is now the matrix element. At low energies, where the structure<br />

of the particles can be neglected, we have already considered weak 0 + → 0 +<br />

transitions caused by charged weak currents.<br />

The matrix element is given by Eq. (11.68), and the differential cross section in<br />

this case is<br />

dσ = − G2F V 2 ud<br />

2π�4c2 E<br />

W dq2 . (11.89)<br />

The total cross section is obtained by integrating over dq 2 . The minimum<br />

squared momentum transfer is −4p 2 ν, the maximum as given by Eq. (11.87) is 0,<br />

and the integration from 0 to −4p 2 ν yields<br />

σtot = 2G2 F V 2 ud<br />

π� 4 c 2<br />

E<br />

W p2 ν<br />

(11.90)<br />

For the case of spin zero considered here, the cross section is modified by a weak<br />

form factor, Fw, and Eq. (11.89) becomes<br />

dσ = − G2F V 2 ud<br />

2π�4c2 E<br />

W |Fw(q 2 )| 2 dq 2 . (11.91)<br />

The weak form factor Fw is predicted by the CVC hypothesis. Feynman and Gell-<br />

Mann postulated that the vector form factors appearing in the electromagnetic and<br />

in the weak currents must have the same form. For our simplified example CVC<br />

states that for the vector interaction<br />

Fw(q 2 )=Fem(q 2 ). (11.92)<br />

No spinless nucleons exist, and the form factor Fem for our specific example cannot<br />

be determined. However, we can assume that Fem has the same form as the form<br />

factors that appear in the nucleon structure. In particular we can identify Fem with<br />

GD as given in Eq. (6.42): The weak cross section then becomes, with Eq. (11.92),<br />

dσ = − G2F V 2 ud<br />

2π�4c2 E<br />

W<br />

dq 2<br />

(1 + |q2 |/q2 . (11.93)<br />

0 )4

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