Subatomic Physics

440 Strong Interactions
isotropic, as displayed in Fig. 14.12(a). (20) At higher energies, however, the behavior
is very different from the one sketched in Fig. 14.11.
A measurement at a neutron energy of 418 MeV is reproduced in
Fig. 14.12(b). (21) The differential cross section displays a pronounced peak in the
backward direction. Such a behavior cannot be understood with an ordinary potential
that leaves the neutron a neutron and the proton a proton.
It is evidence for an exchange force that changes the incoming
neutron into a proton through the exchange of
a charged meson with the target proton. The forwardmoving
neutron now has become a proton, and the recoiling
target proton a neutron. In effect, then, the neutron
is observed in the backward direction after scattering.
The exchange nature of the nucleon–nucleon force
can also be understood simply from the Yukawa meson
exchange theory. As shown in Fig. 14.13, the exchange
of a charged meson transfers the charge from the proton
to the neutron and vice versa, sothatanexchange
force results.
Figure 14.13: Charged-pionexchange
force between a
neutron and proton.
Spin–Orbit Force The existence of a spin–orbit interaction can be seen in scattering
experiments involving either polarized particles or polarized targets. (22) The
idea underlying such experiments can be explained with a simple example, the scattering
of polarized nucleons from a spinless target nucleus, for instance, 4 He or 12 C.
Assume that the nucleon–nucleus force is attractive; it then gives rise to trajectories
as shown in Fig. 14.14(a). Assume further that the two incoming protons are fully
polarized, with spins pointing “up,” perpendicular to the scattering plane. Proton
1, scattered to the right, has an orbital angular momentum L1 with respect to the
nucleus that is pointing “down.” Proton 2, scattered to the left, has its orbital
angular momentum L2 “up.” Assume that the nuclear force consists of two terms,
a central potential, Vc, andaspin–orbit potential of the form VLSL · σ,
V = Vc + VLSL · σ. (14.35)
Figure 14.14(b) implies that the scalar product L · σ has opposite signs for nucleons
1 and 2. Consequently, the total potential V is larger for one nucleon than for the
other, and more polarized nucleons will be scattered to one side than to the other.
20 J. C. Alred, A. H. Armstrong, and L. Rosen, Phys. Rev. 91, 90 (1953).
21 D. V. Bugg, Prog. Part. Nucl. Phys., (D. H. Wilkinson, ed.) 7, 47 (1981).
22 For a nice introduction see H.H. Barshall, Am. Jour. Phys. 35, 119 (1967); for current
issues see Proceedings of the 16th International Spin Physics Symposium and Workshop on Polarized
Electron Sources and Polarimeters, SPIN2004, F. Bradamante, A. Bressan, A. Martin, K.
Aulenbacher eds., World Sci. (2005).