Small-angle scattering 55 Finally, differentiating, we have the effective cross-section cos do. The angle X varies from zero (for p=0) to Xmax (for P = a), where cos 1xmax = 1/n. The total effective cross-section, obtained by integrating do over all angles within the cone Xmax, is, of course, equal to the geometrical cross-section 2 a to a FIG. 21 §20. Small-angle scattering The calculation of the effective cross-section is much simplified if only those collisions are considered for which the impact parameter is large, so that the field U is weak and the angles of deflection are small. The calculation can be carried out in the laboratory system, and the centre-of-mass system need not be used. We take the x-axis in the direction of the initial momentum of the scattered particle M1, and the xy-plane in the plane of scattering. Denoting by P1’ the momentum of the particle after scattering, we evidently have sin 01 = P1y’/P1’. For small deflections, sin 01 may be approximately replaced by 01, and P1’ in the denominator by the initial momentum P1 = MIUoo: (20.1) Next, since Py = Fy, the total increment of momentum in the y-direction is (20.2) The force Fy = - JULY (dU/dr)dr/dy = (dU/dr)y/r. Since the integral (20.2) already contains the small quantity U, it can be calculated, in the same approximation, by assuming that the particle is not deflected at all from its initial path, i.e. that it moves in a straight line y = p with uniform velocity Voo. Thus we put in (20.2) Fy = - (dU/dr)p/r, dt = dx/voo. The result is 56 Collisions Between Particles §20 Finally, we change the integration over x to one over r. Since, for a straight path, r2 = x2+p2, when x varies from - 00 to + 80, r varies from 8 to P and back. The integral over x therefore becomes twice the integral over r from p to 80, and dx = r dr/v(r2-p2). The angle of scattering O1 is thus given byt (20.3) and this is the form of the function 01(p) for small deflections. The effective cross-section for scattering (in the L system) is obtained from (18.8) with 01 instead of X, where sin 01 may now be replaced by A1: (20.4) PROBLEMS PROBLEM 1. Derive formula (20.3) from (18.4). SOLUTION. In order to avoid spurious divergences, we write (18.4) in the form PO and take as the upper limit some large finite quantity R, afterwards taking the value as R 00. Since U is small, we expand the square root in powers of U, and approximately replace rmin by p: dr The first integral tends to 1/11 as R 00. The second integral is integrated by parts, giving = This is equivalent to (20.3). PROBLEM 2. Determine the effective cross-section for small-angle scattering in a field U=a/m(n) 0). t If the above derivation is applied in the C system, the expression obtained for X is the same with m in place of M1, in accordance with the fact that the small angles 01 and X are related by (see (17.4)) 01 = m2x/(m1 +m2). §20 Small-angle scattering 57 SOLUTION. From (20.3) we have dr The substitution p2/r2 = U converts the integral to a beta function, which can be expressed in terms of gamma functions: Expressing P in terms of 01 and substituting in (20.4), we obtain do1. 3 CHAPTER V SMALL OSCILLATIONS $21. Free oscillations in one dimension A VERY common form of motion of mechanical systems is what are called small oscillations of a system about a position of stable equilibrium. We shall consider first of all the simplest case, that of a system with only one degree of freedom. Stable equilibrium corresponds to a position of the system in which its potential energy U(q) is a minimum. A movement away from this position results in the setting up of a force - dU/dq which tends to return the system to equilibrium. Let the equilibrium value of the generalised co-ordinate q be 90. For small deviations from the equilibrium position, it is sufficient to retain the first non-vanishing term in the expansion of the difference U(q) - U(90) in powers of q-qo. In general this is the second-order term: U(q) - U(q0) 112 1k(q-90)2, where k is a positive coefficient, the value of the second derivative U”(q) for q = 90. We shall measure the potential energy from its minimum value, i.e. put U(qo) = 0, and use the symbol x = q-90 (21.1) for the deviation of the co-ordinate from its equilibrium value. Thus U(x) = . (21.2) The kinetic energy of a system with one degree of freedom is in general of the form 1a(q)q2 = 1a(q)x2. In the same approximation, it is sufficient to replace the function a(q) by its value at q = qo. Putting for brevity a(go) = m, we have the following expression for the Lagrangian of a system executing small oscillations in one dimension: L = 1mx2-1kx2. (21.3) The corresponding equation of motion is m+kx=0, (21.4) or w2x=0, (21.5) where w= ((k/m). (21.6) + It should be noticed that m is the mass only if x is the Cartesian co-ordinate. + Such a system is often called a one-dimensional oscillator. 58