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16-disintegration-of-particles

IN many cases the laws of conservation of momentum and energy alone can be used to obtain important results concerning the properties of various mech- anical processes. It should be noted that these properties are independent of the particular type of interaction between the particles involved. Let us consider a "spontaneous" disintegration (that is, one not due to external forces) of a particle into two "constituent parts", i.e. into two other particles which move independently after the disintegration. This process is most simply described in a frame of reference in which the particle is at rest before the disintegration. The law of conservation of momen- tum shows that the sum of the momenta of the two particles formed in the disintegration is then zero; that is, the particles move apart with equal and opposite momenta. The magnitude Po of either momentum is given by the law of conservation of energy: here M1 and m2 are the masses of the particles, E1t and E2i their internal energies, and E the internal energy of the original particle. If € is the "dis- integration energy", i.e. the difference E= E:-E11-E2i, (16.1) which must obviously be positive, then (16.2) which determines Po; here m is the reduced mass of the two particles. The velocities are V10 = Po/m1, V20 = Po/m2. Let us now change to a frame of reference in which the primary particle moves with velocity V before the break-up. This frame is usually called the laboratory system, or L system, in contradistinction to the centre-of-mass system, or C system, in which the total momentum is zero. Let us consider one of the resulting particles, and let V and V0 be its velocities in the L and the C system respectively. Evidently V = V+vo, or V -V = V0, and SO (16.3) where 0 is the angle at which this particle moves relative to the direction of the velocity V. This equation gives the velocity of the particle as a function 41 42 Collisions Between Particles §16 of its direction of motion in the L system. In Fig. 14 the velocity V is repre- sented by a vector drawn to any point on a circle+ of radius vo from a point A at a distance V from the centre. The cases V < vo and V>00 are shown in Figs. 14a, b respectively. In the former case 0 can have any value, but in the latter case the particle can move only forwards, at an angle 0 which does not exceed Omax, given by (16.4) this is the direction of the tangent from the point A to the circle. C V V VO VO max oo oo A V A V (a) V<VO )V>Vo FIG. 14 The relation between the angles 0 and Oo in the L and C systems is evi- dently (Fig. 14) tan (16.5) If this equation is solved for cos Oo, we obtain V 0 (16.6) For vo > V the relation between Oo and 0 is one-to-one (Fig. 14a). The plus sign must be taken in (16.6), so that Oo = 0 when 0 = 0. If vo < V, however, the relation is not one-to-one: for each value of 0 there are two values of Oo, which correspond to vectors V0 drawn from the centre of the circle to the points B and C (Fig. 14b), and are given by the two signs in (16.6). In physical applications we are usually concerned with the disintegration of not one but many similar particles, and this raises the problem of the distribution of the resulting particles in direction, energy, etc. We shall assume that the primary particles are randomly oriented in space, i.e. iso- tropically on average. t More precisely, to any point on a sphere of radius vo, of which Fig. 14 shows a diametral section. §16 Disintegration of particles 43 In the C system, this problem is very easily solved: every resulting particle (of a given kind) has the same energy, and their directions of motion are isotropically distributed. The latter fact depends on the assumption that the primary particles are randomly oriented, and can be expressed by saying that the fraction of particles entering a solid angle element doo is proportional to doo, i.e. equal to doo/4rr. The distribution with respect to the angle Oo is obtained by putting doo = 2m sin Oo doo, i.e. the corresponding fraction is (16.7) The corresponding distributions in the L system are obtained by an appropriate transformation. For example, let us calculate the kinetic energy distribution in the L system. Squaring the equation V = V0 + V, we have 2.2 = V02 + +200 cos Oo, whence d(cos 00) = d(v2)/200V. Using the kinetic energy T = 1mv2, where m is m1 or M2 depending on which kind of particle is under consideration, and substituting in (16.7), we find the re- quired distribution: (1/2mvov) dT. (16.8) The kinetic energy can take values between Tmin = 3m(e0-V)2 and Tmax = 1m(vo+V)2. The particles are, according to (16.8), distributed uniformly over this range. When a particle disintegrates into more than two parts, the laws of con- servation of energy and momentum naturally allow considerably more free- dom as regards the velocities and directions of motion of the resulting particles. In particular, the energies of these particles in the C system do not have determinate values. There is, however, an upper limit to the kinetic energy of any one of the resulting particles. To determine the limit, we consider the system formed by all these particles except the one concerned (whose mass is M1, say), and denote the "internal energy" of that system by Ei'. Then the kinetic energy of the particle M1 is, by (16.1) and (16.2), T10 = 2/2m1 = (M-m1)(E:-E1-EiM where M is the mass of the primary particle. It is evident that T10 has its greatest possible value when E/' is least. For this to be so, all the resulting particles except M1 must be moving with the same velocity. Then Ei is simply the sum of their internal energies, and the difference E;-E-E; is the disintegration energy E. Thus T10,max = (M - M1) E M. (16.9) PROBLEMS PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra- tion into two particles. SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling 010 simply Oo and using formula (16.5) for each of the two particles, we can put 7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then 44 Collisions Between Particles