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