117 lines
5.6 KiB
Markdown
117 lines
5.6 KiB
Markdown
§18
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§18. Scattering
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As already mentioned in §17, a complete calculation of the result of a
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collision between two particles (i.e. the determination of the angle x) requires
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the solution of the equations of motion for the particular law of interaction
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involved.
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We shall first consider the equivalent problem of the deflection of a single
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particle of mass m moving in a field U(r) whose centre is at rest (and is at
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the centre of mass of the two particles in the original problem).
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As has been shown in $14, the path of a particle in a central field is sym-
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metrical about a line from the centre to the nearest point in the orbit (OA
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in Fig. 18). Hence the two asymptotes to the orbit make equal angles (o,
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say) with this line. The angle X through which the particle is deflected as it
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passes the centre is seen from Fig. 18 to be
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X = -200.
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(18.1)
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A
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X
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to
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FIG. 18
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The angle do itself is given, according to (14.7), by
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(M/r2) dr
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(18.2)
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taken between the nearest approach to the centre and infinity. It should be
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recalled that rmin is a zero of the radicand.
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For an infinite motion, such as that considered here, it is convenient to
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use instead of the constants E and M the velocity Voo of the particle at infinity
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and the impact parameter p. The latter is the length of the perpendicular
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from the centre O to the direction of Voo, i.e. the distance at which the particle
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would pass the centre if there were no field of force (Fig. 18). The energy
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and the angular momentum are given in terms of these quantities by
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E = 1mvoo²,
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M = mpVoo,
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(18.3)
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§18
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Scattering
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49
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and formula (18.2) becomes
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dr
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(18.4)
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Together with (18.1), this gives X as a function of p.
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In physical applications we are usually concerned not with the deflection
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of a single particle but with the scattering of a beam of identical particles
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incident with uniform velocity Voo on the scattering centre. The different
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particles in the beam have different impact parameters and are therefore
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scattered through different angles X. Let dN be the number of particles
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scattered per unit time through angles between X and X + dx. This number
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itself is not suitable for describing the scattering process, since it is propor-
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tional to the density of the incident beam. We therefore use the ratio
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do = dN/n,
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(18.5)
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where n is the number of particles passing in unit time through unit area of
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the beam cross-section (the beam being assumed uniform over its cross-
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section). This ratio has the dimensions of area and is called the effective
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scattering cross-section. It is entirely determined by the form of the scattering
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field and is the most important characteristic of the scattering process.
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We shall suppose that the relation between X and P is one-to-one; this is
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so if the angle of scattering is a monotonically decreasing function of the
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impact parameter. In that case, only those particles whose impact parameters
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lie between p(x) and p(x) + dp(x) are scattered at angles between X and
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+ dx. The number of such particles is equal to the product of n and the
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area between two circles of radii P and p+dp, i.e. dN = 2mp dp n. The
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effective cross-section is therefore
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do = 2mp dp.
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(18.6)
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In order to find the dependence of do on the angle of scattering, we need
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only rewrite (18.6) as
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do = 2(x)|dp(x)/dx|dx
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(18.7)
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Here we use the modulus of the derivative dp/dx, since the derivative may
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be (and usually is) negative. t Often do is referred to the solid angle element
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do instead of the plane angle element dx. The solid angle between cones
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with vertical angles X and x+dx is do = 2n sin x dx. Hence we have from
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(18.7)
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do.
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(18.8)
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t If the function p(x) is many-valued, we must obviously take the sum of such expressions
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as (18.7) over all the branches of this function.
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50
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Collisions Between Particles
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§18
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Returning now to the problem of the scattering of a beam of particles, not
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by a fixed centre of force, but by other particles initially at rest, we can say
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that (18.7) gives the effective cross-section as a function of the angle of
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scattering in the centre-of-mass system. To find the corresponding expression
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as a function of the scattering angle 0 in the laboratory system, we must
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express X in (18.7) in terms of 0 by means of formulae (17.4). This gives
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expressions for both the scattering cross-section for the incident beam of
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particles (x in terms of 01) and that for the particles initially at rest (x in terms
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of O2).
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PROBLEMS
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PROBLEM 1. Determine the effective cross-section for scattering of particles from a perfectly
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rigid sphere of radius a (i.e. when the interaction is such that U = 8 for r < a and U = 0
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for r>a).
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SOLUTION. Since a particle moves freely outside the sphere and cannot penetrate into it,
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the path consists of two straight lines symmetrical about the radius to the point where the
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particle strikes the sphere (Fig. 19). It is evident from Fig. 19 that
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a sin to = a sin 1(-x) = a cos 1x.
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A
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to
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p
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&
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FIG. 19
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Substituting in (18.7) or (18.8), we have
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do = 1ma2 sin X do,
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(1)
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i.e. the scattering is isotropic in the C system. On integrating do over all angles, we find that
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the total cross-section o = na2, in accordance with the fact that the "impact area" which the
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particle must strike in order to be scattered is simply the cross-sectional area of the sphere.
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In order to change to the L system, X must be expressed in terms of 01 by (17.4). The
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calculations are entirely similar to those of $16, Problem 2, on account of the formal resemb-
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lance between formulae (17.4) and (16.5). For M1 < m2 (where M1 is the mass of the particle
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and m2 that of the sphere) we have
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do1,
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where do1 = 2n sin 01 d01. If, on the other hand, M2 < M1, then
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For m1 = M2, we have do = a²|cos 01| do1, which can also be obtained directly by sub-
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stituting X = 201 from (17.9) in (1).
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