Elastic Scattering. R = m 1r 1 + m 2 r 2 m 1 + m 2. is the center of mass which is known to move with a constant velocity (see previous lectures):

Transcription

1 Elastic Scattering In this section we will consider a problem of scattering of two particles obeying Newtonian mechanics. The problem of scattering can be viewed as a truncated version of dynamic problem where one is not interested in all the details of motion except for just the asymptotic (at t + ) values of velocities of the two particles, v 1 and v, as functions of their initial (at t ) values of velocities, v 1 and v, and the geometry of the collision. (Later on we will see that the only relevant geometric quantity is the so-called impact parameter equal to the distance between two parallel lines which the two particles follow in the center-of-mass reference frame prior to the collision.) Kinematics of the pair collision We start with the kinematics of the pair collision. That is we establish a number of crucial features of the problem that follow purely from the conservation of the total momentum and energy. These features are especially transparent in the center-of-mass (CM) reference frame. So, our first step is to introduce the CM-frame by Galilean transformation where r i r i R(t) (i = 1, ), (1) R = m 1r 1 + m r () is the center of mass which is known to move with a constant velocity (see previous lectures): v c = Ṙ = m 1v 1 + m v = P = const. (3) In accordance with (1), the initial velocities of the two particles in the CM-frame are where v c1 = v 1 v c = m (v 1 v ) = m v v c = v v c = m 1(v v 1 ) = m 1v = m m 1 v, (4) = m m v, (5) v = v 1 v = v c1 v c (6) is the relative velocity which is one in the same in both reference frames (Galilean transformation does not change relative distances, and, correspondingly, relative velocities), and is the reduced mass. For the momenta of the two particles in the CM-frame we have m = m 1m (7) p c1 = m 1 v c1 = m v, (8) The fact that p c = m v c = m v. (9) p c = p c1 (10) 1

2 is clear from a simple observation that for any mechanical system the total momentum is always zero in the CM-frame. It is this fact that renders the kinematics of the pair collision very simple. By conservation of the total momentum we have On the other hand, the conservation of energy implies p c = p c1. (11) p c1 m 1 + p c m = (p c1 ) m 1 + (p c ) m, (1) which being combined with (10) and (11) immediately yield a simple and very important result: p c1 = p c = p c1 = p c. (13) On the basis of Eqs. (10), (11), and (13) we conclude that in the CM-frame the momentum of each of the two particles has the same absolute value before and after the collision, and just rotates by some angle χ, the same for both particles. Conservation of the angular momentum of relative motion (see previous lectures; note that the relative motion is the same in both reference systems) L = m r ṙ, r = r 1 r (14) requires that the trajectory lie in the plane perpendicular to L. Hence, the rotation of the momenta by the angle χ should take place in this particular plane. It is convenient to introduce a unit vector n 0 pointing along the vector p c1, and write the kinematic relations in terms of n 0 : Dividing momenta by masses, we get the velocities: n 0 = p c1/p c1 = v c1/v c1, (15) p c1 = p c1 n 0 = m vn 0, (16) p c = p c1 n 0 = m vn 0. (17) v c1 = m m 1 vn 0, (18) v c = m m vn 0. (19) Finally, returning back to the original reference fame (by just adding v c to the velocities) we find v 1 v = m m 1 vn 0 + v c = m 1v 1 + m v + m vn 0, (0) = m m vn 0 + v c = m 1v 1 + m v m 1 vn 0. (1) Important special case: m 1 = m, v = 0 This case corresponds to a typical experimental setup for the scattering of two identical particles when one particle is at rest (playing the role of a target ). Now we have v = v 1, v c1 = v 1 / = v/. ()

3 and Eqs. (0)-(1) are reduced to It is then useful to note that v 1 = v + vn 0, (3) v = v vn 0. (4) v 1 + v = v (5) and to represent the kinematic relations geometrically, see Fig. 8. This diagram allows us to easily find the relations for two important angles: the angle θ 1 between v 1 and v 1 and the angle θ between v and v 1. Indeed, since AO = OB = OC = v/, we readily see that θ 1 = χ, θ = π χ, (6) and, in particular, that independently of χ we always have θ 1 + θ = π/. (7) Dynamical part of the problem We have demonstrated that the dynamical part of the problem reduces to finding the angle χ of the rotation of the momenta/velocities in the CM-frame. From the previous analysis of the twobody problem we know that geometric features of the orbits and the angle χ is one of them can be extracted from the form of the interaction potential, given the angular momentum, l = L, and energy, E, of the relative motion (the masses of particle are assumed to be given as well). In the CM-frame, the radius-vectors of the two particles are simply proportional to the radiusvector r of their relative motion: r c1 = m r = m m 1 r, r c = m 1 r = m m r. (8) Hence, the angle of the rotation of the velocity of, say, the first particle, ṙ c1, is precisely the same as the angle of the rotation of the velocity of the relative motion, ṙ. And the latter is trivially related to the total polar angle, θ, swept out by the trajectory from t = to t = + (up to a sign which is not important for our purposes): χ = π θ. (9) Taking into account the reflectional symmetry of the orbit with respect to its perigee, we have θ = θ 0, where θ 0 is the angle swept out by the radius-vector while moving from the perigee to infinity (as well as from infinity to the perigee). Hence, χ = π θ 0. (30) The explicit expression for θ 0 is obtained from the general relation for the polar angle of an orbit as a function of polar radius: θ = ± l m We have to integrate from the perigee radius r p to infinity: dr/r [E Ueff (r)], U eff(r) = U(r) + l m r. (31) θ 0 = l m r p dr/r [E Ueff (r)]. (3) 3

4 Figure 1: Diagram for the special case of m 1 = m, v = 0. 4

5 Now we need to relate the two parameters, E and l, to the initial parameters of the scattering problem. Clearly enough, E = m v = m 1 m (v 1 v ). (33) ( ) The angular momentum l is conveniently related to the so-called impact parameter, b, that is nothing else than the distance between two parallel lines in the CM-frame representing hypothetic trajectories of the two particles in the absence of interaction. From Eq. (14) it is seen that l = m bv = m 1 m b v 1 v ( ). (34) Expressing E and l in terms of v and b, we get θ = ±b dr/r 1 (b/r) U(r)/m v, (35) dr/r θ 0 = b r p 1 (b/r) U(r)/m v, (36) dr/r χ = π b r p 1 (b/r) U(r)/m v. (37) The perigee radius r p r p (v, b) is found from the requirement that the expression in the denominator is equal to zero. Note that the impact parameter b has the same geometric meaning and, correspondingly, the same value in any reference frame where the initial velocities of the two particles are parallel to each other (including the case when one of the two velocities is just zero). This simple observation is very useful for practical purposes. Power-law potentials: Analysis of dimensions. Similarity Here we analyze the scattering problem in the potentials U(r) = γ/r α, with α > 0 and arbitrary sign of γ. We want to introduce proper dimensionless scaling variables revealing the similarities. Since we have already performed such an analysis when considering the motion in attractive power-law potentials, we could simply employ the same variables, ρ = r/r 0 and ε = E/ E min, here. (In the case of repulsive potentials, we could use their attractive counterparts, γ γ, for introducing r 0 and E min.) However, for the scattering problem there exist even more natural scaling variables; we describe them below. In the CM-frame, the scattering is described by the angle χ as a function of five numbers: two dimensional variables, v and b, characterizing the initial conditions; two dimensional parameters, m and γ; and a special dimensionless parameter α characterizing the type of the potential. We have already seen that the shape of the trajectory depends up to a scale of distance, which is not relevant to the angles on just two dimensionless numbers, ε and α. Correspondingly, we understand that the four dimensional parameters, v, b, m, γ can be combined in a single dimensionless variable. Without loss of generality, we can introduce such a variable, β, in the form β = b/b 0, (38) where b 0 b 0 (v, γ, m, α) is a characteristic value of the impact parameter corresponding to χ 1 at given (v, γ, m, α). The physical meaning of the scaling variable β is transparent and directly relevant to the scattering problem: At β 1 the scattering is weak in the sense that χ 1. One can easily 5

6 find an expression for b 0 on the basis of the following reasoning. At b b 0 the scattering is supposed to be weak, with the trajectory close to a straight line. But this is only possible if the potential energy is always negligible as compared to the kinetic energy. The largest absolute value of the potential energy is at the perigee, r p b. Hence, χ 1 Correspondingly, χ 1 By definition, χ 1 means b b 0, and we get m v m v γ b α. (39) γ b α. (40) b 0 ( ) 1 γ α m v (41) Since b 0 is defined up to an arbitrary dimensionless factor of order unity, we can always choose b 0 = ( ) 1 γ α, (4) m v and thus β = b ( m v γ ) 1 α. (43) The parameter b 0 is a natural scale of distance that suggests a dimensionless variable, ρ, for the radius: r = b 0 ρ. (44) Now we take the relation (35) with our potential, θ = ±b and substitute r = b 0 ρ. The result is In particular, θ = ±β dr/r 1 (b/r) γ/m v r α, (45) d ρ/ ρ, σ = γ/ γ = ±1. (46) 1 (β/ ρ) σ/ ρ α d ρ/ ρ θ 0 = ±β ρ p 1 (β/ ρ) σ/ ρ. (47) α (Remember that ρ p is defined as a point where the denominator hits zero.) As we expected, the answer for θ 0 depends only on β and α (apart from the sign of γ which is a discrete variable). It is instructive to establish a relation between two dimensionless variables, ρ and ρ. Clearly, ρ/ρ = r 0 /b 0, and we need to relate b 0 to r 0. The answer is r 0 = [( α)ε] 1/α b 0, (48) and thus ρ = [( α)ε] 1/α ρ. (49) 6

7 One can also directly relate ε and β: ( α)α α/( α) ε = β α/( α) (ε > 0). (50) Problem 17. Derive Eqs. (48) and (50). Coulomb Potential The case of Coulomb potential, U(r) = γ/r, is especially important it corresponds to the scattering of charged particles. The problem can be solved analytically. Eq. (46) now reads θ = ±β d ρ/ ρ, σ = γ/ γ = ±1. (51) 1 (β/ ρ) σ/ ρ The substitution reduces it to θ = ±β dξ 1 (βξ) σξ ρ = 1/ξ (5), σ = γ/ γ = ±1. (53) The new integral is similar to the ones we were dealing with when considering time evolution in the Coulomb potential. It is readily done by the same technique, and the result is θ = ±β arccos β ξ + σ 1 + β + const. (54) The particular value of the free constant of integration is associated with the direction of the x- axis. Without loss of generality, we can set const = 0, which turns out to be a very good choice corresponding to the x-axis pointing at the perigee. We thus get β ρ Noticing that in the case of α = 1 we have = σ β cos θ. (55) β = ε (ε > 0), ρ = ερ, (56) we realize that (55) is nothing else than Eq. (76) of the previous section (at ε > 0), generalized to the repulsive case. The solution to the scattering problem then comes in the form cos θ 0 = σ 1 + β. (57) Problem 18. Show that Eq. (57) implies that: (i) χ(β = ) = 0, (ii) χ(β = 0) = π, (iii) χ(β = 1) = π/. Also, find approximate formula for χ(β) in the case β 1. Hint. At β 1 one can Taylor-expand both sides of Eq. (57) up to the first β-dependent term. 7

8 Finally, the relation between the angle χ and the dimensionless impact parameter β can be written in the form β = cot χ. (58) For the dimensional impact parameter b we thus have b = b 0 cot χ, b 0 = γ m v. (59) Problem 19. Derive Eq. (58). Problem 0. A moving α-particle ( 4 He nucleus) is being scattered by a nucleus of gold, which is at rest. How large should be the velocity of the α-particle for the characteristic impact parameter to be less than 1 Angstrom? Cross Section In this section we elaborate on a language for describing typical experimental set-up when an incident beam of particles of sort 1 is being scattered by a target consisting of particles of sort. All the particles of the incident beam have the same velocity v while the particles of the target are at rest. For the significant scattering events (scattering angles of order unity) to be of two-body nature, we have to require that typical impact parameter for such events, b 0, be much smaller than the interparticle distance in both the beam and the target. These are the conditions of diluteness. If n beam and n target are the number densities of the beam and the target, respectively, then the conditions of diluteness read b 0 n 1/3 beam 1, b 0 n 1/3 target 1. (60) (The typical distance between the particles is the inverse cubic root of the number density.) The conditions of diluteness do not yet guarantee the absence of the multiple scattering, a cascade of successive pair scattering events triggered by the first scattering. The multiple scattering is negligible if the target is thin enough. If d is the width of the target in the direction of the incident beam, then the quantitative criterion for the multiple scattering to be negligible is b 0 (n beam d) 1/ 1, b 0 (n target d) 1/ 1. (61) Problem 1. Argue that Eq. (61) guarantees the absence of multiple scattering. Hint. It is useful to start with realizing that Eq. (61) is the condition of transparency of the target, meaning that the probability of a particle of the incident beam to be significantly scattered is very small. In a realistic experimental situation, there is practically no way to control the impact parameter of each individual collision. Here one has to resort to a statistical description in terms of the flux of particles. The flux, J = n beam v, is the number of particles crossing unit transverse (with respect to velocity) area per unit time. Normally, the width of the incident beam is dramatically larger than the characteristic impact parameter. This means that with a very high accuracy the flux of the incident beam can be treated as spatially homogeneous. The two fluxes of the scattered particles scattered particles of the incident beam and scattered particles of the target are spatially inhomogeneous and, 8

9 generally speaking, anisotropic. The anisotropy comes from the dependence of the scattering angle on the impact parameter. The spatial dependence of the flux of scattered particles has a simple asymptotic structure. (Below we use spherical coordinates, with the origin at the target, and z-axis along the direction of the incident beam.) If we take a small enough solid angle, Ω = sin θ θ ϕ ( θ 1, ϕ 1), then the flux within this solid angle is locally isotropic: the dependence on the angles, if any, is negligibly small due to θ 1, ϕ 1. The dependence of the flux on the distance r from the target is then totally due to the fact that the particles move radially, so that the flux in the given direction (ϕ, θ) has to decrease as 1/r. For this reason for the scattered particles it is more natural to talk of the number of particles per unit solid angle per unit time rather than the flux; the former is r-independent, and is related to the latter by just multiplying by r. Imagine now a particle detector located far enough from the target so that (i) the distance from the target to the detector is much larger than the size of the target and (ii) the solid angle corresponding to the particles being detected is much smaller than unity. Then, for the total number of scattered particles (either of sort 1 or sort ), N, detected per small time interval t we have (below N target stands for the total number of target particles) N J N target Ω t. (6) Here we are saying that N should be directly proportional to (i) the flux of the incident beam, (ii) the number of target particles, (iii) the value of the (small) solid angle, and the (iv) the length of the (small) time interval. For this conclusion it is important that all the scattering events are independent, and that during a small time interval we do not need to take into account the depletion of N target due to scattering. Now we explicitly introduce the proportionality coefficient, s, and write N = s J N target Ω t. (63) By construction, s, is independent of n beam and n target, since the explicit dependence of N on these two quantities is already taken into account by the factors J and N target, respectively. We conclude that s can depend only on ϕ, θ, and v, the dependence coming essentially from the physics of a single pair collision. We then notice that by the symmetry of the central-potential scattering there will be no ϕ-dependence all ϕ s are equivalent, so that s s(θ, v). Once the function s(θ, v) is known, the scattering statistics is completely described by Eq. (63). Our goal now is to extract s s(θ, v) from the properties of a single scattering event. The crucial observation here is the previously-established relation between the angle θ and the impact parameter b (at a given v that we will not write explicitly for the sake of briefness): θ = θ(b). The relation between θ and b implies that the particles scatter into the solid angle sin θ ϕ if and only if the impact parameter b of the pair collision is in the corresponding interval b in the vicinity of the b = b(θ). We thus establish a relation between the solid angle Ω = sin θ θ ϕ and corresponding cross-sectional area σ = b b ϕ b db dθ θ ϕ. (64) For a given scattering event, the scattering into the solid angle sin θ ϕ takes place if and only if the incident particle passes through the area σ in the cross-sectional plane (the plane perpendicular to the velocity). The relation allows us to write N as N = σ J N target t. (65) Comparing Eq. (65) to Eq. (63), and taking into account Eq. (64), we find the desired result s(θ) = b(θ) sin θ 9 db dθ. (66)

10 The quantity dσ = s dω s dϕ sin θ dθ (67) is called differential cross section of the collision. It is the infinitesimal area in the cross-sectional plane responsible for the scattering into the infinitesimal solid angle dω. Often, the quantity s is also referred to as differential cross section. One should remember, however, that this quantity does not have the direct meaning of a cross-sectional area. This fact is quite important when one goes from the laboratory frame to the center-of-mass frame. The quantity dσ remains the same due to its geometrical meaning, and this immediately implies that s must change. Indeed, denoting s = s lab in the laboratory frame and s = s cm in the center-of-mass frame, we have s lab dϕ sin θ lab dθ lab = dσ = s cm dϕ sin χ dχ. (68) Here we use our standard notation χ θ cm and take into account the fact that the angle ϕ is the same in both frames. Since θ lab χ, the quantities s lab and s cm should be different. By the way, we have established an important explicit relation between s lab and s cm : s lab = s cm sin χ sin θ lab dχ dθ = s sin χ cm lab sin θ lab dθ lab dχ 1. (69) Along with differential cross section, one can talk of the cross section σ θ [θ1,θ ] for the scattering into the finite interval θ [θ 1, θ ]. Clearly, σ θ [θ1,θ ] = θ [θ 1,θ ] where b 1 = b(θ 1 ), b = b(θ ). dσ = π 0 dϕ θ θ 1 b s(θ) sin θ dθ = π b 1 b db = π(b b 1), (70) Problem. A homogeneous particle flux J is being scattered by a hard sphere of the radius R. The scattering of each individual particle from the surface of the sphere is governed by the mirror reflection law. Find the function s(θ) for this problem. Make sure that the integral of your differential cross section over the variables (ϕ, θ) yields πr, which is a priori clear from geometric considerations. Rutherford Scattering We are in a position to derive the celebrated Rutherford formula for the differential cross section for scattering of two Coulomb particles (referred to as Rutherford or Coulomb scattering). Differentiating the relation between b and cot χ, Eq. (59), one finds s cm (χ) = b 0 /4 sin 4 χ/, b 0 = γ m v (Rutherford formula). (71) Problem 3. Derive Eq. (71). Now we have to return to the laboratory frame. While Eq. (71) works for both incident and target particles, in the laboratory frame there are two different functions, s (1) lab and s() lab. The two are related to s cm (χ) by Eq. (69). The fact that the target particles are at rest simplifies the relations between the laboratory-frame angles θ 1, θ and χ. Especially simple is the relation for the target particles: θ = π χ 10. (7)

11 With Eq. (7) we get s () lab (θ ) = b 0 cos 3 θ. (73) Problem 4. Derive Eq. (7) and then Eq. (73). The generic relation for s (1) lab in the laboratory frame is rather cumbersome because the of rather complicated relation between θ 1 and χ: tan θ 1 = m sin χ cos χ. (74) There are, however, very important particular cases when the result is quite simple. The first case is when the mass of the target particle is much larger than that of the incident one, m m 1. Now m m 1, θ 1 χ, and we have s (1) lab (θ 1) = ( γ ) 1 m 1 v sin 4 θ 1 / (m 1 m ). (75) Another important case is m 1 = m. We already know that in this case θ 1 = χ/. So, s (1) lab (θ 1) = ( ) γ cos θ 1 m 1 v sin 4 (m 1 = m ). (76) θ 1 Problem 5. Derive Eq. (76). 11