A shadow of the repulsive Rutherford scattering and Hamilton vector
AA shadow of the repulsive Rutherford scatteringand Hamilton vector
D A Shatilov and Z K Silagadze , Novosibirsk State University, Novosibirsk 630090, RussiaE-mail: [email protected] Budker Institute of Nuclear Physics, Novosibirsk 630 090, RussiaE-mail:
Abstract.
The fact that repulsive Rutherford scattering casts a paraboloidal shadowis rarely exploited in introductory mechanics textbooks. Another rarely used const-ruction in such textbooks is the Hamilton vector, a cousin of the more famous Laplace-Runge-Lenz vector. We will show how the latter (Hamilton’s vector) can be used toexplain and clarify the former (paraboloidal shadow). a r X i v : . [ phy s i c s . c l a ss - ph ] J a n
1. Introduction
Recently, in an interesting paper [1], a shadow of the repulsive Rutherford scatteringwas considered in great detail in fixed-target and center-of-mass frames. It was notedthat this shadow, in addition to playing an important role in low-energy ion scatteringspectroscopy, is of great educational value.Despite its educational attractiveness, this simple fact concerning repulsive Cou-lomb orbits was seemingly forgotten. After the boundary parabola equation was derivedfor the envelope of a family of scattered orbits in [2], the problem did not attractattention and disappeared into oblivion, as witnessed by the accidental rediscovery ofparabolic shadow by an undergraduate student in computer simulations [3].In this article, we will show how the Hamilton vector can be used to obtain theparaboloidal shape of the shadow in a simple and transparent way. We hope thatthe publication of [1] will stimulate interest in this neither too trivial nor too difficultproblem, and help it to finally find its way into introductory textbooks on classicalmechanics.
2. Hamilton vector
The Hamilton vector is an extra constant of motion in the Kepler problem, like thewell-known Laplace-Runge-Lenz vector (in fact, the two are simply related) [4]. This“lost sparkling diamond of introductory level mechanics” [5] was well known in the past,but then almost forgotten (see [4, 6, 7, 8] and references therein).A great virtue of the Hamilton vector is that it can be introduced in a simple andnatural way [9]. Let e r = cos φ i + sin φ j , e φ = − sin φ i + cos φ j , (1)be time-dependent polar-system unit vectors. Then (the dot denotes time-derivative)˙ e r = ˙ φ e φ , ˙ e φ = − ˙ φ e r . (2)Using the second equation in (2), we can write Newton’s second law for the Cou-lomb/Kepler problem in the form (in fixed-target frame)˙ V = αµr ˙ e r = − αµr ˙ φ ˙ e φ = dd t (cid:18) − αL z e φ (cid:19) , (3)where α = Z p Z t e π(cid:15) , Z p and Z t being the projectile and target charges, respectively, inunits of the elementary charge e and V is the the target-relative projectile velocity. Atthe last step we have used that L z = µr ˙ φ , z -component of the total angular momentumof the system, is conserved in the central field (like any other components of the angularmomentum). It is important to note that, similar to the traditional formulation of thetwo-body problem, a reduced mass of the system is introduced µ = m p m t m p + m t , (4)where m p and m t are the projectile and target masses in the specified order. Equation(3) indicates that the following vector (the Hamilton vector) is conserved in theCoulomb/Kepler field: H = V + αL z e φ . (5)The physical meaning of the Hamilton vector is especially clear in the case of anattractive potential, when α <
0. Then H = V − V c , (6)where V c = | α | L z e φ is the velocity that the particle would have if it moved in acircular orbit around the center of the field with the same L z component of the angularmomentum. Therefore, the Hamilton vector is a kind of velocity defect (deficit) that isconserved during motion in the Coulomb/Kepler field.Note that the better known Laplace-Runge-Lenz vector A is related to the Hamiltonvector: A = H × L = V × L + α e r . (7)
3. The shape of the shadow
Let us consider some projectile trajectory (in the fixed-target frame) launched at x = −∞ and with an initial velocity V ∞ directed along the x axis. Selected point B at this trajectory has the polar angle φ . When point A on this trajectory movestowards negative infinity on the x -axis, the corresponding unit vector e φ approachesnegative direction of the y axis, and its x -projection tends to zero (see figure 1). Let us Figure 1.
Projectile trajectory along which H ( A ) = H ( B ) for any choice of A and B . write the conservation law of the Hamilton vector in the x and y -projections for points A and B in the limit, when the point A corresponds to x = −∞ : V ∞ = V x − αL z sin φ, − αL z = V y + αL z cos φ, (8)where V = ( V x , V y ) is the projectile velocity at point B . Then V = (cid:18) V ∞ + αL z sin φ (cid:19) + α L z (1 + cos φ ) = V ∞ + 2 α L z + 2 αL z V ∞ sin φ + 2 α L z cos φ. (9)On the other hand, V can be determined from the energy conservation law µV ∞ µV αr , (10)which gives V = V ∞ − αµr , (11)where r is the target-relative distance at point B . Then, substituting (11) into the l.h.s.of (9), we get after a simple algebra1 µr = − ασ (1 + cos φ ) + σV ∞ sin φ, (12)where σ = | L z | (note that L z <
0, because ˙ φ < φ , the shadow boundary is determined by the minimumvalue of r . Therefore, we must choose σ m (inverse angular momentum) in such a waythat dd σ (cid:104) − ασ (1 + cos φ ) + σV ∞ sin φ (cid:105)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) σ = σ m = 0 . (13)This gives σ m = V ∞ sin φ α (1 + cos φ ) . (14)To calculate the minimum value of r for a given φ , we substitute (14) into (12) and get1 µr min = V ∞ sin φ α (1 + cos φ ) = V ∞ α (1 − cos φ ) . (15)Therefore, the boundary of the shadow is given by the equation1 r min = µV ∞ α (1 − cos φ ) , (16)which is the equation of the parabola in polar coordinates.Note that finding the envelope of trajectories from the equation (12) can be putin a general envelope determination context by treating this formula as a condition F ( r, φ, σ ) = 0 and then applying the general rule for the envelope ∂∂σ F ( r, φ, σ ) = 0(more on this in the concluding remarks).So far, we have implicitly assumed that all possible projectile trajectories areconfined to the x − y plane. In the case of three-dimensional flow, due to the axialsymmetry of the problem, the shape of the shadow is a paraboloid obtained by rotatingthe parabola (16) around its axis of symmetry.Introducing the cylindrical coordinates z = r min cos φ , ρ = r min sin φ , equation (16)can be rewritten in the form4 αµV ∞ = (cid:113) ρ + z − z. (17)Solving for z , we obtain the following equation for the shadow paraboloid in cylindricalcoordinates: z = ρ ξ − ξ, (18)where ξ = αmV ∞ . (19)
4. Concluding remarks
As we can see, using the Hamilton vector allows a simple and transparent way to get themain result of [1], the equation (18) for the shape of the shadow. In fact, we have beenusing this problem for some time as an exercise in an introductory mechanics course atNovosibirsk State University [10, 11]. In [11], two more solutions to the problem aregiven.The first solution is based on the fact that the shadow boundary is the envelope ofthe projectile trajectories. A one-parameter family of the projectile trajectories is givenby the equation (for details, see [11]. Note that in [11] the flow falls from the right). F ( r, φ, φ ) = αµV ∞ tan φ r + 1 + cos ( φ − φ )cos φ = 0 , (20)where φ is the polar angle of the symmetry axis of the hyperbolic trajectory. Thenit is known from mathematics (for inquisitive readers: there are some subtleties in themathematical definition of an envelope, see [12, 13, 14]) that the envelope equation isobtained by excluding the parameter φ from the system F ( r, φ, φ ) = 0 ,∂∂φ F ( r, φ, φ ) = 0 . (21)In this way we can obtain the envelope equation1 r min = µV ∞ α (1 − cos φ ) . (22)The sign in front of cos φ can change assuming that the flow falls from the right, forexample, as it is done in [11]. In the present article (as in [1]) it was assumed that theflow falls from the left.In the second solution method, considered in [11], we fix the polar angle φ inthe family (20) and choose the parameter φ so that the corresponding trajectoryminimizes r . This is the same idea that was used in [1] (and in this article), justthe parameterizations in the one-parameter family of projectile trajectories are different.Therefore, we emphasize that the most original contribution relative to [1] is not so muchabout the procedure for obtaining the shadow itself, but rather obtaining trajectories,which is facilitated by the use of the Hamilton vector.The method used in this article can be extended to the interesting case whenthe infalling flux is originated from an emitter at a finite distance from the repulsiveCoulomb potential centre. Such a modification of the problem of the shape of theshadow in repulsive Rutherford scattering is also of considerable pedagogical interest.It represents a physical situation that actually occurs in atomic physics, and allowsstudents to become familiar with such interesting phenomena as the glory and rainboweffects in scattering processes [15].Consider projectile trajectories all starting from some point A on the x -axis at adistance R from the center of the field (see figure 2). We assume that the particles arereleased from the emitter with the same in magnitude initial velocity V . If the initial Figure 2.
The definition of kinematic variables for the projectile trajectory launchedfrom the point A on the x -axis. velocity V makes an angle θ with the x -axis, the conservation equations (8) of theHamilton vector components will be modified in the following way: V cos θ = V x − αL z sin φ, V sin θ − αL z = V y + αL z cos φ. (23)Then V = V + 2 αL z V [ sin ( φ − θ ) − sin θ ] + 2 (cid:18) αL z (cid:19) (1 + cos φ ) , (24)and using the energy conservation law µV αR = µV αr , (25)we derive the following relation1 µr = − ασ (1 + cos φ ) + σV sin ( φ − θ ) = − ασ (1 + cos φ ) + V sin φ (cid:115) σ − µ R V − cos φµR , (26)where we have taken into account that σ = | L z | = µV R sin θ .The equation (26) indicates that the minimum r for a given φ corresponds to σ m such that (cid:115) σ m − µ R V = V sin φ α (1 + cos φ ) . (27)Substituting σ m from (27) into (26), we finally obtain for the shadow boundary theexpression 1 r min = µV ∞ α (1 − A ) (cid:18) − A − A cos φ (cid:19) , (28)where we have introduced a convenient dimensionless parameter [15, 16] A = V ∞ V − αµV R . (29)Since
A > A → Acknowledgments
The work is supported by the Ministry of Education and Science of the RussianFederation.
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