As with the Bohr atom, classical mechanics gave the correct results for alpha-particle scattering by nuclei

The initial state is shown in the diagram at the right. Mass m is moving with velocity **v** on a straight line, labeled "asymptote," approaching another mass m' with which it will interact, perhaps through the electromagnetic force derivable from the potential ZZ'e^{2}/r, or a gravitational force derivable from the potential Gmm'/r, where r is the distance between m and m' (or Z and Z'). The electromagnetic force may be repulsive or attractive, but the gravitational force is always attractive. If there were no interaction, the closest distance between m and m' would be p, the *impact parameter*, and m would continue in the same straight line.

The six degrees of freedom will be resolved into the three coordinates of the center of mass, and the three coordinates of the relative motion. In this article, we shall consider only the relative motion. The effective mass for the relative motion is the reduced mass mm'/(m + m'), which we shall denote simply by m for convenience. The interaction potential is already expressed in terms of the relative distance r. The initial relative velocity v = |**v**| and the impact parameter p are the same in the relative system. The total energy E is constant at mv^{2}/2, and the relative angular momentum is constant at **N** = mvp in the relative motion.

We know from studying planetary orbits that the trajectory under an inverse-square central force will be a conic section with the centre of force in one focus, and in this case the trajectory will be a hyperbola, illustrated in the diagram at the left. If the force is repulsive, then the relative particle will enter on the asymptote at the lower left-hand corner, the centre of force will be located at the focus F, and after collision the relative particle will recede toward the upper left-hand corner, after being scattered through an angle Θ. If the force were attractive, we could consider the relative particle entering at the lower right-hand corner and exiting at the upper right-hand corner instead. In either case, the orbit will be essentially the same, and in solving one problem we solve the other. Let's assume that the force is repulsive in our developments.

To draw a hyperbola, we might start with the asymptotes, crossing at the origin at some angle Θ, or 2θ. Clearly, Θ = π - 2θ is the relation of the angles. This gives us the *shape* of the hyperbola. Then, we choose the location of the vertex V on the x-axis. The other vertex V' is located symmetrically with respect to the origin. The parameter a is the distance of the vertices from the origin. Now we have specified the *size* of the hyperbola. If we draw a vertical tangent at a vertex, it meets the asymptotes at a distance b from the x-axis. This defines a square, shown dotted in the diagram, that completely specifies the hyperbola. The diagonals of the square are of length 2c, and c is the distance from the origin to the foci F and F'. In order to make this assertion, we must specify the curve of the hyperbola precisely, and this may be done by the equation (x/a)^{2} - (y/b)^{2} = 1. Clearly, c^{2} = a^{2} + b^{2}. The eccentricity e of the hyperbola is given by c = ea, or e^{2} = 1 + (b/a)^{2} = 1 + tan^{2}θ. Once we have drawn the asymptotes, we have determined the eccentricity. In our present problem we shall proceed in the opposite direction, finding the eccentricity first and from it the angle between the asymptotes, which will give us the scattering angle. Don't confuse the eccentricity with the electronic charge!

An ellipse is drawn *inside* a box of sides a and b, while a hyperbola is entirely *outside* a similar box. The hyperbola falls into two disconnected *branches*, only one of which is the trajectory. If a line is drawn perpendicular to an asymptote to the focus, it is easy to see that it makes a right triangle congruent to the one of sides a and b, since they share a common vertex angle and a side of length c. Equality of the bases then shows that p = b: the impact parameter is the parameter b. The distance of closest approach will be the distance V'F = a + c = s. Now, c = p/sin θ and a = c cos θ, so s = p(1 + cos θ)/sin θ = p cot(θ/2).

The significant thing about the vertex V' is that we can calculate the velocity there in terms of p, s and the constancy of the angular momentum, and also from the energy, and equate the two results. In fact, v's = vp, so v'/v = tan(θ/2) from angular momentum. Energy gives v^{2} = v'^{2} + 2ZZ'e^{2}/ms, so that (v'/v)^{2} = 1 - (2ZZ'e^{2}/mv^{2})/s = 1 - n/s, where n = 2ZZ'e^{2}/mv^{2}. Therefore, (v'/v)^{2} = tan^{2}(θ/2) = 1 - (n/p)tan(θ/2). Then, n/p = [1 - tan^{2}(θ/2)]/tan(θ/2) = [cot(θ/2) - tan(θ/2)] = 2/tan θ. Therefore, p = (n/2)tan(θ) = (n/2)cot(Θ/2), since θ = π/2 - Θ/2. This equation relates the impact parameter to the scattering angle, and is what we desired to find.

The algebra for this problem is amazingly troublesome, so a careful path must be steered to the answer. We are now past those hazards. To make the problem applicable to gravitation, just replace ZZ'e^{2} by Gmm'. A famous application was to the scattering of alpha particles (Z = 2) from atomic nuclei, which led to the recognition of the atomic nucleus by Ernest Rutherford, after the grueling work of his students Geiger and Marsden, who did the experiments between 1909 and 1913, counting flashes "manually" using a ZnS screen. The scattering was in gold and silver foils, while the velocity of the alpha particles was varied using thin mica sheets. For this application, we want the differential cross section σ(Θ), defined as the projected area for scattering into unit solid angle dΩ at a scattering angle Θ. This solid angle is dΩ = 4πsin(Θ/2)cos(Θ/2)dΘ, where we have put in the half-angles knowing what will be expected of them.

The desired area is -2πpdp, since Θ decreases as p increases. We use p = (n/2)cot(Θ/2), and find σ(Θ)dΩ = 2π(n^{2}/4)cot(Θ/2)csc^{2}(Θ/2)(dΘ/2) = (n^{2}/16)csc^{4}(Θ/2)dΩ. Writing out n, we have σ(Θ) = (ZZ'e^{2}/2mv^{2})^{2}sin^{-4}(Θ/2). This formula was amply verified by experiment. In fact, the size of the nucleus was estimated from the distance of closest approach s = p cot(π/4 - Θ/4) when deviations were noted. It is remarkable that a classical analysis gave the correct result, something that also happened with the Bohr atom. A full quantum-mechanical analysis gives the same answer, but is much more difficult, as in the case of the atom. Even more remarkably, the Born approximation gives the correct answer in the first order of perturbation theory, when it should not be valid!

For the gravitational field, if we consider a light body m and a heavy body m', then p = (2Gm'/v^{2})cot(Θ/2). If v' is the escape velocity from the heavy body of radius r, v'^{2} = 2Gm'/r, then tan(Θ/2) = (v'/v)^{2}(r/p), from which the deflection can be calculated if the initial relative velocity and the impact parameter are known. This has great application in the matter of the so-called "gravitational assist," which is simply an elastic collision with a moving heavy body. See Gravitational Assist for a discussion. For the earth, v' = 11.2 km/s and r = 6371 km. If the initial relative velocity is equal to the escape velocity, and the impact parameter is two earth radii, then the deflection angle is Θ = 53.1°. This applies, in particular, to meteroroids, which may be deflected into the earth's atmosphere.

Imagine the earth moving in its orbit at its usual speed of 29.785 km/s, 2.66 times the escape velocity. There is a small body at rest at a distance of 2 earth radii from the path of the earth. The diagram at the right shows the transformation to the CM system, the origin of which coincides with the earth for all practical purposes. The small body now approaches the earth at rest at a speed of v = 29.785 km/s. The scattering angle under these conditions will be Θ = 8.09°. The distance of closest approach will be s = p tan(θ/2) for the attractive force, instead of p cot(θ/2). This will be 1.86r, so the small body will miss the earth by a good margin as the earth passes by. In the diagram, v' is the velocity after the encounter. Now we subtract the same velocity we added to get to the CM system to find the velocity of the small body in space after the earth goes by. This gives a velocity u, which is easily calculated to be 4.2 km/s, at an angle of 4.04° with the perpendicular to the path of the earth. The earth will, therefore, sweep out a torus as it moves. Those particles within a little more than 1.1 earth radii of the orbit will collide with the earth, and presumably stick to it, while those outside will be sent to distant parts. At the present time, there does not appear to be a great deal of débris at rest around the earth's orbit, but this may not always have been the case.

I. Kaplan, *Nuclear Physics* (Cambridge, MA: Addison-Wesley, 1955), pp. 41-51.

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Composed by J. B. Calvert

Created 21 May 2003

Last revised