A shadow of the repulsive Rutherford scattering and Hamilton vector

Recently, in an interesting paper [1], a shadow of the repulsive Rutherford scattering was considered in great detail in fixed-target and center-of-mass frames. It was noted that this shadow, in addition to playing an important role in low-energy ion scattering spectroscopy, is of great educational value.

Despite its educational attractiveness, this simple fact concerning repulsive Coulomb orbits was seemingly forgotten. After the boundary parabola equation was derived for the envelope of a family of scattered orbits in [2], the problem did not attract attention and disappeared into oblivion, as witnessed by the accidental rediscovery of parabolic shadow by an undergraduate student in computer simulations [3].

In this article, we will show how the Hamilton vector can be used to obtain the paraboloidal shape of the shadow in a simple and transparent way. We hope that the publication of [1] will stimulate interest in this neither too trivial nor too difficult problem, and help it to finally find its way into introductory textbooks on classical mechanics.

The Hamilton vector is an extra constant of motion in the Kepler problem, like the well-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, 6–8] and references therein).

A great virtue of the Hamilton vector is that it can be introduced in a simple and natural way [9]. Let

$$\begin{equation}{\mathbf{e}}_{r}=\mathrm{cos}\phi \mathbf{i}+\mathrm{sin}\phi \mathbf{j},\hspace{25.0pt}{\mathbf{e}}_{\phi }=-\mathrm{sin}\phi \mathbf{i}+\mathrm{cos}\phi \mathbf{j},\end{equation} \tag{ 1 }$$

be time-dependent polar-system unit vectors. Then (the dot denotes time-derivative)

$$\begin{equation}{\dot {\mathbf{e}}}_{r}=\dot {\phi }{\mathbf{e}}_{\phi },\hspace{25.0pt}{\dot {\mathbf{e}}}_{\phi }=-\dot {\phi }{\mathbf{e}}_{r}.\end{equation} \tag{ 2 }$$

Using the second equation in (2), we can write Newton's second law for the Coulomb/Kepler problem in the form (in fixed-target frame)

$$\begin{equation}\dot {\mathbf{V}}=\frac{\alpha }{\mu {r}^{2}}{\dot {\mathbf{e}}}_{r}=-\frac{\alpha }{\mu {r}^{2}\dot {\phi }}{\dot {\mathbf{e}}}_{\phi }=\frac{\mathrm{d}}{\mathrm{d}t}\left(-\frac{\alpha }{{L}_{z}}{\mathbf{e}}_{\phi }\right),\end{equation} \tag{ 3 }$$

where $\alpha =\frac{{Z}_{\mathrm{p}}{Z}_{\mathrm{t}}{e}^{2}}{4\pi {{\epsilon}}_{0}}$, Z_p and Z_t being the projectile and target charges, respectively, in units of the elementary charge e and V is the target-relative projectile velocity. At the last step we have used that ${L}_{z}=\mu {r}^{2}\dot {\phi }$, z-component of the total angular momentum of the system, is conserved in the central field (like any other components of the angular momentum). It is important to note that, similar to the traditional formulation of the two-body problem, a reduced mass of the system is introduced

$$\begin{equation}\mu =\frac{{m}_{\mathrm{p}}{m}_{\mathrm{t}}}{{m}_{\mathrm{p}}+{m}_{\mathrm{t}}},\end{equation} \tag{ 4 }$$

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 the Coulomb/Kepler field:

$$\begin{equation}\mathbf{H}=\mathbf{V}+\frac{\alpha }{{L}_{z}}{\mathbf{e}}_{\phi }.\end{equation} \tag{ 5 }$$

The physical meaning of the Hamilton vector is especially clear in the case of an attractive potential, when α < 0. Then

$$\begin{equation}\mathbf{H}=\mathbf{V}-{\mathbf{V}}_{\mathbf{c}},\end{equation} \tag{ 6 }$$

where ${\mathbf{V}}_{\mathbf{c}}=\frac{\vert \alpha \vert }{{L}_{z}}{\mathbf{e}}_{\phi }$ is the velocity that the particle would have if it moved in a circular orbit around the center of the field with the same L_z component of the angular momentum. Therefore, the Hamilton vector is a kind of velocity defect (deficit) that is conserved during motion in the Coulomb/Kepler field.

Note that the better known Laplace–Runge–Lenz vector A is related to the Hamilton vector:

$$\begin{equation}\mathbf{A}=\mathbf{H}{\times}\mathbf{L}=\mathbf{V}{\times}\mathbf{L}+\alpha {\mathbf{e}}_{r}.\end{equation} \tag{ 7 }$$

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 moves towards negative infinity on the x-axis, the corresponding unit vector e_ϕ approaches negative direction of the y axis, and its x-projection tends to zero (see figure 1). Let us 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 = −∞:

$$\begin{equation}{V}_{\infty }={V}_{x}-\frac{\alpha }{{L}_{z}}\enspace \mathrm{sin}\enspace \phi ,-\frac{\alpha }{{L}_{z}}={V}_{y}+\frac{\alpha }{{L}_{z}}\enspace \mathrm{cos}\enspace \phi ,\end{equation} \tag{ 8 }$$

where V = (V_x , V_y ) is the projectile velocity at point B. Then

$$\begin{align}\hfill {V}^{2}& ={\left({V}_{\infty }+\frac{\alpha }{{L}_{z}}\enspace \mathrm{sin}\enspace \phi \right)}^{2}+\frac{{\alpha }^{2}}{{L}_{z}^{2}}{\left(1+\mathrm{cos}\enspace \phi \right)}^{2}\hfill \\ \hfill & ={V}_{\infty }^{2}+2\frac{{\alpha }^{2}}{{L}_{z}^{2}}+2\frac{\alpha }{{L}_{z}}{V}_{\infty }\enspace \mathrm{sin}\enspace \phi +2\frac{{\alpha }^{2}}{{L}_{z}^{2}}\enspace \mathrm{cos}\enspace \phi .\hfill \end{align} \tag{ 9 }$$

Zoom In Zoom Out Reset image size

On the other hand, V² can be determined from the energy conservation law

$$\begin{equation}\frac{\mu {V}_{\infty }^{2}}{2}=\frac{\mu {V}^{2}}{2}+\frac{\alpha }{r},\end{equation} \tag{ 10 }$$

which gives

$$\begin{equation}{V}^{2}={V}_{\infty }^{2}-2\enspace \frac{\alpha }{\mu r},\end{equation} \tag{ 11 }$$

where r is the target-relative distance at point B. Then, substituting (11) into the lhs of (9), we get after a simple algebra

$$\begin{equation}\frac{1}{\mu r}=-\alpha {\sigma }^{2}\left(1+\mathrm{cos}\enspace \phi \right)+\sigma {V}_{\infty }\enspace \mathrm{sin}\enspace \phi ,\end{equation} \tag{ 12 }$$

where $\sigma =\frac{1}{\vert {L}_{z}\vert }$ (note that L_z < 0, because $\dot {\phi }{< }0$). This formula describes every possible (for different initial parameters) projectile trajectory, which is a hyperbola in polar coordinates.

The rationale in finding shadow shape from this point can be followed as in [1]. Namely, for a given polar angle ϕ, the shadow boundary is determined by the minimum value of r. Therefore, we must choose σ_m (inverse angular momentum) in such a way that

$$\begin{equation}{\left.\frac{\mathrm{d}}{\mathrm{d}\sigma }\left[-\alpha {\sigma }^{2}\left(1+\mathrm{cos}\enspace \phi \right)+\sigma {V}_{\infty }\enspace \mathrm{sin}\enspace \phi \right]\right\vert }_{\sigma ={\sigma }_{\mathrm{m}}}=0.\end{equation} \tag{ 13 }$$

This gives

$$\begin{equation}{\sigma }_{\mathrm{m}}=\frac{{V}_{\infty }\enspace \mathrm{sin}\enspace \phi }{2\alpha \left(1+\mathrm{cos}\enspace \phi \right)}.\end{equation} \tag{ 14 }$$

To calculate the minimum value of r for a given ϕ, we substitute (14) into (12) and get

$$\begin{equation}\frac{1}{\mu {r}_{\mathrm{min}}}=\frac{{V}_{\infty }^{2}\enspace {\mathrm{sin}}^{2}\enspace \phi }{4\alpha \left(1+\mathrm{cos}\enspace \phi \right)}=\frac{{V}_{\infty }^{2}}{4\alpha }\left(1-\mathrm{cos}\enspace \phi \right).\end{equation} \tag{ 15 }$$

Therefore, the boundary of the shadow is given by the equation

$$\begin{equation}\frac{1}{{r}_{\mathrm{min}}}=\frac{\mu {V}_{\infty }^{2}}{4\alpha }\left(1-\mathrm{cos}\enspace \phi \right),\end{equation} \tag{ 16 }$$

which is the equation of the parabola in polar coordinates.

Note that finding the envelope of trajectories from the equation (12) can be put in 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 $\frac{\partial }{\partial \sigma }F\left(r,\phi ,\sigma \right)=0$ (more on this in the concluding remarks).

So far, we have implicitly assumed that all possible projectile trajectories are confined to the x–y plane. In the case of three-dimensional flow, due to the axial symmetry of the problem, the shape of the shadow is a paraboloid obtained by rotating the 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 form

$$\begin{equation}\frac{4\alpha }{\mu {V}_{\infty }^{2}}=\sqrt{{\rho }^{2}+{z}^{2}}-z.\end{equation} \tag{ 17 }$$

Solving for z, we obtain the following equation for the shadow paraboloid in cylindrical coordinates:

$$\begin{equation}z=\frac{{\rho }^{2}}{8\xi }-2\xi ,\end{equation} \tag{ 18 }$$

where

$$\begin{equation}\xi =\frac{\alpha }{m{V}_{\infty }^{2}}.\end{equation} \tag{ 19 }$$

As we can see, using the Hamilton vector allows a simple and transparent way to get the main result of [1], the equation (18) for the shape of the shadow. In fact, we have been using this problem for some time as an exercise in an introductory mechanics course at Novosibirsk State University [10, 11]. In [11], two more solutions to the problem are given.

The first solution is based on the fact that the shadow boundary is the envelope of the projectile trajectories. A one-parameter family of the projectile trajectories is given by the equation (for details, see [11]. Note that in [11] the flow falls from the right).

$$\begin{equation}F\left(r,\phi ,{\phi }_{0}\right)=\frac{\alpha }{\mu {V}_{\infty }^{2}}\enspace \frac{{\mathrm{tan}}^{2}\enspace {\phi }_{0}}{r}+1+\frac{\mathrm{cos}\left(\phi -{\phi }_{0}\right)}{\mathrm{cos}\enspace {\phi }_{0}}=0,\end{equation} \tag{ 20 }$$

where ϕ₀ is the polar angle of the symmetry axis of the hyperbolic trajectory. Then it is known from mathematics (for inquisitive readers: there are some subtleties in the mathematical definition of an envelope, see [12–14]) that the envelope equation is obtained by excluding the parameter ϕ₀ from the system

$$\begin{align}\hfill F\left(r,\phi ,{\phi }_{0}\right)& =0,\hfill \\ \hfill \frac{\partial }{\partial {\phi }_{0}}F\left(r,\phi ,{\phi }_{0}\right)& =0.\hfill \end{align} \tag{ 21 }$$

In this way we can obtain the envelope equation

$$\begin{equation}\frac{1}{{r}_{\mathrm{min}}}=\frac{\mu {V}_{\infty }^{2}}{4\alpha }\left(1-\mathrm{cos}\enspace \phi \right).\end{equation} \tag{ 22 }$$

The sign in front of cos ϕ can change assuming that the flow falls from the right, for example, as it is done in [11]. In the present article (as in [1]) it was assumed that the flow falls from the left.

In the second solution method, considered in [11], we fix the polar angle ϕ in the family (20) and choose the parameter ϕ₀ so that the corresponding trajectory minimizes r. This is the same idea that was used in [1] (and in this article), just the 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 much about 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 when the infalling flux is originated from an emitter at a finite distance from the repulsive Coulomb potential center. Such a modification of the problem of the shape of the shadow in repulsive Rutherford scattering is also of considerable pedagogical interest. It represents a physical situation that actually occurs in atomic physics, and allows students to become familiar with such interesting phenomena as the glory and rainbow effects in scattering processes [15].

Consider projectile trajectories all starting from some point A on the x-axis at a distance R from the center of the field (see figure 2). We assume that the particles are released from the emitter with the same in magnitude initial velocity V₀. If the initial velocity V₀ makes an angle θ with the x-axis, the conservation equation (8) of the Hamilton vector components will be modified in the following way:

$$\begin{equation}{V}_{0}\enspace \mathrm{cos}\enspace \theta ={V}_{x}-\frac{\alpha }{{L}_{z}}\enspace \mathrm{sin}\enspace \phi ,\hspace{25.0pt}{V}_{0}\enspace \mathrm{sin}\enspace \theta -\frac{\alpha }{{L}_{z}}={V}_{y}+\frac{\alpha }{{L}_{z}}\enspace \mathrm{cos}\enspace \phi .\end{equation} \tag{ 23 }$$

Then

$$\begin{equation}{V}^{2}={{V}_{0}}^{2}+2\frac{\alpha }{{L}_{z}}{V}_{0}\left[\mathrm{sin}\left(\phi -\theta \right)-\mathrm{sin}\enspace \theta \right]+2{\left(\frac{\alpha }{{L}_{z}}\right)}^{2}\left(1+\mathrm{cos}\enspace \phi \right),\end{equation} \tag{ 24 }$$

and using the energy conservation law

$$\begin{equation}\frac{\mu {V}_{0}^{2}}{2}+\frac{\alpha }{R}=\frac{\mu {V}^{2}}{2}+\frac{\alpha }{r},\end{equation} \tag{ 25 }$$

we derive the following relation

$$\begin{align}\hfill \frac{1}{\mu r}& =-\alpha {\sigma }^{2}\left(1+\mathrm{cos}\phi \right)+\sigma {V}_{0}\mathrm{sin}\left(\phi -\theta \right)\hfill \\ \hfill & =-\alpha {\sigma }^{2}\left(1+\mathrm{cos}\phi \right)+{V}_{0}\mathrm{sin}\phi \sqrt{{\sigma }^{2}-\frac{1}{{\mu }^{2}{R}^{2}{V}_{0}^{2}}}-\frac{\mathrm{cos}\phi }{\mu R},\hfill \end{align} \tag{ 26 }$$

Zoom In Zoom Out Reset image size

where we have taken into account that $\frac{1}{\sigma }=\vert {L}_{z}\vert =\mu {V}_{0}R\mathrm{sin}\enspace \theta$.

The equation (26) indicates that the minimum r for a given ϕ corresponds to σ_m such that

$$\begin{equation}\sqrt{{\sigma }_{\mathrm{m}}^{2}-\frac{1}{{\mu }^{2}{R}^{2}{V}_{0}^{2}}}=\frac{{V}_{0}\mathrm{sin}\phi }{2\alpha \left(1+\mathrm{cos}\phi \right)}.\end{equation} \tag{ 27 }$$

Substituting σ_m from (27) into (26), we finally obtain for the shadow boundary the expression

$$\begin{equation}\frac{1}{{r}_{\mathrm{min}}}=\frac{\mu {V}_{\infty }^{2}}{4\alpha }\left(1-A\right)\left(1-\frac{1+A}{1-A}\mathrm{cos}\phi \right),\end{equation} \tag{ 28 }$$

where we have introduced a convenient dimensionless parameter [15, 16]

$$\begin{equation}A=\frac{{V}_{\infty }^{2}}{{V}_{0}^{2}}-1=\frac{2\alpha }{\mu {V}_{0}^{2}R}.\end{equation} \tag{ 29 }$$

Since A > 0 equation (28) corresponds to a hyperbola with vertex at the emission point and focus at the force center [15]. As the emission point moves towards infinity, A → 0 and the hyperbola (28) degenerates into the parabola (16).

For an attractive potential, the envelope of Kepler ellipses, all starting from the same point in space, can be obtained in a similar way [16].

The work is supported by the Ministry of Education and Science of the Russian Federation.