A couple of columns back, I strongly suggested that the definition of the scattering cross section be changed to really reflect its probabilisitic nature and that the geometric picture associated with classical scattering be eliminated. The reasons for that were expressed in the column but what I would like to add in this and subsequent columns is that I am not unsympathetic to the traditional view of the scattering cross section. In fact, I’ve had occasions to use this notion quite often in both professional and hobbyist contexts. So, in the interest of completeness, I am going to be devoting the next four columns to notions associated with scattering within the classical context.
I’ll start with the general theory of scattering from a $$1/r$$-potential. This covers both gravitational and electromagnetic forces, the two forces of classical mechanics.
The equation of motion, quite generally, is
\[ \ddot{\vec r} + \frac{\alpha \vec r}{r^3} = 0 \; .\]
where the sign of $$\alpha$$ determines whether the potential is attractive ($$\alpha > 0$$) or repulsive ($$\alpha < 0$$). Gravitational interactions are always attractive and $$\alpha$$ is usually denoted $$\mu$$ but, for generality, we’ll stick with $$\alpha$$. Now this coupled set of ordinary differential equations is readily solved numerically on just about any computer but the generation of lots of numbers doesn’t help the analysis. A better approach is to interrogate the equations of motion for conserved quantities (click here for the same approach within the context of the Maxwell equations).
Before beginning, it is useful to remind ourselves of some basic identities from many-variable calculus. The first is
\[ \frac{d}{dt} r = \frac{d}{dt} \sqrt{ x^2 + y^2 + z^2 } = \frac{1}{2 \sqrt{ x^2 + y^2 + z^2}} \frac{d}{dt} \left( x^2 + y^2 + z^2 \right) \\ = \frac{ x \dot x + y \dot y + z \dot z}{\sqrt{x^2 + y^2 + z^2}} = \frac{ \vec r \cdot \dot{\vec r} }{r} \equiv \frac{\vec r \cdot \vec v}{r} \; .\]
The second one, which follows directly from the first, is
\[ \frac{d}{dt} \left( \frac{1}{r} \right) = -\frac{ \dot {\vec r} \cdot \vec r}{r^3} \; .\]
Now, the first conserved quantity results from dotting the equation of motion with $$\vec v = \dot {\vec r}$$ to get
\[ \ddot {\vec r} \cdot \vec v + \frac{\alpha \vec r \cdot \vec v }{r^3} \; \]
The first term can be written as a total time derivative since
\[ \frac{d}{dt} \left(\frac{1}{2} \dot {\vec r} \cdot \dot {\vec r} \right) = \dot {\vec r} \cdot \ddot {\vec r} \; .\]
The second term is also a total time derivative as seen from the identities listed above. And so we get our first conserved quantity
\[ E = \frac{1}{2} \dot {\vec r} \cdot \dot {\vec r} – \frac{\alpha}{r} \equiv \frac{1}{2} \vec v \cdot \vec v – \frac{\alpha}{r} \; ,\]
which is the energy of the system.
The next conserved quantity is derived by crossing the equation of motion from the left by $$\vec r$$
\[ \vec r \times \ddot{\vec r} + \frac{\alpha \vec r \times \vec r }{r^3} = 0 \; .\]
Because of the central nature of the force, the last term is identically zero, and the remaining term
\[ \vec r \times \ddot{\vec r} = 0 \]
can be written as a total time derivative
\[ \frac{d}{dt} \left( \vec r \times \dot {\vec r} \right) = 0 \; ,\]
resulting in the conservation of angular momentum
\[ \vec L = \vec r \times \dot {\vec r} = \vec r \times \vec v \; .\]
The final conserved quantity is obtained by crossing the equation of motion from the right by the angular momentum
\[ \ddot {\vec r} \times \vec L + \frac{\alpha \vec r \times \vec L}{r^3} = 0 \; .\]
Since the angular momentum is a conserved quantity, the first term can be rewritten as
\[ \ddot {\vec r} \times \vec L = \frac{d}{dt} \left( \dot {\vec r} \times \vec L \right) = \frac{d}{dt} \left( \vec v \times \vec L \right) \; .\]
The best way to simplify the second term is to recognize that
\[ \frac{d}{dt} \frac{\vec r}{r} = \frac{\vec v}{r} – \frac{\vec r(\vec v \cdot \vec r)}{r^3} = \frac{\vec v r^2 – \vec r(\vec v \cdot \vec r)}{r^3} = -\frac{[\vec r \times (\vec r \times \vec v)]}{r^3} = – \frac{\vec r \times \vec L}{r^3} \; ,\]
where the second-to-last step results from the BAC-CAB rule and the last step is the definition of the angular momentum.
After these manipulations are completed, the resulting total time derivative is
\[ \frac{d}{dt}\left( \vec v \times \vec L – \frac{\alpha \vec r}{r} \right) = 0 \; ,\]
from which we see the conserved quantity
\[ \vec v \times \vec L – \frac{\alpha \vec r}{r} = \vec A \; .\]
This latter vector is known as the Laplace-Runge-Lenz vector.
With all these conserved quantities in place, it is relatively easy to predict certain aspects of the scattering trajectories without solving the equations directly. First, the conservation of angular momentum requires the motion to take place in the plane. As a result, plane-polar coordinates can be used to describe the motion. Since the Laplace-Runge-Lenz vector is conserved and is in the plane perpendicular to the angular momentum, it makes an excellent axis against which to measure the movement of the polar angle $$\nu$$.
\[ \vec A \cdot \vec r = A r \cos \nu \]
The variation of the radius in the plane can then be expressed as a function of $$\nu$$ as follows. Expand the dot product between $$\vec A$$ and $$\vec r$$
\[ \vec A \cdot \vec r = \left( \vec v \times \vec L – \frac{\alpha \vec r}{r} \right) \cdot \vec r = (\vec v \times \vec L) \cdot \vec r – \alpha r \; .\]
Next note, because of the cycle property of the triple-scalar product, that
\[ (\vec v \times \vec L) \cdot \vec r = (\vec r \times \vec v) \cdot \vec L = L^2 \; \]
Equating the two expressions gives
\[ A r \cos \nu = L^2 -\alpha r \; , \]
which can be readily solved to give
\[ r = \frac{L^2 / \alpha}{1 + A/\alpha \cos \nu} \;.\]
Now the interpretation of this equation requires a modest amount of care. If $$\alpha > 0$$, then the above form can immediately be recognized as a conic section with
\[ r = \frac{p}{1+e \cos \nu} \; \; \alpha > 0 ;, \]
where the semi-latus rectum $$p = L^2/\alpha$$ and the eccentricity $$e = A/\alpha$$.
If $$\alpha < 0$$, then we need to multiply the numerator and denominator by $$-1$$ to get \[ r = \frac{L^2 / |\alpha|}{A/|\alpha| \cos \nu – 1} \; , \] from which we get \[ r = \frac{p}{e \cos \nu – 1} \; ,\] where $$p = L^2/|\alpha|$$ and $$e = A/|\alpha|$$. Next week we will analyze individual scattering orbits using these conic equations and the conserved quantities.