User:Rmarhatta/sandbox

Introduction
The scattering experiment is one of the important tool to understand the microscopic physics. The motion of a particle or [particles] is governed by a potential energy function. Depending upon the total energy and the nature of the potential function, the motion will be either bound, with the particle trajectories bounded, or free, with one or more particles able to access all of space. Bound motion is either oscillatory, rotational, or vibrational, while free motion is commonly referred to as scattering. Almost everything we know about nuclear and atomic physics has been discovered by scattering experiments, e.g. Rutherford’s discovery of the nucleus, the discovery of sub-atomic particles (such as quarks), etc. In low energy physics, scattering phenomena provide the standard tool to explore solid state systems, e.g. neutron, electron, x-ray scattering, etc. Scattering is important in many physical contexts:
 * 1) 	establishment and maintenance of equilibrium in gases;
 * 2) 	pumping of population inversions in gas lasers;
 * 3) 	effusion and diffusion of gases;
 * 4) 	broadening of emission lines in atomic and molecular spectra;
 * 5) 	absorption of neutrons in a reactor;
 * 6) 	chemical reaction;
 * 7) 	creation and annihilation of fundamental particles at high energy;
 * 8) 	generation and destruction of plasmas, e.g. in the ionosphere;
 * 9) 	formation of structure in solar systems and galaxies.

The Differential and total cross section
Both classical and quantum mechanical scattering phenomena are characterized by the scattering cross section. Imagine a particle with impact parameter $$b$$ and energy E is scattered off a scattering center. It will emerge at some scattering angle $$\theta$$. Now the question is: what is the scattering angle $$\theta$$ for a given impact parameter? Ordinarily scattering angle depends on impact parameter. The different impact parameters lead to different scattering angles.

Let us imagine that particles incident within the ring of area $$d\sigma$$ will scatter into a solid angle $$d\Omega$$. The larger the $$d\sigma$$ the greater the $$d\Omega$$ would be. This angular information is very important; when it can be measured, it provides a much more detailed picture of the scattering process and the interaction force responsible for it. Accordingly, we can write the angle-differential collision cross section (usually referred to simply as the differential cross section) as a ratio of two different kinds of fluxes: the outgoing particle flux, measured as particles per unit solid angle per unit time, divided by the incoming flux, measured in particles per unit area per unit time and the ratio $$d\sigma/d\Omega$$ is called scattering cross section. In terms of impact parameter b, we can write $$d\sigma$$
 * $$d\sigma=2\pi bdb=\frac{d\sigma}{d\Omega}d\Omega$$

The differential cross section is,
 * $$ \frac{d\sigma}{d\Omega}=\frac{b}{sin\theta}\left|\frac{db}{d\theta}\right|$$

The total cross section is then,
 * $$ \sigma=\int_\Omega \frac{d\sigma}{d\Omega}\,d\Omega$$

Let us assume that a point projectile is scattered from a massive slippery spherical target so that it does not recoil. This is the hard sphere scattering with differential cross section $$R_t^2/4$$ and total cross section  $$\pi R_t^2$$. Here $$R_t$$ is the radius of the target. If the projectile has a finite size of radius $$R_p$$ then the total cross section will be simply $$\pi (R_t+R_p)^2$$.

Role of the potential
We explore atom-atom scattering as an example. The calculation of the atom-atom potential energy function is a vast science in and of itself. In fact, extracting the potential function is one of the aims of scattering experiments, since once the potential is known, one can calculate properties of bulk gases (and to some extent liquids). At the moment, our goal is to understand the generic features of the cross section; for this purpose, it is sufficient to understand the generic features of the potential energy function. In general, neutral atoms attract one another at long range and repel one another at short range. Since the atoms are neutral, it might be wondered why they attract at all; the answer is that they are polarizable. Classically, fluctuations in the electron clouds surrounding the nuclei become correlated, so that their instantaneous dipoles are aligned. For two neutral atoms in their ground state, this effect appears in second order in perturbation theory, and consequently the $$1/r^3$$ dipole potential translates into a $$1/r^6$$ atom-atom attractive dispersion potential energy function, often known as the van der Waals potential. In fact, this is the leading term in a series:
 * $$V(r)=\frac{C_6}{r^6}+\frac{C_8}{r^8}+\frac{C_{10}}{r^{10}} .........(1)$$

The coefficient of $$1/r^6$$ is always negative, and this is sometimes factored out; the sign of the higher-order coefficients can be either negative or positive and is usually included with the coefficient. The short-range portion of the potential function is not expressible in a simple form. It originates in the Coulomb repulsion between the electron clouds of the two atoms and also includes a contribution from the "exchange" interaction due to the Pauli exclusion principle that manifests itself when the electron clouds overlap. It is much more repulsive than the long-range portion of the potential is attractive, and there is a longstanding custom of representing it using a $$1/r^{2n}$$ function that often simplifies computation when coupled with an attractive $$1/r^n$$ component:
 * $$V(r)=\frac{C_{2n}}{r^{2n}}-\frac{C_n}{r^n}  .........(2)$$

Equation 2 is often referred to as the Lennard-Jones potential; when n = 6 it is called the Lennard-Jones 6-12 potential. However, the short-range potential is usually better represented by an exponential, leading to the exp-6 potential,
 * $$V(r)=Ae^{-Br}-\frac{C_6}{r^6}$$

The general form of an atom-atom potential function is shown in Figure 2. Note that it is conventional for the zero of energy to coincide with the asymptotic value of the energy. This makes possible a very convenient division between bound states (E < 0) and scattering (E > 0). Much can be learned about the types of possible collisions by studying the features of the effective potential, shown in Figure 2 for different values of b. One price paid for being able to express the motion of two bodies in three dimensions as that of a single body in one dimensions is the dependence of the effective potential energy on the impact parameter, and this expresses itself as the family of curves shown in Figure 2. For $$b$$ = 0 ($$L$$ = 0), $$V_{eff}(r) = V (r)$$. As $$b$$ increases, the effective potential becomes more and more distorted. For $$b$$ = 0, there is one turning point in the trajectory for $$E > 0$$; as soon as $$b > 0$$, however, the positive $$1/r^2$$ centrifugal potential overwhelms any long-range potential for which n > 2 and there are three turning points at low energy. The inner two turning points are those of a bound state, and it is possible classically for a pair of particles to persist indefinitely in such a bound state. With further increases in $$b$$, however, the entire well is lifted up above the $$E$$ = 0 line; at this point the state becomes "quasi-bound". It is unstable with respect to dissociation, although in the absence of a perturbation it could last indefinitely. Quantum mechanical tunneling, however, limits the lifetime of such a state, and even classically such states are fragile and unstable to perturbations from within or outside the system. A further increase in the impact parameter eliminates the attraction altogether, and there is once again a single turning point.

We know that angular momentum $$L =\mu vb$$ is conserved. If we reduce the problem to a one-body problem involving the vector separation $$r$$ of the two bodies, the kinetic energy of the particles is $$K = 1/2\mu v^2=1/2\mu(dr/dt)^2$$. As long as the particles are far away, $$v$$ can be thought of as parallel to $$r$$ and then $$dr/dt$$. But when the two bodies get closer to one another, the direction of $$r$$ as well as its magnitude changes, and then there is another term, even in the absence of an interaction between the two bodies. Since, in the absence of a force, $$r = vt + b,$$ the Pythagorean theorem leads to
 * $$r^2 = v^2t^2 + b^2$$

for straight-line trajectories. Differentiating this equation leads to
 * $$2r\frac{dr}{dt}=2v^2t=2v(r^2-b^2)^{1/2}$$

then
 * $$r^2(\frac{dr}{dt})^2=v^2(r^2-b^2)$$
 * $$1/2\mu (\frac{dr}{dt})^2=1/2\mu v^2(1-b^2/r^2)$$

and hence
 * $$K =1/2\mu v^2=1/2\mu(\frac{dr}{dt})^2+K\frac{b^2}{r^2}$$

This admittedly unusual way of writing the kinetic energy makes it easy to see the division between translational and rotational components. When $$b/r<< 1$$, the first term on the right hand side, due to the changing distance between the two bodies, dominates. When the bodies get close to one another, the second term, due to the relative rotation of the bodies, becomes important. At their point of closest approach, $$r = b$$ and $$dr/dt$$ = 0, and the rotational term is the entire kinetic energy. As long as there is no interaction between the two bodies, the kinetic energy is equal to the total energy, which is a constant of the motion. If there is a potential energy of interaction between the two bodies, equation is modified according to
 * $$E = K + V (r) $$
 * $$1/2\mu v_{rel}^2=1/2\mu(\frac{dr}{dt})^2+E\frac{b^2}{r^2}+V(r)$$

where $$v_{rel}$$ is the asymptotic relative speed of the two bodies. it is now easy to solve for the derivative of $$r$$:
 * $$\frac{dr}{dt}=v_{rel}(1-\frac{V(r)}{E}-\frac{b^2}{r^2})^{1/2}$$

We can rearrange this to calculate t(r):
 * $$t - t_0 =\int_\infty^r\frac{dr'}{(1-\frac{V(r)}{E}-\frac{b^2}{r^2})^{1/2}}$$

Naturally, this integral will be difficult or impossible to solve analytically, in general. But beyond that, inversion of the result to obtain $$r(t)$$ will pose additional problems. More importantly, for the calculation of the classical collision cross section, we don't need to know anything about the time dependence. We need to know about the trajectory.

Deflection Function
We know, the angular momentum is,
 * $$L = \mu r^2\dot\phi$$

where $$\mu$$ is reduced mass along with $$L = \mu v_{rel}b$$, we can write
 * $$\dot\phi=bv_{rel}/r^2$$

After eliminating the time from the equation of motion:
 * $$\frac{d\phi}{dr}=\frac{\dot\phi}{\dot r}=-\frac{b}{r^2}(1-\frac{V(r)}{E}-\frac{b^2}{r^2})^{-1/2}$$

The negative sign indicates that the value of $$\phi$$ decreases as r increases. Now this equation can be integrated, with the result:
 * $$\phi_0 =\int_\infty^{r_0}\frac{d\phi}{dr}dr$$
 * $$\phi_0 =b\int_{r_0}^\infty\frac{dr}{(1-\frac{V(r)}{E}-\frac{b^2}{r^2})^{1/2}}$$

and, defining the deflection angle $$\chi = \pi-2\phi_0$$
 * $$\chi =\pi-2b\int_{r_0}^\infty\frac{dr}{(1-\frac{V(r)}{E}-\frac{b^2}{r^2})^{1/2}}.........(3)$$

The turning point $$r_0$$ is the smallest value that $$r$$ achieves during the collision; it depends on both $$b$$ and $$E$$. Since the collision is symmetric about the turning point owing to the time-reversal symmetry (energy conservation) of the system, the deflection angle is $$\pi-2\phi_0$$. The scattering angle is always positive, but the deflection angle can be negative, so that $$\theta = |\chi|$$ The deflection angle is a function of the impact parameter and the energy, so we can refer to it as the deflection function $$\chi(b;E)$$. The simple relationship between the scattering angle and the deflection angle means that we can rewrite the angle-differential cross section as
 * $$\sigma(\theta,E)=\frac{b}{sin\theta|d\chi/db|}$$

Now we can calculate the deflection function and the differential cross section from its derivative with respect to the impact parameter. There is one remaining ingredient: the potential function $$V(r)$$. For example let us consider hard sphere scattering with radius $$R_t$$. The deflection function is;
 * $$\chi =\pi-2b\int_{R_t}^\infty\frac{dr}{(1-\frac{b^2}{r^2})^{1/2}}$$

The integral is,
 * $$\chi=\pi-2sin^{-1}(b/R_t)$$

This formula is valid only for $$b < R_t$$. The deflection function is most easily graphed by solving for $$b$$; it is shown in Figure 4. Please note that the deflection function is always positive and hence it is equal to the scattering angle. Let see another example of Rutherford Scattering. If the potential has the form $$V (r) =B/r, B>0$$ the integral becomes, after transforming from r to u,
 * $$\chi(E; b) =\pi-2b\int_{0}^{u_c}\frac{du}{r^2(1-\frac{Bu}{E}-b^2u^2)^{1/2}}$$

where $$u_c = 1/r_c$$ is the positive root of the radical in the denominator of the integrand. The solution is
 * $$\chi(E;b) = 2sin^{-1}[\frac{B/E}{(B^2/E^2+4b^2)^{1/2}}]$$

This deflection function is shown in Figure 5 for the choice E/B = 1Å. There are two things to note about Rutherford Scattering. First, the deflection function is special, because instead of overwhelming the potential at long range as it does for n > 2, the centrifugal potential becomes dominant at short range instead. Consequently, the minimum in the effective potential persists for all values of $$b$$. Second, the dominance of the $$1/r$$ potential at long range means that the differential cross section cannot be integrated ( the total cross section for Coulomb scattering is infinite).

Example of atom-atom elastic scattering
To explore the features of the deflection function in the case of the Lennard Jones potential (Equation 2), we note that the potential can be written (sometimes called the Lennard-Jones s-t potential) $$V(r) =\frac{C_s}{r^s}-\frac{C_t}{r^t}$$ We require s > t so that the potential is attractive at long range and repulsive at short range. The potential can be scaled to read
 * $$V (r) =\frac{V_0}{s-t}[\frac{t}{\rho^s}-\frac{s}{\rho^t}]$$

Here $$\rho = r/r_m$$, with $$r_m$$, the minimum value of the potential, given by
 * $$r_m =(\frac{sC_s}{tC_t})^{1/(s-t)}$$

and the well depth $$V_0$$ is given by
 * $$V_0 = (s - t)\frac{C_s}{t}(\frac{sC_s}{tC_t})^{-s/(s-t)}$$

Let us now specialize to the Lennard-Jones 6-12 potential, which is now written
 * $$V (r) = V_0(\frac{1}{\rho^{12}}-\frac{2}{\rho^6})$$

where
 * $$r_m = (\frac{2C_{12}}{C_6})^{1/6}$$

and
 * $$V_0 =\frac{C_6^2}{4C_{12}}$$

The 6-12 potential has a classical turning point for $$L$$ = 0 and $$E$$ = 0 given by
 * $$r_0(E = 0; L = 0) =(\frac{C_{12}}{C_6})^{1/6}$$

The value of $$r_0$$ declines as the energy increases for $$E\ge 0$$, but not rapidly, as the potential rises very rapidly at positive energy. To the potential must be added the centrifugal term to get $$V_{eff}$$ :
 * $$\tilde {V}_{eff}(\rho) =\frac{1}{\rho^{12}}-\frac{2}{\rho^6}+\epsilon\frac{\tilde {b}^2}{\rho^2} ........(4)$$

In the above formula, we have scaled all lengths by $$r_m$$ and all energies by $$V_0$$ to arrive at a universal, dimensionless effective potential for all 6-12 potentials. Equation 4 is shown for a variety of values of angular momentum in Figure 6. In the absence of a potential, the turning point $$r_0 = b$$. This is clear intuitively,and it is also apparent from the centrifugal potential when it is written in the form $$V_{cent} = Eb^2/r^2$$. At energies well below the top of the angular momentum barrier - equivalently, for impact parameters sufficiently large for a given energy - this is the value of the turning point even if the potential is nonzero, since the angular momentum term dominates here. But as the energy increases, the trajectory surmounts the angular momentum barrier and the turning point is determined by the hard wall of the potential. Thinking about this is a little tricky, because the height of the barrier depends on the energy. Moreover, it is not possible to find expressions for the barrier location and height for a realistic potential with both attractive and repulsive components. Nevertheless, we can illuminate the matter by developing a simple approximate model for the centrifugal barrier for the 6-12 potential. Unlike the Coulomb potential, with its $$1/r$$ dependence, the long-range behavior of the 6-12 potential is dominated by the centrifugal potential with its $$1/r^2$$ dependence, and the short-range dependence is dominated by the $$1/r^{12}$$ term. The location and height of the centrifugal barrier for the 6-12 potential cannot be found in closed form, but since the barrier occurs outside $$r_m$$, where the repulsive part proportional to $$1/\rho^{12}$$ is negligible, it is not unreasonable to ignore the repulsive portion of the potential. Assuming that the potential is a pure attractive $$1/\rho^6$$ potential results in the following reduced expressions for the location and height of the centrifugal barrier:
 * $$\rho_B =(\frac{6}{\epsilon\tilde{b}^2})^{1/4}$$

and
 * $$\tilde{V}_{eff}(\rho_B)=\frac{2(\epsilon\tilde{b}^2)^{3/2}}{3\sqrt{6}} ........(5)$$

These approximate locations of the barriers are superimposed on the effective potentials shown in Figure 6. Equation 4 reveals that the height of the barrier at a fixed impact parameter increases faster than linearly in energy, making it difficult to use Figure 6 to gain intuition about the dependence of the barrier height on energy. However, for every energy, there does exist an impact parameter at which the barrier height equals the energy. An estimate of the relationship comes from setting $$\tilde{V}_{eff} = \epsilon$$, with the result
 * $$\tilde{b}_B=\frac{3}{(2\epsilon)^{1/3}}$$

For impact parameters greater than this critical value $$\tilde{b}_B$$, the barrier is too high and the collisional turning point $$r_0$$ is outside the barrier. For impact parameters lower than $$\tilde{b}_B$$, the barrier is surmounted and the turning point occurs at the hard repulsive wall, as shown in Figure 7. At high enough energy, the barrier becomes an inflection point, and there is no abrupt transition between the two regimes. Notice that, for large values of $$\tilde{b}$$, the angular momentum barrier always dominates and $$\rho_0\approx \tilde{b}$$. At small values of $$\tilde{b}$$, the angular momentum barrier is always insignificant, and the turning point is just the nearly energy-independent location of the hard wall. Classically, high energy collisions are defined by the insignificance of the well and centrifugal barrier in influencing the collision, i.e., $$E\gg |V_{eff}(r_e)|$$ and $$ E \gg V_{eff}(r_B)$$. If the collision energy does not substantially exceed the centrifugal barrier, the radial motion of the atoms slows significantly as they approach the barrier and then speeds up again after crossing it before stopping altogether at $$r_0$$. If the collision energy is precisely equal to $$V_{eff} (r_B)$$, the atoms become trapped in an orbiting resonance, and the deflection function diverges. This can be seen by examining Equation 3, since the square root in the denominator is proportional to $$\sqrt{E-V-Eb^2/r^2}=\sqrt{E-V_{eff}}$$ The deflection angle diverges, and the atoms spiral toward an orbital radius of $$\rho_B$$, taking infinite time to do so. The interaction of the short-range, long-range, and centrifugal potentials results in more complexity than one might expect at first in an effectively one dimensional problem. We discuss below their contributions to the deflection function. This deflection function, Equation 3, takes the form in scaled units
 * $$\chi(b;E) = \pi- 2\tilde{b}\int_{\rho_0}^{\infty}\frac{d\rho}{\rho^2\sqrt{1-\tilde{V}_{eff}}}$$

Orbiting Collisions
The phenomenon in which the barrier is exactly matched by the collision energy, is called the orbiting resonance; it occurs when the collision energy $$\epsilon_B =\tilde{V}_{eff} (\rho_B)$$. This phenomenon manifests itself in an infinite deflection, and examples of this can be seen in Figure 9. It is possible to approximate the local maximum in the potential by a parabola in order to obtain an analytical result valid in the vicinity of the orbiting resonance. As L increases, $$\epsilon_B$$ grows, until there comes a particular energy at which there is no longer a potential well, and the well and barrier coalesce into an inflection point. For $$L$$ (and hence $$b$$) greater than this critical value, orbiting resonances are impossible, and there is only one turning point.

Rainbow Scattering
Figure shows two examples of deflection functions for energies at which no orbiting resonance occurs. The singularity is replaced by a minimum. This minimum gives rise to rainbow scattering, whose origin can be understood by reference to Figure 8. The atoms are deflected from their initial trajectories by the rainbow angle, which corresponds to the maximum negative deflection angle. The cross section still contains contributions from larger scattering angles, corresponding to positive deflection angles all the up to $$\pi$$, so the rainbow angle $$\theta_r$$ is not the largest scattering angle. But because $$d\chi/db = 0$$, it does produce a singularity in the cross section as shown in Figure 9. Rainbow scattering is a general phenomenon that results from a pileup of trajectories, or caustic. The familiar atmospheric rainbow is another example, resulting from an extreme scattering angle that occurs after an internal reflection within a raindrop. In fact, Sir George Biddell Airy's 1838 paper giving a wave theory of the atmospheric rainbow was titled "On the intensity of light in the neighborhood of a caustic" 2. The rainbow is due to this pileup of rays; the colors that we associate with the rainbow are a separate phenomenon due to the dispersion in the index of refraction of water. Droplets of a nondispersive medium would still produce a rainbow, but it would be white.

Glory Scattering
As the impact parameter increases, the deflection angle decreases; using the impulse approximation, it is possible to obtain an approximate analytical expression for the deflection function and cross section, accurate in the limit of large impact parameter and small scattering angles. But there is another case that leads to precisely zero scattering angle, known as glory scattering. Figure 8 show examples of this case, which results from a perfect balance between the attractive and repulsive components of the potential, resulting in $$\chi =\theta = 0$$. This gives an additional contribution the deflection function, and hence to the cross section, at small angles. In the quantum mechanical theory of scattering, interference structure arises from interference between these two classes of trajectories. Perhaps, while traveling by air, you have seen a bright circle around the shadow of your plane on the clouds below; this forward-scattered optical phenomenon is known as a glory, and its physical basis is the same as that of glory scattering in the present context.