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## Central force and Scattering, Physics tutorial

:Central force and Scattering:The Generic Central Force ProblemWe first discuss the central force problem in general terms, considering arbitrary radially dependent potential energy functions.

The Equation of Motion:Central force is stated as one which satisfies strong form of Newton's third law. That is, given two particles a and b, force applied by particle a on b is equal and opposite to that exerted by particle b on particle a, and, furthermore, force depends only on separation of two particles and points along vector between two particles.

:Properties of an Isolated Two-Body Central-Force SystemLet us now consider an isolated two-body system interacting via a conservative central force. There are no other forces acting on the bodies.

From concepts of force, momentum, and energy for systems of particles. One can abstract the following facts about isolated two-body system:

0 = d/dtP

^{→}= d/dt(m_{a}r_{a}^{→}+ mbr^{→}_{b})Since P

^{→}is constant, the velocity of the center-of-mass R^{→}= P^{→}/M is fixed and thus the center of-mass system is inertial. We may therefore assume, without loss of generality, that r^{→}a and r^{→}b are coordinates in the center of mass system, where P^{→}vanishes and the center of mass is at the origin:0 = R

^{→}= m_{a}r^{→}_{a}+ m_{b}r^{→}_{b}0 = P

^{→}= m_{a}dr^{→}_{a}+ m_{b}dr^{→}_{b}Defining reduced mass

μ = (m

_{a}m_{b})/(m_{a }+ m_{b})Gives us

r

^{→}_{a}= μ/m_{a}r^{→}_{ab}r^{→}_{b}= μ/m_{b}r^{→}_{ab}Dynamics of two-particle system will be reduced to that of the single particle with mass μ moving in potential U(r

_{ab}). Consider two simple limits immediately:In the limit when m

_{b}is far greater than ma we have μ→m_{a}, r^{→}_{b}→ 0, and r^{→}_{ab}→ r^{→}_{a}. That is, center of mass is fixed on heavier mass and motion is completely of smaller mass. Angular momentum of system is conserved.In limit m

_{a}= m_{b}= m, we have μ = m/2, r^{→}_{a}= r^{→}_{ab}/2 and r^{→}_{b }= r^{→}_{ab}/2As system's center-of-mass has been taken to be at rest at origin, angular momentum comprises only of the internal angular momentum because of motion of two particles about center of mass.

The Lagrangian is L = 1/2μdr

^{→}_{ab}^{2}+ U(r_{ab})Reduction to a one body problem:For simplicity and better understanding of the above equation of motion we discuss the reduction of the two body problem to a mathematically equivalent problem of a single particle moving in one direction. First we reduce it to a one-body problem, and then we reduce the dimensionality. Our original problem has six degrees of freedom, but because of the symmetries in the problem, many of these can be simply separated and solved for. As there are no external forces, we expect the center of mass coordinate to be in uniform motion, and it allows us to use

R→ = Σ

_{α}m_{α}r_{α}^{→}/(Σ_{α}m_{α}) = (m_{1}r_{1}^{→ }+ m_{2}r_{2}^{→})/(m_{1}+ m_{2})Where M=m

_{1}+m_{2}.The kinetic energy is

T = 1/2MdR

^{→2}+ 1/2μdr^{→2}Where μ = m

_{1}m_{2}/(m_{1}+ m_{2}) is known as reduced mass. Therefore kinetic energy is transformed to form for two effective particles of mass M and μ that is neither simpler nor more complicated than it was in original variables.Dynamics of an Isolated Two-Body Central-Force System:Now, explore dynamics using Lagrangian. Conservation of L

^{→}already leads us to expect that one of the Euler-Lagrange equations will be trivial. Explicitly, we have (dropping the ab subscripts now):μd

^{2}r/dt = d/dt[μr^{2}sin^{2}ΦΘ] = 0We make general point that, when writing Euler-Lagrange equations for the multidimensional system, it is a good idea to begin with time derivatives on left side unexpanded until rest has been simplified as they are just time derivatives of canonical momenta and few of them may end up being conserved if right side of corresponding equation disappears, either explicitly or by suitable choice of initial conditions. Φ equation of motion tells us l

_{Φ}= constant, that would have become very unapparent if derivative had been expanded in three terms.1D Lagrangian is not attained just by rewriting 3D Lagrangian by using Φ = 0, Φ

^{2}= 0 and μr^{2}Θ = l_{Θ}one would have got wrong sign for centrifugal term. This difficulty happens as Lagrangian formalism assumes independent variations of different coordinates. Instead, we should just start with effective potential.Effective total energy is E = 1/2μdr

^{2}+ l_{Θ}^{2}/2μr^{2}+ U(r)This is conserved as effective potential is conservative. This effective total energy turns out to be equal to true total energy as Θ kinetic energy is included via l

_{Θ }term.Reduction to one dimension:In the problem under discussion, though, there is additional restriction that potential depends only on magnitude of r

^{→}, that is, on distance between two particles, and not on direction of r^{→}. Therefore we now convert from cartesian to spherical coordinates (r; Θ; Φ), for r^{→}. In terms of Cartesian coordinates (x; y; z)r = (x

^{2}+ y^{2}+z^{2})1/2Θ = cos

^{-1}(z/r)Φ = tan

^{-1}(y/z)x = rsinΘcosΦ, y = rsinΘsinΦ, z = rcosΘ

Plugging in kinetic energy is messy but finally reduces to rather simple form

T = 1/2μ[dxx

^{2}_{1}+ dy^{2 }+ dz^{2}]= 1/2μ[dr

^{2}+ r^{2}dΘ^{2}+ r^{2}sin^{2}ΘΦ^{2}]r sinΘ is distance of particle from the z-axis, so PΦ is just z-component of angular momentum, l

_{z}Of course all component of angular momentum L^{→}= l_{Θ}= r^{→}x p^{→}is conserved, as in our effective one body problem there is no torque about the origin. Thus l_{Θ}is the constant and motion should remain in the plane perpendicular to l_{Θ}and passing through origin as the consequence of fact that r^{→}is perpendicular to l_{Θ}. It simplifies things if we select coordinates so that L^{→}is in z-direction. Then Θ = π/2, dΘ = 0, l_{Θ}= μr^{2}Φ. r equation of motion is thenμd

^{2}r/dr = μrΦ^{2}+ dU/dr = 0 = μr^{2}- lΘ^{2}/μr^{3}+ dU/drThis is the one-dimensional motion of body in the effective potential

U

_{eff}(r) = U(r) + lΘ^{2}/2μr^{2}Therefore we have reduced the two-body three-dimensional problem to one with single degree of freedom, without any addition of the centrifugal barrier term l

_{Θ}^{2}/2μr^{2}to potential.:Kepler's problemJohannes Kepler suggested three laws of planetary motion. There are three statements which described motion of planets in the sun-centered solar system. Kepler's efforts to describe underlying reasons for such motions are no longer accepted; however, actual laws themselves are still considered accurate explanation of motion of any planet and any satellite.

Kepler's three laws of planetary motion can be described as follows:

i)

Kepler's first law:Path of planets about the sun is elliptical in shape, with center of the sun being situated at one focus.- Law of Ellipses.

Using first Lagrange's equation we can prove Kepler's first law that defines that each planet moves in elliptical orbit with sun at one focus

d/dt(∂L/∂dr) - ∂L/∂r = 0 μ(d

^{2}r/dr - rdΘ^{2}) = -∂U(r)/∂r = - k/r^{2}μr

^{2}dΘ = l_{Θ}= d_{Θ}= l_{Θ}u^{2}/μdt = dΘ(μ/lu

^{2})In polar coordinates, we have

r = α/(1 + pcos(Θ - Θ

_{0}))Where α = l

_{Θ}^{2}/kμ and P = √1 + qEl_{Θ}^{2}/k^{2}μThis is the equation for conic sections that explains

p > hyperbola

p = ellipse

p < parabola

Motion of planet is bounded to sun and thus corresponds to case p<1. Furthermore, as m

_{s}is far less than m_{p}1/μ = 1/m

_{p}+ 1/m_{s}≈ 1/m_{p}And R = (m

_{p}r_{p}+ m_{s}r_{s})/(m_{p}+ m_{s}) ≈ [m_{p}/(m_{p}+ m_{s})]r_{p}+ [m_{s}/(m_{p}+ m_{s})]r_{s}≈ r_{s}I.e. centre of mass around coincides with position of sun and r is approximately distance from sun to planet. Thus, we have proved Kepler's first law that defines that every planet moves in elliptical orbit with sun at one focus.

ii)

Kepler's second law:The imaginary line drawn from center of the sun to center of the planet will sweep out equivalent areas in equal intervals of time. - Law of Equal Areas.

Radius vector r sweeps out area dA = 1/2r

^{2}Θ in r^{2}Θ in time dt. Rate at which radius vector sweeps out area is dA/dt = 1/2 r^{2}Θ. Comparing this rate with momentum PΘ we proved thatdA/dt = 1/2r

^{2}dΘ = 1/2μ = constant.This is Kepler's second law that defines that radius vector drawn from sun to the planet explains equal areas in equal times.

iii)

Kepler's third law:Ratio of squares of periods of any two planets is equivalent to ratio of cubes of their average distances from sun. - Law of Harmonies.

Using second Kepler's law dA/dt = 1/2r

^{2}and expression for angular momentum PΘ = μr^{2}Θ to write dA = 1/2μdt. Integrating expression over whole area of the ellipse, we attained A = πab =1/2μT, where T is period of planetary motion and πab is area of ellipse.Using given relation a = α

^{-1}b^{2}to attain πa^{1/2}a^{3/2}= 1/2μT^{2}= a^{3}= l^{2}α^{-1}/(2πμ)^{2}T^{2}. Using the definition α = l^{2}/kμ to finally get a3 = (k/(4π2μ))T^{2}that is Kepler's third law that states that square of period of revolution about sun is proportional to the cube of major axis of orbit.Time Dependence of the Kepler Problem Solutions:Orbit solutions as functions r(Θ). This of course explains much of the dynamics of the problem. But one does indeed often want orbit as the function of time, so we attain that result here.

Period of Elliptical Orbits

We can get period of elliptical orbits by Kepler's second law, Kepler's second law states that

dA/dt = l

_{Θ}/2μArea of ellipse is A =πab, so period is time required to sweep out the area of the ellipse,

T = A/dA/dt = πab/l

_{Θ}/2μWrite this in terms of parameters of orbit p and ε to get Kepler's third law:

T = (2πμP

^{2}/(1 - ε^{2})^{3/2})(1/μ√GMp)= 2π√a

^{3}/GMPeriod depends only on a. Of course, a encodes information about total energy E and angular momentum l

_{Θ}. Implication of Kepler's third law for the solar system is that all orbits must lie on the single T^{2}∝ a^{3}curve as M, dominated by sun, is almost the same for all planets.:Scattering cross sectionsScattering cross-section is effective area that quantifies intrinsic rate a given events occurs during scattering of two particle species (in quantum mechanics, this also includes waves). Conventionally, one of the species is treated as projectile (incident beam), and other species is treated as target (scattering center).

Intuitively, event is said to have cross-section of σ if rate is equal to rate of collisions in equivalent, idealized experiment where:

Initial Conditions:We consider the particle incident on the force center subject to the scattering potential. It is supposed that particle is asymptotically free, having enough energy to be in the unbound orbit. To have unbound orbits, potential energy should go to zero at large r. Energy of system is therefore

E = 1/2μv

_{∞}^{2}Where v

_{∞}is asymptotic particle speed as r → ∞. We will use v_{∞}to parametize initial energy of system, v_{∞}only specifies energy. We should also specify geometry. Only free parameter left is l_{Θ}that we can see specifies geometry as follows. As system is unbound, trajectory should become straight lines at large radii, reflecting incoming and outgoing velocity vectors. For instance, for a 1/r potential, we know that ingoing and outgoing velocity vectors classify asymptotes of hyperbolic orbit.We can compute distance between scattering center and the straight lines where they come closest to scattering center. This distance is stated to be impact parameter, generally related with symbol b. Small line from center to dotted line is impact parameter. Impact parameter is related to l

_{Θ}. We know that l_{Θ}= r^{→}x μdr^{→}. At r= ∞, we know that dr^{→}points in direction along asymptote and r^{→}points to origin, so angle between r^{→}and dr^{→}that we call γ is subtended by impact parameter b. r^{→}is display by line that extend from center to -10 point. So,l

_{Θ}= rμv_{∞}sinγ = μv_{∞}bl

_{Θ }specifies geometry through b, that fixes position of asymptote (v_{∞}) relative to scattering center.The Generic Cross SectionDifferential Cross Section:Incident particles in beam will be scattered into the range of angles depending on input impact parameters (and beam velocity). We state differential scattering cross section, dσ/dΩ (Θ

_{s}, Φ_{s}), via probability of the incident particle being scattered into solid angle dΩ in direction (Θ_{s}, Φ_{s}) where Φs is polar angle estimated from beam axis and Φs is azimuthal angle around beam axis. If dN(Θ_{s}, Φ_{s}) is the number of particles per unit time scattered into the solid angle dΩ at (Θ_{s}, Φ_{s}) we state differential cross section via relation1/Adσ/dΩ (Θ

_{s}, Φ_{s}) dΩ = dN(Θ_{s}, Φ_{s}) /FA:1/r PotentialFor 1/r potential, we can determine differential cross section explicitly as we have explicit relationships between Θ and r.

Total Cross Section:If we compute the total cross section from Rutherford formula there will be an infinite result. This is due to, for a 1/r potential, probability of scattering doesn't decrease adequately quickly with increasing b as effective area of scattering center is infinite. If potential is made to converge more rapidly (e.g., by multiplying be the exponential decay), then finite total cross section is attained.

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