Motion in a central potential

Consider a particle moving in a central potential. 
U = U(r),  F = f(r)(r/r),  f(r) = -dU/dr.
In a central potential the energy E and the angular momentum M are conserved.  The motion is in a plane.  Choose this plane to be the x-y plane.  Then

L = T - U = ½m[(dr/dt)2 + r2(dφ/dt)2] - U(r).
E = T + U = ½m[(dr/dt)2 + r2(dφ/dt)2] + U(r) = constant.
M = pφ = ∂L/∂(dφ/dt) = mr2(dφ/dt) = constant, angular momentum is conserved.

Kepler’s second law:

Areal velocity = dA/dt = ½r(rdφ/dt) = ½r2(dφ/dt) = constant.
The area swept out per unit time is constant for all central potentials.

Equations of motion involving r only:

E = ½m(dr/dt)2 + M2/(2mr2) + U(r) = ½m(dr/dt)2 + Ueff(r),
with Ueff(r) = M2/(2mr2) + U(r).
yields dr/dt = ±[(2/m)(E - Ueff(r))]½.

Lagrange’s equations yield
md2r/dt2 - M/(mr3) = f(r),  or  md2r/dt2 = -dUeff/dr,
with f(r) = -dU/dr.

Equations for the orbit:

dφ = [M/(mr2)]dt,  d/dt = [M/(mr2)] d/dφ.

From
dr/dt = [(2/m)(E - Ueff(r))]½
we obtain
dφ = [M/(mr2)]dr [(2/m)(E - Ueff(r))].
φ = ∫M(dr/r2)/[2m(E - Ueff(r))]½ + φ0,
or, with u = 1/r,
φ = φ0 - ∫Mdu/[2m(E - Ueff(u))]½.

From
md2r/dt2 - M/(mr3) = f(r)
we obtain
(M/r2) d/dφ[(M/(mr2))dr/dφ] - M/(mr3) = f(r),
or
(M2u2/m)(d2u/dφ2 + u) = -f(u),  where  u = 1/r.


The Kepler problem

Let U(r) = -α/r,  Ueff(r) = -α/r + M2/(2mr2),  f(r) = -α/r2.
Then the equation for the orbit yields
φ = cos-1[(M/r - mα/M)/(2mE + m2α2/M2)½] + φ0.
cos(φ - φ0) = (1/e)(p/r - 1),
with p = M2/(mα), e = (1 + 2EM2/(mα2))½,
or
p/r = 1 + e cos(φ - φ0).
This is the equation of a conic section.  Here e is the eccentricity.

E > 0, e > 1       hyperbola
E = 0, e = 1       parabola
E < 0, e < 1       ellipse
E = -mα2/(2M2), e = 0       circle

image

The ellipse and the circle are closed orbits.  For closed orbits we have
rmin = p/(1 + e),  rmax = p/(1 - e),
semi-major axis:  a = ½(rmin + rmax) = p/(1 - e2) = ½α/|E|,
semi-minor axis:  b = a(1 - e2)½ = M/(2m|E|)½.

From
dA/dt = M/(2m),  A = MT/(2m) = πab
we find the period T.
T = 2mπab/M = 2π(m/α)½a3/2.
This is Kepler’s third law.  It only holds for the attractive 1/r potential.

For the gravitational potential V(r) = -GM/r remember:
All orbits with the same semi-major axis have the same total energy per unit mass and the same period.


Two interacting particles

Consider two particles with masses m1 and m2, subject to internal forces
F12 = F12(r2 - r1)/r12,    r12 = |r1 - r2|,     F21 = -F12.
The equations of motion are 
m1(dv1/dt) = F21,  m2(dv2/dt ) = F12  = -F21,  m1dv1/dt + m2dv2/dt = 0.

Define the center of mass (CM) coordinate R = (m1r1 + m2r2)/(m1+ m2).
The center of mass moves with constant velocity.

Define the relative coordinate  r = r1 - r2.  
Then r1 = R + (m2/(m1+m2))r r R - (m1/(m1 + m2))r .
We find that (m1m2/(m1+ m2 ))d2r/dt2 = F21(r), or µd2r/dt2 = F21(r). 
Here µ = m1m2/(m1 + m2) is called the  reduced mass.
The problem of the relative motion of two interacting masses m1and m2 can be solved by solving for the motion of one fictitious particle of reduced mass µ in a central field.

Equation of motion of the fictitious particle:
µd2r/dt2 = F21(r) = F21(r)(r/|r|).

The equations of the motion for the system can be separated into equations for the center of mass motion and equations for the relative motion.  Similarly, the sum of the momenta and sum of the angular momenta can be split into parts pertaining to the center of mass motion and parts pertaining to the relative motion.

In the CM frame of two interacting particles the Lagrangian
L = ½m1(dr1/dt)2 + ½m2(dr2/dt)2 - U(|r1 - r2|)
can be written as
L = ½μ(dr/dt)2 - U(r),  μ = m1m2/(m1 + m2),  r = r1 - r2.
The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass µ moving in a central potential U(r).


Elastic scattering

Consider the scattering of a particle by a central potential. 
We define the differential scattering cross section σ(Ω) = dσ/dΩ through the following expression:

# of particles scattered into the solid angle dΩ per unit time = I σ(Ω) dΩ ,

where I is the intensity of the incident beam, i.e. the number of beam particles per unit area per unit time.
For a central potential σ(Ω) is independent of φ. 
We write σ(Ω) = σ(θ) = dσ/dθ.
The number of particles scattered through an angle between θ and θ + dθ per unit time is
I σ(Ω) dΩ = I σ(θ) 2π sinθ dθ.

We define the impact parameter b through
M = mv0b = b(2mE)½,
where M is the angular momentum and v0 is the incident speed at infinite distance.
Once E and b are fixed, the scattering angle is uniquely determined.
A particle incident with impact parameter between b and b + db will be scattered through an angle between θ and θ + dθ.  We can write

I 2π σ(θ) sinθ dθ = -I 2π b db,
σ(θ) =|(b/sinθ)(db/dθ)|.

In a central potential the motion is in a plane and M and E are constant.
E = ½m(dr/dt)2 + b2E/r2 + U(r),  dr/dt = [(2/m)(E(1 - b2/r2) - U(r))]½.

From
dr/dt = [(2/m)(E(1 - b2/r2) - U(r))]½,  d/dt = [b(2mE)½/(mr2)] d/dφ,
we find
dφ = [b(2mE)½/(mr2)]dr [(2/m)(E(1 - b2/r2) - U(r))].
φ(u) = b∫0udu'/[1 + b2u'2 - U(u')/E]½,
with u = 1/r and φ(z = -∞) = 0.

Let θ be the angle between the incident and the scattered direction and φ0 be the angle between r(z = -∞) and rmin.  Then the equations
φ0 = b∫0umaxdu'/[1 + b2u'2 - U(u')/E]½,
and
E = M2/(2mr2min) + U(rmin) = b2E/r2min + U(rmin) =  b2Eu2max + U(umax)
determine umax.  We have
θ = π - 2φ0 for a repulsive potential,
θ = 2φ0 - π for an attractive potential, (or θ = π - 2φ0, θ < 0).

image

If U(r) = α/r, U(u) = αu, U(umax) = αumax, then
umax = -α/(2b2E) + [α2/(4b4E2) + 1/b2]½ and φ0 = cot-1(α/(2bE)).
cotφ0 = α/(2bE),  cot(θ/2) = cot(π/2 - φ0) = tanφ0 = 1/cotφ0.
b = [α/(2E)] cot(θ/2), db/dθ =  -½[α/(2E)] sin-2(θ/2),
σ(θ) = ¼[α/(2E)]2/sin-4(θ/2).
This is the Rutherford’s formula.


Frame transformations

Let θ be the scattering angle in the lab frame and θ0 be the scattering angle in the CM frame.  The number of particles scattered into a detector is the same in the laboratory and in the CM frame.  Therefore
σ(θ) sinθ dθ = σ(θ0) sinθ00.