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} + r^{2}(dφ/dt)^{2}]
- U(r).

E = T + U = ½m[(dr/dt)^{2} + r^{2}(dφ/dt)^{2}]
+ U(r) = constant.

M = p_{φ} =
∂L/∂(dφ/dt) = mr^{2}(dφ/dt) = constant, angular momentum is
conserved.

Areal velocity = dA/dt =
½r(rdφ/dt) = ½r^{2}(dφ/dt) =
constant.

The area swept out per unit time is constant for **all central potentials**.

E = ½m(dr/dt)^{2} + M^{2}/(2mr^{2})
+ U(r) = ½m(dr/dt)^{2} + U_{eff}(r),

with
U_{eff}(r) = M^{2}/(2mr^{2})
+ U(r).

yields dr/dt = ±[(2/m)(E - U_{eff}(r))]^{½}.

Lagrange's equations yield

md^{2}r/dt^{2} - M^{2}/(mr^{3}) = f(r),
or
md^{2}r/dt^{2} = -dU_{eff}/dr,

with f(r) = -dU/dr.

dφ = [M/(mr^{2})]dt, d/dt =
[M/(mr^{2})] d/dφ.

From

dr/dt = [(2/m)(E - U_{eff}(r))]^{½}

we obtain

dφ = [M/(mr^{2})]dr [(2/m)(E - U_{eff}(r))]^{-½}.

φ = ∫M(dr/r^{2})/[2m(E - U_{eff}(r))]^{½}
+ φ_{0},

or, with u = 1/r,

φ = φ_{0} - ∫Mdu/[2m(E - U_{eff}(u))]^{½}.

From

md^{2}r/dt^{2} - M^{2}/(mr^{3}) = f(r)

we obtain

(M/r^{2}) d/dφ[(M/(mr^{2}))dr/dφ] - M^{2}/(mr^{3}) = f(r),

or

(M^{2}u^{2}/m)(d^{2}u/dφ^{2 }
+ u) = -f(u), where u = 1/r.

Let U(r) = -α/r,
U_{eff}(r) = -α/r
+ M^{2}/(2mr^{2}), f(r) = -α/r^{2}.

Then the equation for the orbit yields

φ = cos^{-1}[(M/r - mα/M)/(2mE +
m^{2}α^{2}/M^{2})^{½}]
+ φ_{0}.

cos(φ - φ_{0})
= (1/e)(p/r - 1),

with p = M^{2}/(mα),
e = (1 + 2EM^{2}/(mα^{2}))^{½},

or

p/r = 1 + e cos(φ -
φ_{0}).

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 = -mα^{2}/(2M^{2}),
e = 0 |
circle |

The ellipse and the circle are **closed orbits**.
For closed orbits we have

r_{min} = p/(1 + e), r_{max} = p/(1 - e),

semi-major axis: a = ½(r_{min} +
r_{max}) = p/(1 - e^{2}) = ½α/|E|,

semi-minor axis: b = a(1 - e^{2})^{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)^{½}a^{3/2}.

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.**

Consider two particles with masses m_{1} and m_{2}, subject to
internal forces

**F**_{12} = F_{12}(**r**_{2
}- **r**_{1})/r_{12}, r_{12 }= |**r**_{1 }-
**r**_{2}|,
**F**_{21 }= -**F**_{12}.

The
equations of motion are

m_{1}(d**v**_{1}/dt) =
**F**_{21},_{
}m_{2}(d**v**_{2}**/**dt) =
**F**_{12 }= -**F**_{21},_{ }m_{1}d**v**_{1}/dt_{
}+ m_{2}d**v**_{2}/dt_{ }= 0.

Define the center of mass (CM)
coordinate **R** = (m_{1}**r**_{1
}+ m_{2}**r**_{2})/(m_{1}+ m_{2}).

The center of mass moves with constant velocity.

Define the relative
coordinate **r** = **r**_{1 }-
**r**_{2}.

Then **r**_{1
}= **R **+ (m_{2}/(m_{1}+m_{2}))**r**,
**r**_{2 }= **R **- (m_{1}/(m_{1 }+ m_{2}))**r
.
**We find that (m

Here μ = m

The problem of the relative motion of two interacting masses m

Equation of motion of
the fictitious particle:

μd^{2}**r**/dt^{2} =
**F**_{21}(r) = F_{21}(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 = ½m_{1}(d**r**_{1}/dt)^{2}
+ ½m_{2}(d**r**_{2}/dt)^{2} - U(|**r**_{1}
- **r**_{2}|)

can be written as

L = ½μ(d**r**/dt)^{2} - U(r),
μ = m_{1}m_{2}/(m_{1} + m_{2}),
**r** = **r**_{1} -
**r**_{2}.

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).

Consider the scattering of a particle by a central
potential.

We define the differential scattering cross section
σ(Ω) = dσ/dΩ through the 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/dΩ.

The number of particles scattered through an angle between
θ and
θ
+ dθ
per unit time is

I σ(θ) 2π sinθ
dθ.

We define the **impact parameter**** b**
through

M = mv_{0}b = b(2mE)^{½},

where M is the angular momentum and v_{0} 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} + b^{2}E/r^{2}
+ U(r), dr/dt = ±[(2/m)(E(1 - b^{2}/r^{2}) - U(r))]^{½}.

From

dr/dt = [(2/m)(E(1 - b^{2}/r^{2}) - U(r))]^{½},
d/dt =
[b(2mE)^{½}/(mr^{2})] d/dφ,

we find the equation for the trajectory.

dφ = [b(2mE)^{½}/(mr^{2})]dr
[(2/m)(E(1 - b^{2}/r^{2}) - U(r))]^{-½}.

φ(u) = b∫_{0}^{u}du'/[1 - b^{2}u'^{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 φ_{0} be the angle between
**r**(z =
-∞)
and **r**_{min}.

Then

φ_{0} = b∫_{0}^{umax}du'/[1
- b^{2}u'^{2} - U(u')/E]^{½},

and

E =
M^{2}/(2mr^{2}_{min})
+ U(r_{min}) = b^{2}E/r^{2}_{min}
+ U(r_{min}) = b^{2}Eu^{2}_{max}
+ U(u_{max})

determines u_{max}. We have

θ
= π - 2φ_{0}
for a repulsive potential,

θ
= 2φ_{0 }- π
for an attractive potential, (or θ = π - 2φ_{0}, θ
< 0).

If U(r) = α/r, U(u) = αu, U(u_{max}) = αu_{max}, then

u_{max} = -α/(2b^{2}E) + [α^{2}/(4b^{4}E^{2})
+ 1/b^{2}]^{½} and φ_{0}
= cot^{-1}(α/(2bE)).

cotφ_{0}
= α/(2bE), cot(θ/2) = cot(π/2 - φ_{0}) = tanφ_{0}
= 1/cotφ_{0}.

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**.

Let θ be the scattering angle in the lab frame and θ_{0} 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(θ_{0})
sinθ_{0} dθ_{0}.