__Definitions__

A target of Aluminum (A = 27) with an aerial density of 1 mg/cm^{2}
is positioned perpendicular to a 0.5 μA beam of deuterons (Z = 1). If the cross
section for producing protons at an angle of 30^{o} with respect to the
beam direction is 15 mb/sr, (1 mb = 10^{-27} cm^{2}), how many
protons per second will be incident on a 1 cm^{2} detector facing the
target at 30^{o} and 10 cm from the target?

Solution:

- Concepts:

The cross section - Reasoning:

We are given the cross section for producing a proton in a single deuteron - Aluminum nucleus collision.

We are asked to find the total number of protons hitting a detector per second. - Details of the calculation:

If we have not one, but many scattering centers we distinguish the following limiting cases:

(a) small beam, big target:

(# of particles scattered into the solid angle dΩ)/s

= [(# of beam particles)/s] * [(# of target particles)/area ] * σ(Ω)dΩ.

(b) big beam, small target:

(# of particles scattered into the solid angle dΩ)/s

= [(# of beam particles)/(area * s)] * (# of target particles) * σ(Ω)dΩ.

In this problem we have case (a).

(# of beam particles)/s = (0.5*10^{-6 }C/s)/(1.6*10^{-19 }C) = 3.125*10^{12}/s.

(# of aluminum atoms)/cm^{2}= (1 mg/cm^{2})/(27*1.6*10^{-27 }kg) = 2.29*10^{19}/cm^{2}.

What is dΩ?

r^{2}dΩ = (10 cm^{2})dΩ = 1 cm^{2}, dΩ = 0.01 sr.

# of protons/s = (3.125*10^{12}/s) * (2.29*10^{19}/cm^{2}) * (15*10^{-27 }cm^{2}) * 0.01 = 1 * 10^{4}/s.

__Hard sphere scattering__

Calculate the differential and total scattering cross section for scattering a particle off a fixed, "hard" sphere of radius R.

Solution:

- Concepts:

The scattering cross section, scattering from a spherically symmetric potential - Reasoning:

The potential has spherical symmetry, V(r) = ∞, r < R; V(r) = 0, r >R. - Details of the calculation:

For scattering off a central potential we have σ(θ) = |(b/sinθ)(db/dθ)|, θ = π - 2φ_{0}.

Referring to the figure above we have

b/R = sinα

θ = π - 2α.

α = (π - θ)/2.

b = Rsin(π/2 - θ/2) = Rcos(θ/2).

db/dθ = (R/2)sin(θ/2).

σ(θ) = (b/sinθ)((R/2)sin(θ/2) = R^{2}cos(θ/2)sin(θ/2)/(2sinθ) = R^{2}/4.

σ_{total}= ∫σ(θ)dΩ, = ∫∫σ(θ)sin(θ)dθdΦ = πR^{2}= total elastic scattering cross section.

Solution:

- Concepts:

Conservation of energy and momentum - Reasoning:

In the elastic collision kinetic energy and momentum are conserved. - Details of the calculation:

Let disk A initially move in the x direction.

The in the CM frame the initial velocity of disk A is 5 m/s**i**and the initial velocity of disk B is -5 m/s**i**.

After the collision disk A has velocity components v_{xA}= -5 m/s cosθ, v_{yA}= 5 m/s sinθ, and disk B has velocity components v_{xB}= 5 m/s cosθ, v_{yB}= -5 m/s sinθ.

In the lab frame v_{A}= 5 m/s (1 - cosθ)**i**+ 5 m/s sinθ**j**,

and v_{B}= 5 m/s (1 + cosθ)**i**- 5 m/s sinθ j.

v_{A}^{2}= 25 m^{2}/s^{2}= 25 m^{2}/s^{2}[(1 - cosθ)^{2}+ sin^{2}θ].

2 - 2cosθ = 1, θ = 60^{o}.

Impact parameter b: b = 2R sin30^{o}= R.

Consider the perfectly elastic scattering of a hard sphere of mass m and
initial velocity v_{0} against a stationary hard sphere of mass M. The
sum of the radii of the two spheres is D.

(a) Compute the energy lost by the small mass in the collision in the lab frame
as a function of the scattering angle θ in the center of mass frame.

(b) Compute the probability P(θ)dθ of scattering through an angle θ in the
center of mass frame, assuming that the particles m in the incident beam are
uniformly distributed across a cross section of the beam. Find the differential
scattering cross section σ(θ), and show that your expression integrates to the
known total hard sphere cross section πD^{2}.

(c) Use the results of parts (a) and (b) to compute the average energy loss per
collision. For what value of m/M does the average energy loss maximize?

Solution:

- Concepts:

Elastic scattering, energy and momentum conservation, frame transformations - Reasoning:

In the elastic collision kinetic energy and momentum are conserved. A frame transformation is necessary because the energy lost by the small mass in the lab frame is given in terms of the scattering angle in the CM frame. - Details of the calculation:

(a) The CM frame moves with respect to the laboratory frame with speed V.

Before the collisions: m(v_{0}- V) = MV. Therefore: V = mv_{0}/((m + M).

In the CM frame m has speed (v_{0}- V) = Mv_{0}/(m + M) = v' before and after the collision.

In the lab frame the speed of m after the collision is v.

v^{2}= v'^{2}+ V^{2}+ 2v'Vcosθ = v_{0}^{2}(m^{2}+ M^{2}+ 2mMcosθ)/(m + M)^{2}

= v_{0}^{2}(1 - 2mM(1 - cosθ)/(m + M)^{2}).

(v^{2}- v_{0}^{2})/v_{0}^{2}= ∆E/E = -2mM(1 - cosθ)/(m + M)^{2}).

∆E = -v_{0}^{2}m^{2}M(1 - cosθ)/(m + M)^{2}) is the energy lost by the small mass in the collision in the lab frame as a function of the scattering angle θ in the center of mass frame.

(b) Let b be the impact parameter.

For particles which collide, P(b)db = 2πbdb/(πD^{2}) is the probability of a collision with impact parameter between b and b + db.

θ = π - 2α. b = Dsinα.

The relationship between the impact parameter and the scattering angle is

b = Dsin(π/2 - θ/2) = Dcos(θ/2), db/dθ = -(D/2)sin(θ/2).

P(θ)dθ = P(b)|db/dθ|dθ = 2b|db/dθ|dθ/D^{2}= cos(θ/2)sin(θ/2)dθ = ½sinθdθ.

∫_{0}^{π}P(θ)dθ = 1.

σ(θ) =|(b/sinθ)(db/dθ)| = ½D^{2}cos(θ/2)sin(θ/2)/sinθ.

σ(θ) 2π sinθ dθ = πD^{2}cos(θ/2)sin(θ/2)dθ = πD^{2}P(θ)dθ.

∫_{0}^{π}σ(θ) 2π sinθ dθ = πD^{2}.

(c) average energy loss = ∫_{0}^{π}∆E(θ) P(θ)dθ

= (½v_{0}^{2}m^{2}M/(m + M)^{2}) ∫_{0}^{π}(1 - cosθ)sinθ dθ = v_{0}^{2}m^{2}M/(m + M)^{2}

= v_{0}^{2}m(m/M)/(m/M + 1)^{2}.

d(x/(x +1)^{2})/dx = 0 for x = 1.

The average energy loss maximizes for m/M = 1. The average energy loss then equals ½ of the initial kinetic energy.

__Scattering by a 1/r ^{n} potential__

A particle of mass m moves under a central repulsive force F(r) = km/r^{3}.
At its distance of closest approach r_{0} it has speed v_{0}.

(a) Find the orbital equation r(θ) for the particle motion, evaluating
constants in terms of r_{0} and v_{0}.

(b) Find the impact parameter and the total angular deflection, assuming the
particle approaches from large r.

(c) Sketch the particle trajectory, indicating the impact parameter and total
deflection calculated in part (b).

Solution:

- Concepts:

Motion in a central potential,**F**= f(r)(**r**/r), f(r) = -dU/dr, impact parameter b (M = mbv_{inf})

For motion under the influence of a central force E and M are conserved.

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

The equation that relates the orbit to the force is

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

We are given the force and are asked to find the equation of the orbit. We are then asked to find a relationship between the impact parameter and the total deflection. - Details of the calculation:

(a) f(u) = kmu^{3}, d^{2}u/dφ^{2 }+ u = -(km^{2}/M^{2})u, d^{2}u/dφ^{2 }+ (1 + km^{2}/M^{2})u = 0,

u = A sin(Bφ + C).

B^{2}= (1 + km^{2}/M^{2}), M = mbv_{inf}= mr_{0}v_{0}, B^{2}= (1 + k/(r_{0}v_{0})^{2}).

Let u = 0 at φ = 0, then C = 0.

At the distance of closest approach φ = φ_{0}, r = r_{0}= r_{min}, u = u_{max}.

1/r_{0}= A sin(Bφ_{0}) = A sin(π/2), Bφ_{0}= π/2, A = 1/r_{0},

1/r = (1/r_{0})sin((1 + k/(r_{0}v_{0})^{2})^{½}φ)

(b) Bφ_{0}= π/2, φ_{0}= π/(2B). We need φ_{0}in terms of the impact parameter b.

From energy conservation we have E = ½mv_{inf}^{2 }= ½mv_{0}^{2 }+ km/2r_{0}^{2},

v_{inf}^{2 }= v_{0}^{2 }+ k/r_{0}^{2}.

b = r_{0}v_{0}/v_{inf}, b^{2}= (r_{0}v_{0})^{2}/(v_{0}^{2 }+ k/r_{0}^{2}) = r_{0}^{2}/B^{2}. b = r_{0}/B is^{ }the impact parameter.

φ_{0}= πb/(2r_{0}). θ = π - 2φ_{0}is the deflection angle.

(c) For example, pick b = r_{0}/2, then φ_{0}= π/4, θ = π/2.

A fixed force center scatters a particle of mass m and initial velocity
**u**_{0}
according to the force law f(r) = k/r^{3}. Determine the
differential scattering cross section.

Solution:

- Concepts:

A particle moving in a central potential

U = U(r),**F**= f(r)(**r**/r), f(r) = (-∂U/∂r). U = k/(2r^{2}). E > 0 for scattering. - Reasoning:

We are asked to find the differential scattering cross section for a particle being scattered by a central force. - Details of the calculation:

f_{r}= -∂U/∂r = k/r^{3}.

σ(θ) = |(b/sinθ)(db/dθ)| θ = π - 2φ_{0}if k > 0.

We need to find θ as a function of b.

U = k/(2r^{2}), U(u) = ku^{2}/2, u = 1/r.

φ_{0}= ∫_{0}^{u_max }bdu/(1 - b^{2}u^{2}- ku^{2}/(2E))^{½}

for motion in a central potential.

u_{max}= (E/(b^{2}E + k/2))^{½}.

Let k > 0, repulsive potential:

φ_{0}= b(b^{2}+ k/(2E))^{-½}∫_{0}^{1 }du'/(1 - u'^{2})^{½}, u' = (b^{2}+ ku/(2E))^{½}.

φ_{0}= b(b^{2}+ k/(2E))^{-½}sin^{-1}(1) = (bπ/2)/(b^{2}+ k/(2E))^{½}

since ∫dx/(a^{2}- x^{2})^{½}= sin^{-1}(x/a).

θ = π - πb/(b^{2}+ k/(2E))^{½}, (π - θ)^{2}= (πb)^{2}/(b^{2}+ k/(2E)),

b^{2}= [k/(2E)](π - θ)^{2}/[θ(2π - θ)].

db^{2}/dθ = 2bdb/dθ = -(k/E)π^{2}(π - θ)/[θ^{2}(2π - θ)^{2}].

σ(θ) = (k/(2E))π^{2}(π - θ)/[θ^{2}(2π - θ)^{2}sinθ].

Let k < 0, attractive potential:

If b^{2}E < |k/2| we find no solution for u_{max}, there is no turning point, the particle spirals into the origin. Its angular velocity increases beyond any limit, such that mr^{2}dφ/dt = M stays constant.

If b^{2}E > |k/2| we have φ_{0}= (bπ/2)/(b^{2}+ (k/(2E)))^{½}> π/2.

φ_{0}may be much larger than 2π, the particle may turn around the force center several times.

Two particles of mass m are moving in the x-y plane. The inter-particle
potential energy is U(**r**_{1 }-
**r**_{2}) = U_{0}/|**r**_{1}
- **r**_{2}|^{2} with U_{0} > 0. The initial
conditions are**r**_{1}(t = 0) = (-∞, -y_{0}),
**r**_{2}(t
= 0) = (∞, y_{0}), and **p**_{1}(t = 0) = (p_{0}, 0),
**p**_{2}(t = 0) = (-p_{0}, 0),

with p_{0} > 0.
What is the distance of closest approach of the two particles?

Solution:

- Concepts:

Motion in a central potential. - Reasoning:

The problem of the relative motion of two interacting masses m_{1}and m_{2 }can be solved by solving for the motion of one fictitious particle of reduced mass μ in a central field. - Details of the calculation:

For motion in a central field energy E and angular momentum M are conserved.

E = p_{0}^{2}/2μ = p_{0}^{2}/m, M = 2y_{0}p_{0}.

At the distance of closest approach: E = p_{c}^{2}/m + U_{0}/r_{c}^{2}, M = r_{c}p_{c}. Here r_{c}= |**r**_{1}-**r**_{2}|_{c}.

Then p_{c}= M/r_{c}, E = M^{2}/(mr_{c}^{2}) + U_{0}/r_{c}^{2}= 4y_{0}^{2}p_{0}^{2}/(mr_{c}^{2}) + U_{0}/r_{c}^{2}= p_{0}^{2}/m.

r_{c}^{2}= 4y_{0}^{2}+ mU_{0}/p_{0}^{2}.

The distance of closest approach is r_{c}= (4y_{0}^{2}+ mU_{0}/p_{0}^{2})^{1/2}.

A point like comet of mass m approaches a sun with mass M and Radius R with
speed v_{∞} > 0. What is the total cross section for the comet to
crash on the sun?

Solution:

- Concepts:

Motion in a central potential.

Reasoning:

The comet has a hyperbolic orbit. Energy E and angular momentum L are conserved, the motion is in a plane.

L = mbv_{∞}= mr_{c}v_{c}, where r_{c}is the distance of closest approach, and v_{c}is the speed at r_{c}.

The potential energy of the comet is U(r) = -GMm/r. If r_{c}< R, the comet will crash onto the sun.

We find the relationship between r_{c}and b, and find the largest impact parameter b_{max}for which r_{c}< R.

The cross section for crashing onto the sun is σ = πb_{max}^{2}. - Details of the calculation:

For motion in a central field energy E and angular momentum L are conserved.

E = ½mv_{∞}^{2}= ½mv_{c}^{2}- GMm/r_{c}. v_{c}^{2}= v_{∞}^{2}+ 2GM/r_{c}.

L = mbv_{∞}= mr_{c}v_{c}, v_{∞}= r_{c}v_{c}/b.

b = r_{c}v_{c}/v_{∞}= r_{c}(1 + 2GM/(r_{c}v_{∞}^{2}))^{½}. b_{max}= R(1 + 2GM/(Rv_{∞}^{2}))^{½}.

σ = πR^{2}(1 + 2GM/(Rv_{∞}^{2}))^{½}.

In frame B particles with speed v_{B} << c are scattered uniformly in
all directions. Frame B moves with velocity v_{0} **k** (v_{0}
<< c) with respect to frame A.

(a) If the scattering angle of a particle in frame B is θ_{B},
what is the corresponding scattering angle θ_{A} in frame A, and what is
the speed of the particle v_{A}?

(b) Given that dσ/dΩ_{B} = K = constant, derive an
expression for dσ/dΩ_{A} in terms of v_{0}, v_{B},
v_{A}, and θ_{A}.

(c) Show that when v_{0} > v_{B}, there is a maximum angle
θ_{A_max}, which is given by sinθ_{A_max} = v_{B}/v_{0}.

(d) What happens to dσ/dΩ_{A} near θ_{A_max}?

Solution:

- Concepts:

Frame transformations - Reasoning:

Using the figure and trigonometry, we can find v_{A}and θ_{A}in terms of v_{0}, v_{B}, and θ_{B}. - Details of the calculation:

(a)**v**_{A}=**v**_{0}+**v**_{B}.

v_{A}cosθ_{A}= v_{B}cosθ_{B}+ v_{0}, v_{A}sinθ_{A}= v_{B}sinθ_{B}, or tanθ_{A}= sinθ_{B}/(cosθ_{B}+ γ).

Here γ = v_{0}/v_{B}.

θ_{A}= tan^{-1}(sinθ_{B}/(cosθ_{B}+ γ)).

v_{A}^{2}= v_{0}^{2}+ v'_{B}^{2}+2v_{0}v_{B}cosθ_{B}.

(b) σ(θ_{A},φ_{A})dΩ_{A}= σ(θ_{B},φ_{B})dΩ.

σ(θ_{A},φ_{A}) sinθ_{A}dθ_{A}= σ(θ_{B},φ_{B}) sinθ_{B}dθ_{B}. (The angle φ is the same in the CM and laboratory frame.)

We can therefore write

σ(θ_{A},φ_{A}) = σ(θ_{B},φ_{B}) sinθ_{B}dθ_{B}/(sinθ_{A}dθ_{A}) = σ(θ_{B},φ_{B}) dcosθ_{B}/dcosθ_{A}.

We have tanθ_{A}= sinθ_{B}/(cosθ_{B}+ γ). Somehow we have to use this to evaluate dcosθ_{B}/dcosθ_{A}.

Use tanθ_{A}= (1/cos^{2}θ_{A}- 1)^{½}= sinθ_{B}/(cosθ_{B}+ γ).

Let x = cosθ_{B}and y = cosθ_{A}.

tanθ_{A}= (1/y^{2}- 1)^{½}= (1 - y^{2})^{½}/y = (1 - x^{2})^{½}/(x + γ).

y^{2}= (x + γ)^{2}/(1 + 2xγ + γ^{2}).

dy/dx = (1 + γx)/1 + γ^{2}+ 2γx)^{3/2}.

This then yields

σ(θ_{A},φ_{A}) = σ(θ_{B},φ_{B}) (1 + γ^{2}+ 2γcosθ_{B})^{3/2}/|1 + γcosθ_{B}|.

(c) When v_{0}> v_{B}, γ = v_{0}/v_{B}> 1.

d(tanθ_{A})/dθ_{B}= (cos^{2}θ_{B}+ γcosθ_{B}+ sin^{2}θ_{B})/(cosθ_{B}+ γ)^{2 }= (1 + γcosθ_{B})/(cosθ_{B}+ γ)^{2}= 0 --> cosθ_{B}= -1/γ.

tanθ_{A}has a maximum when cosθ_{B}= -1/γ.

tanθ_{A-max}= (1 - 1/γ^{2})^{½}/(γ - 1/γ) = (1/γ)/(1 - 1/γ^{2})^{½}.

1 + cot^{2}θ_{A-max}= 1/sin^{2}θ_{A-max}= γ^{2}.^{ }sinθ_{A-max}= 1/γ.

We can also just work from the figure. θ_{A}has its maximum value when v_{A}is tangential to the circle. Then the angle between v_{A}and v_{B}is a right angle, and^{ }sinθ_{A-max}= v_{B}/v_{0}= 1/γ.

(d) |1 + γcosθ_{B}| --> 0, therefore σ(θ_{A},φ_{A}) --> ∞.