Elastic Collision and Rotation Angle

Consider an elastic collision (namely a collision where the particles involved do not change their internal state) of an incident particle of mass m₁, momentum p₁, and energy ε₁ (see notation below), with a particle of mass m₂ at rest. Let the final energies be ἓ₁ and ἓ₂ and the final momenta be  and  (all of this in the laboratory frame).

a) In the center of mass frame (CMF), denote by p₁₀ and p₂₀ the incoming momenta of the two particles and by ε₁₀, ε₂₀ their energies.

One has

 (i)

where p₀ is a vector. In the center of mass frame the collision rotates the direction of the momenta. Let the outgoing momenta and energies be  and the rotation angle be χ. From conservation of energy and momentum, what can you tell about   About ἓ₁₀ and ἓ₂₀?

b) From the energy and momentum conservation laws,

 (ii)

show that

 (iii)

c) Evaluate the first and third terms of the left-hand side of equation (3) in the laboratory frame. Evaluate the second term in the CMF in terms of p₀ and χ (and masses). Now use equation (3) to find an expression for (ἓ - ε) in terms of p₀ and χ (and masses).

d) Find an expression for p₀ in terms of ε₁ (and masses) by evaluating p₁ . p₂ both in the laboratory and CMF.

e) Give ἓ₁ and ἓ₂ in terms of ε₁, ε_2, and χ.

f) Consider the case for maximal energy transfer. What is the value of χ? For this case find the ratio of the final kinetic energy to the incident kinetic energy for the incoming particle (in the laboratory frame).

Notation: Here we use

SOLUTION

a) Express conservation of energy as

(1)

Conservation of momentum gives

(2)

From the information given and (2)

(3)

These equations may be rearranged to yield

(4)

Substituting (1) into (4) yields

(5)

Adding or subtracting (1) from (5) results in

(6)

and by (3),

(7)

b) Inspection of (iii) seems to indicate the way to proceed, since  does not appear. Subtract  from both sides of (ii) and square:

(8)

c) The instructions in (c) exploit the invariance of a product of 4-vectors under a Lorentz transformation, here from the laboratory frame to the CMF. In the laboratory frame

So we have

In the CMF

where

Substitution into (8) yields

Now, ε²₁₀ = p²₀ + m²₁, so

(9)

d) From (c), p₁ . p₂ in the laboratory frame equals ε₁m₂, and p₁ . p₂ in the CMF equals ε₁₀ε₂₀ + p²₀, so

(10)

Rearrange and square (10):

(11)

yielding

(12)

If the incident mass has no kinetic energy, in the CMF it should have no momentum, as seen in (12). If m₂/m₁ → 0, then the frame of particle 1 is the CMF, so p₀ again should equal 0.

e) From (9) and (12) we have

Conservation of energy gives

So

f) We maximize  by letting χ = π. So

For m₁ = m₂, the final kinetic energy of the incident particle equals 0.