Cross section for momentum transfer
Definition of cross section for momentum transfer
σm =
2π
0
dϕ
π
0
(1 − cos θ)I(v, θ) sin θdθ (1)
Reminder, the total cross section is
σ =
2π
0
dϕ
π
0
I(v, θ) sin θdθ (2)
Where does the factor 1 − cos θ come from??
Average change of velocity
∆v = v − v =
1
σ
2π
0
dϕ
π
0
∆vI(v, θ) sin θdθ (3)
If an electron interacts with a heavy ion, its speed does not change
too much:
v ≈ v (4)
We split the velocity after the collision into a velocity component
that is parallel and perpendicular to the velocity before the collision:
v = v⊥ + v (5)
Also, the angle between the velocities in the scattering angle θ:
v · v = vv cos θ = v2
cos θ (6)
Then:
v − v = v⊥ + v − v = v⊥ + v(1 − cos θ) (7)
Average change of velocity
∆v = v⊥ + v(1 − cos θ) = v⊥(θ, ϕ) + v (1 − cos θ) (8)
For azimuthally symmetric potetnials, cross section does not depend
on ϕ I = I(θ) and the perpendicular velocity will be distrubted
symmetrically. For every vector of perpendicular velocity there will
one vector with the same magnitude but opposite direction. The
velocity integral through ϕ of the perpendicular velocity will be zero:
v⊥ = 0 (9)
Also see wikipedia pages for the particular integrals.
Last remarks
∆v = v (1 − cos θ) = (10)
v
σ
2π
π
0
(1 − cos θ)I(v, θ) sin θdθ = (11)
v
σm
σ
(12)
Relation between cross section can be written:
σm = σ (1 − cos θ) (13)