Isotropic scattering probability — normalization condition An electron in a metal suffers a collision. If the probability of it being scattered into the solid angle between polar angles θ and θ + dθ (integrated over azimuth) is given by 2π P(θ) sinθ dθ, what normalization condition must this distribution satisfy over θ ∈ [0, π]?

Introduction / Context:
In transport theory for metals and plasmas, scattering probabilities over directions must integrate to unity. The function P(θ) describes how likely a carrier is deflected by an angle θ; the factor sinθ dθ accounts for the spherical surface element, and 2π integrates out the azimuthal angle for axisymmetric scattering.

Given Data / Assumptions:

• Scattering distribution depends only on polar angle θ, not on azimuth.
• Element of solid angle: dΩ = sinθ dθ dφ.
• Azimuth φ runs from 0 to 2π.

Concept / Approach:
Any probability density over solid angle must integrate to 1 over the full sphere. Since azimuthal symmetry is assumed, integration over φ yields a factor 2π, leaving the polar integral with sinθ dθ. The normalization condition imposes a single scalar equation on P(θ).

Step-by-Step Solution:
Start with total probability = ∫ dΩ P(θ) = 1.Write dΩ = sinθ dθ dφ, integrate φ from 0 to 2π → factor 2π.Hence, ∫₀^π 2π P(θ) sinθ dθ = 1.This ensures the distribution is properly normalized.

Verification / Alternative check:
For isotropic scattering, P(θ) is constant. Setting P(θ) = C gives 2π C ∫₀^π sinθ dθ = 2π C * 2 = 4π C = 1 ⇒ C = 1/(4π), which is the familiar uniform distribution over a sphere.

Why Other Options Are Wrong:
Equalities without the integral are meaningless for a differential element. Values 0, 2, or −1 violate probability axioms. The last option incorrectly integrates θ over φ’s range.

Common Pitfalls:
Forgetting the sinθ Jacobian; mixing up the roles of θ and φ; omitting limits.

Final Answer:
∫₀^π 2π P(θ) sinθ dθ = 1