Scattering 1 Classical scattering of a charged particle (Rutherford ...

E = √ 1 ∞∫2π−∞E(t) = √ 1 ∞∫2πdt e iωt E(t)−∞dt e −iωt EThe energy radiated is the time integral of the power. Therefore;dWdΩ= 12πµc∞∫−∞dt∞∫−∞dω∞∫−∞dω ′ ⃗ E∗ · ⃗E e i(ω′ −ω)tInterchange the time and frequency integration and differentiate with respect to ω.dWdΩ= 12µc |E(ω)|2 R 2The above is the energy radiated per unit solid angle. The equation has been multiplied bythe differential area element, R 2 dΩ and a factor of (1/2) is included so that we define ω > 0always. The radiation field (Gaussian units) is;E(t) = (q/c)[ ˆN × (ˆn − β) ⃗ × ˙⃗ βκ 3 ] = (e/cκ) d Rdt [ˆn × (ˆn × β)] ⃗The right side of the energy equation has the form G ∗ G.G = 1(4µc) 1/2 E(ω)R = 1(8πµc) 1/2∞∫−∞dt e iωt [ˆn × (ˆn − ⃗ β) × ˙⃗ βκ 3 ]Change to present time units, κt, and approximate the time in the phase component byt ′ ≈ t − ˆn · ⃗r/c for large r. Integrate by parts over the time, and then put this value for Gback into the equation for the differential power above to obtain;d 2 Idω dΩ = q2 ω 216π 2 c 3 |∞∫−∞dt[ˆn × (ˆn × ⃗ β)] e iω(t−ˆn·⃗r/c) | 2In a dielectric medium the velocity of the EM wave is c → c/n = c/ √ ǫ with n the index ofrefraction. Assume straight line motion so that ⃗r = ⃗ V t .d 2 Idω dΩ = µq2 cω 216π 2 |ˆn × V ⃗ ∞∫(ω 2 /2π) dt e iωt(1−ˆn·⃗r/nc) | 2−∞The integral is a δ function and results in the expression for the intensity of the Cherenkovradiation (Gaussian units);d 2 Idω dΩ = q2 β 2nc sin2 (θ)|δ(1 − (β/n) cos(θ)| 2The above δ function gives the Cherenkov angle as previously obtained. As an example,glass has an index of refraction of 1.5. The Cherenkov angle is then;18