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

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oth energy and momentum, write;⃗p I = ⃗p F + ⃗p γE I = E F + E γWhen combining the equationsM 2 γ = E 2 γ − p 2 γ = 2[M 2 + p I p F cos(θ) − E I E F ]In the above M is the mass <strong>of</strong> the <strong>particle</strong> and θ is the angle between the incident andoutgoing momentum. Now M 2 γ < 0 so that M γ is imaginary. However, in some cases it isuseful to think that this photon has some <strong>of</strong> the the properties <strong>of</strong> a real photon. Indeed inQM such a virtual <strong>particle</strong> can live within the uncertainty relation ∆E ∆t ≈ . For laternotation we let this photon have energy ω = E I − E F and momentum ⃗q = ⃗ P I − ⃗p F . ThenM 2 γ = ω2 − q 2 < 014 <strong>Scattering</strong> <strong>of</strong> virtual quantaThis technique is an approximation to the interaction <strong>of</strong> <strong>charged</strong> <strong>particle</strong>s, by viewing thefield as a collection <strong>of</strong> virtual photons. The approximation works for photons that are almost“on the mass shell”, ie virtual photons that have mass nearly zero.Consider the <strong>scattering</strong> <strong>of</strong> a <strong>charged</strong> <strong>particle</strong> from another charge system. The figure 12 isa schematic <strong>of</strong> the model to be described. Indeed the figure illustrates how bremsstrahlungcould be described as the scatering <strong>of</strong> a virtual photon into one <strong>of</strong> real mass. Generally wewould have a light <strong>particle</strong> <strong>scattering</strong> from a heavier one, but we use a frame in which theheavier mass is in motion and the lighter mass is at rest so that the motion <strong>of</strong> the heaviermass can be assumed to move with constant velocity during the collision. The lighter<strong>particle</strong> recoils and emitts radiation. The relativistic factor for the movement <strong>of</strong> the chargeQ is γ ≫ 1. Now if γ ≫ 1 the E and B fields are almost transverse. The fields at point P are;E 1 = Q4πǫE 2 = Q4πǫγvt(b 2 + (γvt) 2 ) 3/2γb(b 2 + (γvt) 2 ) 3/2⃗B = ⃗ V × ⃗ EThe equation for the ⃗ B is due to the fact there is no magnetic field in the rest frame <strong>of</strong> Q.The E field is just the static field <strong>of</strong> a charge Q a distance r ′ away from the point, Q. Apply21

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