Mechanics Scattering Problem, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Scattering Problem, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity! Mechanics Physics 151 Lecture 7 Scattering Problem (Chapter 3) What We Did Last Time Discussed Central Force Problems Problem is reduced to one equation Analyzed qualitative behaviors Unbounded, bounded, and circular orbits Condition for stable circular orbits Defined orbit equation and solved it for the Kepler problem Conic orbits depending on E 2 3 ( ) lmr f r mr = + 2 2( ) ( ) 2 lV r V r mr ′ = + 2 2 2 1 21 1 cos( )mk El r l mk θ θ ⎛ ⎞ ′= + + −⎜ ⎟⎜ ⎟ ⎝ ⎠ Why Scattering Problem? Physical “observations” are scattering phenomena Photons scattered by an object Seeing Electrons scattered by an object Electron microscope Experiments on microscopic objects Electron-nucleus scattering to probe nuclear structure Neutrino-electron scattering to measure neutrino energy Classical description fails with such targets Still a good approximation in many cases Classical framework of describing scattering used in QM as well – and it’s more intuitive to understand Lousy Shooter Model Imagine shooting bullets at a small target Suppose you have very poor aim Bullets spread uniformly Number of bullets / area / time = intensity I Number of hits will be proportional to the target size hitsN I σ= ⋅ Target cross section (m2) Intensity (bullets/m2/sec) Hit frequency (bullets/sec) Spherical Target Imagine the target is a solid sphere We want to know which direction the bullets ricochet Number of bullets ricocheting into solid angle dΩ around a direction Ω is Concentrate on the scattering angle Θ Target is round = has rotational symmetry Number of bullets between Θ and Θ + dΘ is Θ ( )N I dσ= ΩΩ Differential crosssection (m2/str) ( ) ( )sind d dσ σ φΩ = Θ Θ ΘΩ 2 0 ( )sin ( )2 sinN d I d I d π φ σ σ π= Θ Θ Θ = Θ Θ Θ∫ Central Force Scattering Now consider scattering by general central force How does Θ relate to s? We need to know the shape of the orbit at large r Look at the orbit equation Angular momentum l is related to s by If we assume V(r) → 0 as r → ∞ ( ) sin s ds d σ Θ = Θ Θ 0 0sinl rp spθ= × = =0r p Θ s 0p 2 2 2 (1/ ) 0d u m dV uu d l duθ + + = 2 0 2 pE T m = = 0 2l sp s mE= = Central Force Scattering One can in principle solve this equation and get r → ∞ at θ = Θ Then we can calculate 2 2 2 1 (1/ ) 0 2 d u dV uu d s E duθ + + = ( , , )u u s Eθ= ( , , ) 0u s EΘ = Solve ( , )s s E= Θ ( , ) sin s dsE d σ Θ = Θ Θ Let’s look at the orbit we already know Inverse-square force Orbit equation in terms of the impact parameter s and the energy E Inverse Square Force Consider a repulsive 1/r2 force Equation and solution same as Kepler problem Just flip the sign of k Radius > 0 2( ) kf r r = ( ) kV r r = Ex: electrostatic force between two like-sign charged particles 2 2 2 1 21 1 cos( )mk El r l mk θ θ ⎛ ⎞ ′= − + + −⎜ ⎟⎜ ⎟ ⎝ ⎠ 2 2 21 1El mk ε = + > Eccentricity Hyperbola Rutherford Scattering Before Rutherford’s discovery Electron was known to exist in matter Positive charge must exist in atoms, but the distribution was unknown Measurement of σ(Θ) showed Positive charge of +Ze is in one particle e.g. Z particles of +e each would give Discovery of atomic nuclei 22 4 2 1 1( ) 4 2 sin ZZ e E σ Θ ′⎛ ⎞ Θ = ⎜ ⎟ ⎝ ⎠ 22 4 2 1 4 2 sin Z Z e E Θ ′⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Total Cross Section Integrating Rutherford cross section gives Because electrostatic force is long-range No matter how large is the impact parameter s, the particle still gets slightly deflected Reality: electrostatic field is shielded by the electrons around the nucleus Finite cross section 22 40 24 22 1 2 30 2 1 1( ) 2 sin 4 2 sin (sin )2 2 sin T ZZ ed d E dZZ e E π π σ σ π π Θ Θ Θ ′⎛ ⎞ = = Θ Θ⎜ ⎟ ⎝ ⎠ ′⎛ ⎞ =⎜ ⎝ ⎠ ∞= ⎟ ∫ ∫ ∫ Ω Ω Rainbow Scattering Equation for σ(Θ) assumes that s(Θ) is single-valued True for Rutherford scattering, but not always If s(Θ) is not monotonous At maximum Θ = Θm Called rainbow scattering ( ) sin s ds d σ Θ = Θ Θ Θ 2s 1s mΘ ( ) sin i i i s ds d σ Θ = Θ Θ∑ Sum up for possible s’s 0d ds Θ = ( )σ Θ = ∞ ( )σ Θ Rainbow Minimum of Θ Sharp peak of intensity I(Θ) Reflection observed only at Θmin This depends on n, which depends slightly on λ This is really how rainbow is created Θ s R minΘ Θ minΘ ( )I Θ ∞ mins r > rmin r < rmin minΘ From Physics 15c Attractive Force Repulsive force can only scatter by Attractive force can do more If the potential and the energy are just right, particle can make multiple turns before emerging Called spiraling or orbiting 0 π< Θ < r ( )V r′ E Orbiting region: E – V’ is small r varies slowly Summary Discussed scattering problem Foundation for all experimental particle physics Defined and calculated cross sections Differential cross section Rutherford scattering Done with central force problems Next: Rigid Bodies ( ) sin s ds d σ Θ = Θ Θ hitsN I σ= ⋅