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Coulomb Scattering. Hyperbolic Orbits  The trajectory from an inverse square force forms a conic section. e < 1 ellipsee < 1 ellipse e =1 parabolae =1.

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Presentation on theme: "Coulomb Scattering. Hyperbolic Orbits  The trajectory from an inverse square force forms a conic section. e < 1 ellipsee < 1 ellipse e =1 parabolae =1."— Presentation transcript:

1 Coulomb Scattering

2 Hyperbolic Orbits  The trajectory from an inverse square force forms a conic section. e < 1 ellipsee < 1 ellipse e =1 parabolae =1 parabola e >1 hyperbola.e >1 hyperbola.  The force center is at a focus. attractive focus  r repulsive focus aae 

3 Reoriented View  Orient the incident axis to horizontal.  Scattering mass forms a hyperbolic trajectory. attractive focus  r repulsive focus aae  h m v0v0 x b 

4 Impact Parameter  The potential is defined at infinity.  The impact parameter would be closest approach for no force. Compare to angular momentumCompare to angular momentum h m v0v0 x b 

5 Rutherford Cross Section  Scattering cross section is based on the potential. Impact parameter Differential impact parameter  The result is the Rutherford scattering cross section b  

6 Lunar Miss Problem  A spaceship of mass m moving with velocity v 0 approaches the Moon.  The impact parameter is b.  The velocity v 0 is perpendicular to the orbital velocity V of the moon. x b m   Show that if the spaceship passes behind the Moon it gains kinetic energy as it leaves the Moon. M

7 Lunar Frame  The moon is much more massive. Lunar frame is CMLunar frame is CM Incident and final energies same in CMIncident and final energies same in CM m1m1 v0v0 x b  b M

8 Energy Boost  In the Moon’s frame the velocity components come from the scattering angle.  In the observer’s frame the velocity is boosted by the moon. next


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