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One can set up a two-dimensional scattering theory, which could be applied to puck projectiles sliding on an ice rink and colliding with various...
One can set up a two-dimensional scattering theory, which could be applied to puck projectiles sliding on an ice rink and colliding with various target obstacles. The cross section (sigma) would be the effective width of a target and the differential cross section d(sigma)/d(theta) would give the number of projectiles scattered in the angle d(theta). a) Show that the two dimensional analog is d(sigma)/d(theta)=absolute value of (db/dtheta). Note that in two dimensional scattering it is convenient to let theta range from -pi to pi. b)Now consider the scattering of a small projectile off a hard "sphere"(actually a hard disk) of radius R pinned down to the ice. Find the differential cross section. (Note that in two dimensions, hard "sphere" scattering is not isotropic.) c)By integrating your answer to part b, show that the total cross section is 2R as expected.
