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Mech9Sols

Mech9Sols - Physics 105B Problem Set 9 June 3 2008 Jeff...

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Physics 105B Problem Set 9 June 3, 2008 Je ff Schonert: schonert (at) physics.ucsb.edu Taylor 15.80 Refer to the equation of motion listed in problem 15.79: F = γ m a + ( F · v ) v /c 2 Since the magnetic force is perpendicular to both v and B , this means that F · v = 0 We also know that magnetic forces cannot change the speed of a particle; they can only change its direction. Therefore, v is a constant, meaning that γ is a constant as well. Plugging these into the above force equation, we get that γ m ˙ v = e ( v × B ) This is the same as the non-relativistic equation except now m has been replaced by γ m . So the physics of the particle is basically the same, with this minor mass modification. Therefore, if v is initially orthogonal to B , then it will stay orthogonal. This perpendicular force will result in circular motion with radius r = γ mv eB = p eB Taylor 15.87 Let the pion initially be traveling along the x -axis, so that the two photons are traveling in the xy -plane. Initial 4-momentum is denoted p π and the two final 4-momenta are p 1 and p 2 . Conservation of momentum requires that p π = p 1 + p 2 . Because the pion has no momentum in the y -direction, its 4-momentum is ( E π /c, p π , 0 , 0) , meaning that the y -momenta of the two photons must add up to zero in order to have conservation of momentum. Therefore, p y 1 = -

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Mech9Sols - Physics 105B Problem Set 9 June 3 2008 Jeff...

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