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410ahw4soln - HW#4Solutions Phys410Fall 2011 Prof Ted...

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HW#4—Solutions —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ [email protected] 7.49 ( charged particle in uniform magnetic field — polar gauge ) (a) Just evaluate. (b) A = 1 2 ˆ φ so v · A = 1 2 2 ˙ φ , hence L = 1 2 m ( ˙ ρ 2 + ρ 2 ˙ φ 2 + ˙ z 2 ) + 1 2 qBρ 2 ˙ φ. (1) The EL eqns are m ¨ ρ = ˙ φ 2 + qBρ ˙ φ (2) p φ = 2 ˙ φ + 1 2 qBρ 2 = const. (3) m ¨ z = 0 (4) (c) The z component of velocity is constant in all solutions. If ρ is constant then m ˙ φ + qB = 0, so ˙ φ = - qB/m. (5) All orbits, regardless of radius, have this same angular frequency, the cyclotron fre- quency . The general orbit is a spiral that is counterclockwise looking in the positive z -direction, with constant velocity in the z direction, and angular velocity qB/m . (d) A calculation shows the Hamiltonian is equal to the kinetic energy: H = 1 2 m ( ˙ ρ 2 + ρ 2 ˙ φ 2 + ˙ z 2 ) . (6) (In general, H = ˙ q i ∂L/∂ ˙ q i - L , so terms in L independent of the velocity appear in H with the opposite sign, terms linear in the velocity do not appear, and terms quadratic in the velocity appear unchanged between L and H .) (e) p φ in (3) is the angular momentum about the z axis. (f) For orbits of constant ρ in the xy - plane, the energy is E = 1 2 2 ˙ φ 2 = 1 2 ( qBρ ) 2 /m . For these orbits, using (5) the angular momentum is p φ = - 1 2 qBρ 2 , so E = - ( qB/m ) p φ . (g) Dimensional analysis: [ m ] = M , [ qB ] = [ MT - 1 ], [ p φ ] = ML 2 T - 1 , and energy = ML 2 T - 2 , so the only way to make energy is by the combination ( qB/m ) p φ . (h) If p φ = - n ¯ h is quantized (the sign of the allowed integers n is the sign of qB ), then E n = n ¯ hqB/m . [Actually there is a zero-point energy in quantum mechanics, and E n = ( n + 1 2 hqB/m .] 7.51 ( pendulum with constrained Cartesian coordinates ) L = 1 2 m ( ˙ x 2 + ˙ y 2 ) + mgy , with y
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