Exercises with Magnetic Monopoles - Kurt Nalty

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or so, effectively ruling out colocated charge as the origin of electron spin at the large scale. Let’s look at these numbers from another point of view. From the conserved magnitude of the duality rotated charge, we have c 2 q 2 e + q 2 m = [( 2.9989 · 10 8 m/s ) · (1.6 · 10 −19 C )] 2 = 2.3 · 10 −21 Amp 2 m 2 For the conserved angular momentum, we have L = µq eq m = 1 4π 2¯h ( ) 1 µq e q m = 4π 2¯h = 2h q e q m = 2h µ cq e q m = 2hc µ = 2(1.325 · 10−33 J s)(2.9989 · 10 8 m/s) 4π10 −7 H/m = 3.16 · 10 −19 Amp 2 m 2 Before proceding, we note that the ratio cq e q m /(c 2 q 2 e + q 2 m) = 137.04, the (inverse) fine structure constant. Returning to the discussion, the point of closest approach for the hyperbola xy = K 2 to the origin is K √ 2. With our values above, the angular momentum hyperbola never intersects the duality circle by a factor of 137, effectively ruling out any model of the electron as a simple colocated electric and magnetic charge combo. <strong>Magnetic</strong> Monopole and Charge Trajectories While we can rule out a colocated electric and real magnetic charge model for the electron, we have not excluded composite, separated assemblies of electric and magnetic charge, nor have we addressed imaginary magnetic charge, which will be a separate note. 20
Electron Motion in a Radial <strong>Magnetic</strong> Field The force law for a pure electron in an electromagnetic field is ⃗F = q e ( ⃗E + ⃗v × ⃗ B ) A pure magnetic field can do no work. Force is at right angles to motion, and results only in a change of direction. Consequently, the speed is constant, and the velocity and acceleration are always orthogonal. To describe the motion of the electron, I will use the Frenet-Serret approach, and identify the curvature and torsion formulas for the electron in the radial magnetic field. I assume that Newtonian mass and acceleration still apply in this scenario. ⃗B = µq m ⃗r 4π r 3 ⃗F = q e (⃗v × B ⃗ ) = µq eq m ⃗v × ⃗r 4π r 3 F ⃗a = ⃗ m = µq eq m 1 ⃗v × ⃗r 4π m r 3 We see above that ⃗a is normal to velocity, and see below that speed will be constant. ⃗a · ⃗v = 0 = 1 d (⃗v · ⃗v) 2 dt = 1 dv 2 2 dt v 2 = const The magnitude of curvature in three dimensions is ⃗κ = ⃗a × ⃗v v 3 av sin θ κ = v 3 = a for ṽ ⊥ ã v 2 21
Page 1 and 2: Exercises with Magnetic Monopoles K
Page 3 and 4: Field Angular Momentum due to Charg
Page 5 and 6: ⃗∇(ρ e ρ m ) −1 . Each form
Page 7 and 8: We now write the field angular mome
Page 9 and 10: Looking at the radial component in
Page 11 and 12: ∫ ρ=∞ ρ=0 ρ 2 ∫ ρ=∞ (
Page 13 and 14: So, x = r cos θ sin φ y = r sin
Page 15 and 16: For the zone R/2 < r < ∞, we have
Page 17 and 18: Figure 1: tan −1 (4rR/(4r 2 − R
Page 19: if the ratio of magnetic to electri
Page 23 and 24: τ = = = = [ µq e q m ⃗a × ⃗r
Page 25 and 26: ρ = r = mv 4πr2 µq e q m r = µq
Page 27 and 28: Without loss of generality, I can p
Page 29: [5] Alexander Russell, “The Magne

Exercises with Magnetic Monopoles - Kurt Nalty

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