Exercises with Magnetic Monopoles - Kurt Nalty

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This expression has zero divergence, away from the sources. We show this most easily using the gradient definitions for the electric and magnetic fields. ⃗∇ · ⃗∇ · ρ e = √ x 2 + y 2 + (z − R/2) 2 ρ m = √ x 2 + y 2 + (z + R/2) 2 ⃗E = − q ( ) e ∇ 4πɛ ⃗ 1 ρ e ⃗B = − µq ( ) m ∇ 4π ⃗ 1 ρ [ ( m ) ( 1 ⃗∇ × ∇ 16π 2 ρ ⃗ 1 e ɛµ S ⃗ = µq )] eq m ρ m = µq [ ( ) ( )] eq m ∇ 16π ⃗ 1 · ⃗∇ × ∇ 2 ρ ⃗ 1 e ρ m = µq [ ( ) ( ( )) eq m 1 ⃗∇ · 16π 2 ρ ⃗∇ 1 × ⃗∇ m ρ e = 0 since curl of gradient is zero ( ɛµ S ⃗ ) ( ɛµ S ⃗ ) ( ) ( ( ))] − ∇ ⃗ 1 · ρ ⃗∇ 1 × ⃗∇ e ρ m Our next challenge, having shown zero divergence for this expression, is to recast this as the curl of some other function. (This is motivated by the identity that the divergence of a curl is zero.) Using the relationship We see that if ⃗ A = ⃗ ∇g, ⃗∇ × (f ⃗ A) = f( ⃗ ∇ × ⃗ A) − ⃗ A × ( ⃗ ∇f) ⃗∇ × (f ⃗ ∇g) = f( ⃗ ∇ × ⃗ ∇g) − ⃗ ∇g × ( ⃗ ∇f) We easily see two nice answers. = ⃗ ∇f × ⃗ ∇g ⃗∇f × ⃗ ∇g = ⃗ ∇ × (f ⃗ ∇g) = − ⃗ ∇ × (g ⃗ ∇f) Given that the curl of a gradient is zero, we have a family of answers, just like gauge transformations. The two answers shown above differ by 4
⃗∇(ρ e ρ m ) −1 . Each form favors one of the two charge types. I prefer to use a version which is more symmetric (actually anti-symmetric) <strong>with</strong> respect to the charges. Thus, I introduce ⃗α = 1 2 ( = ( 1 ρ e ⃗ ∇ ( 1 ρ m ) − 1 ρ m ⃗ ∇ ( 1 ρ e )) x 2 + y 2 + z 2 + ( R 2 ( ) 2 ) x 2 + y 2 + z 2 + ( R 2 ⃗r × ⃗α = − ρ 3 eρ 3 m ( ( R ) ) 2 2 − x 2 − y 2 − z 2 ⃗r · ⃗α = ρ 3 eρ 3 m ( ( R ) 2 2 − r 2) Rz 2 = ρ 3 eρ 3 m ( ) ( ) ⃗∇ × ⃗α = ∇ ⃗ 1 × ∇ ρ ⃗ 1 e ρ m R ⃗a 2 z − zR (x⃗a x + y⃗a y + z⃗a z ) ρ 3 eρ 3 m ) ) 2 Rz 2 R 2 ρ⃗a θ We see that ⃗α is to momentum density as ⃗ A is to magnetic field density ⃗B. Consequently, I’ll call ⃗α the momentum density vector potential. So, going back to our momentum density, we have ɛµ ⃗ S = µq eq m 16π 2 yR⃗a x − xR⃗a y ( x2 + y 2 + (z − R/2) 2) 3/2 ( x2 + y 2 + (z + R/2) 2) 3/2 The momentum density circulates around the z axis, so we expect our total linear momentum to be zero. Our integral for total momentum is ∫∫∫ ⃗M = ɛµ Sdxdydz ⃗ = µq ∫∫∫ eq m yR⃗a x − xR⃗a y 16π 2 ( x2 + y 2 + (z − R/2) 2) 3/2 ( x2 + y 2 + (z + R/2) 2) dxdydz 3/2 5
Page 1 and 2: Exercises with Magnetic Monopoles K
Page 3: Field Angular Momentum due to Charg
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 and 20: if the ratio of magnetic to electri
Page 21 and 22: Electron Motion in a Radial Magneti
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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