r is an unknown function we expand it in a taylors

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Unformatted text preview: (θ ) dr dθ dϕ ⋅ dS z cos(θ ) 4π r 2 The partial current J z− can now be written as: J − z ∫ dN = Σ = dS z ∞ π / 2 2π e 4π ∫ ∫ ∫ s 0 0 φ ( r ) sin(θ )cos(θ )dr dθ dϕ −Σt r (13) 0 Since ϕ ( r ) is an unknown function, we expand it in a Taylor’s series assuming it varies slowly with position: ∂φ ∂φ ∂φ φ ( r ) = φ0 + x +y +z +L ∂x 0 ∂y 0 ∂z 0 Writing x, y, and z in spherical coordinates: φ ( r ) = φ0 + rsin(θ )cos(ϕ ) ∂φ ∂φ ∂φ + rsin(θ )sin(ϕ ) + r cos(θ ) ∂x 0 ∂y 0 ∂z (14) 0 Substituting Eq. 14 into Eq. 13, we get, Σ J=s 4π − z ∞ π 2 2π ∫∫ 00 ⎡ ∂φ ∂φ ∂φ ⎤ e −Σt r ⎢φ0 + rsin(θ )cos(ϕ ) + rsin(θ )sin(ϕ ) + r cos(θ ) ⎥ sin(θ ) cos(θ )drdθ dϕ ∫ ∂x 0 ∂y 0 ∂z 0 ⎦ 0 ⎣ (15) The terms containing cos(ϕ ) and sin(ϕ ) integrate to zero over the interval ϕ ∈ [0, 2π ] , thus: J z− = Σs 4π ∞ π / 2 2π ∫ ∫ ∫e 0 0 ⎡ ∂φ ⎤ ⎢φ0 + r cos(θ ) ∂z ⎥ sin(θ ) cos(θ ) dr dθ dϕ 0⎦ ⎣ −Σt r 0 (16) The first term can be evaluated as: ∞ π / 2 2π Σ I1 = s φ0 ∫ 4π 0 ∫ ∫e 0 −Σt r sin(θ ) cos(θ ) dr dθ dϕ 0 = 11 Σs φ0 ⋅ 2π ⋅ ⋅ 4π Σt 2 = 1 Σs ϕ0 4 Σt The second term also becomes: Σ ∂φ I2 = s 4π ∂z ∞ π / 2 2π ∫ ∫ ∫ re 00 0 −Σt r sin(θ ) cos2 (θ ) dr dθ dϕ 0 = 11 Σ s ∂φ ⋅ 2π ⋅ 2 ⋅ 4π ∂z 0 Σt 3 = 1 Σ s ∂φ 6 Σt2 ∂z 0 Thus Eq. 16 becomes: J z− = Similarly: 1 Σs 1 Σ s ∂φ φ0 + 4 Σt 6 Σ t2 ∂z (17) 0 J z+ = 1 Σs 1 Σ s ∂φ φ0 − 4 Σt 6 Σt2 ∂z (18) 0 Substituting Eqs. 17 and 18 into Eq. 9, we can write: Jx = − Jy = − Jz = − 1 Σ s ∂φ...
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