leaves the expressions (5.17)and (5. 18)unchanged. This is called a gauge transformation, and the choice of a certain r alters the specification of v. Ae. Thus we may begin with the Coulomb gauge as our baseline, and allow any alteration of Ae according to(5. 20) vA by v Vr= v2r Once V. Ae is specified, the relationship between the potentials and the current J can be found by substitution of (5.17) and(5. 18) into Ampere's law. At this point e assume media that are linear, homogeneous, isotropic, and described by the time- invariant parameters A, e, and o. Writing J=J+oE we have V×(V×A)=J (5.22) Taking the divergence of both sides of (5. 22 )we get 0=V·J Then, by substitution from the continuity equation and use of (5. 19)along with VVoe= V-pe we obtain (p'+eVe)=-ov2oe For a lossless medium this reduces to Vφe=-p/∈ (5.24) pe r, 0= p(r,) av (5.25) 4丌∈R We can obtain an equation for Ae by expanding the left-hand side of (5. 22) to get v(v- A)-v2A=uJ'-op--ouVp a-A at 2-kea vae VA J+σμ一+aμVφe+∈V中e under the Coulomb gauge. For lossless media this becomes vA-H∈a2=-H+∈aV中 (5.27) Observe that the left-hand side of (5.27) is solenoidal(since the Laplacian term came from the curl-curl, and V. Ae=0), while the right-hand side contains a general vector field J and a lamellar term. We might expect the Voe term to cancel the lamellar portion of J, and this does happen 91. By(5.12) and the continuity equation we can write the lamellar component of the current as i(,)=-p/VJ(r,t) J -v/pr,t) Thus(5.27)becomes (5.28) ②2001leaves the expressions (5.17) and (5.18) unchanged. This is called a gauge transformation, and the choice of a certain  alters the specification of ∇ · Ae. Thus we may begin with the Coulomb gauge as our baseline, and allowany alteration of Ae according to (5.20) as long as we augment ∇ · Ae by ∇·∇ = ∇2. Once ∇ · Ae is specified, the relationship between the potentials and the current J can be found by substitution of (5.17) and (5.18) into Ampere’s law. At this point we assume media that are linear, homogeneous, isotropic, and described by the time￾invariant parameters µ, , and σ. Writing J = Ji + σE we have 1 µ ∇ × (∇ × Ae) = Ji − σ ∂Ae ∂t − σ∇φe − ∂2Ae ∂t 2 − ∂ ∂t ∇φe. (5.22) Taking the divergence of both sides of (5.22) we get 0 =∇· Ji − σ ∂ ∂t ∇ · A − σ∇·∇φe − ∂2 ∂t 2 ∇ · Ae − ∂ ∂t ∇·∇φe. (5.23) Then, by substitution from the continuity equation and use of (5.19) along with ∇·∇φe = ∇2φe we obtain ∂ ∂t  ρi + ∇2 φe = −σ∇2 φe. For a lossless medium this reduces to ∇2 φe = −ρi / (5.24) and we have φe(r, t) =  V ρi (r , t) 4πR dV . (5.25) We can obtain an equation for Ae by expanding the left-hand side of (5.22) to get ∇ (∇ · Ae) − ∇2 Ae = µJi − σµ ∂Ae ∂t − σµ∇φe − µ ∂2Ae ∂t 2 − µ ∂ ∂t ∇φe, (5.26) hence ∇2 Ae − µ ∂2Ae ∂t 2 = −µJi + σµ ∂Ae ∂t + σµ∇φe + µ ∂ ∂t ∇φe under the Coulomb gauge. For lossless media this becomes ∇2 Ae − µ ∂2Ae ∂t 2 = −µJi + µ ∂ ∂t ∇φe. (5.27) Observe that the left-hand side of (5.27) is solenoidal (since the Laplacian term came from the curl-curl, and ∇ · Ae = 0), while the right-hand side contains a general vector field Ji and a lamellar term. We might expect the ∇φe term to cancel the lamellar portion of Ji , and this does happen [91]. By (5.12) and the continuity equation we can write the lamellar component of the current as Ji l(r, t) = −∇  V ∇ · Ji (r , t) 4π R dV = ∂ ∂t ∇  V ρi (r , t) 4π R dV = ∂ ∂t ∇φe. Thus (5.27) becomes ∇2 Ae − µ ∂2Ae ∂t 2 = −µJi s. (5.28)