above. Computing the derivatives 2(E(r)(k2-0)=V(VE(r)e(-0)+E(r)Ve(k0) VE(r)e(kz-wt)+VE(r).Ve +VE(r)·Ve(k2-0)+E(r) (E(r)e(k-)=E(r)e(- →(v2-e2)(E(r)(-)=VE(rl4k-)+vEre(=0 02 +E(r)(v2-e2 ei(kiz-wt) The derivatives of the exponential can be simplified, assuming a particular alue for k 2(kz-ut) k4+epw)e?(kz-wt) →(2-cP)k=0=0k=√甲=Vo0 The second term in the expression above can also be simplified Ve VE(r aE(r). aE(r).aE(r →VE(r)Ve dE(r) Canceling the remaining exponential factors in the Maxwell equation OE VE+2k As mentioned above, The variation of VE is slow with respect to e his we can neglect a2 with respect to k a2. Hence we can neglect the component of the V2 operator, although the variation in the x and y components is still significant ay 2above. Computing the derivatives, ∇2 ￾ E(r)e ı(kz−ωt)
= ∇ ￾ ∇E(r)e ı(kz−ωt) + E(r)∇e ı(kz−ωt)
= ∇2E(r)e ı(kz−ωt) + ∇E(r) · ∇e ı(kz−ωt) + ∇E(r) · ∇e ı(kz−ωt) + E(r)∇2 e ı(kz−ωt) ∂ 2 ∂t2 ￾ E(r)e ı(kz−ωt)
= E(r) ∂ 2 ∂t2 e ı(kz−ωt) ⇒  ∇2 − µ ∂ 2 ∂t2  ￾ E(r)e ı(kz−ωt)
= ∇2E(r)e ı(kz−ωt) + 2∇E(r)∇e ı(kz−ωt) + E(r)  ∇2 − µ ∂ 2 ∂t2  e ı(kz−ωt) The derivatives of the exponential can be simplified, assuming a particular value for k,  ∇2 − µ ∂ 2 ∂t2  e ı(kz−ωt) = ￾ −k 2 + µω2
e ı(kz−ωt) ⇒  ∇2 − µ ∂ 2 ∂t2  e ı(kz−ωt) = 0 if k = ω √ µ = ω c r µ 0µ0 = ω vph The second term in the expression above can also be simplified, ∇e ı(kz−ωt) = ıkeı(kz−ωt) zˆ ∇E(r) = ∂E(r) ∂x xˆ + ∂E(r) ∂y yˆ ∂E(r) ∂z zˆ ⇒ ∇E(r) · ∇e ı(kz−ωt) = ık ∂E(r) ∂z e ı(kz−ωt) Canceling the remaining exponential factors in the Maxwell equation, ∇2E + 2ık ∂E ∂z = 0 As mentioned above, The variation of ∇2E is slow with respect to e ıkz . This we can neglect ∂ 2E ∂z2 with respect to k ∂E ∂z . Hence we can neglect the ∂ 2 ∂z2 component of the ∇2 operator, although the variation in the xˆ and yˆ components is still significant,  ∂ 2 ∂x2 + ∂ 2 ∂y2 + 2ık ∂ ∂z E(r) = 0 (1) 7