We assume the material constants are symmetric about some plane, say z=0. Then (x,y,-3)=∈(x,y,z), H(x,y,-2)=(x,y,z), o(x, y, -z)=o(x,y, z) That is, with respect to z the material constants are even functions. We further assume that the boundaries and boundary conditions, which guarantee uniqueness of solution,are also symmetric about the z=0 plane. Then we define two cases of reflection symmetry. Conditions for even symmetry. We claim that if the sources obey Jm(x, y,2)=-J (x,y,23)=引(x,y,-z),Jm(x,y,z) J(x,y,z)=-/2(x,y,-2),Jm2(x,y,)=Jm2(x,y,-z), then the fields obey E(x, y, z)=E(x, y, -z), Hr(x, y,z)=-H(r, y, -z) Ey(x, y, z)=Ey(, y,-2), Hy(r, y,2)=-H,(, y,-2) E2(x,y,z)=-E2(x,y,-) H2(x,y,x)=H2(x,y,-) The electric field shares the symmetry of the electric source: components parallel to the z=0 plane are even in z, and the component perpendicular is odd. The magnetic field shares the symmetry of the magnetic source: components parallel to the z=0 plane are dd in z, and the component perpendicular is even We can verify our claim by showing that the symmetric fields and sources obey Maxwell's equations. At an arbitrary point z=a>0 equation(5. 1) requires E-21|=-4=2|-m,= By the assumed symmetry condition on source and material constant we get de +Jiz= If our claim holds regarding the field behavior, then dey dHr and we have de- =k=2 +J So this component of Faradays law is satisfied. With similar reasoning we can show that the symmetric sources and fields satisfy (5. 2)-(5.6)as well ②2001We assume the material constants are symmetric about some plane, say z = 0. Then (x, y, −z) = (x, y,z), µ(x, y, −z) = µ(x, y,z), σ(x, y, −z) = σ(x, y,z). That is, with respect to z the material constants are even functions. We further assume that the boundaries and boundary conditions, which guarantee uniqueness of solution, are also symmetric about the z = 0 plane. Then we define two cases of reflection symmetry. Conditions for even symmetry. We claim that if the sources obey J i x (x, y,z) = J i x (x, y, −z), J i mx (x, y,z) = −J i mx (x, y, −z), J i y (x, y,z) = J i y (x, y, −z), J i my (x, y,z) = −J i my (x, y, −z), J i z (x, y,z) = −J i z (x, y, −z), J i mz(x, y,z) = J i mz(x, y, −z), then the fields obey Ex (x, y,z) = Ex (x, y, −z), Hx (x, y,z) = −Hx (x, y, −z), Ey (x, y,z) = Ey (x, y, −z), Hy (x, y,z) = −Hy (x, y, −z), Ez(x, y,z) = −Ez(x, y, −z), Hz(x, y,z) = Hz(x, y, −z). The electric field shares the symmetry of the electric source: components parallel to the z = 0 plane are even in z, and the component perpendicular is odd. The magnetic field shares the symmetry of the magnetic source: components parallel to the z = 0 plane are odd in z, and the component perpendicular is even. We can verify our claim by showing that the symmetric fields and sources obey Maxwell’s equations. At an arbitrary point z = a > 0 equation (5.1) requires ∂Ez ∂y z=a − ∂Ey ∂z z=a = −µ|z=a ∂ Hx ∂t z=a − J i mx |z=a. By the assumed symmetry condition on source and material constant we get ∂Ez ∂y z=a − ∂Ey ∂z z=a = −µ|z=−a ∂ Hx ∂t z=a + J i mx |z=−a. If our claim holds regarding the field behavior, then ∂Ez ∂y z=−a = −∂Ez ∂y z=a , ∂Ey ∂z z=−a = −∂Ey ∂z z=a , ∂ Hx ∂t z=−a = −∂ Hx ∂t z=a , and we have −∂Ez ∂y z=−a + ∂Ey ∂z z=−a = µ|z=−a ∂ Hx ∂t z=−a + J i mx |z=−a. So this component of Faraday’s lawis satisfied. With similar reasoning we can showthat the symmetric sources and fields satisfy (5.2)–(5.6) as well.