Chapter 5 Field decompositions and the EM potentials 5.1 Spatial symmetry decompositions Spatial symmetry can often be exploited to solve electromagnetics problems. For analytic solutions, symmetry can be used to reduce the number of boundary conditions that must be applied. For computer solutions the storage requirements can be reduced. Typical symmetries include rotation about a point or axis, and reflection through a plane, along an axis, or through a point. We shall consider the common case of reflection through a plane. Reflections through the origin and through an axis will be treated in the exercises Note that spatial symmetry decompositions may be applied even if the ields possess no spatial symmetry. As long as the boundaries and material media are symmetric, the sources and fields may be decomposed into constituents that individually mimic the symmetry of the environment 5.1.1 Planar field symmetry Consider a region of space consisting of linear, isotropic, time-invariant media having laterial parameters E(r), u(r), and o(r). The electromagnetic fields(E, H)within this region are related to their impressed sources (, Jm)and their secondary sources J '=gE through Maxwells curl equations der dEy dHr der dE dey aEr dHz m Hy dEx at +oEr+J ahr dh. dey +oe+ aH aHr ae ②2001
Chapter 5 Field decompositions and the EM potentials 5.1 Spatial symmetry decompositions Spatial symmetry can often be exploited to solve electromagnetics problems. For analytic solutions, symmetry can be used to reduce the number of boundary conditions that must be applied. For computer solutions the storage requirements can be reduced. Typical symmetries include rotation about a point or axis, and reflection through a plane, along an axis, or through a point. We shall consider the common case of reflection through a plane. Reflections through the origin and through an axis will be treated in the exercises. Note that spatial symmetry decompositions may be applied even if the sources and fields possess no spatial symmetry. As long as the boundaries and material media are symmetric, the sources and fields may be decomposed into constituents that individually mimic the symmetry of the environment. 5.1.1 Planar field symmetry Consider a region of space consisting of linear, isotropic, time-invariant media having material parameters (r), µ(r), and σ(r). The electromagnetic fields (E, H) within this region are related to their impressed sources (Ji , Ji m) and their secondary sources Js = σE through Maxwell’s curl equations: ∂Ez ∂y − ∂Ey ∂z = −µ ∂ Hx ∂t − J i mx , (5.1) ∂Ex ∂z − ∂Ez ∂x = −µ ∂ Hy ∂t − J i my , (5.2) ∂Ey ∂x − ∂Ex ∂y = −µ ∂ Hz ∂t − J i mz, (5.3) ∂ Hz ∂y − ∂ Hy ∂z = ∂Ex ∂t + σ Ex + J i x , (5.4) ∂ Hx ∂z − ∂ Hz ∂x = ∂Ey ∂t + σ Ey + J i y , (5.5) ∂ Hy ∂x − ∂ Hx ∂y = ∂Ez ∂t + σ Ez + J i z . (5.6)
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 ②2001
We 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.
Conditions for odd symmetry. We can also show that if the sources obey Ji(x,y, z)=-(r, y,-2), Jm(x, y, z)=Jm(x, y, -z) Jy(x,y,2)=-x, y,-z), my(x, y, z)=Jmy (x, J(x,y,3)=J(x,y,-z),m:(x,y,x)=-/m2(x,y,-z) then the fields obey E(x, y, 2)=-ECx, y, -z), H,(r, y, z)=H(x, y, -z) Ey(r, H,(x,y,z)=H,(x,y,-x) E2(x,y,x)=E2(x,y,-) H2(x,y,z)=-H2(x,y,-x) Again the electric field has the same symmetry as the electric source. However, in this case components parallel to the z=0 plane are odd in z and the component perpendicular is even. Similarly, the magnetic field has the same symmetry as the magnetic source. Here components parallel to the z=0 plane are even in z and the component perpendicular is odd Field symmetries and the concept of source images. In the case of odd symmetr the electric field parallel to the z=0 plane is an odd function of z. If we assume that the field is also continuous across this plane then the electric field tangential to z=0 must vanish: the condition required at the surface of a perfect electric conductor(PEC) We may regard the problem of sources above a perfect conductor in the z=0 plane equivalent to the problem of sources odd about this plane, as long as the sources in both cases are identical for z>0. We refer to the source in the region z 0. Thus the image source (', Jm)obeys I(x,, -z)=-J(, y, z), JA(x, y, -2)=Jm (x, y, z) J!(x,y,-)=(x,y,z),Jm2(x That is, parallel components of electric current image in the opposite direction, and the perpendicular component images in the same direction; parallel components of the magnetic current image in the same direction, while the perpendicular component images in the opposite direction. In the case of even symmetry, the magnetic field parallel to the z=0 plane is odd, d thus the magnetic field tangential to the z=0 plane must be zero. We therefore have an equivalence between the problem of a source above a plane of perfect magnetic conductor(PMC) and the problem of sources even about that plane. In this case we identify image sources that obey Jl(x,y, -z)=Ji(x, y, z), Jm (x, y, -z)=-Jmi(, y, z) J(x, y, -z)=J(x, y, z), Jmy J2(x,y,-2)=-12(x,y,z),m2 Jm(x, y, z) Parallel components of electric current image in the same direction, and the perpendicular component images in the opposite direction; parallel components of magnetic current image in the opposite direction, and the perpendicular component images in the same direction In the case of odd symmetry, we sometimes say that an"electric wall"exists at z=0 The term"magnetic wall"can be used in the case of even symmetry. These terms are particularly common in the description of waveguide fields ②2001
Conditions for odd symmetry. We can also showthat 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) = −Ei x (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). Again the electric field has the same symmetry as the electric source. However, in this case components parallel to the z = 0 plane are odd in z and the component perpendicular is even. Similarly, the magnetic field has the same symmetry as the magnetic source. Here components parallel to the z = 0 plane are even in z and the component perpendicular is odd. Field symmetries and the concept of source images. In the case of odd symmetry the electric field parallel to the z = 0 plane is an odd function of z. If we assume that the field is also continuous across this plane, then the electric field tangential to z = 0 must vanish: the condition required at the surface of a perfect electric conductor (PEC). We may regard the problem of sources above a perfect conductor in the z = 0 plane as equivalent to the problem of sources odd about this plane, as long as the sources in both cases are identical for z > 0. We refer to the source in the region z 0. Thus the image source (JI , JI m) obeys 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). That is, parallel components of electric current image in the opposite direction, and the perpendicular component images in the same direction; parallel components of the magnetic current image in the same direction, while the perpendicular component images in the opposite direction. In the case of even symmetry, the magnetic field parallel to the z = 0 plane is odd, and thus the magnetic field tangential to the z = 0 plane must be zero. We therefore have an equivalence between the problem of a source above a plane of perfect magnetic conductor (PMC) and the problem of sources even about that plane. In this case we identify image sources that 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). Parallel components of electric current image in the same direction, and the perpendicular component images in the opposite direction; parallel components of magnetic current image in the opposite direction, and the perpendicular component images in the same direction. In the case of odd symmetry, we sometimes say that an “electric wall” exists at z = 0. The term “magnetic wall” can be used in the case of even symmetry. These terms are particularly common in the description of waveguide fields
etric field deco ition. Field sym source distributions through a symmetry decomposition of the sources and fields. Con sider the general impressed source distributions (J, Jm). The source set #(xy,z)=5[H(x,y,2)+f(x,y,-) J(,y,z)=5[(x,,2)+(x,y,-) J,(x, y, z) 2 [(x,y,z) Jm(x, y, z) 2 Jm.(,y,z) I(x, y, z) (x,y,z)-my(x,y,-2) learly of even symmetric type while the source set (a,y,)=1 [(x,y,x)-J(x,y,-2 1 J(x [(x,y.z)-(x,y,-2) (x,y,x)=2(x,y.3)+(,y-], Jm(,y, z)=5Jm(x, y, z)+Jmx(x, y, -2) m(,y,x)=5[m(x,y,x)+ J is of the odd symmetric type. SinceJ'=J+J and Jm= Jin +Jm, we can decompose any source into constituents having, respectively, even and odd symmetry with respect to a plane. The source with even symmetry produces an even field set, while the source with odd symmetry produces an odd field set. The total field is the sum of the fields Planar symmetry for frequency-domain fields. The symmetry conditions intro- duced above for the time-domain fields also hold for the frequency-domain fields. Because both the conductivity and permittivity must be even functions, we combine their effects and require the complex permittivity to be even. Otherwise the field symmetries and urce decompositions are identical Example of symmetry decomposition: line source between conducting planes. Consider a z-directed electric line source lo located at y =h, x=0 between conducting planes at y=td, d>h. The material between the plates has permeability i(o) and complex permittivity E(o). We decompose the source into one of even symmetric type ith line sources lo/2 located at y= th, and one of odd symmetric type with a line ②2001
Symmetric field decomposition. Field symmetries may be applied to arbitrary source distributions through a symmetry decomposition of the sources and fields. Consider the general impressed source distributions (Ji , Ji m). The source set J ie x (x, y,z) = 1 2 J i x (x, y,z) + J i x (x, y, −z) , J ie y (x, y,z) = 1 2 J i y (x, y,z) + J i y (x, y, −z) , J ie z (x, y,z) = 1 2 J i z (x, y,z) − J i z (x, y, −z) , J ie mx (x, y,z) = 1 2 J i mx (x, y,z) − J i mx (x, y, −z) , J ie my (x, y,z) = 1 2 J i my (x, y,z) − J i my (x, y, −z) , J ie mz(x, y,z) = 1 2 J i mz(x, y,z) + J i mz(x, y, −z) , is clearly of even symmetric type while the source set J io x (x, y,z) = 1 2 J i x (x, y,z) − J i x (x, y, −z) , J io y (x, y,z) = 1 2 J i y (x, y,z) − J i y (x, y, −z) , J io z (x, y,z) = 1 2 J i z (x, y,z) + J i z (x, y, −z) , J io mx (x, y,z) = 1 2 J i mx (x, y,z) + J i mx (x, y, −z) , J io my (x, y,z) = 1 2 J i my (x, y,z) + J i my (x, y, −z) , J io mz(x, y,z) = 1 2 J i mz(x, y,z) − J i mz(x, y, −z) , is of the odd symmetric type. Since Ji = Jie + Jio and Ji m = Jie m + Jio m , we can decompose any source into constituents having, respectively, even and odd symmetry with respect to a plane. The source with even symmetry produces an even field set, while the source with odd symmetry produces an odd field set. The total field is the sum of the fields from each field set. Planar symmetry for frequency-domain fields. The symmetry conditions introduced above for the time-domain fields also hold for the frequency-domain fields. Because both the conductivity and permittivity must be even functions, we combine their effects and require the complex permittivity to be even. Otherwise the field symmetries and source decompositions are identical. Example of symmetry decomposition: line source between conducting planes. Consider a z-directed electric line source ˜I0 located at y = h, x = 0 between conducting planes at y = ±d, d > h. The material between the plates has permeability µ(ω) ˜ and complex permittivity ˜ c(ω). We decompose the source into one of even symmetric type with line sources ˜I0/2 located at y = ±h, and one of odd symmetric type with a line
source Io/2 located at y= h and a line source -lo/2 located at y= -h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results to find the solution to the original problem For the even-symmetric case, we begin by using(4.407) to represent the field EI (,)=-x+A For y> h this becomes E(x, y, o= 2 cos k The secondary(scattered) field consists of waves propagating in both the ty-directions +jΔ [A*(r, o)- /k, '+A(kx,oeky ]e ik dkr.(5.7) ∞+j△ The impressed field is even about y =0. Since the total field E,= E;+Es must be even in y(E is parallel to the plane y=0), the scattered field must also be even. Thus, A+=a and the total field is for y > h E 2A(kr, a)cos ky,'so lo(o)2 cos ky h -jk, e -jk, dks Now the electric field must obey the boundary condition E,=0 at y=+d. How since E is even the satisfaction of this condition at y= d automatically implie atisfaction at y=-d. So we set 2A*(, o)cos kyd-a lo(o)2 cos k,h and invoke the Fourier integral theorem to get AT(Kx, o)=op lo(o) cos kyh The total field for this case is E ∞+ For the odd-symmetric case the impressed field is El (x, y, o) 一 hksi dkx ②2001
source ˜I0/2 located at y = h and a line source −˜I0/2 located at y = −h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results to find the solution to the original problem. For the even-symmetric case, we begin by using (4.407) to represent the impressed field: E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| + e− jky |y+h| 2ky e− jkx x dkx . For y > h this becomes E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j 2 cos kyh 2ky e− jky y e− jkx x dkx . The secondary (scattered) field consists of waves propagating in both the ±y-directions: E˜ s z(x, y,ω) = 1 2π ∞+ j −∞+ j A+(kx ,ω)e− jky y + A−(kx ,ω)e jky y e− jkx x dkx . (5.7) The impressed field is even about y = 0. Since the total field Ez = Ei z + Es z must be even in y (Ez is parallel to the plane y = 0), the scattered field must also be even. Thus, A+ = A− and the total field is for y > h E˜ z(x, y,ω) = 1 2π ∞+ j −∞+ j 2A+(kx ,ω) cos ky y − ωµ˜ ˜I0(ω) 2 2 cos kyh 2ky e− jky y e− jkx x dkx . Nowthe electric field must obey the boundary condition E˜ z = 0 at y = ±d. However, since E˜ z is even the satisfaction of this condition at y = d automatically implies its satisfaction at y = −d. So we set 1 2π ∞+ j −∞+ j 2A+(kx ,ω) cos kyd − ωµ˜ ˜I0(ω) 2 2 cos kyh 2ky e− jkyd e− jkx x dkx = 0 and invoke the Fourier integral theorem to get A+(kx ,ω) = ωµ˜ ˜I0(ω) 2 cos kyh 2ky e− jkyd cos kyd . The total field for this case is E˜ z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| + e− jky |y+h| 2ky − − 2 cos kyh 2ky e− jkyd cos kyd cos ky y e− jkx x dkx . For the odd-symmetric case the impressed field is E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| − e− jky |y+h| 2ky e− jkx x dkx ,
hich fo h E 2j sin k,h The scattered field has the form of (5.7)but must be odd. Thus A+=-A-and the total field for y h E2( 2JA(kx, a)sin kyy -ou lo()2j sin k, Setting E, =0 at z=d and solving for A+ we find that the total field for this case is +j△ , (. y, 0)=2/[ jk ly-hl 2 i sin k、he jkr sin k, d Adding the fields for the two cases we find that +j△ oulo(o) -jkyIy-hl y dk (5.8) which is a superposition of impressed and scattered fields 5.2 Solenoidal-lamellar decomposition We now discuss the decomposition of a general vector field into a lamellar comp aving zero curl and a solenoidal component having zero divergence. This is known as a Helmholtz decomposition. If V is any vector field then we wish to write where Vs and Vi are the solenoidal and lamellar components of v. Formulas expressing these components in terms of V are obtained as follows. We first write Vs in terms of a vector potential”Aas v=V×A. This is possible by virtue of(B 49). Similarly, we write VI in terms of a"scalar potential v=Vφ ②2001
which for y > h is E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j 2 j sin kyh 2ky e− jky y e− jkx x dkx . The scattered field has the form of (5.7) but must be odd. Thus A+ = −A− and the total field for y > h is E˜ z(x, y,ω) = 1 2π ∞+ j −∞+ j 2 j A+(kx ,ω)sin ky y − ωµ˜ ˜I0(ω) 2 2 j sin kyh 2ky e− jky y e− jkx x dkx . Setting E˜ z = 0 at z = d and solving for A+ we find that the total field for this case is E˜ z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| − e− jky |y+h| 2ky − − 2 j sin kyh 2ky e− jkyd sin kyd sin ky y e− jkx x dkx . Adding the fields for the two cases we find that E˜ z(x, y,ω) = −ωµ˜ ˜I0(ω) 2π ∞+ j −∞+ j e− jky |y−h| 2ky e− jkx x dkx + +ωµ˜ ˜I0(ω) 2π ∞+ j −∞+ j cos kyh cos ky y cos kyd + j sin kyh sin ky y sin kyd e− jkyd 2ky e− jkx x dkx , (5.8) which is a superposition of impressed and scattered fields. 5.2 Solenoidal–lamellar decomposition We nowdiscuss the decomposition of a general vector field into a lamellar component having zero curl and a solenoidal component having zero divergence. This is known as a Helmholtz decomposition. If V is any vector field then we wish to write V = Vs + Vl, (5.9) where Vs and Vl are the solenoidal and lamellar components of V. Formulas expressing these components in terms of V are obtained as follows. We first write Vs in terms of a “vector potential” A as Vs =∇× A. (5.10) This is possible by virtue of (B.49). Similarly, we write Vl in terms of a “scalar potential” φ as Vl = ∇φ. (5.11)
To obtain a formula for VI we take the divergence of(5. 9 )and use(.11)to get v=vvt=V.Vφ=Vφ The result may be regarded as Poissons equation for the unknown This equation is solved in Chapter 3. By (3.61)we have V.V(r) 4丌R where R= r-r, and we have V. V(r v1(r)=-V Similarly, a formula for Vs can be obtained by taking the curl of (5.9)to get VxV=VXV Substituting(. 10)we have xv=V×(V×A)=V(·A)-V2A We may choose any value we wish for V. A, since this does not alter Vs=VxA (We discuss such "gauge transformations"in greater detail later in this chapter. )With V.A=O we obtain VxV=V-A This is Poissons equation for each rectangular component of A; therefore V'xV(r) 4丌R and we have V' x V(r) V,(r=VX Summing the results we obtain the Helmholtz decomposition V=V+V dV′+V 4T R (5.13) Identification of the electromagnetic potentials. Let us write the electromagnetic fields as a general superposition of solenoidal and lamellar components E=V×AE+VφE (5.14) B=V×AB+Vφ (5.15) One possible form of the potentials AE, AB, E, and B appears in(5.13).However, because e and B are related by Maxwells equations, the potentials should be related to the sources. We can determine the explicit relationship by substituting(5. 14) and (5.15) ②2001
To obtain a formula for Vl we take the divergence of (5.9) and use (5.11) to get ∇ · V =∇· Vl =∇·∇φ = ∇2 φ. The result, ∇2 φ =∇· V, may be regarded as Poisson’s equation for the unknown φ. This equation is solved in Chapter 3. By (3.61) we have φ(r) = − V ∇ · V(r ) 4π R dV , where R = |r − r |, and we have Vl(r) = −∇ V ∇ · V(r ) 4π R dV . (5.12) Similarly, a formula for Vs can be obtained by taking the curl of (5.9) to get ∇ × V =∇× Vs. Substituting (5.10) we have ∇ × V =∇× (∇ × A) = ∇(∇ · A) − ∇2 A. We may choose any value we wish for ∇ · A, since this does not alter Vs =∇× A. (We discuss such “gauge transformations” in greater detail later in this chapter.) With ∇ · A = 0 we obtain −∇ × V = ∇2 A. This is Poisson’s equation for each rectangular component of A; therefore A(r) = V ∇ × V(r ) 4π R dV , and we have Vs(r) =∇× V ∇ × V(r ) 4π R dV . Summing the results we obtain the Helmholtz decomposition V = Vl + Vs = −∇ V ∇ · V(r ) 4π R dV +∇× V ∇ × V(r ) 4π R dV . (5.13) Identification of the electromagnetic potentials. Let us write the electromagnetic fields as a general superposition of solenoidal and lamellar components: E =∇× AE + ∇φE , (5.14) B =∇× AB + ∇φB. (5.15) One possible form of the potentials AE , AB, φE , and φB appears in (5.13). However, because E and B are related by Maxwell’s equations, the potentials should be related to the sources. We can determine the explicit relationship by substituting (5.14) and (5.15)
into Ampere's and Faradays laws. It is most convenient to analyze the relationship Ising superposition of the cases for which Jm =0 and J=0. With Jm=0 Faraday's law V×E (5.16) Since v xe is solenoidal that B=0, which is equivalent to the auxiliary Maxwell equation V.B= 0. Now, substitution of (5. 14)and(5.15)into(5.16)gives Using V×(VφE)=0 and combining the terms we get da V×v×AE+ dAg + ve Substitution into(. 14)gives +VφE+V] Combining the two gradient functions together, we see that we can write both e and B in terms of two potentials dA E= Vφe, B=V×A (518) where the negative sign on the gradient term is introduced by convention. Gauge transformations and the Coulomb gauge. We pay a price for the simplicity of using only two potentials to represent E and B. While V x ae is definitely solenoidal, Ae itself may not be: because of this (5.17) may not be a decomposition into solenoidal and lamellar components. However, a corollary of the Helmholtz theorem states that a vector field is uniquely specified only when both its curl and divergence are specified. Here there is an ambiguity in the representation of E and B; we may remove this ambiguity and define Ae uniquely by requiring that Then Ae is solenoidal and the decomposition (5.17) is solenoidal-lamellar. This require- ment on Ae is called the Coulomb gauge. The ambiguity implied by the non-uniqueness of V. Ae can also be expressed by the observation that a transformation of the type A→A+Vr, (5.20) (5.21) ②2001
into Ampere’s and Faraday’s laws. It is most convenient to analyze the relationships using superposition of the cases for which Jm = 0 and J = 0. With Jm = 0 Faraday’s lawis ∇ × E = −∂B ∂t . (5.16) Since ∇ × E is solenoidal, B must be solenoidal and thus ∇φB = 0. This implies that φB = 0, which is equivalent to the auxiliary Maxwell equation ∇ · B = 0. Now , substitution of (5.14) and (5.15) into (5.16) gives ∇ × [∇ × AE + ∇φE ] = − ∂ ∂t [∇ × AB] . Using ∇ × (∇φE ) = 0 and combining the terms we get ∇ × ∇ × AE + ∂AB ∂t = 0, hence ∇ × AE = −∂AB ∂t + ∇ξ. Substitution into (5.14) gives E = −∂AB ∂t + [∇φE + ∇ξ ] . Combining the two gradient functions together, we see that we can write both E and B in terms of two potentials: E = −∂Ae ∂t − ∇φe, (5.17) B =∇× Ae, (5.18) where the negative sign on the gradient term is introduced by convention. Gauge transformations and the Coulomb gauge. We pay a price for the simplicity of using only two potentials to represent E and B. While ∇ × Ae is definitely solenoidal, Ae itself may not be: because of this (5.17) may not be a decomposition into solenoidal and lamellar components. However, a corollary of the Helmholtz theorem states that a vector field is uniquely specified only when both its curl and divergence are specified. Here there is an ambiguity in the representation of E and B; we may remove this ambiguity and define Ae uniquely by requiring that ∇ · Ae = 0. (5.19) Then Ae is solenoidal and the decomposition (5.17) is solenoidal–lamellar. This requirement on Ae is called the Coulomb gauge. The ambiguity implied by the non-uniqueness of ∇ · Ae can also be expressed by the observation that a transformation of the type Ae → Ae + ∇, (5.20) φe → φe − ∂ ∂t , (5.21)
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) ②2001
leaves 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 timeinvariant 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)
Therefore the vector potential Ae, which describes the solenoidal portion of both E and B, is found from just the solenoidal portion of the current. On the other hand, the scalar potential, which describes the lamellar portion of E, is found from p' which arises from V- J, the lamellar portion of the current From the perspective of field computation, we see that the introduction of potential functions has reoriented the solution process from dealing with two coupled first-order partial differential equations(Maxwell's equations), to two uncoupled second-order equa- tions(the potential equations(5. 24)and(5.28). The decoupling of the equations is often worth the added complexity of dealing with potentials, and, in fact, is the solution tech nique of choice in such areas as radiation and guided waves. It is worth pausing for a moment to examine the form of these equations. We see that the scalar potential obeys Poisson's equation with the solution(5.25), while the vector potential obeys the wave equation. As a wave, the vector potential must propagate aw from the source with finite velocity. However, the solution for the scalar potential (5.25)shows no such behavior. In fact, any change to the charge distribution instantaneously permeates all of space. This apparent violation of Einsteins postulate shows that we must be careful hen interpreting the physical meaning of the potentials. Once the computations (5.17) and(5. 18)are undertaken, we find that both e and B behave as waves, and thus propa gate at finite velocity. Mathematically, the conundrum can be resolved by realizing that individually the solenoidal and lamellar components of current must occupy all of space even if their sum, the actual current J, is localized [911 The Lorentz gauge. a different choice of gauge condition can allow both the vector and scalar potentials to act as waves. In this case e may be written as a sum of two terms: one purely solenoidal, and the other a superposition of lamellar and solenoidal Let us examine the effect of choosing the lorentz gauge (5.29) Substituting this expression into(5.26)we find that the gradient terms cancel, giving (5.30) a2A (531) and(5.23)become For lossy media we have obtained a second-order differential equation for Ae, but e must be found through the somewhat cumbersome relation(5.29). For lossless media e coupled Maxwell equations have been decoupled into two second-order equations, one involving Ae and one involving e. Both(5.31)and(5.32) are wave equations, with J as the source for Ae and p' as the source for e. Thus the expected finite-velocity wave nature of the electromagnetic fields is also manifested in each of the potential functions The drawback is that, even though we can still use(5.17)and(5.18), the expression for E is no longer a decomposition into solenoidal and lamellar components. Nevertheless, the choice of the Lorentz gauge is very popular in the study of radiated and guided waves ②2001
Therefore the vector potential Ae, which describes the solenoidal portion of both E and B, is found from just the solenoidal portion of the current. On the other hand, the scalar potential, which describes the lamellar portion of E, is found from ρi which arises from ∇ · Ji , the lamellar portion of the current. From the perspective of field computation, we see that the introduction of potential functions has reoriented the solution process from dealing with two coupled first-order partial differential equations (Maxwell’s equations), to two uncoupled second-order equations (the potential equations (5.24) and (5.28)). The decoupling of the equations is often worth the added complexity of dealing with potentials, and, in fact, is the solution technique of choice in such areas as radiation and guided waves. It is worth pausing for a moment to examine the form of these equations. We see that the scalar potential obeys Poisson’s equation with the solution (5.25), while the vector potential obeys the wave equation. As a wave, the vector potential must propagate away from the source with finite velocity. However, the solution for the scalar potential (5.25) shows no such behavior. In fact, any change to the charge distribution instantaneously permeates all of space. This apparent violation of Einstein’s postulate shows that we must be careful when interpreting the physical meaning of the potentials. Once the computations (5.17) and (5.18) are undertaken, we find that both E and B behave as waves, and thus propagate at finite velocity. Mathematically, the conundrum can be resolved by realizing that individually the solenoidal and lamellar components of current must occupy all of space, even if their sum, the actual current Ji , is localized [91]. The Lorentz gauge. A different choice of gauge condition can allowboth the vector and scalar potentials to act as waves. In this case E may be written as a sum of two terms: one purely solenoidal, and the other a superposition of lamellar and solenoidal parts. Let us examine the effect of choosing the Lorentz gauge ∇ · Ae = −µ ∂φe ∂t − µσφe. (5.29) Substituting this expression into (5.26) we find that the gradient terms cancel, giving ∇2 Ae − µσ ∂Ae ∂t − µ ∂2Ae ∂t 2 = −µJi . (5.30) For lossless media ∇2 Ae − µ ∂2Ae ∂t 2 = −µJi , (5.31) and (5.23) becomes ∇2 φe − µ ∂2φe ∂t 2 = −ρi . (5.32) For lossy media we have obtained a second-order differential equation for Ae, but φe must be found through the somewhat cumbersome relation (5.29). For lossless media the coupled Maxwell equations have been decoupled into two second-order equations, one involving Ae and one involving φe. Both (5.31) and (5.32) are wave equations, with Ji as the source for Ae and ρi as the source for φe. Thus the expected finite-velocity wave nature of the electromagnetic fields is also manifested in each of the potential functions. The drawback is that, even though we can still use (5.17) and (5.18), the expression for E is no longer a decomposition into solenoidal and lamellar components. Nevertheless, the choice of the Lorentz gauge is very popular in the study of radiated and guided waves.
