MIT OpenCourseWare http://ocw.mit.edu Electromagnetic Field Theory: A Problem Solving Approach For any use or distribution of this textbook, please cite as follows: Markus Zahn, Electromagnetic Field Theory: A Problem Solving Approach. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-NonCommercial-Share Alike. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.chapter 9radiation664 RadiationIn low-frequency electric circuits and along transmissionlines, power is guided from a source to a load along highlyconducting wires with the fields predominantly confined tothe region around the wires. At very high frequencies thesewires become antennas as this power can radiate away intospace without the need of any guiding structure.9-1 THE RETARDED POTENTIALS9-1-1 Nonhomogeneous Wave EquationsMaxwell's equations in complete generality are0BVxE= (1)ataDVxH=J +-(2)atV B = 0 (3)V-D=pf (4)In our development we will use the following vector iden-titiesVx (V V) =O (5)V -(VxA) = 0 (6)Vx (Vx A) = V(V -A) - V2 A (7)where A and V can be any functions but in particular will bethe magnetic vector potential and electric scalar potential,respectively.Because in (3) the magnetic field has no divergence, theidentity in (6) allows us to again define the vector potential Aas we had for quasi-statics in Section 5-4:B=VxA (8)so that Faraday's law in (1) can be rewritten as/ A\Vx(E+-=0Oat\Ot/"The Retarded Potentials 665Then (5) tells us that any curl-free vector can be written as thegradient of a scalar so that (9) becomesaAE+-= -VV (10)atwhere we introduce the negative sign on the right-hand sideso that V becomes the electric potential in a static situationwhen A is independent of time. We solve (10) for the electricfield and with (8) rewrite (2) for linear dielectric media (D =eE, B = H):1 vaV aA1 1Vx(VxA)= p•Jt+ 1-Va, c=--at-(11)The vector identity of (7) allows us to reduce (11) toS 1 Vi 1 a2AV2V - cA+-•--2jC2 at2---t (12)Thus far, we have only specified the curl of A in (8). TheHelmholtz theorem discussed in Section 5-4-1 told us that touniquely specify the vector potential we must also specify thedivergence of A. This is called setting the gauge. Examining(12) we see that if we set1 avV 1A=A (13)c atthe middle term on the left-hand side of (12) becomes zero sothat the resulting relation between A and J, is the non-homogeneous vector wave equation:V2A c2A= -Jrf (14)The condition of (13) is called the Lorentz gauge. Note thatfor static conditions, V A = 0, which is the value also pickedin Section 5-4-2 for the magneto-quasi-static field. With (14)we can solve for A when the current distribution J1 is givenand then use (13) to solve for V. The scalar potential can alsobe found directly by using (10) in Gauss's law of (4) asVV+ a(VA) = -P (15)at EThe second term can be put in terms of V by using theLorentz gauge condition of (13) to yield the scalar waveequation:1 a2V -p_-a (16)C at (666 Radiation-Note again that for static situations this relation reduces toPoisson's equation, the governing equation for the quasi-staticelectric potential.9-1-2 Solutions to the Wave EquationWe see that the three scalar equations of (14) (one equationfor each vector component) and that of (16) are in the sameform. If we can thus find the general solution to any one ofthese equations, we know the general solution to all of them.As we had earlier proceeded for quasi-static fields, we willfind the solution to (16) for a point charge source. Then thesolution for any charge distribution is obtained using super-position by integrating the solution for a point charge over allincremental charge elements.In particular, consider a stationary point charge at r = 0that is an arbitrary function of time Q(t). By symmetry, theresulting potential can only be a function of r so that (16)becomes1 1 a2I9 V-4rr ---y= 0, r>O (17)where the right-hand side is zero because the charge densityis zero everywhere except at r=O. By multiplying (17)through by r and realizing thatI a ,aV a'rr =---(r V ) (18)we rewrite (17) as a homogeneous wave equation in the vari-able (rV):.a' I a'(rV)--;P (rV)= 0 (19)ar c atwhich we know from Section 7-3-2 has solutionsrV=f(t-)1+f-_) (20)We throw out the negatively traveling wave solution as thereare no sources for r >0 so that all waves emanate radiallyoutward from the point charge at r =0. The arbitraryfunction f+. is evaluated by realizing that as r -0 there can beno propagation delay effects so that the potential shouldapproach the quasi-static Coulomb potential of a pointcharge:Q(Q) (2t1lim V= Q( f+(t) (21)r-.o 4irer 41sRadiationfrom Point Dipoles 667The potential due to a point charge is then obtained from(20) and (21) replacing time t with the retarded time t-rlc:Q(t -r/c)V(r, t) = (22)4rerThe potential at time t depends not on the present value ofcharge but on the charge value a propagation time r/c earlierwhen the wave now received was launched.The potential due to an arbitrary volume distribution ofcharge pf(t) is obtained by replacing Q(t) with the differentialcharge element p1(t) dV and integrating over the volume ofcharge:pf(t -rqplc)V(r, t)= chare ( t -rc) dV (23)where rQp is the distance between the charge as a source atpoint Q and the field point at P.The vector potential in (14) is in the same direction as thecurrent density Jf.The solution for A can be directly obtainedfrom (23) realizing that each component of A obeys the sameequation as (16) if we replace pIle by l&J1:A(r, t) = V (24)faIl current 41rQp9-2 RADIATION FROM POINT DIPOLES9-2-1 The Electric DipoleThe simplest building block for a transmitting antenna isthat of a uniform current flowing along a conductor ofincremental length dl as shown in Figure 9-1. We assume thatthis current varies sinusoidally with time asi(t)=Re (fej) (1)Because the current is discontinuous at the ends, charge mustbe deposited there being of opposite sign at each end [q(t)=Re (Q e•)]:dq dli(t)= • I= ijoC), z = ±-(2)dt 2This forms an electric dipole with momentp= q dl i, (3)If we can find the potentials and fields from this simpleelement, the solution for any current distribution is easilyfound by superposition.668 Radiation2A Ap =Qd~i,Figure 9-1 A point dipole antenna is composed of a very short uniformly distributedcurrent-carrying wire. Because the current is discontinuous at the ends,
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