3. Electromagnetic Waves II

Transcription

1 Lectue 3 - Electomagnetic Waves II 9 3. Electomagnetic Waves II Last time, we discussed the following. 1. The popagation of an EM wave though a macoscopic media: We discussed how the wave inteacts with the media and how all of the details of inte-atomic and atom-em wave inteactions can be descibed by a constitutive elation. Examples we discussed last time included the change of the speed of the popagation, eflection and efaction, gain and loss, and tunneling though a thin slab.. Macoscopic Maxwell s equations: We deived the macoscopic Maxwell s equations. These equations can descibe all the EM inteaction with macoscopic media with a linea dimension > 1 nm. The constitutive elation can be measued fom expeiments. It can also be calculated by taking into account all the micoscopic inteactions. Fo example, to the fist ode, the dielectic constant of a homogeneous medium can be detemined by the polaization vecto P which is the total dipole moment in the medium. We will show you how this can be done late in the class when we discuss micoscopic inteaction between light and mattes Geneation of EM waves ε = ε + E (3.1) P/ obseve Antenna basics In the discussions so fa, we have only studied the behavio of a given EM wave (e.g. a plane wave) and its inteaction with macoscopic media but we have not discussed how the EM wave is geneated. We will see in this section how a time-vaiant cuent can geneate an EM wave. We will solve the wave equation with the inclusion of the cuent as follows. E k E = iωµ J (3.) Once the electic field is obtained, we can calculate the magnetic field by: EECS 598- Nanophotonics and Nanoscale Fabication Winte 6, P.C.Ku

2 Lectue 3 - Electomagnetic Waves II 1 E H = iωµ (3.3) The solution to (3.) can be witten as follows fo any obseve at outside the distibution egion. E ( ) = iωµ G (, ') J ( ') d ' whee G is a dyadic Geen s function which satisfies G k G = Iδ ( ') (3.4) (3.5) I is an identity matix ( I = xx ˆˆ+ yy ˆˆ + zz ˆˆ = ˆˆ+ ˆˆ θθ + ˆˆ ϕϕ ). You can easily veify by inseting (3.4) into the LHS of (3.) and using (3.5), you get the RHS of (3.). Note also that because is outside the, we can intechange and the volume integal. Please note all the diffeential opeatos below including (3.5) act on unless othewise specified. To solve G, fist we notice that vecto identity ( A) = G = G+ G. If we take the divegence of (3.5) and use the, we get: k G = δ ( ') Substituting (3.6) back into (3.5), we have: + k G = I + δ ( ') k ( ) We can veify that G now can be witten in tems of a scala function: G = I + g k whee the scala function g satisfy: To solve g in (3.9), we notice that g(, ') + k g(, ') = δ ( ') ( ) should depend only on ' (3.6) (3.7) (3.8) (3.9) but not the absolute location of '. Theefoe we can abitaily set must be spheical symmetic aound the oigin. (3.9) becomes: ' at oigin. Afte the choice of ', we can easily see the solution fo g The solution to (3.1) is: d g() dg() + + k g( ) = δ ( ) (3.1) d d e g () = C (3.11) Dyadic G is a diect poduct of two vectos. Fo example, if G = AB, its index notation becomes G = AB. In matix notation, the diect poduct of two vectos can be epesented by a 3x3 matix. ij i j EECS 598- Nanophotonics and Nanoscale Fabication Winte 6, P.C.Ku

3 Lectue 3 - Electomagnetic Waves II 11 To detemine the constant C, we integate (3.9) ove a volume including the oigin and let the volume go to zeo: gd = 1 C = Combining (3.4), (3.8), (3.11) and (3.1), we have: The magnetic field is (fom (3.3) and (3.13)): E H = iωµ V S 1 dg g nds = π = d ˆ 4 1 = δ ik k (3.1) e E ( ) = iωµ I+ J ( ') d' (3.13) 4 π = ik e J ( ') d' 4 π (3.14) Befoe we poceed, we have to emembe that all the quantities in (3.13) and (3.14) ae in the fequency i t domain. We have dopped thei e ω i t dependence. If J ( ') e ω is a static, ω = and k = Geneal popeties of nea field and fa field Depending on the distance between the obseve and the, we can study two exteme cases, the nea field ( k 1 ) and the fa field ( k 1 ). In the fa field, we have ˆ '. The electic field becomes: e E i I J e d ik ' ( ) = ωµ + ( ') ' k (3.15) The integal in (3.15) esults in a function that depends only on θ and ϕ. We can define a vecto cuent moment as: ik ' f ( θϕ, ) = J( ') e d' (3.16) In the fa field egion, we only keep tems on the ode of 1/k and neglect all the highe ode tems. Using: ˆ 1 1 = ˆ + θ + ˆ ϕ θ sin θ ϕ (3.17) (3.15) becomes: e e ( ˆˆ) ( ˆ ˆ θ ϕ ) E ( ) = iωµ I f iωµ θ f ϕ f = + (3.18) EECS 598- Nanophotonics and Nanoscale Fabication Winte 6, P.C.Ku

4 Lectue 3 - Electomagnetic Waves II 1 This is an outgoing wave with a spheical wave font. The electic field is pependicula to the popagation diection. At lage distance, the wave can be appoximately by a plane wave. In the nea field, we have = and the obseve is at a distance many times smalle ik e ik than the wavelength fom the. (3.13) becomes: J( ') E ( ) = iωµ I+ d' (3.19) 4 k π This is a quasi-static field because of the absence of the oscillating exponential tem. Note the field is not i t tuly static since thee is an implicit time hamonic facto e ω. Similaly, the magnetic field is: J( ') H ( ) = d' (3.) 4 π Because in nea field egion, we have k 1, the contibution fom the nd tem in the paenthesis of (3.19) dominates and the magnetic field can usually be neglected (because magnetic field gets diffeentiation once while the electic field gets diffeentiation twice). We will see an example in the following when we discuss the dipole adiato. But we notice that if magnetic field can be neglected, the solution of fields satisfies the electostatic equation o the Poisson equation: E = (3.1) O in tems of the potential φ : φ = (3.) Dipole adiation The most fundamental antenna is a Hetzian dipole which consists of a cuent-caying wie with an infinitesimal length l: J ( ') = zil ˆ δ ( ') (3.3) Substituting (3.3) into (3.13), we get: ik e E ( ) = iωµ I+ zil ˆ δ ( ') d' 4 k π e = iωµ I + zil ˆ k e = iωµ Il zˆ + k z (3.4) Now we make the coodinate tansfomation to the spheical coodinate by ˆ 1 1 = ˆ + θ + ˆ ϕ θ sinθ ϕ e 1 e = ik cos θ z zˆ = ˆ cosθ ˆ θsinθ (3.5) EECS 598- Nanophotonics and Nanoscale Fabication Winte 6, P.C.Ku

5 Lectue 3 - Electomagnetic Waves II 13 (3.4) becomes: e i ( ) ˆ i cos ˆ i i E = iωµ Il θ + + θ sinθ 1+ + k k k k (3.6) Nea and fa fields fo dipole adiatos In the fa field, (3.6) educes to: e E ( ) = iωµ Il ˆ θsinθ (3.7) The wavefont is a spheical outgoing wave and adiation patten consists of two side lobes with no electic field along the z axis. The polaization of the electic field is pependicula to the diection of the popagation. The magnetic field can be calculated by (3.3) to be: e H( ) = ikil ˆ sin ϕ θ (3.8) In the nea field, (3.6) educes to: The field fades away quickly with and has the ˆ component. 1 ( ) ˆ i cos ˆ i E = iωµ Il θ + θ sinθ k k 3 iil 1 = ˆcosθ ˆ + θsinθ ωε Radiation fom a Moving Chage in contast to fa field dependence of (3.9) 1. Note the field is quasi-static The dipole adiato is one special type of adiation s. Since the dipole usually consists of lots of oscillating (o moving) chages. It is inteesting to study the adiation fom a single moving chage. It has applications fo example in paticle detectos (using Cheenkov adiation) and synchoton adiation. The chage density of a moving chage with a tajectoy '( t) is: ρ( ', t) = qδ ( ' '( t)) (3.3) The cuent density is theefoe: d '( t) J t = q δ t dt To solve the electic field with (3.13), we need to convet J ( ', t) ( ', ) ( ' '( )) (3.31) to the fequency domain. To do that, we use the Fouie tansfom as follows. d '( t) i( ωt k ') J ( k, ω) = d ' dtq δ( ' '( t)) e dt (3.3) d '( t) i( ωt k '( t)) = dtq e dt EECS 598- Nanophotonics and Nanoscale Fabication Winte 6, P.C.Ku

6 Lectue 3 - Electomagnetic Waves II 14 We conside two cases. In the fist case, the paticle is moving at a constant velocity along the z-axis (i.e. no acceleation.) (3.3) becomes: J ( k, ω) = πqv δ ω k v ( ) Fom (3.33), the electic field will have tems with wavevectos given by (3.33) k = ω /( v cos θ ) (3.34) whee θ is the angle fom the z-axis to the popagation diection. Remembe that the wavevecto itself needs to satisfy the Maxwell s equations, i.e. k = ωn/ c. We have: c c v = > (3.35) ncosθ n That is in ode to geneate adiation fom a chage moving at a constant velocity, the velocity has to be geate than the speed of light in the media being consideed. The adiation geneated in such a way is called the Cheenkov adiation. That s why usually a moving chage without any acceleation does not adiate. The second case we will conside is a chage moving with acceleation. Because of the acceleation, the integal in (3.3) will have tems that can geneate EM waves. It is geneally had to evaluate such an integal. But in summay an acceleated chage adiates. EECS 598- Nanophotonics and Nanoscale Fabication Winte 6, P.C.Ku