Question 1: Some algebra

Transcription

1 October 13, 017 Cornell University, Department of Physics PHYS 337, Advance E&M, HW # 6, due: 10/4/017, 11:15 AM Question 1: Some algebra 1. Prove the vector identity used in lecture to derive the energy conservation law for EM fields: E ( B) B ( E) = (E B). (1) It is probably simplest to use index notation where (a b) i = ɛ ijk a j b k. I will use the shorthand i = x i. In index notation, the LHS of the equation reads E i ɛ ijk j B k B i ɛ ijk j E k = ɛ ijk j (E i B k ) ɛ ijk B k j E i B i ɛ ijk j E k (product rule) = i (ɛ jik E j B k ) ɛ kji B i j E k B i ɛ ijk j E k (relabelling dummy indices) = i (ɛ ijk E j B k ) + ɛ ijk B i j E k B i ɛ ijk j E k (antisymmetry of ɛ ijk ) = i (ɛ ijk E j B k ) which is the RHS of the equation in index notation. (). Consider a plane wave where E(x, t) = E 0 cos( k r ωt) (3) and show that E = 0 implies that k E 0 = 0. Here you may find it simpler to use the complex representation of the wave and use index notation. In index notation, i E i = i [Re(E 0i e i(k jx j ωt) )] = Re[ik i E 0i e i(k jx j ωt) ] = Re[k i E 0i e i(k jx j ωt+π/) ] = k i E 0i cos(k j x j ωt + π/) (4) = k i E 0i sin(k j x j ωt) = 0 Therefore, we see that k i E 0i = 0, or k E 0 = 0. 1

2 Question : Time-averaged Poynting vectors Solve Q. 5-7 in Heald and Marion. As discussed in Pg. 178 of H&M, for two complex oscillatory functions F = F 0 e iωt and G = G 0 e iωt, where F 0 and G 0 are complex amplitudes, the time-averaged product is given by Re(F )Re(G) = 1 Re(F 0G 0) = 1 Re(F 0 G 0 ) (5) We now apply it to Poynting vectors, defined in free space by (a) In this case, S 4π Re( E) Re( B) (6) E = [ E 0 1 ˆx + E 0 e iα ŷ ] e ikz e iωt B = [ E 0 e iαˆx + E 0 1ŷ ] e ikz e iωt (7) Therefore, S 8π Re [ (E1 0 ˆx + Ee 0 iα ŷ)e ikz ( Ee 0 iαˆx + E1ŷ) 0 e ikz] 8π Re [ (E1 0 ˆx + Ee 0 iα ŷ) ( Ee 0 iαˆx + E1ŷ) ] 0 8π Re [ (E1) 0 ẑ + (E) 0 ẑ ] (8) [ (E 0 8π 1 ) + (E) 0 ] ẑ Given the expression of the Poynting vector for a single plane wave in H&M (5.51), we see that our expression is indeed the sum of Poynting vectors of the two waves separately. The reason for this is that the polarisations of the two wave modes are orthogonal, so while they superpose, they do not interfere in that the amplitude of E in the x-direction (due to Wave 1) is unmodified by the addition of Wave. (b) In this case, Therefore, E = [ E E 0 3e iα] e ikz ˆxe iωt B = [ E E 0 3e iα] e ikz ŷe iωt (9) S 8π Re [ (E1 0 + E3e 0 iα )e ikz ˆx (E1 0 + E3e 0 iα ) e ikz ŷ ] 8π Re [ (E1 0 + E3e 0 iα )(E1 0 + E3e 0 iα ) ] ẑ 8π Re [ (E1) 0 + (E3) 0 + E1E cos α ] ẑ [ (E 0 8π 1 ) + (E3) 0 + E1E cos α ] ẑ (10)

3 The results are different from Part (a) because the two wave modes are polarised in the same direction, and so they interfere when superposed. (c) In this case, Therefore, E = [ E 0 1e ikz + E 0 1e ikz] ˆxe iωt = E 0 1 cos(kz)ˆxe iωt B = [ E 0 1e ikz E 0 1e ikz] ŷe iωt = ie 0 1 sin(kz)ŷe iωt (11) S 8π Re [ (E 0 1 cos(kz))ˆx (ie 0 1 sin(kz)) ŷ ] 8π Re [ 4i(E 0 1) cos(kz) sin(kz) ] ẑ = 0 This is not surprising, because we are dealing with standing waves which are nonpropagating. (This becomes obvious when you take the real parts of E and B). We want to work out the energy density. Electric portion : Magnetic portion : 1 8 Re( E) = 1 (E0 1) cos (kz) cos (ωt) 1 8 Re( B) = 1 (E0 1) cos (kz) sin (ωt) We see that they oscillate in space and time with the same magnitude, but are out of phase. (1) (13) Question 3: Polarization Consider a general plane wave traveling in the z direction, that we write as E(x, t) = E xˆx exp[i(kz ωt + ϕ x )] + E y ŷ exp[i(kz ωt + ϕ y )], (14) where E x and E y are real, and taking the real part is implicit. We can always rotate the reference frame such that ( ) ( ) ( ) x cos θ sin θ x (15) y sin θ cos θ y We usually call this rotation a basis change. We can also use translation, that is t t + c t, x i x i + c i (16) An alternative way to write it is to work in the circular basis where we write the wave as E(x, t) = E L ˆL exp[i(kz ωt + ϕl )] + E R ˆR exp[i(kz ωt + ϕr )] (17) 3

4 where ˆR = ˆx + iŷ, ˆL = ˆx iŷ. (18) and E L and E R are real. The circular polarization is defined as P circular = E R E L ER +. (19) E L and is a measure of how circularly polarised the EM wave is. For example P circular = ±1 means the light is completely RH/LH circularly polarised. If we don t care about the handedness, just how circularly polarised the wave is, we can use a different definition P circular = P circular (0) The goal of this question is to explore an analogous definition of linear polarisation, i.e. a measure of how linearly polarised the EM wave is. For instance, one naive defintion we start with is given by P linear = E x Ey. (1) Ex + Ey 1. Start with the wave specified in Eq. (14). Use translation to eliminate ϕ y. Explain why you cannot also simultanously eliminate ϕ x? Assume ϕ y = 0 from now on. Do a translation so that z = z ϕ y /k () kz ωt + ϕ y = kz ωt kz ωt + ϕ x = kz ωt + ϕ x ϕ y (3) We see that we can get rid of ϕ y, but in general it is not possible to remove ϕ x at the same time. Note that you can also do a time translation, or a combination of both.. Again starting with the same wave, now do a basis rotation by angle θ and find the new E x and E y in terms of the old E x, E y, ϕ x and θ. (You don t have to find the new ϕ x and ϕ y.) Under a basis rotation, the complex amplitudes rotate among each other, so we have ( ) ( ) ( ) ( ) E x exp[iϕ x] cos θ sin θ E x exp[iϕ x ] E x exp[iϕ x ] cos θ + E y sin θ = = E y exp[iϕ y] sin θ cos θ E y exp[iϕ y ] E x exp[iϕ x ] sin θ + E y cos θ (4) 4

5 so we have (pulling out the magnitude of the complex amplitude) E x = (E x cos ϕ x cos θ + E y sin θ) + (E x sin ϕ x cos θ) (5) E y = (E x cos ϕ x sin θ E y cos θ) + (E x sin ϕ x sin θ) Note that if you have chosen a passive rotation, you might get a sign difference in your θ, which is perfectly fine. 3. Show that P linear depends on θ and explain why that fact means it is not a good definition of linear polarisation. (Feel free to use Mathematica to calculate P linear if the algebra gets too painful. The final expression is relatively simple-looking.) Just do a direct substitution. After some simplification, we get P linear = (E x E y) cos(θ) + E x E y cos(ϕ x ) sin(θ) E x + E y This is clearly θ-dependent. Therefore, it is not a good definition. E.g. we could have started with something completely linearly polarised, e.g. E y = 0, so P linear = 1. After a θ = 45 basis rotation, we get P linear = 0 even though we have not physically done anything to the wave. (6) 4. To get rid of the θ dependence, we now redefine the linear polarization to be the maximum possible value of P linear (θ) as you vary θ, and denote it by P linear. P linear as an expression of E x, E y and ϕ x. Find Since P linear is of the form a cos(θ)+b sin(θ), the maximum value poasible is a + b, which in this instance gives (after some simplification) P linear = 1 4E xe y sin ϕ x (E x + E y) (7) 5. Find E L, E R and hence P circular in terms of E x, E y, and ϕ x. (Again feel free to use Mathematica.) We first note that ˆx = ˆL + ˆR ŷ = i ˆL ˆR (8) 5

6 Therefore, we can rewrite the original equation as ˆL + ˆR E(x, t) = E x exp[i(kz ωt + ϕ x )] + E y i ˆL ˆR exp[i(kz ωt)] ( ) Ex e iϕx + ie y E = x e ˆL iϕx ie y + ˆR exp[i(kz ωt)] (9) Let s write the complex amplitudes in polar form E E L exp[iϕ L ] = x cos ϕ x + (E x sin ϕ x + E y ) E E R exp[iϕ R ] = x cos ϕ x + (E x sin ϕ x E y ) E L, E R, ϕ L and ϕ R can now be read off. We also have [ ( exp i tan 1 Ex sin ϕ x + E y E x cos ϕ x [ ( exp i tan 1 Ex sin ϕ x E y E x cos ϕ x )] )] (30) P circular = E R E L E R + E L = E x cos ϕ x + (E x sin ϕ x E y ) E x cos ϕ x (E x sin ϕ x + E y ) E x cos ϕ x + (E x sin ϕ x E y ) + E x cos ϕ x + (E x sin ϕ x + E y ) = E xe y sin ϕ x E x + E y (31) 6. Now that we have the expressions for P linear and P circular in terms of E x, E y and ϕ x, show that ( P linear ) + ( P circular ) = 1 (3) and explain the meaning of it. What you have shown is that the polarization is just another way to describe the phase shift between the two components of the wave. ( P linear ) + ( P circular ) = 1 is very obvious from direct substitution. This tells us that light can be linearly or circularly polarised, or somewhere in-between. The polarisation is something truly physical and not an artifact of a basis choice. 6