4. (12) Write out an equation for Poynting s theorem in differential form. Explain in words what each term means physically.

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1 Eletrodynamis I Exam 3 - Part A - Closed Book KSU 205/2/8 Name Eletrodynami Sore = 24 / 24 points Instrutions: Use SI units. Where appropriate, define all variables or symbols you use, in words. Try to tell about the physis involved, more than the mathematis, if possible.. 0) What expression gives the time-dependent eletri field in terms of salar and vetor potentials? Ex, t) = Φ t A 2. 2) In the presene of any time-dependent soures ρx,t) and Jx,t), what equation is obeyed by the salar potential when using the Lorenz gauge? It is a wave equation for salar potential Φx, t), driven by the harge density ρx, t) saled by ; there would be a similar equation for the vetor potential Ax, t): 2 Φ A 2 2 t 2 Φ = ρx, t)/. t 2A = µ 0Jx, t). 3. 0) Consider a 3D wave equation, 2 Ψ 2 2 t 2 Ψ = 4πfx,t), where fx,t) is the soure that drives some waves Ψx, t). Write out the retarded spae- and time-dependent Green s funtion for this equation. G + x, t;x, t ) = x x δ t t + x ) x This produes a response at time t from a soure at time t, related by t = t + x x that is, the response omes after the propagation time for the waves to arrive at the observer s position. 4. 2) Write out an equation for Poynting s theorem in differential form. Explain in words what eah term means physially. u t + S + J E = 0. where u is EM energy density, S = E H is the Poynting vetor, and J and E are urrent density and eletri field. The first term is the inrease in EM field energy in a volume element. The seond term is the flux of EM energy into that volume element. The third term is the mehanial work done on harges in that volume by the eletri field. Poynting s theorem demonstrates that total mehanial plus EM energy is onserved. 5. 2) How does L em = 2 d 3 r r E H) transform under spae inversion? Time inversion? Sine E r) = Er) odd), and H r) = Hr) even), while r itself reverses odd), L em is EVEN under spatial inversion. Sine E is even while H is odd, their produt is ODD under time reversal and so is L em. These are the usual properties for angular momentum.

2 6. 2) A plane EM wave is traveling in the x-diretion in a medium with µ = µ 0 and ɛ = 9. With linearly polarized Ex,t) = E 0 ẑ exp[ikx ωt)] write an expression for Bx,t) in this wave. Faraday s Law gives the relation between E and B, for harmoni time dependene / t iω), and with ik, and dispersion relation k = ω ɛµ, E + B t = ik E iωb = 0 or B = ω k E = ɛµ ˆn E where ˆn = k/k = ˆx. The ross produt ˆn E ˆx ẑ = ŷ, and ɛµ = 3 µ 0 = 3/. We get Bx, t) = 3 E 0 ŷ exp[ikx ωt)]. 7. 2) A plane wave travels in the x-diretion: Ex,t) = E 0 ŷ + iẑ)exp[ikx ωt)]. Looking into the wave at a fixed point in spae, in whih diretion does the eletri field vetor rotate lokwise or outerlokwise)? Whih irular polarization is this right or left)? Taking a fixed spae point x = 0), and looking at the time-dependene, use the implied real part to get the omponents of E: E = Re {E 0 ŷ + iẑ)exp iωt)} = E 0 [ŷ osωt) + ẑ sinωt)] For small times t > 0 the ẑ omponent inreases positively, and the rotation of the wave towards the observer is ounterlokwise. Pointing your left thumb bak towards the soure, your left 4 fingers rotate in the ounterlokwise sense. This is left irular polarization. 8. 0) When light undergoes total internal refletion at an interfae between two optial media with indexes n inident side) and n refration side), what is required of the inident angle θ? Use Snell s Law, n sinθ = n sin θ with θ 90 determines the ritial angle. For TIR, the inident angle must be greater than the ritial angle, θ > θ, with θ defined by θ = sin n /n), requiring n < n. 9. 0) Write an expression for the dieletri funtion ɛω) in a plasma. ɛω) = ω2 p ω 2 where the squared plasma frequeny ω p depends on the total volume density of free harges n, by ω 2 p = ne2 m where m is the arrier mass usually, the eletron mass). 0. 0) What does ɛω) imply for EM waves of low frequeny traveling in a plasma? If ω < ω p, then the dieletri funtion beomes negative, ɛ < 0. The wave vetor k = ω ɛµ beomes pure imaginary, leading to strong damping of the waves entering a plasma, over a short distane. Waves with frequenies below ω p do not propagate through a plasma. ) Part A Sore = 0 /0 2

3 Eletrodynamis I Exam 3 - Part B - Open Book KSU 205/2/8 Name Instrutions: Use SI units. Please show the details of your derivations here. Explain your reasoning for full redit. Open-book only, no notes.. 48) Consider the 3D wave equation for an EM field omponent ψx,t), 2 ψ 2 2 t2φ = 4πfx,t). It is being driven by a soure funtion fx,t ) = δx )δy )δt ), whih represents a flash of a line soure along the z -axis at time t = 0. You know that the retarded 3D Green s funtion for this wave equation is G + x,t;x,t ) = x x δ t t + x ) x. Consider the EM response in 2D around the line soure at a distane ρ = x 2 + y 2 from that line soure. a) 2) Write the general integral expression for the signal ψx,t) in terms of an integration over the soure funtion fx,t ). Do not yet evaluate it. + ψx, t) = d 3 x dt G + x, t;x, t )fx, t ) + = d 3 x = dt x x δ t t + x x d 3 x x x δx )δy ) + ) δx )δy )δt ) dt δ t t + x x ) δt ). b) 2) Write out the result after only the integration over soure time t is performed. Using the last delta funtion to fore only the point t = 0 to ontribute, in the other delta funtion, ψx, t) = d 3 x x x δx )δy )δ t x ) x. ) 24) Now do the integrations over soure point x, inluding over z. Determine the final signal ψx, t) as a funtion in the form ψρ, t). Is there some restrition between the observation time t and the radial distane ρ from the soure? Hint: Due to the symmetry, the integration over z might be easier to do as an integration over soure to observer distane R = x x. What is the restrition on R? It simplifies expressions to write in terms of R = x x. But the delta funtions in x and y fore x = y = 0, so the only oordinate left to really integrate over is z. An observation point is at a radial distane ρ from the z-axis. The signal one gets depends only on ρ and the observer time t: + ψρ, t) = dz R δ t R ), R = ρ 2 + z ) 2. 3

4 One an see there are two points on the z axis that satisfy the delta funtion: z = ± R 2 ρ 2, but only if R > ρ. If we hange to R as the variable of integration, both points from the z ontribute equally, so take twie the ontribution of the one at positive z. First do δt R/) = δt R) hek that dimensions are orret). The we use R dr = z dz. The integration now beomes ψρ, t) = 2 ρ dr R z R δt R) = 2 0 dr ΘR ρ) δt R) R2 ρ2 The Heaviside step funtion enfores R > ρ. The delta funtion piks off R = t only, giving ψρ, t) = 2Θt ρ) 2 t 2 ρ 2. It is lear that the step funtion does not allow a signal to arrive at the observer until the minimum propagation time t = ρ/ has elapsed. The solution obeys ausality! 4

5 2. 44) A plane wave of intensity I 0 = 48.0 kw/m 2 and some wave vetor of magnitude k is traveling in the +ẑ diretion, with its eletri field vetor polarized along +ˆx. Jakson s Eq. 6.2), d dt P meh + P field ) α + S T αβ ) n β da = 0 suggests that T αβ is the flux per unit area per unit time) of α omponent of field momentum in the β diretion. The omponents of the Maxwell stress tensor are [ T αβ = E α E β + 2 B α B β E E + 2 B B ) ] δ αβ. 2 β a) 6) For the given plane wave, determine how T zz ) is related to the z-omponent of the eletromagneti momentum density, g = E H)/ 2. Assuming oskz ωt) time dependene, k = ω µ 0 ẑ, and ˆn = k/k = ẑ, we an take the eletri and magneti fields of the plane wave as E = E 0ˆxoskz ωt), B = µ 0ˆn E = E 0 ŷ oskz ωt). The amplitude of magneti indution is B 0 = E 0 /. Other omponents of E and B are zero. Now using α = β = z, we get the relevant omponent of the stress tensor, [ T zz = Ez Bz 2 ] E B0 2 ) os 2 kz ωt) = E0 2 os2 kz ωt). Compare the momentum density, using 2 = µ 0, g z = 2 E H) z = 2 E B y x = µ 0 2 E E 0 0 os 2 kz ωt) = µ 0 E2 0 os2 kz ωt). Comparing the results show that T zz = g z. In other words, T zz really is the flux of momentum per area per time in the waves. That is the same as the pressure fore per area) in the waves if they are absorbed by some surfae. Multiplying the momentum volume density g z by the speed of light onverts it into the flux of momentum, or pressure in the z diretion. Note: the solution an also be found by doing the time averaging of harmoni fields, inluding a omplex onjugate on one quantity of a quadrati pair, and a fator of /2. b) 6) If the given wave is inident on a perfetly onduting mirror at normal inidene, determine the radiation pressure on the mirror, in terms of I 0 and in N m 2. If the waves are refleted from a perfet ondutor, the wave momentum is reversed, whih reates a pressure equal to twie the momentum flux. The intensity is the time averaged Poynting vetor S = E H, whih is half of its peak value, Then the pressure produed by total refletion is The number is E 0 I = S ˆn = 2 E 0H 0 = 2 E 0 = E2 0. µ 0 2µ 0 E H P = 2 g z = 2 2 = 2I. P = )2 W m m s = 3.20 N m 2. 5

6 ) 2) How is the answer to part b) hanged if the angle of inidene is θ = 60? If the waves are inident at angle θ, then the omponent of momentum perpendiular to the surfae is proportional to osθ. In addition, the waves are inident on an area that is inreased by the fator / osθ. The pressure is thus hanged by the fator os 2 θ, P = 2 g z osθ / osθ = 2I os2 θ. For the ase osθ = os60 = 2, the pressure is redued to 4 P = 0.80 N m 2. of that at normal inidene, then You ould also arrive at this result by finding T nn where n refers to the omponent perpendiular to the refleting surfae. 6

7 3. 48) The imaginary part of a dieletri funtion is known to be ɛ I ω) = γω ω 2 + γ 2. a) 6) Apply the Kramers-Kronig relations to obtain the real part of ɛω). Use the simplest form, with the prinipal valued integral, losing the ontour in the upper half plane: ɛ R ω) = π P dz ɛ Iz) z ω = π P dz z ω = [ ] π [πi resω) + 2πi resiγ)] = i γω ω 2 + γ 2 + 2γ iγ iγ ω)2iγ) ɛ R ω) = + γ2 ω 2 + γ 2 γz z 2 + γ 2 = γ 2 ω 2 + γ 2 b) 6) Sketh the real and imaginary parts of ɛω) vs. ω, and identify the regions of normal and anomalous dispersion. ) 6) Determine the omplex funtion ɛω) and loate its poles in the omplex ω-plane. Do its poles our in the region assoiated with ausal response? Adding the real and imaginary parts give easily ɛω) = ɛ Rω) + i ɛ I ω) = + γ2 ω 2 + γ 2 + i ɛω) = + iγω iγ) ω 2 + γ 2 = + iγ ω + iγ. γω ω 2 + γ 2 The Only pole is at ω = iγ, in the lower half plane. Thus, ɛω) is analyti in the upper half plane, as required by ausality. Part B Sore = 40 /40 7