Physics 14 Midterm Exam Solutions Winter 017 1. A linearly polarized electromagnetic wave, polarized in the ˆx direction, is traveling in the ẑ-direction in a dielectric medium of refractive index n 1. The wave is normally reflected from the surface of a conductor of conductivity σ (the conductor occupies the x y plane). Assume that µ = µ 0 for both the dielectric and the conductor. (a) Find the phase change undergone by the electric field vector of the wave after reflection, assuming the refractive index of the conductor of n = n 1 (1+iζ), where ζ > 0. This problem is a simplified version of Jackson, problem 7.4, whose solution was given in Solution Set 1. But it is simple enough to derive the necessary results from Section 7.3 of Jackson directly. Without loss of generality, we define the z-axis to lie along the direction of the incoming wave and the x-axis to lie along the polarization vector of the incoming wave. Thus, we can take the incoming, transmitted and reflected waves to be, E = ˆxe i(kz ωt), E = E 0ˆxei(k z ωt), E = E 0ˆxe i(kz+ωt), (1) respectively, where the corresponding indices of refraction are ǫi µ i ǫi n i = =, for i = 1,, () ǫ 0 µ 0 ǫ 0 and k = n 1 ω/c, k = n ω/c. InthenotationofSection7.3ofJackson, thecaseofnormalincidenceofthewavecorresponds to k = k = n = ẑ, k = ẑ, = ˆx, E 0 = ˆx and E 0 = ˆx. Using the last two equations of eq. (7.37) of Jackson, it then follows that +E 0 = E 0, n 1 ( E 0) = n E 0. Eliminating, we can immediately solve for E 0 /, E 0 = n 1 n n 1 +n = iζ +iζ = ζ(ζ +i) ζ +4 = after using n = n 1 (1+iζ), where the phase α is given by 1 eiα, (3) tanα = Im(E 0 /) Re(E 0/ ) = ζ. (4) Note that the complex number E 0 / given in eq. (3) lies in the third quadrant of the complex plane. Consequently, eq. (4) yields α = tan 1 (/ζ) π, where the principal value of the arctangent satisfies 0 tan 1 (/ζ) < π for ζ > 0. 1 In the standard convention, we take π < α π as the principal value of the argument of E 0 /.
Using the reflected wave given in eq. (1), E = E 0ˆxe i(kz+ωt) = ˆxe iα e i(kz+ωt). Hence, the relative phase of the incident and reflected wave at the interface (z = 0) is φ rel = α = tan 1 (/ζ) π. (5) (b) How is ζ related to the conductivity σ in the limit of high frequency (i.e., in the limit of ω σ/ǫ 0 )? The indices of refraction are given in eq. (). Following the solution to Jackson, problem 7.4, which is given in Solution Set 1, we identify ǫ = ǫ 1 + iσ ω. In the limit of ω σ/ǫ 0, we can approximate, ǫ ǫ1 n = n 1 (1+iζ) = = + iσ ǫ 0 ǫ 0 ωǫ 0 Thus, we identify ζ = σ ωǫ 1. ǫ1 ǫ 0 1+ iσ. ωǫ 1. A classical point magnetic dipole moment µ at rest has a vector potential (in gaussian units), which is given by A = µ r r 3 and no scalar potential (Φ = 0). Suppose that the point magnetic dipole moment µ now moves with velocity v. (a) Compute and compare the interaction energy between the moving magnetic dipole and static external fields E and B and the interaction energy computed in the rest frame of the magnetic moment. Explain why the two results agree (or disagree). Assume that v c and keep only terms up to O(v/c). Following part (b) of problem 11.7 of Jackson, we define K to be the rest frame of the magnetic dipole with dipole moment µ. In frame K, the magnetic dipole (or equivalently frame K ) is moving at velocity v = βc. In the solutions to problem 11.7, we showed that to order O(v/c), the magnetic dipole moment in frame K is µ = µ. In addition, there is also an electric dipole moment in frame K given by p = β µ. The relative phase φ rel is defined modulo π, so feel free replace π with π in eq. (5) if so inclined.
To compute the interaction energies, we need expressions for the electric and magnetic fields in both reference frames. Using eq. (11.149) of Jackson, E = γ( E + β B) γ γ +1 β( β E), B = γ( B β E) γ γ +1 β( β B). Keeping only terms up to O(v/c), we have E = E + β v B +O, B = B β v E +O. c In frame K, the interaction energy is cf. eqs.(4.4) and (5.7) of Jackson: U = p E µ B = ( β v µ) E µ B +O. Thefirst termontherighthandsideabovecanberewrittenwiththehelpofvector identities, ( β µ) E = ( µ β) E = µ ( β E). c c Hence, the interaction energy in frame K is given by U = µ ( B β v E)+O c. (6) In frame K, the interaction energy is given by U = µ B = µ ( B β v E)+O, where we have used µ = µ and B = B β E, where terms of order O(v /c ) have been neglected. The two computations agree to first order in v/c. The interaction energy is not a Lorentz invariant. Indeed, we expect that U = γu. But if we drop terms of O(v /c ), then γ 1 in which case U U as obtained above. c (b) Compute the exact expression for the scalar potential generated by a point magnetic dipole µ moving with velocity v. (Do not assume that v c.) Express your answer in terms of the angle between ˆn and v, where ˆn is a unit vector pointing from the magnetic dipole to the point at which the scalar potential is measured. Thefour-vectorpotentialisA µ = (Φ; A)transformsjustlikeanyfour-vectorcf.eq.(11.19) of Jackson, Φ = γ(φ β A), A = A+ (γ 1) β ( β A) β γβφ.
x (x,y,z) O x vt R ψ µ z Figure 1: A magnetic dipole is moving at constant velocity v = βc in the z-direction as seen from reference frame K. The origin of the laboratory frame K is denoted by O. The angle ψ is defined so that ˆv ˆR = cosψ. By convention, we take 0 ψ π. Note that the specific choice of x, y and z axes is arbitrary and not used in the solution to this problem. The inverse transformation is obtained simply by changing the sign of β, Φ = γ(φ + β A ), A = A + (γ 1) β ( β A ) β +γβφ. (7) In the rest frame, we have a magnetic dipole moment but no electric dipole moment, which corresponds to A = µ r r 3, Φ = 0, (8) where r r. The coordinate r in frame K is related to the coordinate r in frame K by eq. (11.19) of Jackson, r (γ 1) = r + ( β β r) β γβct, (9) where t is the time measured in frame K. The moving magnetic dipole as seen from the laboratory frame K is shown in Figure 1. It is convenient to define R to be the vector in frame K that points from the location of the magnetic dipole at time t to the location of the observer. It follows that R = r vt, (10) where R Rˆn and ˆn is the unit vector that points in the direction of R. Using eq. (9), it follows that r = R+ (γ 1) ( β β R) β. Note that β ( µ r ) = β ( µ R), since β ( µ β) = 0. Hence, eqs. (7) and (8) yield Φ = γ β A = γ β ( µ r ) r 3 = γ β ( µ R) r 3. (11)
The last step involves relating r to R R. In eq. (13) of the class handout entitled, The electromagnetic fields of a uniformly moving charge, I showed that r 3 = γ 3 R 3 (1 β sin ψ) 3/, (1) where ψ is the angle between v and R as seen in frame K (cf. Figure 1). Inserting eq. (1) into the denominator of eq. (11) and using v = c β and R = Rˆn, we end up with Φ = c v ( µ ˆn) γ R (1 β sin ψ) 3/. Notethatwe have expressed our answer intermsofthemagnetic dipolemoment asmeasured in the rest frame K. As discussed in the solution to Jackson problem 11.7 (in Solution Set ), the exact relationship between µ and µ is somewhat complicated, although as noted in part (a) above, the two dipole moments coincide if terms of O(v /c ) are dropped. 3. A magnetic dipole m undergoes precessional motion with angular frequency ω and angle ϑ 0 with respect to the z-axis as shown below. That is, the time-dependence of the azimuthal angle is ϕ 0 (t) = ϕ 0 ωt. z Electromagnetic radiation is emitted by the precessing dipole. (a) Write out an explicit expression for the time-dependent magnetic dipole vector m in terms of its magnitude m 0, the angles ϑ 0 and ϕ 0 and the time t. Show that m consists of the sum of a time-dependent term and a time-independent term. Verify that the time-dependent term can be written as Re( µe iωt ), for some suitably chosen complex vector µ. Using the figure shown above, the magnetic dipole moment vector is given by: m = m 0 ˆxsinϑ 0 cos(ϕ 0 ωt)+ŷsinϑ 0 sin(ϕ 0 ωt)+ẑcosϑ 0 = Re m 0 sinϑ 0 e iϕ 0 (ˆx iŷ)e iωt +m 0 cosϑ 0 ẑ. (13) Thus, we can write the time-dependent term of m as Re( µe iωt ), where µ = m 0 sinϕ 0 e iϕ 0 (ˆx iŷ). (14)
(b) Compute the angular distribution of the time-averaged radiated power, with respect to the z-axis defined in the above figure. The angular distribution of the time-averaged power is given by eq. (9.1) of Jackson in SI units, dp dω = 1 Re r ˆn E H. The magnetic and electric fields of the magnetic dipole are given by eqs. (9.35) and and (9.36) of Jackson. Keeping only the leading terms of O(1/r), we see that H = 1 Z 0 E ˆn, where Z 0 = µ 0 /ǫ 0 is the impedance of free space. It follows that ˆn E H = 1 Z 0ˆn E ( E ˆn) = 1 Z 0 E E ˆn = 1 Z 0 E, since E ˆn = 0 (due to the transverse nature of electromagnetic radiation). Hence, dp dω = r Z 0 E, (15) where the leading O(1/r) term of eq. (9.36) of Jackson, applied to the complex magnetic moment vector µ, yields E = Z 0 4π k ( n µ) eikr r. (16) Inserting this result into eq. (15), we end up with dp dω = Z 0 3π k4 ˆn µ. (17) The squared magnitude of the cross product above is easily computed, ˆn µ = (ˆn µ) (ˆn µ ) = µ ˆn µ, since ˆn is a unit vector. Explicitly, µ is given by eq. (14) and Hence, it follows that. ˆn = sinθcosφ ˆx+sinθsinφŷ +cosθẑ. µ = m 0 sin ϑ 0, ˆn µ = m 0 sinϑ 0 sinθ, and ˆn µ = m 0sin ϑ 0 ( sin θ) = m 0sin ϑ 0 (1+cos θ).
Thus, the angular distribution of the time-averaged radiated power is given by 3 dp dω = Z 0m 0sin ϑ 0 k 4 (1+cos θ). (18) 3π An alternative technique for computing the angular distribution of the time-averaged radiated power which makes use of the real physical magnetic dipole moment is given in the Appendix to these exam solutions. (c) Compute the total power radiated. Integrating eq. (18) over solid angles, dω(1+cos θ) = π Hence, 1 1 P = Z 0m 0k 4 sin ϑ 0 6π (1+cos θ)dcosθ = 16π 3. (19). (0) (d) What is the polarization of the radiation measured by an observer located along the positive z-axis far from the precessing dipole? How would your answer change if the observer were located in the x y plane? The polarization is determined from the electric field given in eq. (16). Thus, we must evaluate ˆn µ, ˆx ŷ ẑ ˆn µ = det sinθcosφ sinθsinφ cosθ m 0 sinϑ 0 e iϕ 0 im 0 sinϑ 0 e iϕ 0 0 = im 0 e iϕ 0 sinϑ 0 cosθ(ˆx iŷ) im 0 sinϑ 0 sinθe i(ϕ 0 φ)ẑ. (1) The polarization depends on the location of the observer. If the observer is located on the positive z-axis then θ = 0. In this case, ˆn = ẑ and E ˆx iŷ, which corresponds to right-circularly polarized light cf. p. 300 of Jackson. If the observer is located in the x y plane, the θ = 1 π. In this case, ˆn = ˆxcosφ + ŷsinφ and E ẑ, which corresponds to linearly polarized light in the z-direction. REMARK: If θ = π, then ˆn = ẑ and E ˆx iŷ, which corresponds to left-circularly polarized light. For any other value of θ 0, 1 π or π, the radiation is elliptically polarized. 3 To obtain the angular distribution of the time-averaged radiated power in gaussian units, one must replace Z 0 4π/c and m 0 m 0 c in eq. (18).
APPENDIX: An alternative method for solving problem 3(b) Instead of evaluating eq. (17), which requires the complex magnetic moment µ given in eq. (14), one can instead employ the result of problem 9.7(a) of Jackson, dp(t) dω = Z 0 16π c 4.. m ˆn, ().. where m d m/dt,and misthetime-dependentmagneticdipolemomentgivenineq.(13). Note that eq. () yields the time dependent power distribution, so to recover the results obtained in problem 1(b), we must time-average over one cycle. For convenience, we rewrite eq. (13) here: m = m 0 ˆxsinϑ 0 cos(ϕ 0 ωt)+ŷsinϑ 0 sin(ϕ 0 ωt)+ẑcosϑ 0 Taking two time derivatives, we obtain:.. m = m 0 ω ˆxsinϑ 0 cos(ϕ 0 ωt)+ŷsinϑ 0 sin(ϕ 0 ωt). (3) Next, we compute the square of the cross product, after using the fact that ˆn is a unit vector, Using eqs. (3) and (4), it follows that.. m ˆn =.. m.. m (ˆn.. m), ˆn = ˆxsinθcosφ+ŷsinθsinφ+ẑcosθ. (4).... m m = m 0 ω4 sin ϑ 0, and.. ˆn m = m 0 ω sinϑ 0 sinθ cosφcos(ϕ ωt)+sinφsin(ϕ 0 ωt) = m 0 ω sinϑ 0 sinθcos(ωt ϕ 0 +φ). Hence,.. m ˆn = m 0 ω4 sin ϑ 1 sin θcos (ωt ϕ 0 +φ). Inserting the above result into eq. () and using ω = kc, we end up with dp(t) dω = Z 0m 0 sin ϑ k 4 1 sin θcos (ωt ϕ 16π 0 +φ). (5) Time-averaging over one cycle, cos (ωt ϕ 0 +φ) = 1. Since 1 1 sin θ = 1 (1+cos θ), we recover eq. (18). One can also check that the total power obtained by integrating eq. (5) over solid angles is time-independent and coincides with eq. (0).