ECE 6340 Intermediate EM Waves Fall 2013 Prof. David R. Jackson Dept. of ECE Notes 6 1 Power Dissipated by Current Work given to a collection of electric charges moving in an electric field: W q E E This could be conduction current or impressed current. vS E Note: The magnetic field never does any work: q v B dr v J vv S 0 v t q v S Power dissipated per unit volume: Pd v E v v E J E S t t W Note: We assume no increase in kinetic energy, so the power goes to heat. 2 Power Dissipated by Current (cont.) Power dissipated per unit volume for ohmic current (we assume here a simple linear media that obeys Ohm’s law): Pd J c E E E E E E Note: E E 2 2 Pd E 2 E 2 E E 2 From this we can also write the power generated per unit volume due to an impressed source current: i Ps J E 3 Poynting Theorem: Time-Domain E M H J B t D t From these we obtain H E M H H E H J E E B t D t Subtract, and use the following vector identity: H E E H E H 4 Poynting Theorem: Time-Domain (cont.) We then have E H M H J E H B t E D t J J E i Now let M M i so that E H J i E E M 2 i H E D t H B t 5 Poynting Theorem: Time-Domain (cont.) Next, use D E E E t t and E D E (simple linear media) E E E 2E t t t 2 1 E E D E E t 2 t t 2 Hence And similarly, 1 H H t 2 t B 2 6 Poynting Theorem: Time-Domain (cont.) S EH Define We then have S J E E M i 2 i 1 1 2 H E H t 2 2 2 Next, integrate throughout V and use the divergence theorem: S S nˆ dS V i i J E M H dV E dV 2 V 1 1 2 E H t V 2 2 2 dV Note: We are assuming the volume to be stationary (not moving) here. 7 Poynting Theorem: Time-Domain (cont.) S S nˆ dS i i J E M H dV E dV 2 V V 1 1 2 E H t V 2 2 2 dV Interpretation: We 1 E dV sto re d e le ctric e n e rg y 1 H 2 2 V Wm 2 2 dV sto re d m a g n e tic e n e rg y V Pd E dV d issip a te d p o w e r 2 V Ps i J E M i H dV so u rce p o w e r V Pf S nˆ dS p o w e r flo w in g o u t o f S S (see next figure) 8 Poynting Theorem: Time-Domain (cont.) S S nˆ dS V i i J E M H dV E dV 2 V P f Ps Pd t 1 1 2 E H t V 2 2 2 dV We Wm or Ps Pd Pf t We Wm 9 Poynting Theorem: Time-Domain (cont.) Ps Pd Pf t We Wm Power flow out of surface E , , H source 10 Poynting Theorem: Note on Interpretation S EH Does the Poynting vector really represent local power flow? S J E E M i Consider: 2 i 2 S S A A Note that 1 1 2 H E H t 2 2 A rb itra ry v e cto r fu n ctio n S S This “new Poynting vector” is equally valid! They both give same TOTAL power flowing out of the volume, but different local power flow. 11 Note on Interpretation (cont.) Another “dilemma” A static point charge is sitting next to a bar magnet S EH 0 N Is there really power flowing in space? q Note: There certainly must be zero net power out of any closed surface: S Ps Pd Pf t We Wm 12 Note on Interpretation (cont.) Bottom line: We always get the correct result if we assume that the Poynting vector represents local power flow. Because… In a practical measurement, all we can ever really measure is the power flowing through a closed surface. Comment: At high frequency, where the power flow can be visualized as "photons" moving in space, it becomes more plausible to think of local power flow. In such situations, the Poynting vector has always given the correct result that agrees with measurements. 13 Complex Poynting Theorem E M j H i Frequency domain: H J j c E i Hence (generalized linear media) H E M H j H i * * E H EJ 2 * i* j c E * 2 Subtract and use the following vector identity: A B B A A B EH * H E E H * * Hence EH * i * M H E J i* j H 2 j c E * 2 14 Complex Poynting Theorem (cont.) S Define 1 EH * (complex Poynting vector) 2 Note: R e S 1 Re E H 2 * EH S Then S EJ 2 1 i* i M H * 2 1 j * c E 2 H 2 Next use c c j c j c j c * c Next, collect real and imaginary parts in the last term on the RHS. 15 Complex Poynting Theorem (cont.) S 2 EJ EJ 1 i* i M H * 1 2 j c E 2 H 2 1 2 c E 2 H 2 or S 1 2 i* i M H * 1 2 j c E 4 2 1 4 H 2 1 c E 2 2 H 2 Next, integrate over a volume V and apply the divergence theorem: S S nˆ dS V 1 2 E J 1 c E 2 V i* 2 i M H 1 2 * H 1 dV 2 j c E 4 V 2 2 1 4 H 2 dV dV 16 Complex Poynting Theorem (cont.) Final form of complex Poynting theorem: 2 E J 1 V i* i M H * dV S nˆ dS S 1 2 j H 4 V 1 c E 2 V 2 2 1 4 1 2 2 c E dV H 2 dV 17 Complex Poynting Theorem (cont.) Interpretation of Ps : Ps V R e Ps 2 1 i* EJ E J i i M H M i * H dV dV V Ps We therefore identify that Ps com plex source pow er 18 Complex Poynting Theorem (cont.) Interpretation of Pf : Pf S R e Pf 1 2 E H EH * nˆ dS nˆ d S S Pf We therefore identify that P f co m p le x p o w e r flo w in g o u t o f S 19 Complex Poynting Theorem (cont.) Interpretation of energy terms: 1 4 c E 2 1 * c E E 2 2 1 2 1 c 2 1 2 Hence 1 Re E E c E E 1 4 c E 2 dV V Similarly, 1 4 H V * 1 2 Note: The real-part operator may be added here since it has no effect. c E Note: We know this result represents stored energy for simple linear media, so we assume it is true for generalized linear media. 2 1 2 E c 2 dV We V 2 dV 1 2 c H 2 dV Wm V 20 Complex Poynting Theorem (cont.) Interpretation of dissipation terms: 1 2 c E 2 1 * c R e E E 2 c E E c E c j 2 real constant For simple linear media: c Hence 1 2 V 2 c E dV E 2 dV Pd e This is the time-average power dissipated due to conducting losses. We assume the same interpretation holds for generalized linear media. V 21 Complex Poynting Theorem (cont.) Interpretation of dissipation terms (cont.): Similarly, 1 2 H 2 dV Pd m V This is the time-average power dissipated due to magnetic losses. 22 Complex Poynting Theorem (cont.) Summary of Final Form V 2 1 EJ i* i M H * dV S nˆ dS S 1 c E 2 V 2 1 2 j H 4 V 1 H 2 2 Ps P f Pd j 2 Wm We 1 4 2 dV 2 c E dV 23 Complex Poynting Theorem (cont.) Ps P f Pd j 2 Wm We Pabs We can write this as Ps P f Pabs where we have defined a complex power absorbed Pabs : R e Pabs Pd Im Pabs 2 Wm We 24 Complex Poynting Theorem (cont.) Ps P f Pabs This is a conservation statement for complex power. Im Pabs 2 Wm We R e Pabs Pd VARs Watts Pf E Complex power flow out of surface VARS consumed Power (watts) consumed H c, Source 25 Example L I + V - Ideal inductor Denote: Pabs com plex pow er absorbed W abs jQ abs Calculate Pabs using circuit theory, and verify that the result is consistent with the complex Poynting theorem. Note: Wabs = 0 (lossless element) 26 Example (cont.) L I + V - Pabs 1 V I * 2 1 2 1 2 1 Z I I * j L I I j L I Ideal inductor 2 * Note: Pabs jQ abs 2 27 Example (cont.) Q abs Im Pabs 1 L I 2 2 1 * L Re I I 2 L i 2 Wm 2 1 2 2 Li 2 28 Example (cont.) Since there is no stored electric energy in the inductor, we can write Q abs 2 Wm We Hence, the circuit-theory result is consistent with the complex Poynting theorem. Note: The inductor absorbs positive VARS. 29 Example We use the complex Poynting theorem to examine the input impedance of an antenna. Pf Js Iin V0 + - S Antenna Model: Iin V0 + - Pin 1 1 Zin=Rin+j Xin 2 2 V 0 I in * Z in I in 1 2 Z in I in I in* 2 30 Example (cont.) so Z in 2 Pin I in Real part: 2 R in 2 R e Pin I in 2 R e Pin Pin Prad (no losses in vacuum surrounding antenna) Note: The far-field Poynting vector is much easier to calculate than the nearfield Poynting vector. R e Prad Hence R in 2 I in 2 R e S nˆ dS S 31 Example (cont.) R in 2 I in 2 R e S nˆ dS S In the far field (r ) Im S 0 (This follows from plane-wave properties in the far field.) Hence R in 2 I in 2 S nˆ dS S 32 Example (cont.) 2 Imaginary part: X in I in 2 Im Pin where Im Pin Im Prad 2 Wm We Hence X in 2 I in 2 Im S nˆ dS S Wm , We 33 Example (cont.) X in 2 I in 2 2 W We m Hence X in 4 I in 2 1 4 0 H V 2 1 4 2 0 E dV However, it would be very difficult to calculate the input impedance using this formula! 34 Dispersive Material The permittivity and permeability are now functions of frequency: (A lossy dispersive media is assumed here. It is assumed that the fields are defined over a fairly narrow band of frequencies.) The formulas for stored electric and magnetic energy now become: We 1 Wm 1 E 4 4 H 2 2 Note: The stored energy should always be positive, even if the permittivity or permeability become negative. Reference: J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 263). 35 Momentum Density Vector The electromagnetic field has a momentum density (momentum per volume): p D B In free space: p 0 0 E H or p 1 c 2 S [(kg m/s)/m3] = [kg/(s m2)] Reference: J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 263). 36 Momentum Density Vector (cont.) Photon: From physics, we have a relation between the energy E and the momentum p of a single photon. E pc E hf h 6.626068 10 34 [J s] (Planck’s constant) Photons moving: Calculation of power flow: Sz E nergy A t E A c t N c t p A t Ec N p pc N p 2 pz c pN c S S z zˆ 2 p Np photons per unit volume 2 Hence v A pz 1 c 2 Sz This is consistent with the previous momentum formula. 37 Momentum Density Vector (cont.) Example: Find the time-average force on a 1 [m2] mirror illuminated by normally incident sunlight, having a power density of 1 [kW/m2]. A = 1 m2 pz Fz inc 2 1 c inc pz 2 Sz c Ac t t 1 inc 2 2 z (1000) inc z p Ac 2000 1 2 2 (1000) 1 c c c F z 6.671 10 6 [N ] 38 Solar Sail Sunjammer Project NASA has awarded $20 million to L’Garde, Inc. (Tustin, CA), a maker of “inflatable space structures,” to develop a solar sail, which will rely on the pressure of sunlight to move through space when it takes its first flight as soon as 2014. A smaller version of L'Garde's solar sail unfurled in a vacuum chamber in Ohio in 2005. This one was about 3,400 square feet, a quarter the size of the sail the company plans to loft into space as soon as 2014. Photo courtesy NASA and L'Garde. http://www.lgarde.com 39 Solar Sail (cont.) http://www.lgarde.com 40 Maxwell Stress Tensor This gives us the stress (vector force per unit area) on an object, from knowledge of the fields on the surface of the object. References: D. J. Griffiths, Introduction to Electrodynamics, Prentice-Hall, 1989. J. D. Jackson, Classical Electrodynamics, Wiley, 1998. J. Schwinger, L. L. DeRaad, Jr., K. A. Milton, and W.-Y. Tsai, Classical Electrodynamics, Perseus, 1998. 41 Maxwell Stress Tensor T xx T T yx T zx T xy T yy T zy T xz T yz T zz 1 Tij 0 EE 0 H i H j ij 0 E i j 2 2 1 2 0 H 2 i, j x, y , z (for vacuum) 1, i j ij 0, i j 42 Maxwell Stress Tensor (cont.) Momentum equation: T nˆ dS F S Total flow rate of momentum given to object from EM field d dt p dV V Total force on object (rate of change of mechanical momentum) Rate of change of electromagnetic momentum inside of region F p 1 c 2 EH 43 Maxwell Stress Tensor (cont.) T nˆ dS F S d dt p dV V In many practical cases the time-average of the last term (the rate of change of electromagnetic momentum inside of region) is zero: No fields inside the body Sinusoidal steady-state fields In this case we have F T nˆ dS S The Maxwell stress tensor (matrix) is then interpreted as the stress (vector force per unit area) on the surface of the body. 44 Maxwell Stress Tensor (cont.) Example: Find the time-average force on a 1 m2 mirror illuminated by normally incident sunlight, having a power density of 1 [kW/m2]. x nˆ zˆ TEMz wave A = 1 m2 S F z zˆ T nˆ T xx F T nˆ T yx T zx T xy T yy T zy Note: The fields are assumed to be zero on the back side of the mirror, and no fields are inside the mirror. T xz 0 T yz 0 T zz 1 F z T zz z T xz T yz T zz 45 Maxwell Stress Tensor (cont.) x nˆ zˆ A = 1 m2 TEMz wave T zz 1 0 Ez Ez 0 H z H z 0 E 2 1 0 E 2 2 1 2 0 H 2 2 z 1 2 0 H 2 inc inc y US Assume E Ex xˆ , H H y yˆ Ex /H 0 46 Maxwell Stress Tensor (cont.) 1 1 2 F z 0Ex 0 H 2 2 1 2 Fz 1 2 0H 0 H 2 y 2 y 1 * 0 Re H y H y 2 2 1 0 Re H 2 2 1 4 0 H y (PEC mirror) 2 y 1 1 2 y Fz 1 1 4 4 0 H y 2 0 2 H inc y 2 (The tangential magnetic field doubles at the shorting plate.) 2 47 Maxwell Stress Tensor (cont.) Fz 1 4 0 2 H inc y 2 0 H inc y 2 Next, use inc Sz Fz 0 H inc y inc inc y 1 2 inc inc * Ex H y 1 2 inc Hy 2 0 2 0 2 S z 4 10 Ex H inc 7 / 0 2 1000 / 376.7303 F z 6.671 10 6 [N ] 48 Foster's Theorem Consider a lossless system with a port that leads into it: Z in jX in Lossless system (source free) Sp c Coaxial port PEC enclosure z dX in d 0 R. E. Collin, Field Theory of Guided Waves, IEEE Press, Piscataway, NJ, 1991. 49 Foster's Theorem (cont.) The same holds for the input susceptance: 1 Z in Yin 1 jX in jB in X in dX in d 1 B in 1 dB in B in d 2 dB in d 0 50 Foster's Theorem (cont.) Start with the following vector identity: A B B A A B Hence we have (applying twice, for two different choices of the (vectors) * * H H E * H E E * * E* E E H H H Add these last two equations together: 51 Foster's Theorem (cont.) * * H * H E E H * H E E * * E E H H Using Maxwell's equations for a source-free region, * * H * H E E H * H j H E * * E E H j E 52 Foster's Theorem (cont.) * * H * H E E H * H j H E * * E E H j E Using Maxwell's equations again, and the chain rule, we have: H E * * H E * * j E * j H j * j E * H * j j E * H * We then have 53 Foster's Theorem (cont.) * * H * H E E H * E * E j j H E j * * H E * H j H j j E cancels Simplifying, we have * * H E * * E H E j E H j H or * * H E E H j E 2 H 2 54 Foster's Theorem (cont.) * * H E E H j E 2 H 2 Applying the divergence theorem, S * * H E H nˆ dS j 4 We Wm E Therefore, Im S * * H E H nˆ dS 0 E 55 Foster's Theorem (cont.) Im S * * H E H nˆ dS 0 E The tangential electric field is only nonzero at the port. Hence we have: * * H E Im E H nˆ dS 0 Sp Assume that the electric field (voltage) at the port is fixed (not changing with frequency). Then we have E * 0 Z in jX in Lossless system (source free) c Sp Coaxial port z 56 Foster's Theorem (cont.) * H Im E Sp Then nˆ dS 0 At the coaxial port: Z in jX in Lossless system (source free) E ˆ E c nˆ H ˆ H Sp zˆ Coaxial port z Hence: * H Im E dS 0 Sp or Im Sp E H * dS 0 (since the electric field is fixed and not changing with frequency) 57 Foster's Theorem (cont.) Im E H dS 0 Sp Im 2 Pz 0 Z in jX in Lossless system (source free) c Im V I 0 * Im V * z 2 Yin 0 * Iz Sp Coaxial port z V Y in * Im Yin 0 58 Foster's Theorem (cont.) * Im Yin 0 Im Yin 0 B in 0 X in 0 59 Foster's Theorem (cont.) Example: Short-circuited transmission line Zin Z0 ZL = 0 l Z in jZ 0 tan l LC Xin inductive π/2 3π/2 5π/2 l capacitive 60