Electricity and Magnetism-Poynting Vector and its Applications

Topics: 􀂄 􀂄 􀂄 􀂄 8.022 (E&M) – Lecture 21 Energy and momentum carried by EM waves Poynting vector Transmission lines Scattering of light and sunset demo… 2 Last time 􀂄 􀂄 Solution of wave equation combination of plane waves: 􀂄 Properties 􀂄 􀂄 i􀂄 􀂄 􀂄 0 􀂄 Circular: when the vector E0 irclime 􀂄 ii􀂄 2 2 2 2 1 ∂∇ = ∂ 􀁇􀁇 􀁇 EE c t ˆ( )fr ckt±􀁇 0 0); )E E B tω ω= − = − 􀁇 􀁇􀁇 􀁇 􀁇 􀁇􀁇 􀁇i i G. Sciolla – MIT 8.022 – Lecture 21 Solution of Maxwell’s equations in vacuum can be expressed as linear of plane waves: They travel at the speed of light //to k (wave vector) E, B and k are always perpendcular to each other Amplitude of E and B are the same in cgs Polarization of EM waves Linear: when the direction of Eis constant in time describes a ce over tElliptcal: all the situatons in between these 2 cases Today we will complete the study of these properties… sin( sin( kr t B kr 1G. Sciolla – MIT 8.022 – Lecture 21 . 3 EM Energy 􀂄 EM radiation carries energy 􀂄 􀂄 How does this energy propagate? 􀂄 . V Surface A E B 􀁇 􀁇 V V 1 ( )8 1Tl(8 energy EEvolume EE π π = + = = +∫ ∫ 􀁇􀁇 i i 􀁇􀁇 i i G. Sciolla – MIT 8.022 – Lecture 21 Obvious if you think about the fact that is the light from the sun that keeps us warm… Consider a volume V of surface A containing E and B Energy density: u= ota energy: U udV )dV BB BB 􀁇􀁇 􀁇􀁇 4 􀂄 iime? V V V 1 1( (8 4 1 1and B= c c ( B E4 ( ) -( ) ( U E BEE E B t t t t B E t t U c E B t EB E B B π π π ∂ ∂ ∂ ∂ = + = + ∂ ∂ ∂ ∂ ∂ ∂∇× ∇× ∂ ∂ ∂ ⇒ = ∂ ∇ × = + ∫ ∫ ∫ 􀁇 􀁇􀁇􀁇 􀁇 􀁇 i i i i 􀁇 􀁇􀁇 􀁇􀁇 􀁇 􀁇􀁇􀁇 􀁇 i i 􀁇 􀁇 􀁇 􀁇􀁇 􀁇 􀁇 i i i ( ) ) as B4 B4 V V E U c E S t cS E π π ∂ =− ∇ ×⇒ ≡− ∇ ∂ ≡ × ∫ ∫ 􀁇􀁇 􀁇 i i 􀁇 The Poynting vector How does total derivatve change over t)dV )dV Remembering that in vacuum: E=-)dV Remembering that BB ∇× −∇× ∇× ∇× 􀁇􀁇 􀁇􀁇 􀁇 􀁇 where we defined the dV dV Poynting vector 􀁇 􀁇 􀁇 􀁇 25 􀂄 Given: 􀃆 l􀂄 􀃆 􀂄 ing vector: 􀂄 􀂄 Remember that 􀂄 dV () es SAV U Sda t U AS t ∂ =− =−∂ → ∂ Φ ∂∫ ∫ 􀁇 􀁇􀁇 i 􀁇 􀁇i B4 cS Eπ≡ × 􀁇 2 0 0 0 ˆ× = 􀁇 􀁇 􀁇 E B E k A Sda∫ 􀁇 􀁇i G. Sciolla – MIT 8.022 – Lecture 21 Interpretation of Poynting vector The rate of change of EM energy in the volume V is given by the fux of the Poynting vector S through the surface A Minus sign: dA points outward U increases when S is opposite to dA Interpretation of Poyntpoints in the direction of the EM energy flow The flux of S through a surface gives the power through A Stok=− ∇ 􀁇 􀁇 Power through A: Poynting vector: dimensional analysis 􀂄 What are the units of the Poynting vector? 􀁇 􀁇 􀁇 ⎤ cgs⎣⎦ ⎢⎣4π⎦ cBE =cE⎡⎤=⎡cE ×B ⎥=⎡⎤⎡ ⎤⎡ ⎤ ⎡⎤⎡ ⎤ 2S ⎣⎦⎣ ⎦⎣ ⎦ ⎣⎦⎣ ⎦ Lenght c⎡⎤=⎣⎦ Time1 􀁇􀁇 􀁇􀁇Fro m u= (EE +B B) ⇒⎡⎤2 = Energyi E8π i ⎣⎦ Volume 􀁇 ⇒⎡⎤= Lenght Energy = Energy PowerS = ⎣⎦ Time Volume Time Area Area 􀂄 Expected if the flux of S is the power through area A 􀂄 In cgs: [S]=erg s-1 cm-2 􀂄 NB: Magnitude of S is know as Intensity I 􀂄 Intense source of radiations emit a lot of power per unit area G. Sciolla –MIT 8.022 – Lecture 21 6 3 8.022 – Lecture 21 7 Applications: plane waves 􀂄 􀂄 􀂄 􀂄 􀃆 􀂄 i14 Hz) 􀃆 all that matters is 0 0 ˆ) ˆ) ω ω ⎧= −⎪⎨ = −⎪⎩ 􀁇 􀁇 E E B B 2 2 0 ˆE si)4 4 c cS B E kz wπ π≡ × = − 􀁇 2 2 0 1 1 u= ( ) )8 4EE E kz tωπ π+ = −i i ˆS uc uck=⇒ = 􀁇 􀁇 J vρ= 􀁇 􀁇 2 2 0 0 ˆ S= ;8 8 ck cE I Eπ π = 􀁇 G. Sciolla – MIT Consider a linearly polarized plane wave: Poynting vector associated with it: This can be compared to the energy density of the wave: This is similar to another way to show that S tells us about the flow of energy! Usually the oscillation is very fast (e.g.: vsible~10the average energy density and intensity : cos( cos( kz t x kz t y n ( t k 􀁇 􀁇 sin ( BB 􀁇􀁇 􀁇􀁇 8 Application 2: Dipole radiation 􀂄 􀂄 λ=2π/k 􀂄 􀂄 􀂄 2: 2 2 2 2 ( ) ˆ ( ) ˆ ω ωθ θ ω ωθ φ ⎧ − =⎪⎪⎨ −⎪= ⎪⎩ 􀁇 􀁇 p tE c r p tB c r 2 4 2 2 3 2 4 2 2 23 1 si) ˆB sin4 4 ˆsin8 c kr tS E p r c r pS r rc ωω θπ π ω θπ − = ⇒ = 􀁇 􀁇􀁇 􀁇 + -z G. Sciolla – MIT 8.022 – Lecture 21 Radiation emitted by a dipole oriented along the z axis in spherical coordinates: NB: this solution is only valid for r >>This is the Radiation propagates radially, some angular dependence too Poynting vector: NB: Poynting vector (and I) falls as 1/rthis should be intuitive. Why? sinsin sinsin kr kr n ( ×= This is 8.07 stuff: just trust me for the moment 49 Dipole radiation: cont. 􀂄 􀂄 􀂄 􀂄 􀂄 Why? S falls as 1/r2, area increases as r2 􀃆 + -z 4 2 2 23 2 4 2 22 3 23 0 0 4 2 3 30 ( ˆsin8 ˆsin : sin8 4 sin 3 3 R R U pSda a t da R U p R d d t U pd t c π π π ω θπ θφ ω φπ ωθθ ∂ = = ∂ = ∂ = ∂ ∂ = ⇒ = ∂ ∫ ∫ ∫ ∫ ∫ 􀁇 􀁇 􀁇i i 􀁇 G. Sciolla – MIT 8.022 – Lecture 21 Draw a sphere of radius R around the dipole centered in origin: NB: R >> d Compute power radiated through the sphere: NB: power through sphere does not depend on R Power through S (flux through S) is constant: Energy is conserved Since Since Larmor formula) rdrc d r Rc θθ 10 Application 3: capacitor 􀂄 i􀂄 􀂄 i􀂄 􀂄 charging up! 2 2 4 4ˆ ˆ 2 ˆ( ) π φ =− =− = 􀁇 􀁇 Q QE z zA a IrBr ca 2 2 4 42 2ˆˆ ˆB = ( )4 4 c c r rS E z r aca a φπ π π = × = × − 􀁇 􀁇􀁇 G. Sciolla – MIT 8.022 – Lecture 21 The Poynting vector applies to ANY situation in which both E and B appear, not just when we have radiaton Example: charging capacitor Calculate Poyntng vector: NB: what is important here is the direction of S: S points into the center of the capacitor as it should: the plates are From generalized Ampere law: Q IIQ 5Momentum carried by EM wave 􀂄 Since EM carry energy it’s not surprising that they carry momentum as well 2 􀁇2 2 24p c + m c 􀂄 In relativity, E and p are related by E = 􀂄 For EM radiation, m=0: U2 􀁇2 2E = p c ⇒ p= c 􀂄 Remember that 􀁇 Power Energy ⇒ S 􀁇 Energy/c = Momentum S== = Area Time Area c Time Area Time Area 􀂄 Dimensional analysis will also tell us that: 􀁇 S Momentum Force = == Pressure c Time Area Area 􀃆 Radiation exerts pressure Demo G. Sciolla –MIT 8.022 – Lecture 21 11 Summary on Poynting vector 􀂄 Energy flux: Energy /area /unit time 􀂄 Energy density u: Energy /unit volume 􀂄 Momentum flux: Momentum /area /unit time 􀂄 Momentum density: Momentum/unit volume EnergyMomentum Flux X/(Area sec) S ≡ 􀁇 B4 c Eπ × 􀁇􀁇 S c 􀁇 (same as pressure) Density X/Volume S c 􀁇 2 S c 􀁇 G. Sciolla –MIT 8.022 – Lecture 21 12 6 MIT 8.022 – Lecture 21 G. Sciolla – 13 Transmission line 􀂄 Transmission line = a pair of (twisted) cables used to transmit a signal 􀂄 Current flows in one direction on one cable and comes back on the other cable 􀂄 If terminated correctly, Z is purely real: Z~Rtermination 􀂄 Find R when capacitance per unit length=C’ and inductance per unit length=L’ 􀂄 In theory: 􀂄 In practice infinite sum of infinitesimal elements C and L: 􀂄 Calculate Z of the last piece and impose that it’s purely real. . … R 1 2 2 2 1 ' ' ' 1 1 ' . Ignoring term with LC (s L L' R= = Cm C'all): impose eq R i L RLC RZ i i C i L RR i RC i RC iL LCR R R i CR ω ωω ω ω ω ω ω ω − − +⎛ ⎞=+ + = + = =⎜ ⎟ + +⎝ ⎠ − + = + ⇒ L’ C’ … RR … G. Sciolla – MIT 8.022 – Lecture 21 14 Transmission line (2) 􀂄 What happens when transmission line is terminated correctly? 􀂄 Z is purely real: Z~Rtermination 􀃆 Z is a constant of the cable: 􀂄 Z does not depend on how long the cable is! 􀂄 If 􀃆 Z will depend on how long the cable is and on the frequency of the signal 􀃆 Distortions of the signal! 􀂄 Example of transmission line: coaxial cable, a pair of conducting tubes nested in one another 􀂄 Homework: prove that for a cylindrical coaxial cable Z=2 ln(b/a) /c and the velocity of propagation is c. 􀂄 Typical Rtermination: 50 Ohm R L'/:C≠ 715 Transmission line: demo 􀂄 􀂄 liµµsec 􀂄 􀂄 􀂄 iil􀂄 􀃆 􀂄 What happens if: 􀂄 l􀂄 l􀂄 Ω Ch1 Ch2 Pulse generator G. Sciolla – MIT 8.022 – Lecture 21 Coaxial cable (127.4 m long) Puse generator: pulse duraton 0.1 sec, period 20 Simultaneously send pulse from pulse generator (splitter) to Ch 1 of scope to transmission lne (back and forth and dspay on Ch 2) Measure speed of propagation: Time difference: 656 ns v=L/T~2/3c Open: signa will bounce back but nasty reflections Short: signa will be reversed on the same cable, nothing on the other cable If I terminate it with 50resistor: signal comes back on return cable with no reflections Scope Trigger Out Add terminals here 16 (Logical􀂄 directions 􀂄 􀂄 􀂄 We look up in x direction 􀂄 What kind of light do we see? G. Sciolla – MIT 8.022 – Lecture 21 Scattering of light ly this topic belongs to last lecture, but we did not have time…) When we send light into a medium, the light is scattered in many Example: light from Sun (unpolarized) passing through atmosphere Propagation of light //z 817 􀂄 //z 􀂄 polarization //x 􀃆 􀂄 l􀂄 􀂄 polari􀂄 􀂄 ion di􀂄 G. Sciolla – MIT 8.022 – Lecture 21 Scattering of light (2) Since light propagates //z: no polarization We measure the light (with our eyes!) along the x direction: no The light we see must be polarized along the y direction This is actually not really true because the light scatters mutiple times, but it suggests the general tendency What if the put a giant polaroid in front of the Sun? Scattered light would be more intense in direction perpendicular to zation direction Rotating the polaroid would allow us to change intensity of the light: Max intensity when polarizatrection is //y axis Dark when polarization direction is //x axis Scattering of light (3) 􀂄 How is light scattered? 􀂄 Light hits a molecule; the E shakes the molecule’s charges with frequency w; the molecule re-radiates the light often changing the direction 􀃆 changes in polarization 􀂄 Are all frequencies scattered in the same way? 􀂄 Electric fields of scattered radiation depend on acceleration of (dipole) charges 2 Escatee∝∂ d ∝ω2 if dipole moment of the shaken molecule goes as d~cos ωttrd ∂t224 −4 􀂄 Intensity of scattered radiation: I ∝ E ∝ω ∝λscattered 􀂄 Since λ red~ 2 λ blue􀃆 Blue is scattered 16 times more than red 􀂄 This explains why the sky is blue during the day and why it’s red at sunset G. Sciolla –MIT 8.022 – Lecture 21 18 9 19 Summary and outlook 􀂄 􀂄 􀂄 􀂄 􀂄 Scattering of light 􀂄 􀂄 􀂄 G. Sciolla – MIT 8.022 – Lecture 21 Today: Energy and momentum carried by EM waves Poynting vector and some of their applications Transmission lines What happens at sunset? Next Thursday: Magnetic fields through matter? Or review problems? 20 Sunset experiment 􀂄 􀂄 􀂄 Add Na2S2O335H2O (Na thiosulfate) 􀂄 lii􀂄 􀂄 iili􀂄 wall G. Sciolla – MIT 8.022 – Lecture 21 Solution of distilled water and salt. Unpolarized light is shining through it to the wall Lights starts scattering: fog; light on the wal becomes red frst and then dark as all the light is scattered toward the audence (as in sunset) What happened? Chemcal reaction creates bgger and bigger moecules that scatter more and more lght. Blue light is scattered first. Red makes it for a while but eventually scatters too. NB: light is polarized! 1021 Sugar solution experiment (T8) 􀂄 􀂄 iiλ􀂄 wall G. Sciolla – MIT 8.022 – Lecture 21 Light goes through a polarizer and then through an optically active sugar solution The first polarizer creates a linearly polarzed wave, overlap of right-handed and left-handed circularly polarized waves which propagate at different speeds in the solution. This causes linear polarization directon to change slowly. Since the effect depends on , different colors are rotated differently. The second polarizer check polarization direction at exit Polarizer Polarizer 11