Today in Physics 218: Back to Electromagnetic Waves

Today in Physics 218: back to electromagnetic waves

Impedance Plane electromagnetic waves Energy and momentum in plane electromagnetic waves Radiation pressure

Artist’s conception of a “solar sail:” a spacecraft propelled by solar radiation pressure. (Benjamin Diedrich, Caltech.)

30 January 2004 Physics 218, Spring 2004 1 Reflection, transmission and impedance

Consider a simple transverse wave again: −ω fzt(),,= Aeikx( t) ∂∂ff−−ωω for which ==−ikAeikx() t,. iω Ae ikx() t ∂∂zt Consider again the force diagram for the string: f ∂f String Fz()=≅ Tsinθθ T tan = T f ∂z θ T ⎛⎞kT∂∂∂f ff =TZ⎜⎟ −ω =− ≡− . T θ ′ ⎝⎠∂∂∂tvtt

ZTv== Tµ is called the impedance. z z zz+ δ

30 January 2004 Physics 218, Spring 2004 2 Reflection, transmission and impedance (continued) In these terms, the two halves of our string have different impedances. We can case the reflected and transmitted wave amplitudes in this form: TT − vv−− Z Z ZZ AA==21 2 1 AA = 12 . RITT I I vv21+++ ZZ 12 ZZ21 2Z1 Similarly,AATI= . ZZ12+ Changes in impedance are associated with reflection. What does the impedance “mean”? Consider the power carried by the wave past some point z:

30 January 2004 Physics 218, Spring 2004 3 Reflection, transmission and impedance (continued) ddA P =⋅=⋅()FFA ()since the wave is transverse dt dt ∂∂ff⎛⎞v ⎛⎞ ∂ f1 2 =−FFvFTFff =−⎜⎟ − =− f ⎜⎟ − = f, ∂∂tzTzZ⎝⎠ ⎝⎠ ∂ 22 ⎛∂∂ff ⎞T ⎛⎞ ∂ f ⎛⎞ ∂ f or===⎜TZ ⎟ ⎜⎟ ⎜⎟ . ⎝∂∂ztvt ⎠ ⎝⎠ ∂ ⎝⎠ ∂ t 2 1 2 ⎛⎞∂f Compare PFZ == f ⎜⎟ to some familiar electrical quantities: Zt⎝⎠∂

1122⎛⎞ 2 2 P== V RI()DC , P = Re⎜⎟ V = Re( ZI ) () AC . RZ⎝⎠

30 January 2004 Physics 218, Spring 2004 4 Back to electromagnetic waves

With strings providing an introduction, we will now return to electromagnetic waves, which you will recall involve fields that, in vacuum, obey classical wave equations: 11∂∂22E B ∇=22EB,. ∇= ct22∂∂ ct 22 As with strings, the simplest solutions are the sinusoidal ones, called plane waves: −−(kzωω t) −−( kz t) Er(),,,,te== E00 Br() te B where E 00 and B are complex constants. First, a few remarks about the nature of electromagnetic plane waves in vacuum are in order:

30 January 2004 Physics 218, Spring 2004 5 Relationship of E and B

Because the divergences of the fields are both zero in vacuum, we have, for instance, ∂ −ω —◊ E= Eeikz() t No xy or dependence ∂z 0z () ikz− tω ==⇒=ikE00zz e00. E Similarly, B 0 z = 0. Electromagnetic waves in vacuum have no field components along their direction of propagation; they’re transverse, just like waves on a string. We learn something from the curls, too: ∂Ey ∂E —=-¥ Exy ˆ + x ˆ All other terms vanish ∂∂zz

30 January 2004 Physics 218, Spring 2004 6 Relationship of E and B (continued) ikz( −−ωω t) ikz( t) x — ¥ E =−ikE00yx exyˆˆ + ikE e ∂ ω 1 B i ikz()− t E0 E =− = B0e 0x ctω∂ c ω φ i ikz()− t = Be0x xˆω c φ′ () k ikz− t z + ikB0y e yˆ. E0y ω B0y But kc = ω , so B0x EB==−,, E B 00xy 0 y 0 x y B0 which has interesting implications for the relative magnitude, phase, and orientation of the fields:

30 January 2004 Physics 218, Spring 2004 7 Relationship of E and B (continued) the field amplitudes have the same magnitude, as 22 22 EEEBBB000000=+=+=xy yx ; are perpendicular to one another, as EB00yyE ⎛⎞π φφ=′ = =− φφ0x =− = + tan , tan tan tan⎜⎟ ; EBE000xxy⎝⎠2 and are in phase, as we see by combining the last three results: BzE00= ˆ ¥ . In MKS, we’d get EB 00 = c , but the fields would still be perpendicular and in phase.

30 January 2004 Physics 218, Spring 2004 8 Relationship of E and B (continued)

x Snapshot of the fields – in E principle impossible to do, E0 but if you could this is what you’d see)

y B0

B z

30 January 2004 Physics 218, Spring 2004 9 Energy and momentum in electromagnetic waves

More generally, if the wave propagates in a direction other than z (i.e. the wavenumber is a vector, k), then i(kr⋅−ω t) ˆ E ==EBkE.0e , ¥ Taking E and B to be the real parts of the plane-wave fields E and B , we get the following for the energy density, and energy flux: 22 2 1 2 2 EB 2 B uEB=+==()==0E in MKS. 844πππ0 ccˆ SEBEkE==¥¥¥()Use Product Rule #2: 44 ππ ε cccE2 Last one =−kEEˆˆˆˆ()◊◊ EkE= k =cu k . () µsame in MKS. 44ππ4 30 January 2004 Physics 218, Spring 2004 10

π Energy and momentum in electromagnetic waves (continued) For the momentum density, 2 S EuLast one gkk==ˆˆ = . c2 4πcc same in MKS. So, unsurprisingly, an electromagnetic wave carries energy and momentum in its direction of motion.

In the case of light, the periods of oscillation are so short that time averages of these quantities is useful. Recall that the average over a period for the square of a cosine or a sine is 2πω ωωωω 1 cos222ttdtt≡== cos sin . 22∫ π 0 30 January 2004 Physics 218, Spring 2004 11 Flashback (part e, problem 7.54)

Remember the orthonormality of sines and cosines? 22ππ ∫∫cosmx cos nxdx== sin mx sin nxdx πδmn , 00 2 π ∫ cosmx sinωω nxdxω= 0 , so 0 ππ 2 2 11 cos222ttdtxdx=== cos cos , 222∫∫ 00 and, similarly, πω 1 sin2 ωωωttt== and cos sin 0 . 2 π

30 January 2004 Physics 218, Spring 2004 12 Energy and momentum in electromagnetic waves ππ(continued) π Thus, E2 EE22 ut==00cos2 ()kr◊ −=ω . 44 8 Similarly, cE2 Sk= cu ˆˆˆ=≡0 kkI , I = intensity 8π u E2 gk==ˆˆ0 k. cc8 π

30 January 2004 Physics 218, Spring 2004 13 Radiation pressure

When light falls encounters a surface bounding a medium, it can be absorbed or reflect, in either case imparting its momentum to the medium and thus exerting a pressure. The momentum per unit area crossing an area A during a time ∆t is ∆=pgct ∆ . If the medium is a perfect absorber, the pressure is therefore 11dp∆ p E2 I Pcgu======0 . abs Adt A∆ t8π c If the medium’s surface is a perfect reflector, the light rebounds elastically and the pressure is twice as high: E2 PP==2.0 abs ref 4π 30 January 2004 Physics 218, Spring 2004 14 Example: problem 9.10

The intensity of sunlight hitting the Earth is about 1300 W/m2. If sunlight strikes a perfect absorber, what pressure does it exert? How about a perfect reflector? What fraction of atmospheric pressure does this amount to? In cgs units, that’s 1.3¥ 106-1-2 erg sec cm : I P ==4.3¥ 10−5-2 dyne cm . abs c It’s twice that much for a reflector. Either way it’s not much compared to atmospheric pressure,

6-211Pabs − Patm ==1.013¥¥ 10 dyne cm ; 4.3 10 . Patm

30 January 2004 Physics 218, Spring 2004 15