Today in Physics 218: back to electromagnetic waves

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1 Today in Physics 18: 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 004 Physics 18, Spring 004 1

2 Reflection, transmission and impedance Consider a simple transverse wave again: ikx, ( ωt f z t = Ae, f for which z ( ikx ( ωt f ikx, ( ωt = ikae = iω Ae t Consider again the force diagram for the string: f Ff ( z = Tsinθ Ttanθ = T z k f T f f = T = Z ω t v t t f String θ T θ T Z = T v = Tµ is called the impedance z z+ δ z z 30 January 004 Physics 18, Spring 004

3 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: T T v v1 Z Z1 Z1 Z A R = A I = A I = A I v T T + v1 + Z1 + Z Z Z1 Z Similarly, A 1 T = A I Z + Z 1 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 004 Physics 18, Spring 004 3

4 Reflection, transmission and impedance (continued d d P = = dt dt ( F F ( since the wave is transverse f f v f 1 = F = F v = F T = F t z T z Z f f f f f f T f f or = T = = Z z t v t t 1 f Compare P = Ff = Z to some familiar electrical quantities: Z t, 1 1 ( ( P = V = RI DC, P = Re V Re ZI = ( AC R Z 30 January 004 Physics 18, Spring 004 4

5 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: 1 E 1 E B B =, = c t c t As with strings, the simplest solutions are the sinusoidal ones, called plane waves: ( kz ωt ( kz ωt E ( r, t = E 0e, B ( r, t = B 0e, where E 0 and B 0 are complex constants First, a few remarks about the nature of electromagnetic plane waves in vacuum are in order: 30 January 004 Physics 18, Spring 004 5

6 Relationship of E and B Because the divergences of the fields are both zero in vacuum, we have, for instance, ikz ( ωt E = E 0ze No x or y dependence z ikz ( ωt = ike e = 0 E = 0 0z 0z Similarly, B 0z = 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: E E y E ˆ x + ˆ z x = - z y All other terms vanish 30 January 004 Physics 18, Spring 004 6

7 Relationship of E and B (continued ikz ( ωt ˆ ikz ( ωt E = ike e x+ ike e yˆ 0y 0x 1 B iω ikz ( ωt = = B E 0 E 0e 0x c t c φ iω ikz ( ωt = B 0xe xˆ c φ ikz ( ωt + ikb 0ye yˆ B E 0y 0y But k = ω c, so B 0x E 0x = B 0y, E 0y = B 0x, y B 0 which has interesting implications for the relative magnitude, phase, and orientation of the fields: x k z 30 January 004 Physics 18, Spring 004 7

8 Relationship of E and B (continued the field amplitudes have the same magnitude, as E E E B B B 0 = 0x + 0y = 0y + 0x = 0 ; are perpendicular to one another, as E 0y B 0y E 0x π tan φ =, tanφ = = = tanφ = tan φ+ ; E B E 0x 0x 0y and are in phase, as we see by combining the last three results: B 0 = zˆ E 0 In MKS, we d get E 0 = c B 0, but the fields would still be perpendicular and in phase 30 January 004 Physics 18, Spring 004 8

9 Relationship of E and B (continued x E 0 E Snapshot of the fields in principle impossible to do, but if you could this is what you d see y B 0 B z 30 January 004 Physics 18, Spring 004 9

10 Energy and momentum in electromagnetic waves More generally, if the wave propagates in a direction other than z (ie the wavenumber is a vector, k, then i( kr ωt E = E 0e, B = kˆ E 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: 1 ( E B B u= E + B = = = ε0e = in MKS 8π 4π 4π µ 0 c c S= E B = E ( kˆ E Use Product Rule #: 4π 4π c ˆ c ( ( ˆ ce = k E E E k E = kˆ = cukˆ Last one 4π 4π 4π same in MKS 30 January 004 Physics 18, Spring

11 Energy and momentum in electromagnetic waves (continued For the momentum density, S E ˆ u g = = k = kˆ c 4πc c Last one 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 πω ω 1 cos ωt cos ωtdt = = sin ωt π 0 30 January 004 Physics 18, Spring

12 Flashback (part e, problem 754 Remember the orthonormality of sines and cosines? π π 0 0 π 0 cosmx cosnxdx = sin mx sin nxdx = πδ, cosmx sin nxdx = 0, so πω π ω 1 1 cos ωt = cos ωtdt = cos xdx =, π π 0 0 and, similarly, 1 30 January 004 Physics 18, Spring mn sin ωt = and cosωtsinωt = 0

13 Energy and momentum in electromagnetic waves (continued Thus, Similarly, u 0 E0 E E = = cos ( k r ωt = 4π 4π 8π 0 ˆ ce S = cuk = kˆ Ikˆ, I = 8π u ˆ E0 g = k = kˆ c 8π c intensity 30 January 004 Physics 18, Spring

14 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 p = g c t If the medium is a perfect absorber, the pressure is therefore 1 dp 1 p E0 I Pabs = = = c g = = u= Adt A t 8π c If the medium s surface is a perfect reflector, the light rebounds elastically and the pressure is twice as high: E0 Pabs = Pref = 4π 30 January 004 Physics 18, Spring

15 Example: problem 910 The intensity of sunlight hitting the Earth is about 1300 W/m 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 erg sec cm : I 5 - Pabs = = dyne cm c It s twice that much for a reflector Either way it s not much compared to atmospheric pressure, P atm dyne cm - ; Pabs = = P atm 30 January 004 Physics 18, Spring

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