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Radiation Pressure CLASSICAL CONCEPT REVIEW 26 Radiation Pressure Classical Description It was first pointed out by Maxwell in 1871 that electromagnetic (EM) radiation would exert pressure (force per unit area) on surfaces, a theoretical prediction verified experimentally by Lebedev in 1900 and by Nichols and Hull in 1901. The pressure results from the momentum carried by the EM wave, whose energy transport per unit s time per unit area is given by the Poynting vector S : s 1 s s S = E * B RP-1 m0 where Es and Bs are the electric and magnetic fields, respectively, and m is the perme- 0 s ability of vacuum (see Figure RP-1). The wave propagates in the direction of S , whose SI units are W m2. Because the size of both Es and Bs varies harmonically in time, the time average of the magnitude of the Poynting vector is > 1 S = Emax Bmax RP-2 2m0 where Emax and Bmax are the maximum8 9 magnitudes of the fields. In a vacuum the mag- s s 2 s nitudes of E and B are related by E cB, so 〈S〉 E . Thus, although the values of E s s and B vary both positive and negative with time, the value of S is always positive or zero (refer to Figure RP-1 and use the right-hand rule for the cross product). Because the EM wave transports energy, it carries momentum sp and exerts a s s s force FRP = dp dt on any surface the wave encounters. The force FRP that results from the radiation pressure acting on an area A depends on whether the wave is > x z s y RP-1 The electric field (blue arrows) and magnetic field (red arrows) are orthogonal transverse waves. s s s Their cross product, the Poynting vector S , is thus perpendicular to both E and B . The EM wave s travels in the direction of S . Tipler: Modern Physics 6/e Perm fig.: CCR47, New fig.: RP-01 74 First Draft: 2011-10-10 2nd Pass: 2011-10-24 TIPLER_CCR.indd 74 23/11/11 5:46 PM Classical Concept Review 26 75 reflected or absorbed. Referring to Figure RP-2a, if the incident EM wave is com- s s pletely absorbed, the force FRP is in the direction of S and has magnitude S A F cos absorption RP-3a RP = c 8 9 in which case the magnitude of the pressure due1 to radiation2 incident on the area A is FRP S PRP = = cos RP-3b A c s 8 9 If the EM wave is totally reflected, FRP must be perpendicular to the surface of the area A since in reflection there is no net force parallel to the surface. The magni- tude of the force is then 2 S A F cos2 reflection RP-4a RP = c 8 9 and the magnitude of the pressure due to radiation1 falling on2 the area A is FRP 2 S 2 PRP = = cos RP-4b A c (To order to confirm that the units on S c 8are9 those of pressure, pascals, note that J 1 N # m s N * = * = = Pa). s # m2 m s s # m2 m m2 8 9 > > (a) EM wave θ θ reflected wave Area A (edge on) FRP(absorption) FRP(reflection) (b) photon z photon dΩ θ θ dΩ s RP-2 (a) Forces FRP due to radiation pressure on the area A. Fs (green vector) results from total RP s dA absorption of the EM wave, FRP (red vector) from total reflection. (b) Photons impinging on a perfectly reflecting surface dA at an angle of incidence within a solid angle cone d. The photons reflect at the same angle into the same solid angle cone d. TIPLER_CCR.indd 75 23/11/11 5:46 PM Tipler: Modern Physics 6/e Perm fig.: CCR48, New fig.: RP-02 First Draft: 2011-10-10 2nd Pass: 2011-10-24 76 Classical Concept Review 26 Quantum Description As was pointed out in the discussion following Equation 2-31, although photons are massless, they carry momentum of magnitude p = E c. Thus, photons with wave- lengths between l and l 1 dl that are incident on a perfectly reflecting surface of area dA within a solid angle cone d at an angle (see> Figure RP-2b) experience a change in the z component of their momentum given by dpl dl = pref,z - pinc,z dl 1E cos 2 E cos l l dl = c - - c 2c E cos a b d l dl = c 2 u l dl dA dt cos2 d RP-5 = c where u(l) dl is given by Planck’s1 law,2 Equation 3-18. Dividing dpl by dA and dt gives dpl = dpl dt dA = dFl dA RP-6 dt dA 1 > 2 > > where dpl dt = dFl is the force exerted on the photons by the area dA. Newton’s third law reaction to that force, -d pl dt = -dFl , is the force exerted on the area dA by the photons;> the minus sign simply indicates that the force is in the 2z direction. For our discussion here we can ignore> the minus sign. The magnitude of the pressure Pl exerted on the area dA by the photons with wavelengths between l and l 1 dl incident at all angles with z 0 is then dFl 2 P l dl = dl = u l dlcos2 d 3 dA 3 c z 0 z 0 1 2 7 7 1 2 2p p 2 2 2 = > u l dl cos sin d d c 3 30 0 1 2 4p = u l dl RP-7 3c In the case of a gas in a container,1 the2 pressure of the gas exists throughout the con- tainer, not just at the walls. In an isotropic radiation field, the radiation analog of the gas-filled container, radiation pressure exists everywhere in the field, not just on the surface dA. The 2 in Equation RP-5 arose because of the reflection of the photons from the perfectly reflecting surface dA. If we replace the reflecting surface with a mathematical surface, the factor 2 disappears and photons impinge on and pass through the surface dA from both above and below. In that case the magnitude of the radiation pressure due to photons with wavelengths between l and l 1 dl impinging on the area dA becomes TIPLER_CCR.indd 76 23/11/11 5:46 PM Classical Concept Review 26 77 2p 1 p P l dl = u l dl cos2 sin d d c 3 30 0 1 2 4p 1 2 = u l dl RP-8 3c The total radiation pressure Prad 1is 2then 4p 4sT 4 1 Prad = P l dl = u l dl = = U RP-9 3 3c 3 3c 3 0 0 1 2 1 2 where U is the total energy density of the radiation. (To confirm that the units of U are J N # m N pressure, note that = = = Pa.) m3 m3 m2 EXAMPLE RP-1 Radiation Pressure from Sunlight What is the magnitude of the radiation pressure produced by sunlight at Earth’s distance from the Sun? How does it compare with atmospheric pressure? SOLUTION The power per unit area R of sunlight arriving at the top of Earth’s atmosphere, called the solar constant, is 1.36 * 103 W m2. It is related to the energy density U by Equation 3-6: > 1 4 R = cU 1 U = R 4 c Substituting U into Equation RP-9 yields 4 4 1.36 * 103 W m2 Prad = R = 3c 3 3.00 108 m s 1 * > 2 -6 Prad = 6.04 * 10 1 Pa > 2 For comparison, about 99 percent of Earth’s atmosphere lies below 35 km. At that altitude atmospheric pressure is about 103 Pa. The radiation pressure of sunlight there is about nine orders of magnitude smaller. TIPLER_CCR.indd 77 23/11/11 5:46 PM.
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