Download CCR 26: 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
time per unit area is given by the Poynting vector Ss:
1 s
s
RP-1
Ss =
E * B
m0
s and B
s are the electric and magnetic fields, respectively, and m0 is the permewhere E
ability of vacuum (see Figure RP-1). The wave propagates in the direction of Ss,
s and B
s varies harmonically in
whose SI units are W>m2. Because the size of both E
time, the time average of the magnitude of the Poynting vector is
1
E B RP-2
2m0 max max
where Emax and Bmax are the maximum magnitudes of the fields. In a vacuum the mags and B
s are related by E  cB, so 〈S〉  E2. Thus, although the values of E
s
nitudes 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 s
p and exerts a
sRP = dp
sRP that results
s >dt on any surface the wave encounters. The force F
force F
from the radiation pressure acting on an area A depends on whether the wave is
8S 9 =
x
s
z
y
RP-1 The electric field (blue arrows) and magnetic field (red arrows) are orthogonal transverse waves.
s , is thus perpendicular to both E
s and B
s . The EM wave
Their cross product, the Poynting vector S
s.
travels in the direction of S
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Tipler: Modern Physics 6/e
Perm fig.: CCR47, New fig.: RP-01
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Classical Concept Review 26
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reflected or absorbed. Referring to Figure RP-2a, if the incident EM wave is comsRP is in the direction of Ss and has magnitude
pletely absorbed, the force F
8S 9 A
cos  1absorption2
RP-3a
c
in which case the magnitude of the pressure due to radiation incident on the area A is
FRP =
8S 9
FRP
=
cos 
RP-3b
c
A
sRP must be perpendicular to the surface of
If the EM wave is totally reflected, F
the area A since in reflection there is no net force parallel to the surface. The magnitude of the force is then
PRP =
2 8S 9 A
cos2  1reflection2
c
and the magnitude of the pressure due to radiation falling on the area A is
FRP =
RP-4a
2 8S 9
FRP
=
cos2 
RP-4b
c
A
(To order to confirm that the units on 8 S 9 >c are those of pressure, pascals, note that
J
1
N#m
s
N
= 2 = Pa).
*
= # 2 *
2
#
m
m>s
s m
s m
m
PRP =
EM
wave
(a)
θ
θ
reflected
wave
Area A
(edge on)
FRP(absorption)
FRP(reflection)
(b)
photon
dΩ
photon
z
θ
θ
dΩ
dA
sRP due to radiation pressure on
RP-2 (a) Forces F
s
the area A. FRP (green vector) results from total
sRP (red vector) from
absorption of the EM wave, F
total reflection. (b) Photons impinging on a
perfectly reflecting surface dA at an angle of
incidence  within a solid angle cone dV. The
photons reflect at the same angle  into the same
solid angle cone dV.
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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 wavelengths between l and l 1 dl that are incident on a perfectly reflecting surface of
area dA within a solid angle cone dV at an angle  (see Figure RP-2b) experience a
change in the z component of their momentum given by
dpl dl = 1pref,z - pinc,z 2 dl
E l cos 
E l cos 
- ab d dl
c
c
= c
=
2E l cos 
dl
c
2
u1l2 dl dA dt cos2  d
RP-5
c
where u(l) dl is given by Planck’s law, Equation 3-18. Dividing dpl by dA and dt
gives
=
dpl
= 1dpl >dt2 >dA = dFl >dA
dt dA
RP-6
where dpl >dt = dFl is the force exerted on the photons by the area dA. Newton’s
third law reaction to that force, - dpl >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
P1l2 dl = 3
z70
2
= 3
c
dFl
dl = 3
dA
2p
0
z70
3
2
u1l2 dlcos2  d
c
p>2
u1l2 dl cos2  sin  d d
0
4p
u1l2 dl
RP-7
3c
In the case of a gas in a container, the pressure of the gas exists throughout the container, 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
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Classical Concept Review 26
1
P1l2 dl = 3
c
2p
0
2
3 u1l2 dl cos  sin  d d
p
0
4p
u1l2 dl
3c
The total radiation pressure Prad is then
=
4p
4sT 4
1
Prad = 3 P1l2 dl =
u1l2
dl
=
= U
3
3c
3c
3

77
RP-8

0
RP-9
0
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 3 =
= 2 = Pa.)
3
m
m
m
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
cU
4
Substituting U into Equation RP-9 yields
R =
1
U =
4
R
c
411.36 * 103 W>m2 2
4
R =
3c
313.00 * 108 m>s2
Prad =
Prad = 6.04 * 10-6 Pa
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.
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