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: i kx −ωt ) f ( z , t ) = Ae ( , ∂f ∂f i ( kx −ωt ) i kx −ωt ) for which = ikAe = −iω Ae ( , ∂z ∂t Consider again the force diagram for the string: ∂f Ff ( z ) = T sin θ ≅ T tan θ = T ∂z ∂f T ∂f ⎛ k ⎞ ∂f = T ⎜− ⎟ =− ≡ −Z . ∂t v ∂t ⎝ ω ⎠ ∂t
Z = T v = T µ is called the impedance. 30 January 2004
Physics 218, Spring 2004
f
.
String
θ T
T
θ′
z z +δ z
z 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: T T − v2 − v1 Z2 Z1 Z1 − Z2 AR = AI = AI = AI . T T v2 + v1 Z1 + Z2 + Z2 Z1 2 Z1 AI . Similarly, AT = Z1 + Z2 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) d dA P = ( F ⋅ A) = F ⋅ ( since the wave is transverse ) dt dt ∂f ∂f ⎞ 1 2 v ⎛ ⎛ ∂f ⎞ = −F f = −Ff ⎜ − v ⎟ = − Ff ⎜ −T ⎟ = Ff , ∂t ∂z ⎠ Z T ⎝ ⎝ ∂z ⎠ 2
2
⎛ ∂f ⎞ ∂f T ⎛ ∂f ⎞ ⎛ ∂f ⎞ = ⎜T ⎟ = ⎜ ⎟ = Z⎜ ⎟ . or ⎝ ∂z ⎠ ∂t v ⎝ ∂t ⎠ ⎝ ∂t ⎠ 2 ∂ f 1 ⎛ ⎞ Compare P = Ff2 = Z ⎜ ⎟ to some familiar electrical Z ⎝ ∂t ⎠ quantities: 1 2 P = V = RI 2 R 30 January 2004
( )
⎛ 1 2⎞ ( DC ) , P = Re ⎜ V ⎟ = Re ZI 2 ⎝Z ⎠ Physics 218, Spring 2004
( AC ) . 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: ∇2 E =
1 ∂2 E c
2
∂t
2
, ∇2 B =
1 ∂2B c
2
∂t
2
.
As with strings, the simplest solutions are the sinusoidal ones, called plane waves: − kz −ωt ) − kz −ωt ) E r , t = E e ( , B r , t = B e ( ,
(
)
0
(
)
0
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 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, ∂ i( kz −ωt ) — ◊ E = E0 z e No x or y dependence ∂z i kz −ωt ) = ikE e ( = 0 ⇒ E = 0. 0z
0z
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: ∂E y ∂ E — ¥ E = xˆ + x yˆ All other terms vanish ∂z ∂z 30 January 2004
Physics 218, Spring 2004
6
Relationship of E and B (continued) x
i kz −ωt ) ˆ i kz −ωt ) ˆ x + ikE 0 x e ( y — ¥ E = −ikE 0 y e (
1 ∂B iω i( kz −ωt ) =− = B0 e c ∂t c iω i( kz −ωt ) xˆ = B0 x e c i kz −ωt ) ˆ y. + ikB e (
E 0
φ φ′
B E0y 0y
0y
But k = ω c , so E 0 x = B 0 y , E 0 y = − B 0 x
,
E 0x
y
k B 0x
z
B 0
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
E 0 = E02x + E02y = B02y + B02x = B 0 ; are perpendicular to one another, as E 0 y B 0 y E 0 x π⎞ ⎛ =− = − tan φ = tan ⎜ φ + ⎟ ; tan φ = , tan φ ′ = 2⎠ E0 x B0 x E0 y ⎝ 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 2004
Physics 218, Spring 2004
8
Relationship of E and B (continued) x
E 0
y
E
Snapshot of the fields – in principle impossible to do, but if you could this is what you’d see)
B 0 B
30 January 2004
z
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 ( k⋅r −ωt ) , B = kˆ ¥ E . E=E e 0
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: 2 2 2 B 1 E B u= E 2 + B2 = in MKS. = = ε 0 E2 = 8π 4π 4π µ0 c c Use Product Rule #2: S= E¥B = E ¥ kˆ ¥ E 4π 4π 2 Last one c ˆ c cE ˆ ˆ ˆ = k (E ◊ E) − E k◊E = k = cuk . same in MKS. 4π 4π 4π
(
)
(
)
( )
30 January 2004
Physics 218, Spring 2004
10
Energy and momentum in electromagnetic waves (continued) For the momentum density, E2 ˆ u ˆ Last one g= 2 = k= k . same in MKS. 4π c c c So, unsurprisingly, an electromagnetic wave carries energy and momentum in its direction of motion. S
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 cos 2 ωt ≡ 30 January 2004
ω 2π
2π ω
∫
cos 2 ωtdt =
0
Physics 218, Spring 2004
1 = sin 2 ωt 2
. 11
Flashback (part e, problem 7.54) Remember the orthonormality of sines and cosines? 2π
2π
∫ cos mx cos nxdx = ∫ sin mx sin nxdx = πδ mn
0 2π
,
0
∫ cos mx sin nxdx = 0
, so
0
ω cos ω t = 2π 2
and, similarly,
2π ω
∫ 0
1 sin ω t = 2 2
30 January 2004
1 cos ω tdt = 2π 2
and
2π
∫ 0
1 cos xdx = 2 2
,
cos ω t sin ω t = 0 .
Physics 218, Spring 2004
12
Energy and momentum in electromagnetic waves (continued) Thus, Similarly,
E2 u = 4π
2 E02 E = cos2 ( k ◊ r − ωt ) = 0 4π 8π
2 cE S = cukˆ = 0 kˆ ≡ Ikˆ 8π
,
.
I = intensity
E02 ˆ uˆ g = k = k . 8π c c
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 ∆p = g c ∆t . If the medium is a perfect absorber, the pressure is therefore E02 1 dp 1 ∆p I = =c g = =u= Pabs = . A dt A ∆t 8π c If the medium’s surface is a perfect reflector, the light rebounds elastically and the pressure is twice as high: E02 Pabs = 2 Pref = . 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 erg sec-1 cm -2 :
I Pabs = = 4.3 ¥ 10 −5 dyne cm -2 c
.
It’s twice that much for a reflector. Either way it’s not much compared to atmospheric pressure, Patm 30 January 2004
Pabs = 1.013 ¥ 10 dyne cm ; = 4.3 ¥ 10 −11. Patm 6
-2
Physics 218, Spring 2004
15
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: i kx −ωt ) f ( z , t ) = Ae ( , ∂f ∂f i ( kx −ωt ) i kx −ωt ) for which = ikAe = −iω Ae ( , ∂z ∂t Consider again the force diagram for the string: ∂f Ff ( z ) = T sin θ ≅ T tan θ = T ∂z ∂f T ∂f ⎛ k ⎞ ∂f = T ⎜− ⎟ =− ≡ −Z . ∂t v ∂t ⎝ ω ⎠ ∂t
Z = T v = T µ is called the impedance. 30 January 2004
Physics 218, Spring 2004
f
.
String
θ T
T
θ′
z z +δ z
z 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: T T − v2 − v1 Z2 Z1 Z1 − Z2 AR = AI = AI = AI . T T v2 + v1 Z1 + Z2 + Z2 Z1 2 Z1 AI . Similarly, AT = Z1 + Z2 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) d dA P = ( F ⋅ A) = F ⋅ ( since the wave is transverse ) dt dt ∂f ∂f ⎞ 1 2 v ⎛ ⎛ ∂f ⎞ = −F f = −Ff ⎜ − v ⎟ = − Ff ⎜ −T ⎟ = Ff , ∂t ∂z ⎠ Z T ⎝ ⎝ ∂z ⎠ 2
2
⎛ ∂f ⎞ ∂f T ⎛ ∂f ⎞ ⎛ ∂f ⎞ = ⎜T ⎟ = ⎜ ⎟ = Z⎜ ⎟ . or ⎝ ∂z ⎠ ∂t v ⎝ ∂t ⎠ ⎝ ∂t ⎠ 2 ∂ f 1 ⎛ ⎞ Compare P = Ff2 = Z ⎜ ⎟ to some familiar electrical Z ⎝ ∂t ⎠ quantities: 1 2 P = V = RI 2 R 30 January 2004
( )
⎛ 1 2⎞ ( DC ) , P = Re ⎜ V ⎟ = Re ZI 2 ⎝Z ⎠ Physics 218, Spring 2004
( AC ) . 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: ∇2 E =
1 ∂2 E c
2
∂t
2
, ∇2 B =
1 ∂2B c
2
∂t
2
.
As with strings, the simplest solutions are the sinusoidal ones, called plane waves: − kz −ωt ) − kz −ωt ) E r , t = E e ( , B r , t = B e ( ,
(
)
0
(
)
0
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 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, ∂ i( kz −ωt ) — ◊ E = E0 z e No x or y dependence ∂z i kz −ωt ) = ikE e ( = 0 ⇒ E = 0. 0z
0z
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: ∂E y ∂ E — ¥ E = xˆ + x yˆ All other terms vanish ∂z ∂z 30 January 2004
Physics 218, Spring 2004
6
Relationship of E and B (continued) x
i kz −ωt ) ˆ i kz −ωt ) ˆ x + ikE 0 x e ( y — ¥ E = −ikE 0 y e (
1 ∂B iω i( kz −ωt ) =− = B0 e c ∂t c iω i( kz −ωt ) xˆ = B0 x e c i kz −ωt ) ˆ y. + ikB e (
E 0
φ φ′
B E0y 0y
0y
But k = ω c , so E 0 x = B 0 y , E 0 y = − B 0 x
,
E 0x
y
k B 0x
z
B 0
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
E 0 = E02x + E02y = B02y + B02x = B 0 ; are perpendicular to one another, as E 0 y B 0 y E 0 x π⎞ ⎛ =− = − tan φ = tan ⎜ φ + ⎟ ; tan φ = , tan φ ′ = 2⎠ E0 x B0 x E0 y ⎝ 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 2004
Physics 218, Spring 2004
8
Relationship of E and B (continued) x
E 0
y
E
Snapshot of the fields – in principle impossible to do, but if you could this is what you’d see)
B 0 B
30 January 2004
z
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 ( k⋅r −ωt ) , B = kˆ ¥ E . E=E e 0
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: 2 2 2 B 1 E B u= E 2 + B2 = in MKS. = = ε 0 E2 = 8π 4π 4π µ0 c c Use Product Rule #2: S= E¥B = E ¥ kˆ ¥ E 4π 4π 2 Last one c ˆ c cE ˆ ˆ ˆ = k (E ◊ E) − E k◊E = k = cuk . same in MKS. 4π 4π 4π
(
)
(
)
( )
30 January 2004
Physics 218, Spring 2004
10
Energy and momentum in electromagnetic waves (continued) For the momentum density, E2 ˆ u ˆ Last one g= 2 = k= k . same in MKS. 4π c c c So, unsurprisingly, an electromagnetic wave carries energy and momentum in its direction of motion. S
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 cos 2 ωt ≡ 30 January 2004
ω 2π
2π ω
∫
cos 2 ωtdt =
0
Physics 218, Spring 2004
1 = sin 2 ωt 2
. 11
Flashback (part e, problem 7.54) Remember the orthonormality of sines and cosines? 2π
2π
∫ cos mx cos nxdx = ∫ sin mx sin nxdx = πδ mn
0 2π
,
0
∫ cos mx sin nxdx = 0
, so
0
ω cos ω t = 2π 2
and, similarly,
2π ω
∫ 0
1 sin ω t = 2 2
30 January 2004
1 cos ω tdt = 2π 2
and
2π
∫ 0
1 cos xdx = 2 2
,
cos ω t sin ω t = 0 .
Physics 218, Spring 2004
12
Energy and momentum in electromagnetic waves (continued) Thus, Similarly,
E2 u = 4π
2 E02 E = cos2 ( k ◊ r − ωt ) = 0 4π 8π
2 cE S = cukˆ = 0 kˆ ≡ Ikˆ 8π
,
.
I = intensity
E02 ˆ uˆ g = k = k . 8π c c
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 ∆p = g c ∆t . If the medium is a perfect absorber, the pressure is therefore E02 1 dp 1 ∆p I = =c g = =u= Pabs = . A dt A ∆t 8π c If the medium’s surface is a perfect reflector, the light rebounds elastically and the pressure is twice as high: E02 Pabs = 2 Pref = . 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 erg sec-1 cm -2 :
I Pabs = = 4.3 ¥ 10 −5 dyne cm -2 c
.
It’s twice that much for a reflector. Either way it’s not much compared to atmospheric pressure, Patm 30 January 2004
Pabs = 1.013 ¥ 10 dyne cm ; = 4.3 ¥ 10 −11. Patm 6
-2
Physics 218, Spring 2004
15
