is a molecular property. The corresponding property of a continuum is called the polarization: 1 ∑ pi [ ] Coul-m [ ] Coul = P ( x ) ≡ lim V = 3 V → 0 all i m m2 in V
(63)
where the sum is over all dipoles in the region whose volume is V. This can be thought of as representing the local concentration of dipoles at position x. Substituting either (58) or (62) into (63) leads to P ( x= )
1 n p= a E (x) ∑ i V all i V in V
e0 χe
P ( x ) =e0 χe E ( x )
(64)
where n/V is the number of dipoles in V per unit volume. Thus the polarization is expected to be proportional to the local electric field. After factoring out ε0, the remaining proportionality constant, denoted χe, is dimensionless and is called the electric susceptibility of the medium. χe is the continuum counterpart of the molecular property α.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
31
Fall, 2014
Lecture #5 begins here Regardless of which type of polarization we have, we still have to explain how polarization gives rise to a monopole charge density (since a dipole consists of equal amounts of negative and positive charges and therefore no net charge). Indeed dipoles induced completely inside or completely outside some mathematical region do not change the charge of the region; only dipoles induced near the boundaries of the region can change the charge. Consider a region in space completely enclosed by a mathematical boundary (part of which is denoted by the vertical line in the sketch at right) which contains hydrogen atoms (denoted by the circles). Consider two representative hydrogen atoms very near the border: one just inside and the other just outside the region. Because the electron’s charge is smeared out over a spherical shell, some of the negative charge lies inside the region and the rest lies outside, for both atoms. But if the centers lie at equal distances on either side of the interface (as shown), when we add up the charges from both atoms, we have a total of one electron and one proton inside the region, giving no net charge to the system. After the application of an electric field pointing to the right, the electron clouds are pulled the the left and become distorted. We assume the center of mass (i.e. the proton) does not shift. We still have one proton inside and one outside, but less than half of each electron now resides inside our region. Thus the region inside has lost negative charge and has thus acquired a net positive charge. This shows how polarization can cause an initially neutral region to acquire a net charge. For the record, notice that polarization parallel to the interface does not result in a shift of charge across the interface: slightly more than half of one electron cloud and slightly less than half of the other reside inside. A total of one negative and one positive charge reside inside, leaving the interior electrically neutral. Only the normal component of polarization shifts charge. The shift of the electron cloud, together with the number of clouds per unit volume, is quantified by the polarization vector field P. The amount of charge crossing the surface as a result of polarization is calculated from
and this represents the quantity of positive charge displaced
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
32
Fall, 2014
across the interface in the direction of n, as a result of the material becoming polarized. Of course, dotting by the unit normal vector yields the normal component of the polarization. Integrating over some closed surface A leads to
∫ n.P da = {
}
net positive charge lost by system
A
If charges are leaving the system, which was initially electrically neutral, then the remaining system must be charged. We define the local density of charged induced by polarization as ρp = polarization charge density ρp is also sometimes called the bound charge density since it represents electronic charges
which are bound to the nucleus of the atom containing the electrons. In terms of this variable, the change in charge of the system due to polarization is
∫ ρ p dV =− ∫ n.P da =− ∫ ∇.P dV
V
A
(65)
V
Applying the Divergence Theorem for arbitrary V eventually leads to ρ p = −∇.P
(66)
Thus the material must be nonuniformly polarized before a monopole charge is induced by polarization. In addition to the charges induced by polarization of the material, we have the original charges which generated the electric field. We shall refer to these “original” charges as the “free” charge and denote their density as ρf = “free” charge density ρ = ρf + ρp = total charge density
(67)
Gauss’s law (50) (written with the permittivity of vacuum) is still valid, provided we include both types of charges (free and polarization): ρ f + ρp ∇.E = ε0
Substituting (66) for the polarization charge: ρ f − ∇.P ∇.E = ε0
Multiplying through by the universal constant ε0 and combining the two divergences:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
33
Fall, 2014
∇.( ε0 E + P ) = ∇.D = ρf ((
(68)
D
Where D is called the electric displacement: Usually we drop the subscript “f ”. This is the first of Maxwell’s equations. Substituting (64), the displacement becomes D =+ ee eE (1 χe ) e0 E = r 0E =
(69)
εr ≡ 1 + χe
where
is called the dielectric constant (dimensionless) or relative permittivity of the material. Moreover ε ≡ εr ε0
is called the permittivity of the dielectric material.♥ Assuming ε does not vary with position,♦ the ε can be factored out of the divergence ∇.D = ∇.( εE ) = ε∇.E
ρf ∇.E = ε
and Gauss’s law becomes
(70)
Most problems in electrostatics are solved in terms of the electrostatic potential ψ(x) instead of the electric field E(x). Substituting (51) into (70) (and dropping the subscript “f ”), Gauss’s law becomes:
∇2 ψ = −
ρ ε
(71)
which is also called Poisson’s equation . Compared to (56), the permittivity of vacuum ε0 has been replaced by the permittivity of the material ε. Finally let’s return to the integral form of Gauss’s law given by (49): in vacuum:
∫
A
E.n= da
1 ρ dV ε0 ∫ V
This form was derived for charges in a vacuum. Inside a dielectric material, we need to replace ρ by (67) as we did in deriving the general form of the differential form of Gauss’s law: ♥
(69) is only one form this constitutive equation might take. See mention of birefringence and dichroism on page 4. ♦More generally, at very high electric field strength, ε depends on E (see page 4). Since E can depend on position, ε cannot always be assumed constant. A more general form of (70) is (68). Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
34
∫
in dielectric material:
E.n= da
A
1 ε0
Fall, 2014
∫ (ρ f
+ ρ p ) dV
V
Substituting (65) for the polarization charge:
∫
A
1 E.n da = ε0
∫ρf
dV +
V
1 ε0
∫ ρ p dV
V n.P da
∫
A
Multiplying through by ε0 and combining the two surface integrals leads to + P ).n da =∫ ρ f 0E ∫ (ε( ( A
dV
V
D
∫ D.n da= ∫ ρ dV
which is usually written as
A
(72)
V
where we have dropped the subscript “f ”. Boundary Conditions at a Charged Interface Reference: Jackson, 3rd Ed., p16-19 Consider a charged interface between two different dielectric media. To obtain an appropriate boundary condition for such an interface, we apply (72) to a “pill-box” straddling the interface. The two phases on either side of the interface often have different permittivities εI and εII. Now let’s evaluate the quantities on either side of the above equation for the “pill-box” shown above. Let the area of the faces be Af and the thickness of the pill-box be l (sum of lI in Phase I and lII in Phase II). Applying the Mean Value Theorem, we can write the integrals as I II side I II Cl =sA f + ρ A f l I + ρ A f l II D.n A f + D.n A f + D.n q ∫ D.n da A
where the carets denote the appropriate “mean” values and σ is any surface charge density (coul/m2) localized on the boundary. The surface integral was decomposed into three contributions: the two plane faces and the curved side wall of the cylinder; C is the perimeter of the face and Cl is the area of the side wall. Next, we will let the thickness l of the pill box shrink to zero. Three of the five terms vanish (those proportional to l ), leaving Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
35
Fall, 2014
I II D.n + D.n = σ
l → 0:
after dividing out Af. If we now let Af become sufficiently small, we can replace the mean values by the point values. Finally noting that nI = –nII ≡ n1, we obtain
(D1 − D2 ).n1 = σ
Af → 0:
(73)
where n1 is the unit normal pointing into medium 1 and σ is the surface charge density (coul/m2). This implies that a discontinuity in the normal component of E may occur at an interface. To get a relation involving the tangential components of E, we exploit the conservative nature of the electric field. Recall Theorem III:♣ the existence of a scalar field, ψ(x), such that: E = –∇ψ
∫ E.dr = 0
means that
C
for all closed contours, C.♠ For example, let's choose the contour to be a thin rectangle straddling the interface and lying in the xy plane, where the yz plane is the charged interface. On the vertical paths dr = +eydy on the right side (direction is upward) and dr = -eydy on the left side (direction is downward). On the horizontal path, dr = -exdx on the top (direction is toward left) and dr = +exdx on the bottom (direction is toward right). Applying the Mean Value Theorem to evaluating the contour integral:
ε1
ε2
y C
∆x
∆y
x
II II I E.e y ∆y + E.( −e x ) ∆x II + E.( −e x ) ∆x I
(
+ E. −e y
)
I
I II ∆y + E.e x ∆x I + E.e x ∆x II =0
Letting ∆x →0:
(
II E.e y + E. −e y
)
I
0 =
after the ∆y has been cancelled out.
♣ See page 37
of Vector Fields.pdf under the “Appendix” menu item of Blackboard. While this contour integral vanishes in electrostatics problems, it does not vanish when oscillating magnetic fields are present [see (I.16) on p17 of Jackson]. We will revisit this b.c. after we present magnetostatics. ♠
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712 Letting ∆y →0:
36
Fall, 2014
E2y = E1y
Similiarly, we could have choosen our contour to lie in the xz plane, and show that: E2z = E1z Thus we have proved that the tangential components are continuous across a charged interface. We can write this more compactly as a single vector equation:
(E1 − E2 ) × n1 = 0
(74)
(73) and (74) are the two equations relating the normal and tangential components of the electric field on either side of a charged interface. They serve as boundary conditions in problems on electrostatics.
Permittivities of dielectrics are usually determined experimentally by measuring capacitance. In this section, we are going to analyze the simplest geometry for a capacitor: two parallel plate electrodes sandwiching the dielectric material (see figure at right). This problem will also be helpful when we discuss Maxwell’s “displacement current” arising from an oscillating electric field.
Potential
1
Example: Parallel Plate Capacitor
2
3
V
0
εp
ε
εp
δ
x I The electrodes consist of two metallic plates of area A σ23 σ12 separated a distance δ filled with the material of interest V 0 whose permittivity is denoted ε. A battery or other d.c. power supply is used to apply a voltage V between the electrodes and the electric current I is measured as a function of time. Upon applying the voltage, a brief spike in current is observed, which rapidly decays to zero as the capacitor becomes charged. This charge is assumed to accumulate at the interface between the metal and the dielectric: for the polarity shown in the figure, positive charge accumulates at the left interface (x = 0) while negative charge accumulates at the right interface (x = δ). The total charge passed during charging of the capacitor is
∞
Q=
∫ I ( t ) dt
0
so the charge densities arising at equilibrium on either interface are σ23 = −σ12 =
Q A
(75)
Typically Q is found to be proportional to the applied voltage V and the ratio is defined as
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
37 C =
Fall, 2014
coul Q =[ ] =[ ] farad volt V
(76)
which is called the capacitance of the capacitor. The SI unit of capacitance is the farad which is defined as 1 coul/volt. To relate these observations to the permittivity of the fluid, we need a model for the electrostatics of a capacitor. Either Gauss’s law (70) or Poisson’s equation (71) can be used to solve for the potential profile and (73) can be used as the boundary condition. The electrodes and wires of the circuit are usually constructed of a highly conductive material (e.g. a metal). whose conductivity is virtually infinite (compared to that of the dielectric material between the plates). Then (after the transient) all of the voltage will drop across the dielectric (as shown in the figure as the voltage profile in blue) and no electric field remains inside the metal. The carriers of charge in the metal are electrons, which are not bound to the nucleus of the metal atom of a good conductor. Conversely, electrons in dielectric materials remain tightly bound to their nucleus and are immobile. For this problem, we expect the space charge density ρ which appears in (70) or (71) is zero everywhere. Inside the dielectric material, there are no carriers of charge and therefore no free charge. While electrons act as charge carriers in a metal, the virtually infinite conductivity of metals implies that at steady state any excess charge (here “excess” means excess of electrons compared to protons) will migrate to the boundary of the metal and the dielectric. In other words, any nonzero electric field creates an infinite current which tends to rapidly restore local electroneutrality. So good conductors are usually modeled as having no excess charges in their interior as well as no electric field in the interior. When the y- and z-dimensions of the parallel plates are large compared to the x-separation distance δ, any electric field in the dielectric can be expected to have only an x-component, which can depend only on x. So our solution is expected to have the form E = Ex ( x ) e x
Substituting this form into (70) yields the simple result:
dE x =0 dx
or
E x ( x ) = const
But the electric field can be different in each of the three regions and we denote these three values as Ex1, Ex2 and Ex3. At an interface between two media, the discontinuity in the normal component of the electric field is related to the surface charge density and the dielectric constant of the two media through b.c. (73) at x = 0:
( ε1E1 − ε2 E2 ).n1 = σ12
(77)
where n1 is a unit normal pointing into media “1”. Applying this to the left interface, n1 = –ex:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
38
Fall, 2014
( ε p E1 − εE2 ).( −ε x ) = −ε p E x1 + εE x 2 = σ12 ((
(78)
0
To apply (77) this to the right interface, we replace “1” by “3” and note that n3 = +ex:
( ε p E3 − εE2 ).(ε x ) = εpE x3 − εE x 2 = σ23
(79)
0
Taking the electric fields to be zero inside the metallic electrodes, (78) and (79) imply: σ12 = −σ23 = εE2
(80)
The electric field inside the dielectric media is minus the slope of the potential profile: dψ V Ex 2 = − = dx d
Combining (75), (76) (80), (81):
= C
(81)
σ12 A εA = δ V
(82)
Thus measurement of the capacitance of a dielectric material in a capacitor whose geometric dimensions A and δ are known allows the value of the permittivity ε to be calculated. Lecture #6 begins here Some Experimental Data If we use a parallel plate capacitor to measure the dielectric constant of water, we find e water =e 78 0 ε is also related to the speed of light: see the wave equation (13). The ratio of the speed of light
in vacuum c0 to the speed of light in some other medium c is defined as the refractive index of the material n≡
µε c0 == c µ0 ε0
ε ≥1 ε0
(83)
Most transparent media have the same value of magnetic permeability µ so this property cancels out. Eq. (83) applies at frequencies for which absorption of light is negligible. When absorption occurs, ε and n are complex numbers and their relationship also involves the absorption coefficient. Now measurements of refractive index for visible light reveal Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
39 nwater = 1.33
compared to
Fall, 2014
e water = e0
= 78 8.8
Why the big difference? It turns out that ε, n and α are all frequency dependent properties. e water =e 78 0 is based on a static measurement (the electric field is not oscillating), whereas nwater = 1.33 is based on measurement done at optical frequencies. Polarization of a Dielectric Sphere Suppose we have a dielectric particle of one phase dispersed in a second phase. The analog of (61) [ α e = 4πe0 R3 ] can be deduced by calculating the apparent dipole moment induced in a spherical particle of radius R and permittivity εs by a uniform electric field externally applied in the surrounding phase (e.g. water) having permittivity εf. Assuming both phases are dielectrics the result turns out to be (see Hwk2) es − e f a particle = 4pe f R3 e s + 2e f
(84)
3V e f
The important result is that the polarizability of small particles is proportional to their volume. While this was also true of (61), the proportionality constant is different. This difference arises because the electron cloud of an H atom was treated as a good conductor whereas the particles (and the surrounding fluid) are treated as insulators. In particular, different electrical properties show up in the derivations (see Hwk2). Clausius-Mosotti Equation Now ε is a property of a continuum whereas α in (61) is a property of single molecules or atoms. The two properties are related by the Clausius-Mosotti equation (also called the Lorentz-Lorenz formula): = α
3ε0 ε − ε0 3ε0 n 2 − 1 = N ε + 2ε 0 N n2 + 2
(85)
where N is the number of molecules per unit volume; the second relation results from substitutuing (83). Comparing (85) with (84), we find the expressions are very similar: εf has been replaced by ε0 and the particle volume V has been replaced by 1/N. Eq. (85) works well not only for gases but also condensed phases. To see this, let’s look at some experimental measurements taken with liquids and gases. These experimental measurements are tabulated in terms of the molar refractivity A, defined as N N A n2 − 1 cm3 = A≡ A α =[ ] N n2 + 2 3e0 mole
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
40
Fall, 2014
where NA is Avogadro’s number. All tables taken are from Born & Wolf.♣ The first table (just below) shows the molar refractivity of air at various pressures. Increasing the pressure increases the number concentration N of gas molecules as well as the refractive index n.
Although the concentration N of gas molecules varies by two orders of magnitude, the molar refractivity is virtually constant. The second table compares A for various liquids with A for the vapor of the liquid:
Notice that A measured for liquid or vapor of the same compound are virtually equal. This means that the polarizability of molecules does not depend upon the state (vapor or liquid). The third table looks at mixtures of surfuric acid and water in various proportions: ♣
th
Born & Wolf, Principles of Optics, 7 ed., Cambridge U Press (1999).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
41
Fall, 2014
The right-most column contains the A for the mixture inferred from measurements of refractive index. The column just to the left represents A calculated as the weighted average of the A’s for the pure components; the weighting factor is the molecular number fraction. The agreement between these two columns is remarkable in all proportions. This suggests that each molecule contributes independently of the others to the refractive index of the solution. Finally we compare atomic refractivity with molecular refractivity:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
42
Fall, 2014
The right-most column in Table 2.5 is obtained by adding the reflectivity for each atom (from Table 2.4) in the molecule. For example, the value of 6.74 for HCl is obtained by adding 1.02 for H to 5.72 for Cl. These calculated values are compared to the A’s inferred from measurements of refractive index (next to last column). Once again, the agreement is remarkable (the largest error is 11% for CS2). This suggests that each atom of a molecule contributes independently of the others. Comment: Values of the molar refractivity will depend on wavelength (color) of the light used for refractive index measurements. By using visible light (light scource = sodium D lamp), the refractivities reported in the tables above correspond to electronic polarization of the atoms and molecules as described in the section on page 27 above. At lower (sub-microwave) frequencies, polar molecules like H2O or CS2 exhibit much higher α’s owing to orientational polarization arising from orientation of permanent dipole with any applied electric field. Many of the comparisons made above would fail for sub-microwave frequencies.
Magnetostatics of Continua Ref.: Jackson, 3rd ed., Chapt 5. In electrostatics we saw that a small test charge q when placed in a nearly uniform electric field E experiences a force given by F = qE
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
43
Fall, 2014
The quantity q represents a monopole of charge. One important difference between electrostatics and magnetostatics is that magnetic monopoles do not exist. However the application of a magnetic field can induce a permanent magnetic dipole in a few materials (e.g. iron and nickel). Placing a small magnetic dipole m in a magnetic field B results in a torque N being applied to the dipole: N = m×B Of course, electric dipoles do exist and placing an electric dipole p in and electric field E also results in the application of a torque to the dipole: N = p×E The direction of the torque is such that (if the dipole is free to move) the dipole will rotate toward alignment with the electric field. Note that if p is in the same direction as E there will be no torque. Magnetic phenomena have been studied for at least as long as electric phenomena. Perhaps the earliest observations were of bits of iron (e.g. nails) being attracted to a particular naturally occurring rock which today is called lodestone (see picture at right). Lodestone is magnetite, a mixture of ferrous and ferric oxides with Pieces of lodestone the chemical formula Fe3O4. suspended so they can turn will rotate in the earth’s natural magnetic field. This serves as the basis for the compass used by ancient mariners. The word “lode” means “course” (as in a ship’s course) in Middle English. We generally first learn about magnetic fields by playing with permanent magnets. The field lines shown at left are for an ideal cylindrical magnet (axis of cylinder is horizontal). One end of the cylinder (constructed of iron or nickel) is the north pole (analogous to positive charge) and the other end is the south pole (analogous to negative charge). If the strength of the magnetic field is large enough, then iron filings will have a magnetic dipole induced in each particle and rod-shaped particles will tend to align with the field lines or be attracted to either end of the magnet. Relatively few materials have magnetic properties like iron or nickel. However, we will see that magnetic fields play an important role in the propagation of light even in vacuum (or air).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
44
Fall, 2014
Laws of Ampere, Biot and Savart Another way to generate a magnetic field (without using a permanent magnet) is to drive an electric current through a wire, as shown in the figure at right. If current I (units are coul/sec or amp) flows through an infinite straight wire, then the magnetic induction field B at some point x a distance r from the wire (measured normal to wire) has a magnitude given by
µ I B (x) = 0 2πr ( )
(86)
B r
and the direction of B is normal to the plane containing the wire and the point x. The sense (recall there are always two unit normals to any plane) is given by the right-hand rule (see figure at right): with the thumb of the right hand pointing in the direction of positive current flow, the curled fingers point in the direction of B. The proportionality constant (in SI units) is µ0 = 1.256637×10-6 N/amp2
and is called the magnetic permeability of vacuum. (86) is called the Biot-Savart law (1820). The closest electrostatic analog to the magnetic field produced by sending current through a long, straight wire is the electric field produced by a uniform charge density on a long, straight line (shown at right). Application of (52) to this line charge yields an electric field whose magnitude is σ E (x) = 2πε0 r ( )
(87) E
E r
+ σ + + + + + +
E
+ + + +
where σ is the charge density (coul/m) along the line. Notice the similarity between (86) and (87): B ∝ I/r wheras E ∝ σ/r. Also notice the differences: E points radially outward from the wire while B curls around the wire. The proportionality constant µ0 plays a role similar to ε0 although one appears in the numerator and one in the denominator. Early work on the induction of magnetic fields by electric current can be attributed to Oersted (1819), Biot and Savart (1820) and Ampere (1820-1825). All of these authors performed experiments measuring the force which arises between two wires carrying current. For two long (length L) parallel wires, separated by a distance d≪L, carrying currents I1 and I2, the force per unit length turns out to be
F µ0 I1 I 2 = L 2π d Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
45
Fall, 2014
While this force has important applications for design of electric motors, it is not directly relevant to the study of light, so we will not say anymore about it. We are however quite interested in the magnetic field B induced by either wire. So what is meant by B, which is also called the magnetic flux density or magnetic induction? B is the magnetic analog of the electric field E: just like E is the force per unit of electric charge, B is the force per unit of magnetic monopole (if such a thing existed). But magnetic monopoles do not exist. Indeed this makes the study of magnetostatics more complicated than electrostatics. And this is why magnetostatics developed later in history than electrostatics. Just like charges can generate an electric field, electrical current can generate a magnetic field. But the vectorial nature of the two vector fields is quite different as is suggested by comparing the field lines (drawn in blue in either of the two figures above). The magnetic field lines are concentric circles lying entirely in planes normal to the wire; the circles are also concentric with the wire. On the other hand, the electric field lines radiate out normally from the line of charges. Since magnetic monopoles does not exist, the closest analogy to Coulomb’s law is to consider the magnetic field generated by a differential element of thin wire (thin compared to the distance x) through which current is flowing at a rate I amps or coul/s. The differential contribution to the magnetic field at x is given by µ I dr × x dB ( x ) = 0 4π x 3
dr
(88)
Indeed (86) can be obtained by integrating (88) over the length of a long, straight wire. (88) is the analog to (46): Coulomb’s law in vector form applied to a single point charge q located at the origin for the position vector x: E (x) =
single point charge:
1 qx 4πε0 x 3
Notice that both equations predict the field decays as x–2 with I dr playing the role of q. Integrating (88) over an arbitrary distribution of electrical currents (i.e. all currents in the universe) gives
B (x) =
where
µ0 4π
x − x′
∫ J ( x′) × x − x′ 3 dx′
V∞
3
(89)
dV
J = local current density (amp/m2)
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712 and
46
Fall, 2014
= J dV J= awire dr J= awire dr I dr I
where awire is the cross-sectional area of the wire carrying the current. From the above equation, you can see that the integrand of (89) is identical to (88). (89) is analogous to Coulomb’s law summed over all charges in the universe, as given by (52): arbitrary distributon:
1 4πε0
E (x) =
x − x′
∫ ρ ( x′) x − x′ 3 dx′ 3
(52)
dV
V∞
Thus the electric current density J (a vector) plays the same role in magnetic phenonena as the charge density ρ (a scalar) plays in electric phenomena. Clearly the analogy is not one-to-one since the (tensor) rank of these two quantities is different. Like (52), (89) is not the most useful form for the relationship between B and J. In the same way we transformed (52) into (55), we can also transform (89). Recall (53) and (54): x − x′ x − x′
=
3
∂ 1 ∂x x − x ′
1 = ∇ x − x′
Multiplying by a constant J(x' ) (i.e. independent of x): J ( x′) ×
x − x′ 1 = J ( x′) × ∇ 3 x − x′ x − x′
(90)
Next we apply a vector identity:♦ ∇×(su) = s(∇×u) + (∇s)×u where u is an arbitrary vector field and s is an arbitrary scalar field. In (90) we take u = J and = s 1 x − x ′ . Applying (90): 0 (( J ( x′ ) 1 ∇ × J ( x′) −∇× J ( x ′ ) × ∇= ′ (( x − x′ x− x′ x − x (( (((((( u ×∇s = − ( ∇s )× u
s ( ∇× u )
(91)
∇× ( su )
It is important to note that ∇ in (90) is ∂/∂x — not ∂/∂x'. Then the vector J(x' ) can be treated like a constant and its curl vanishes. Using (90) and (91), we can rewrite (89) as ♦
This is entry E.3 in Vector Identities.pdf under Supplemental Files on Blackboard
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
47
µ B (x) = − 0 ∇ × 4π
Fall, 2014
∫ V∞
J ( x′) 3 d x′ x − x ′ dV
(92)
after interchanging the order of curl (with respect to x) and integration (over x'). Recall vector identity C6: the divergence of the curl of any vector field vanishes identically: 0 ∇.B =
(93)
which is the analog of ∇×E = 0 in electrostatics. (93) is the second of Maxwell’s equations. Similarly using vector field theory, we can take the curl of (92) and integrate by parts to obtain
µ ∇ × B = µ0 J + 0 ∇ 4π
∫
V∞
∇ ′.J ( x ′ ) 3 d x′ x − x′
where ∇' denotes ∂/∂x'., ∇ × B = µ0 J
and for ∇.J = 0 (steady state):
(94)
which is the analog of ∇.E =ρ ε0 in electrostatics. The reason that ∇.J = 0 is described as “steady state” can be seen by recalling that conservation of free charges more generally requires: ∂ρ + ∇.J = 0 ∂t
At steady state, the time derivative is negligible, leaving ∇.J = 0. So (94) is like the differential form of Gauss’s law. The corresponding analog of the integral form of Gauss’s law can be obtained by dotting both sides of (94) by the outward pointing unit normal n and integrating over some open surface A:
∫ ( ∇ × B ).n da = µ0 ∫ J.n da A
A
Applying Stokes Theorem to the left integral, we get
∫ B.dr =
µ0 I
(95)
C
where I is the total current passing across the surface A and C is a closed contour forming the edge of the open surface A. (95) is like the integral form of Gauss’s law and is called Ampere’s
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
48
Fall, 2014
Law.♦ If all points on C are the same distance r from some thin wire carrying the current through A, then the integral above is just the magnitude B times the circumference 2πr of the circular countour and the above integral reduces to (86). Lecture #7 begins here The following table summarized the analogy between electrostatics and magnetostatics in situations where the magnetic field is induced by an electric current (rather than a permanent magnetic dipole). Name of Relationship
Electrostatics
Coulomb’s law (single point charge) Coulomb’s law (continuous distribution)
1 4πε0
∫
µ I dr × x dB ( x ) = 0 4π x 3
1 qx 4πε0 x 3
E (x) =
E (x) =
Magnetostatics
x − x′
ρ ( x′)
x − x′
V∞
3
3 ′ d x= B (x) dV
∫
J ( x′) ×
V∞
x − x′
d 3 x′ 3 x − x ′ dV
= J dV J= awire dr J= awire dr I dr
dq = ρ dV
monopole charge
µ0 4π
I
irrotational vs. solenoidal scalar potential vs. vector potential
Gauss’s law (integral form)
Gauss’s law (differential form)
∇×E = 0
1 ψ (x) = 4πε0
∫
E = –∇ψ
implies
E.= n da
A
∫ V∞
ρ ( x′) 3 d x′ x − x′
1 ε0
∫
ρ dV
V q
ρ ∇.E = ε0
0 implies ∇.B =
µ A (x) = − 0 4π
∫ V∞
∫ B.dr =
B = ∇×A
J ( x′) 3 d x′ x − x ′ dV
µ0 I
C
∇ × B = µ0 J
Magnetic Force on Moving Charge Recall from (88) that a steady current (through a wire for example) can generate a magnetic field: ♦
Ampère, André Marie, 1775-1836, French physicist, mathematician, and natural philosopher.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
49
Fall, 2014
µ I dr × x dB ( x ) = 0 4π x 3
(88)
Instead of a wire in which continuous electrical current flows, suppose I have a point charge q moving at some velocity v. Replacing ♥ I dr → dq v
C m s
C
m s
µ qv × x B (x) = 0 4π x 3
in (88) gives
Thus a moving point charge also generates a magnetic field. In the figure at right, the upward velocity acting on a positive charge will induce a magnetic field on the right which points into the page (indicated by ⊗) and a magnetic field on the left which point out of the page (indicated by ). If instead we externally apply a uniform magnetic field B and then move a point charge through it, the point charge will feel a force which is given by Fm= qv × B
(96)
This is called the Lorentz force. The action of the magnetic field B pointing into the page on the positive charge q moving upward is a force pushing the charge to the left (see figure at right). If B is uniform, this Lorentz force will cause the charge to move in a circular orbit in a plane orthogonal to B (see figure at right). Applying Newton’s second law:
ma = F
or
dv m = qv × B dt
(97)
Let’s adopt cylindrical coordinates and align B in the direction of –ez: B = –Bez According the figure above, we expect the trajectory of the particle to be given by ♥
We can drop the “d” from dq if we are dealing with a true “point” charge, meaning that the physical dimensions of the charge are infinitesimal compared to x . Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
50
Fall, 2014
r , θ ( t ) , z = r (t ) = 0 = re r [ θ ( t ) ] and its velocity is♣
v (t ) =
(98)
de dr dr d θ = = r r Ω = r Ωeθ [ θ ( t )] dt d θ dt dθ Ω
eθ
Substituting this result into Newton’s law (97): mr Ω
de q d q = qr ΩB eq × ( −e z ) ( ((( dq dt −e r
−e r Ω
Clearly our guess (98) for the trajectory r(t) satisfies Newton’s law for the force given by (96) [i.e. both sides of Newton’s law correspond to a vector which points in the same direction]. Cancelling out the common terms, we can solve for the orbit frequency Ω: Ω=
qB m
(99)
One important application of the Lorentz force is the cyclotron, in which charged particles (e.g. proton or other ion) are injected into a uniform magnetic field and then accelerated to very high velocities by periodically applying a voltage. The high energy particles can be used for “atom smashing” or treatment of some cancers. This is illustrated in the following example. Example: Calculate the orbit radius and velocity of protons having a kinetic energy of 106 eV exposed to a uniform magnetic field having a strength of 1 Tesla.♠ Solution: A million electron-volts corresponds to a kinetic energy of 6 J ( −19 E= 106 eV = 1.602 × 10−13 J 10 1.602 × 10 C ) = C Assuming this is all kinetic energy, we can calculate the linear velocity of the protons: E = 1 mv 2 2
Substituting the mass of one proton (m = 1.673×10-27 kg) and E from above, we obtain
= v
♣ This
2E m = 1.384 × 107 m s
vector calculus is discussed at greater length on p17 of Vector Fields.pdf on BlackBoard. a big magnetic field. For comparison, the earth’s magnetic field is about 50 µT.
♠ One Tesla is
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
51
Fall, 2014
which is about 5% of the speed of light. The orbit frequency is given by (99) as (substituting q = e): Ω =
rad eB = 9.58 × 107 = 15.24 MHz s m
Finally the orbit radius R can be calculated from = R
v = 0.145 m Ω
Of course, a point charge q also experiences a force in an applied electric field E: Fe = qE which is the electrostatic analog of (96). The total force on a moving point charge is F= q ( E + v × B )
(100)
Faraday’s Disk Generator The sketch at right shows Faraday’s disk, which was the first manmade mechanical generator of electricity. A copper disk (D) is rotated with the hand crank. You also see the edge view of a permanent (horse-shoe shaped) magnet (A) with its north and south poles straddling the disk. This rotation caused an electrical current to flow radially in the disk between the axle of the disk (connected to post B) and a sliding spring (m) in contact with the outer edge of the disk (and connected to the post B’). The rotation of the disk causes free electrons in the metal to undergo solid-body rotation in the θ-direction. The magnet is generating a magnetic flux in the z-direction. When the electrons move in the θ-direction between the poles of the magnet, the magnetic field Bz exerts a Lorentz force on the charges which acts in the r-direction which creates the current. Magnetization Induced in Materials So far we have presented the physics of steady currents in wires inducing a magnetic field in the vacuum outside the wire. If the wire is instead embedded in some material, the magnetic field can induce a magnetic dipole in atoms and molecules of the material, just like an electric field can induce an electric dipole. The analog of the polarization (the concentration of electric dipoles) defined by (63) is Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
52
Fall, 2014
1 M ( x ) ≡ lim ∑ mi V → 0 V all i in V
[analogous to P(x)]
where mi is the magnetic dipole moment induced in molecule i and M(x) is the corresponding continuum variable called the magnetization, which represents the concentration of magnetic dipoles at some location x. Just like a nonuniform polarization generates a polarization charge density ρp of some material given by (66), nonuniform magnetization of the material general an induced current density in the material: JM = ∇ × M
[analogous to ρ p = −∇.P ]
This induced current density needs to be added to the “free” current density J in (94):
∇ × B = µ0 ( J + J M ) = µ0 ( J + ∇ × M ) Dividing both sides by µ0 and then combining the two curls: 1 B − M =J ∇× µ0
or
∇× H =J
(101)
H
where
= H
1 B−M µ0
[analogous to D = ε0 E + P ]
is called the magnetic field. To complete the description, we need a constitutive equation relating H and B. For isotropic nonmagnetic as well as diamagnetic and paramagnetic substances, a simple linear relationship can be used B = µH
(102)
where µ is the magnetic permeability of the material. Typically µ/µ0 differs from unity by only a few parts in 105 with µ>µ0 for paramagnetic substances and µ R 0
(116)♣
♣
(116) is an approximation: there is likely to be a gradual transition from Ez = V/δ to Ez = 0 in the vicinity of r = R. The width of this transition is on the order of δ, so Ez from r = R–δ to r = Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
61
Fall, 2014
Then the z-component of displacement current for 0 ≤ z ≤ δ is given by (115): ∂Dz ( ωε0V0 δ ) cos ωt = ∂t 0
for r < R for r > R
Since J = 0 throughout the region between the disks, according to our generalized form of Ampere’s law (110), the z-component of the curl of the magnetic field is given by
( ∇ × H= )z
∂Dz = ∂t
( ωε0V0 δ ) cos ωt 0
for r < R for r > R
(117)
The other components of the curl must vanish everywhere, since there are no displacement currents (or electric fields) in those directions:
( ∇ × H ) r = ( ∇ × H )θ = 0
(118)
Also we can invoke another of Maxwell’s equations: (93) ∇.B = µ0 ( ∇.H ) = 0
or
∇.H = 0
(119)
Recalling the vector form of the vector field which arises when a steady current is passed through a wire (see figure depicting field lines for B on page 44), we look for a solution in cylinderical to this differential equation for H which has the following form in cylindrical coordinates: H = H θ ( r , t ) eθ
(120)
This form automatically satisfies (118) and (119). Substituting (120) into (117) gives: ( ωε V δ ) cos ωt 1 ∂ ( rH θ ) = 0 0 r ∂r 0
As a boundary condition, we take
for r < R for r > R
(121)
H θ= t) 0 ( r 0,=
Then (121) can be integrated to obtain
( H θ )capacitor
( ωε0V0 δ ) ( r 2 ) cos ωt = ( ωε0V0 δ ) ( R 2 )( R r ) cos ωt
for r < R for r > R
(122)
R+δ is probably affected. But as in the limit δ/R → 0, (116) becomes an increasingly good approximation. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
62
Fall, 2014
For a comparison, we calculate the magnetic field generated by the wire carrying current to this capacitor: substituting (114) into (86) and dividing by µ0 to get Hθ:
= ( H θ )wire
wewe R 2V0 I 1 A = πR 2 0 AV0 = Bθ = → 0 cos wt cos wt µ0 2πr 2πr δ 2r δ
which is identical to (122) for r > R. Below we plot (122) along with Ez given by (116).
This red curve is the same shape of magnetic field profile which would be obtained for a steady current I through a wire of radius R; of course then the blue curve would represent a plot of the dimensionless current density Jz rather than the dimensionless oscillating electric field strength. Indeed, recall from (86) that the magnetic field (outside the wire) induced by a steady current inside a wire decays like r–1; this is also the decay reported by (122) outside the parallel plates (r > R). So we have just seen from Ampere’s equation (110) how an oscillating electric field can induce a oscillating magnetic field. But this new oscillating magnetic field will also induce an oscillating electric field according to Faraday’s equation (107). Recall that B = µ0H
( ωµ0 ε0V0 δ ) ( r 2 ) cos ωt Substituting (120): Bθ ( r , t ) = ( ωµ0 ε0V0 δ ) ( R 2 )( R r ) cos ωt
for r < R for r > R
Differentiating with respect to time:
( (
) )
2 for r < R ∂ −ω µ0 ε0V0 δ ( r 2 ) sin ωt Bθ ( r , t ) = ∂t −ω2 µ0 ε0V0 δ ( R 2 )( R r ) sin ωt for r > R
Substituting into (107):
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
63
( (
Fall, 2014
) )
ω2 µ ε V δ ( r 2 ) sin ωt 0 0 0 ( ∇ × E )θ = (( ω2 µ0 ε0V0 δ ( R 2 )( R r ) sin ωt ∂Er ∂E z − ∂z
for r < R for r > R
(123)
∂r
The remaining two components of the curl are zero. For the record, the curl of the electric field which generated this magnetic field was zero. Therefore the original electric field does not satisfy this equation. The electric field which does satisfy this equation represents a perturbation to the original electric field. This perturbation to the electric field also has to satisfy Coulomb’s law (68) which for a vacuum (D = ε0E), free of any space charges, means ∇.E = 0
(124)
Let’s seek a solution to (123) and (124) which has the form E = Ez ( r , t ) e z
which automatically satisfies (124) as well as makes the r and z components of ∇×E zero. Integrating (123) with respect to r, taking Ez = 0 at r = 0:
( (
)( )(
) )
2 2 for r < R −ω µ0 ε0V0 δ r 4 sin ωt ∆E z ( r , t ) = −ω2 µ0 ε0V0 δ R 2 4 1 + ln ( r R ) sin ωt for r > R
We denoted this result as ∆Ez to distinguish it from Ez. Let’s have a closer look: at r=R we have
ω2 R 2 µ0 ε0 V0 ∆E z ( R, t ) = − sin ωt δ 4 compared to the original electric field obtained by substituting (113) into (116):
E= z (t )
V0 sin ωt δ
After dividing out the common factor, we have
( ωR ) 2 ω2 R 2 µ0 ε0 ∆E z ( R, t ) = − = − 4 Ez ( t ) 4c 2 where
c=
Copyright© 2014 by Dennis C. Prieve
1 m0 ε0
= 3 × 108
(125)
m s
PDF generated November 30, 2014
06-712
64
Fall, 2014
is the speed of light in vacuum. The ratio in (125) is infinitesimally small for any reasonable frequency or R. For example, a large R would be 1 cm and a large frequency would by 1 MHz, which cooresponds to
( ωR )2 1 (10−2 m )( 2π × 106 s −1 ) = 1.1 × 10−8 = 2 m 4 4c 3 × 108 s 2
which is indeed a very small ratio. Lecture #9 begins here
Nature of Waves Reference: Lipson et al., Chapt. 2 Recall that in the absence of any free charges and for linear isotropic materials, Maxwell’s equations imply the wave equation (13) 2
∇ E=
µε 1
∂2E ∂ 2t
and
2
∇ H=
µε 1
c2
∂2H ∂ 2t
(126)
c2
In physics, a wave is a disturbance which travels through space carrying energy with no net transfer of mass. Examples include sound waves and waves on the surface of water. To understand why (126) is called the wave equation and to see why the speed of the wave is associated with the coefficient of the time-derivative, consider the one-dimensional, scalar form
∂2 f ∂x 2
=
1 ∂2 f
(127)
v2 ∂ 2t
D’Alembert♦ deduced a general solution of this PDE: f ( x, t ) = F ( x − vt ) + G ( x + vt )
(128)
where F and G are any two functions. By “general solution” I mean that all solutions to (127) have the form of (128). There is no analog to this statement for parabolic PDE’s like the diffusion equation.
♦
Jean le Rond d’Alembert (1717-1783), French mathematician, physicist, philosopher and music theorist. The wave equation is sometimes referred to as d’Alembert’s equation. Pierre Simon Laplace was his student. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
65
Fall, 2014
Partial Proof: Instead of x and t as the independent variables, let’s define two new independent variables: η = x + vt
and
x = x − vt
f ( x, t )= g ( η, x )
and a new dependent variable:
Then repeatedly applying the Chain Rule of partial differentiation, the wave equation (127) becomes ∂2 g =0 ∂η∂ξ
which can be solved by direct integration. If we integrate first with respect to x: ∂g = A ( η) ∂η
and then integrate with respect to η, we obtain one particular solution g ( η, ξ= )
∫ A ( η)d η=
G ( η)
On the other hand, if we integrate first with respect to η and then with respect to x, we obtain a second particular solution g ( η, ξ= )
)d = ∫ B ( ξξ
F (ξ)
The general solution is the arbitrary linear combination of these two solutions: g ( η, ξ= ) F ( ξ ) + G ( η)
Transforming back into the original set of variables, we obtain (128). Travelling vs. Standing Waves The particular solution f ( x= , t ) F ( x − vt ) is a wave travelling in the +x direction at speed v. This can be seen in the following example in which f is taken as the cosine function:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
66
Fall, 2014
f(x,t) = cos[pi*(x-vt)] 2
f(x,t)
1 0
0 −1 −2
−4
−2
2
0
4
x
The dashed curve gives the function evaluated at t = 0, while the solid curve gives the function evaluated at a slightly later time (e.g. t = 0.2 with v = 1). The solid curve can be seen to be shifted slightly to the right of the dashed curve; thus the wave is travelling to the right. On the other hand, f ( x= , t ) G ( x + vt ) is a wave travelling in the –x direction at speed v. This is illustrated in the following graph in which f is taken as the same cosine function as in the previous example:
f(x,t) = cos[pi*(x+vt)]) 2
f(x,t)
1 0
0
−1 −2
−4
−2
0
2
4
x Once again t = 0 for the dashed curve and t = 0.2 for the solid curve with v = 1. Now the solid curve can be seen to moving to the left. If we average together two periodic travelling waves, which travel in opposite directions and have equal amplitudes and equal frequencies, we get a standing wave. In the graph below, we have averaged the two travelling waves above:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
67
Fall, 2014
f(x,t) = 0.5*{cos[pi*(x-vt)] +cos[pi*(x+vt)]} 2
f(x,t)
1 0
0
−1 −2
−4
−2
0
2
4
x Notice that the maxima and minima remain stationary, but the function oscillates in place. Transport of Energy by Waves In the intro of this section, we suggested that waves transport energy without any net mass transfer. Let’s demonstrate that now by means of a simple example: a vibrating string. Consider a flexible string, stretched under tension τ newtons between horizontal fixed endpoints at x = –L and x = +L. The vertical deflection of the string is given by y(x,t) which satisfies the wave equation:♠
t
∂2 y
∂2 y = σ ∂x 2 ∂ 2t
where σ is the mass of the string per unit length. Comparing with (127), we conclude that the velocity of the wave will be
v=
τσ
To illustrate that energy is being transported in the x-direction, we pick a nonperiodic solution to the wave equation. One particular nonperiodic solution is a Gaussian waveform:
( x − vt )2 y= ( x, t ) A exp − w2 where w is the half-width of the Gaussian and A is the amplitude of the wave. Notice that this solution has the form of d’Alembert’s solution F(x–vt) and represents a wave travelling at speed v in the +x direction. For A = 1, v = 1 and w = 1, the graph below shows the wave at two different t’s:
♠
The derivation is provided in Sect. 19.1 of Greenberg, which is available under “Supporting Files” in Blackboard. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
68
Fall, 2014
The transverse movement of the string in the y-direction represents kinetic energy; in particular
( x − vt )2 KE 1 ∂y 2 ( x − vt )2 A2 σ = 2t exp −2 per unit = 4 2 ∂ 2 t w w length is the kinetic energy per unit length of string, as a function of x and t (recall that σ is the mass of string per unit length). If we integrate over the entire string we obtain the total kinetic energy in the string at any time: ∞
2
1 ∂y 2 π A2 t E= ∫ 2 σ ∂t dx =4w
(129)
−∞
Recall that τ represents the tension in the string (newtons) — not time; so this result is independent of time t. This energy is localized in a subset of the string (having a length of a few w) whose center is moving in the +x direction at speed v. Although the energy E is being transported in the +x direction, the string itself is not moving the x direction. Instead, the string vibrates in the y direction. Finally, notice that energy of the wave is proportional to A2, where A is the amplitude of the oscillating wave function. This is true for all waves, including light waves. Transverse vs. Longitudinal Waves In the previous example of the vibrating string, the vibrations are in the y-direction whereas the wave is propagating in the x-direction. This is an example of a transverse wave. Some waves correspond to vibration in the same direction as propagation. These are longitudinal waves. Sound is an example of a longitudinal wave. Consider a sound speaker which is oscillating in the x-direction at a frequency of ω in air whose mass density ρ depends on the local pressure
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
69
Fall, 2014
p. Treating air as an ideal (but compressible) fluid, Euler’s equation of motion in one-dimension becomes♥
∂ 2 vx ∂t 2 where
= c
= c2
1
(∂ρ ∂p )
∂ 2 vx ∂x 2 ≈ 300
m s
S
is the speed of sound in the air. The subscript S on the partial derivative denotes that it is evaluated along a path on which the entropy S of the gas is held fixed (an ideal fluid is defined as one whose deformations are isentropic). One solution of this equation of motion is ω = vx ( x, t ) U sin ( x − ct ) c
where U is the amplitude of velocity oscillations in the speaker. This is a wave which propagates in the +x direction at speed c — this is the same direction as the movement in the speaker; hence this is a longitudinal wave. Linearly Polarized Plane Waves As we will see now, light is a transverse wave which satisfies the wave equation (126). One particular solution of (126) is a linearly polarized plane wave
(
E= ( r, t ) E0 cos ωt − k.r
)
(130)
where ω is the angular frequency (radians/sec) of oscillation, r is the position vector and k is a vector pointing in the direction of propagation of the wave. This is the type of light which is produced by a HeNe (ruby red) laser such as a laser pointer. To facilitate comparison with d’Alembert’s solution (128) of the wave equation, let’s write this vector equation in terms of its scalar components in Cartesian coordinates. Let’s align the x-axis with the wave vector: k = ke x and we align the y-axis with E0:
E0 = E0 e y
The vector E0 must be orthogonal to k, as we have just arranged. Then the ♥
See pages 38-40 of my Fluids notes from 2008. These notes can be found in the “Lectures” menu of Blackboard. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
70
Fall, 2014
y-component of (130) becomes E y ( x, y= , z , t ) E0 cos ( ωt − kx )
and
Ex = Ez = 0
(131)
The colors in the plot at right represent Ey values calculated from this equation at one particular t. To deduce the velocity of propagation, we ask how does x need to vary with t such that the argument of the cosine remains constant (as it would if we rode along like a surfer on the crest of any wave). Requiring the argument to be constant on a crest: ωt − kx =const, say = 0
or = x
ω = t vt k
Thus we conclude that the wave is propagating in the +x direction at a speed given by v=
ω k
This is analogous to the particular solution F(x – vt) in (128). Linear, Circular and Elliptic Polarization (130) is called linearly polarized because the electric field at a given position r undergoes a periodic linear trajectory with time. To see this, consider the time-dependence of E at a particular point in space, say at r = 0.
E = ( 0, t ) E0 cos ( ωt )
(132)
This is a vector with a fixed direction whose magnitude varies with time. The head of such a vector would trace out a linear trajectory with time, which is why such a wave is said to be “linearly polarized.” There are also circularly polarized waves, in which the electric field traces a circular orbit with time. Such a wave can be obtained by adding two linearly polarized plane waves of the form of (132) provided that E0 for the two waves are 1. equal in magnitude 2. orthogonal in direction, and
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
71
Fall, 2014
3. the two plane waves are out of phase by 90° (π/2) Such a wave can be represented by:
(
= E ( r, t ) E0 e y cos ( ωt − kx ) + E0 e z cos ωt − kx − π 2
)
(133)
Notice that this expression for the electric field is also a particular solution of the wave equation (126). If the magnitudes of the two waves being summed are not equal but they are still 90º out of phase and orthogonal in direction, then their summation produces an elliptically polarized wave. One example is E ( r, t ) = e y cos ( ωt − kx ) + 0.2e z cos ωt − kx − π 2 E0
(
)
(132) is called a plane wave because, at a particular t, the surfaces of constant E are planes; in particular, surfaces having E = E0 are wave fronts. For plane waves, these fronts are planar and the planes are orthogonal to the wave vector k and are spaced one wavelength apart. To see this, note that
= k.r kr cos θ Consider a plane which is orthogonal to k. All points r on this plane have the same value for rcosθ (and that value is the length of the red line in the figure above): as θ increases, r increases but cosθ decreases in such a way that their product remains constant. Since k.r = const on this plane, so is E. Lecture #10 begins here Monochromatic vs. Polychromatic Waves Our LPPW in (131) has a single frequency ω = kv and a single wavelength l = 2π/k. = E y ( x, t ) E0 cos ( kx − ωt )
(131)
Such light is said to be monochromatic. Light produced by the sun (and most lamps) consists of a wide spectrum of wavelengths (or frequencies). Each wavelength has a particular solution of the form of (131). Since the wave equation (126) is linear, a linear combination of contributions like (131) — one for each wavelength — is also a particular solution of the wave equation: = E y ( x, t )
( k j x − ω j t ) ∑ E j cos k j ( x − v j t ) ∑ E j cos= j
Copyright© 2014 by Dennis C. Prieve
(134)
j
PDF generated November 30, 2014
06-712
72
Fall, 2014
where we have substituted ωj = kjvj. Such a solution is polychromatic. It turns out that any periodic shape (even a discontinuous square wave) for the initial waveform Ey(x,0) can be expressed as a sum like (134), which is a Fourier series representation. If this polychromatic wave is propagating through vacuum, then all wavelengths travel at the same speed vj = c: = E y ( x, t )
∑ E j cos k j ( x − ct )
(135)
j
Under these conditions, the shape of the wave remains unchanged (with changes in t) as the wave propagates in the +x direction at speed c; only the center of the waveform shifts with t — its shape does not change with t. (135) is an example of a non-dispersive wave. On the other hand, if this polychromatic wave propagates through some dielectric media like air or water, then the refractive index of the material varies with the wavelength or frequency of the light: n=
c v
or
vj =
c n (ω j )
Then different wavelengths travel at different speeds and the shape of the wave changes as it propagates. This produces a dispersive wave. Attenuated and Evanescent Waves Below in Absorption of Light Energy by a Conductor, we will see that absorption of light energy by some materials causes the amplitude of the oscillations to decrease as the light propagates through the material. Mathematically, this can be captured by writing (131) ♠ = E y E0 cos ( kx − ωt )
(131)
using complex exponential notation (see also Representation of Oscillations as Complex Exponentials on page 8): = E y E0 exp i ( kx − ωt )
(136)
Suppose we take the wave number k to be a complex number, having both real and imaginary parts: k kr + iki = ♠
I have changed the order of the difference in the argument of the cosine (i.e. multiplied the argument by –1). This has no effect on the value of cosine (which is an even function of its argument) but this avoids having to deal with negative numbers in what follows. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
73
Fall, 2014
Then (136) becomes
{
(
}
)
= E y E0 exp i ( kr + ik = E0 exp i 2 ki x exp i ( kr x − ωt ) i ) x − ωt = E0 e− ki x exp i ( kr x − ωt ) Taking the real part of both sides: = Re ( E y ) E0 e− ki x cos ( kr x − ωt )
(137)
Now the amplitude of the oscillations decays exponentially as the wave propagates in the xdirection. As light energy is absorbed by the material, the remaining energy of the wave (proportional to Ey2) decays exponentially with x. The imaginary part ki is also known as the absorption coefficient. We will also encounter situations in which the wave number is purely imaginary. Taking kr = 0 in (137) yields: = Re ( E y ) E0 e− ki x cos ( ωt )
Now the wave doesn’t oscillate in x but simply decays exponentially. evanescent wave.
This is called an
The Diffusion Equation The 1D wave represented by (136) is a particular solution — not only of the wave equation, but also — of the diffusion equation:
∂θ ∂2θ =D ∂t ∂x 2
(138)
where θ(x,t) might represent solute concentration in an unsteady diffusion problem (then D is the diffusion coefficient of the solute), temperature in an unsteady heat-transfer problem (then D is the thermal diffusivity) or the x-component of velocity in an unsteady diffusion of momentum (then D is the kinematic viscosity). Let look for a particular solution to (138) which has the form of (136): = θ ( x.t ) A exp i ( kx − ωt )
(139)
After evaluating the partial derivatives in (138), we have = −iω A exp i ( kx − ωt ) D ( ik ) A exp i ( kx − ωt ) 2
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
74 k2 =
Solving for k2:
−iω D
k= (1 + i )
Assuming ω and D are real:
Fall, 2014
ω 2D
Substituting this back into (139) and taking the real part of both sides: ω ω Re {θ (= x, t )} A exp − x − ωt cos x 2D 2D
(140)
which is a wave propagating in the +x direction whose amplitude decays expontially with x. One physical problem which has this solution is the profile of vy(x,t) next to an infinite plate at x=0 which oscillates back and forth in the y-direction at frequency ω. This oscillation produces ymomentum which diffuses in the x-direction. Although the differential equation has a different form, (140) has the same form as (137) which represents light propagating through an absorbing media. Both absorption of light and viscous dissipation of momentum represent irreversible losses of energy from the wave, which lead to the decaying oscillations. Spherical and Cylindrical Waves So far our study of solutions to the vector wave equation
∇2 E =
µε 1
∂2E
(141)
∂ 2t
c2
has been restricted to plane waves. 1-D particular solutions to (142) are also available in spherical and cylindrical coordinates. In cylindrical coordinates, Eθ(r,t) or Ez(r,t) satisfies 1 ∂ ( rEθ ) 1 ∂ 2 Eθ = r ∂r c 2 ∂t 2 A particular solution can be found having the form
= Eθ ( r , t ) A ( r ) exp [i ( kr − ωt )] where
Copyright© 2014 by Dennis C. Prieve
a A(r ) = 0 r
(143) (144)
PDF generated November 30, 2014
06-712
75
Fall, 2014
(143) and (144) is a cylindrical wave. The wavefronts are cylinders which are moving radially away from the axis at a speed c = ω/k. The amplitude decays with r because as the energy radiates out, the same total energy is being spread over a larger area. In spherical coordinates, Eθ(r,t) or Eφ(r,t) are given by
(
)
2 2 1 ∂ r Eθ 1 ∂ Eθ = ∂r r2 c 2 ∂t 2
A particular solution can again be found having the form of (143), but now the coefficient function is different:
a A(r ) = 0 r
(145)
(143) and (145) is a spherical wave. The wavefronts are spheres which are moving radially away from the origin at a speed c = ω/k.
Energy Flux Reference: Jackson, 3rd ed., Sect. 6.7 Let’s again recall Maxwell’s equations:
∇.D = ρ
Coulomb’s law:
(68)
Maxwell’s generalization of Ampere’s law: Ampere-Maxwell’s law:
Faraday’s law of induction: and
∇ × H= J + ∇×E = −
∂D ∂t
(110)
∂B ∂t
(107)
∇.B = 0
(93)
Besides Maxwell’s equations, the only remaining fundamental equation is the Lorentz force equation, which gives the force on a particle having, charge q and velocity v, experiencing electric and magnetic fields given by E and B, respectively: F= q ( E + v × B )
(146)
Recall that the rate of doing work is force times velocity, or in vector form
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
76
Fall, 2014
dW = F.v = qE.v + q ( v × B ).v (( dt
(147)
0
where the second equality was obtained by substituting (146). The magnetic field B does no work on the particle because it exerts a force which is orthogonal to v (recall the cross product of two vectors is orthogonal to both). If instead of a single point charge q, we have a continuous distribution of current density given by J(r,t), then in a differential volume element dV having charge dq, the following change of variables can be made: dq v = J dV
(148)
In other words, a charge dq moving at velocity v is like a current moving in the direction of v. This proves the equality in direction of either side; to justify the equality of the magnitude, consider the units of either side, which are the same: coul
m [ = ] coul m3 s s-m 2
The rate of doing work on the contents of a finite volume V, having distributed current density J and electric field E is obtained by substituting (148) into the right-hand side of (147) and summing over differential volume elements in V:
dW = ∫ E.J dV dt V
Next, we apply the Ampere-Maxwell law (110) to express J in terms of other quantities:
dW = dt Identity C.3 requires
∂D
∫ E.( ∇ × H ) − E. ∂t dV
(149)
V
) b.( ∇ × a ) − a.( ∇ × b ) ∇.( a × b=
where a and b are arbitrary vector fields. Taking a = E and b = Η and solving the equation above for a.(∇×b):
) H.( ∇ × E ) − ∇.( E × H ) E.( ∇ × H = Next, we use Faraday’s law (107) to eliminate ∇×E: E.( ∇ × H ) = − H.
∂B − ∇.( E × H ) ∂t
(150)
Substituting (150) into (149):
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
77
dW = dt
∂B
∫ −H. ∂t
V
Fall, 2014
∂D − ∇.( E × H ) − E. dV ∂t
(151)
The time derivatives in the integral above turn out to be internal energy terms in an energy balance. Here we take a “time out” to determine the expression representing the local electrostatic energy density corresponding to a particular local electric field E(r) and a particular local displacement vector D(r) = ε0E + P, representing the local state of polarization P(r). Lecture #11 begins here Time Out: Electrostatic Energy Density in a Simple Dielectric Recalling that electrostatic potential ψ(r) represents the electrostatic energy per unit charge, you might be tempted to say that the local energy content is simply ρ(r)ψ(r)dV, but this is not quite right. As charge builds up, the potential ψ changes, so we need to integrate over the changes. Suppose at some intermediate stage of charging, we have charge density ρ(r) and potential ψ(r). Consider the incremental addition of a differential change δρ in the local charge density ρ. This process requires work δW and will produce a small perturbation δψ in the local potential. We use δ’s to denote the changes in each quantity: ρ → ρ + δρ,
W → W + δW
and
ψ → ψ + δψ
So to change by δρ the charge density ρ of a differential volume element dV, the work required is dδW = dq ψ = (δρ dV) ψ This addition of charge also perturbs the local potential by dψ. To obtain the total work, we need to integrate the above over the changes occurring throughout the universe V∞. The incremental work required can be calculated from
dW=
) ψ ( ) dV ∫ dρ ( ρρ
(152)
V∞
where ψ(r) in this integral is the potential arising from the pre-existing charges (before δρ was added). After the addition of this increment of charge, Coulomb’s law for continua (68) requires after:
∇.( D + δD ) = ρ + δρ
whereas before the addition, Coulomb’s law gave before: Subtracting the two:
Copyright© 2014 by Dennis C. Prieve
∇.D = ρ
∇.( δD ) = δρ
(153)
PDF generated November 30, 2014
06-712
78
Fall, 2014
Clearly this change in charge density causes a change δD in the displacement vector, which causes a change δE = –∇δψ in the electric field and therefore a change δψ in the electrostatic potential. With (153), the integrand of (152) becomes
( δρ ) ψ =
ψ∇.( δD )
∇.( sv ) = ∇s.v + s ( ∇.v )
Vector identity C.1 requires
where s and v are arbitrary scalar and vector fields, respectively. Substituting s = ψ and v = δD and solving for ∇.v:
( δρ ) ψ =
. ∇.( ψ δD ) −∇ψ δD +E
The integral of (152) becomes
∫ ∇.( ψ dD ) dV + ∫ E.dDdV
dW=
V∞
(((( ∫ n.( ψdD ) da = 0
V∞
A∞
where we have applied the Divergence Theorem to the first integral and A∞ is the outer boundary of the universe whose interior is V∞. In most problems, the charges are localized in some small subset of this region. When you are very far away from this charged region (e.g. on the boundary of the universe A∞), the potential ψ→0 and δD→0; indeed, far from the charges is usually chosen as the reference state for potential. So the first of the two integrals can be taken as zero, leaving
dW=
∫ E.dDdV V∞
Now there are two differentials in the above expression: δD (arising from δρ) and dV. To calculate the total work required to assemble all the charges in our system, we have to integrate again = D D= D D
W =∫ dW =
∫ ∫ E.dDdV = ∫
∫
dV
E.dD
(154)
= D 0= V∞ V∞ D 0
When E and D are linearly related (recall D = εE for many materials), then we can write E.δD = E. ( εδE ) = ε ( E.δE )
(155)
To simply this last expression, consider applying the chain rule to the derivative of the dot product of E with itself: Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
79
Fall, 2014
δ ( E.E ) =δE.E + E.δE =2E.δE
But the order of the dot product makes no difference, so the two terms on the right-hand side of the first equation are equal, which leads to the second equation above. With this substitution (155) becomes E.δD=
1 εδ ( E.E ) 2
After substituting this last expression into (154), the inner of the double integrals above involves an exact differential, which can be easily integrated: = D D= D D
∫
= E E
1 1 1 1 εδ ( E.E ) = ε ∫ δ ( E.E ) = ε ( E.E ) = ( E.D ) 2 2 2 2
E.δD = ∫
= D 0= D 0
= E 0
where E and D in this last expression are the final values after charging. Then when we integrate over all the increments in charge required to build up the final state of charge in our system, the total work required is W =
Thus we identify
= uE
1 2
∫ E.DdV V∞
e 2 1 energy = E.D = E [ ] 2 2 vol
(156)
as the local electrostatic internal energy per unit volume. The analogous quantity for magnetic fields is♠ energy 1= = uM B.H [ ] 2 vol
(157)
Time In: Now let’s return to our energy balance represented by (151):
dW = dt
∂B
∫ −H. ∂t
V
∂D − ∇.( E × H ) − E. dV ∂t
(151)
Integrating the sum of these two energies (156) and (157) over the volume of our region V gives the internal energy of the region:
= U
1 ( E.D + B.H ) dV 2∫ V
♠ see Jackson, 3th
ed, Sect. 5.16. In particular, see (5.148) on p213.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
80
Fall, 2014
When the linear constitutive equations (D = εE and B = µH) are used, the time derivative of this integral can be written as dU 1 d = dt 2 dt
) dV ∫ ( εE.E + µH.H= V
1 ∂ ( εE.E + µH.H ) dV 2 ∫ ∂t V
∂E ∂H ∂H 1 ∂E .H + µH. dV ε .E + εE. + µ ∫ ∂t ∂t (( ∂t (((( ∂ t 2 (((( V D B ∂ ∂ 2E. 2 H. ∂t ∂t ∂B ∂D = ∫ E. + H. dV ∂t ∂t =
V
Substituting this result into (151):
dW dU = − − ∇.( E × H ) dV dt dt ∫ V
or
dU dW + ∫ ∇.( E × H ) dV = − dt dt V
Applying the Divergence Theorem to the remaining volume integral dU dt
dW + ∫ n.( E × H ) da = − dt A accumulation (((( rate of of internal energy
net rate of energy transported out of system
wor done on system
This result can be considered an energy balance for the arbitrary system represented by the region V enclosed by surface A.♣ This energy balance is called Poynting’s Theorem (1884).♥ Given the form of the surface integral and its interpretation as the rate of transport across the surface element da, we can conclude that Π ≡ E× H
(158)
must represent the energy flux by electromagnetic waves. It is called the Poynting vector:
♣
In this energy balance, we have only considered electromagnetic energies. There are other forms which internal energy can take (e.g. thermal: dU = Cv dT) and other forms for work (e.g. pV work). ♥ John Henry Poynting (1852-1914), English physicist known primarily for deducing the formula for calculating energy flux from electromagnetic radiation. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
81
Fall, 2014
Let’s try to calculate the Poynting vector for the linearly polarized plane wave given by (130).
(
)
= E ( r, t ) E0 exp i ωt − k.r
(130)
The electric field at r = 0 is E(0,t) = E0 eiωt and Re(E) = E0 cosωt whereas the magnetic field at the same point is calculated from (17): H =
1 1 1 k×E = k × E0 eiωt and Re ( H ) = ( k × E0 ) cos ωt µω µω µω
In computing the Poynting vector using complex exponential notation, we should use the real parts of E and H.* Substituting the real parts into (158):
= Π
1 E0 × ( k × E0 ) (( µω
2 cos= ωt
1 2 2 E0 k cos ωt ( ( µω
(159)
cos ( 2ωt ) +1
points in direction of H
((((
2
points in direction of k
Notice that Π is a vector pointing the direction of k: the direction of propagation of the wave. This is consistent with our understanding that energy is transported in the direction of light travel. Moreover, the magnitude is proportional to the square of the amplitude of the electric field; this is one of the general characteristics of waves [recall (129) and note that the energy is proportional to the square of A, the amplitude of the wave]. The magnitude of this vector also oscillates with time, but at twice the frequency of the electric or magnetic fields. If we average the energy flux over one period we obtain: 2π
Π ≡
2π
ω
E0 k ∫ Π (t ) dt = 2µω ω
1
1
2
(160)
0
The magnitude of this vector can also be written as half the product of the amplitudes of E and H:
* As along as we only perform linear operations, we can simply drop the imaginary part (introduced for mathematical convenience) after the operation and obtain the same result as if the same operation had been performed on the real part of the original quantity. However, this cross product of two complex quantities represents a nonlinear operation (like squaring). See Stratton, p137. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
82 = Π
Fall, 2014
E02 1 E0 1 = E0 = E = H [ 2 0 0 mω 2mω 2
]
NA J = 2 C m m −s
(161)
H0
And if we keep track of the units, we see that this does have units of energy flux, as advertised. Time-Average of Products of Harmonic Functions E and H are harmonic functions of time, meaning that their time dependence is sinusoidal with a single frequency ω. Time averages of products of such complex functions can be computed more easily by replacing one of the functions by its complex conjugate and then dividing the result by two. For example, if the Poynting vector is calculated from (158) Π ≡ E× H
(158)
then the time-average in (160) can be computed directly from Π =
1 E× H* 2
(162)
where H* is the complex conjugate of H. If E is given by (130):
(
)
= − k.r E0 eiωt exp ( −ik.r ) E ( r, t ) E0 exp i ωt =
(130)
then H can be calculated from (17): H =
1 1 k×E = k × E0 eiωt exp ( −ik.r ) µω µω
The complex conjugate is computed by replacing i by –i: = H*
1 k × E0 e−iωt exp ( ik.r ) µω
(163)
Subsituting (130) and (163) into (162):
Π =
1 1 E0 eiωt exp ( −ik.r ) × k × E0 e−iωt exp ( ik.r ) 2 µω
E02 1 = E0 × ( k × E = k ) 0 2µω 2µω
(164)
where the cross products were computed as in (159). The result is identical with (160).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
83
Fall, 2014
Lecture #12 begins here
Revisited: Reflection of Light from a Sharp Boundary Reference: see §5.4 of Lipson. Let’s now return to the first problem considered in this course: reflection of light off a plane interface between two phases: see chapter entitled Reflection of Light from a Sharp Boundary starting on page 9. At right is a unit circle in the plane of incidence defining the four unit vectors: ni, nr and nt give the directions of the incident, reflected and transmitted waves while n is normal to the interface. Conservation of Energy Is the total energy of the two daughter rays equal to the energy of the incident ray? It should be. The magnitude of the energy flux is given by (161). Let’s call this qj: ω vj µ =µ0 k=
= qj
kE 2j 2µω
→
E 2j 2µ0 v j
(165)
where the j refers to the wave: j = i, r, or t. Recall that in heat transfer, heat (energy) flux is a vector, pointing in the direction the heat is transported. We can make this quantity a vector simply by multiplying it by a unit vector in the direction of transport:
qi =
Ei2 ni 2µ0 vi
Conservation of energy requires♣
d ∫ u dV = − ∫ n ⋅ q da dt V A accumulation
♣ See Prob.
net rate in
7 of Hwk 3 in 06-703.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
84
Fall, 2014
where u is the internal energy per unit volume, V is the arbitrary 3D system we have chosen to perform a balance on and A is the surface of this region. For our system, let’s choose the cylindrical pillbox shown at right. In the limit that the thickness l of the pillbox shinks to zero, its volume V also shrinks to zero (although the area Af is finite) so that the volume integral vanishes in the above equation, leaving
lim
l →0
0 ∫ n ⋅ q da = A
where n is a unit vector which is everywhere normal to the local differential surface area element da. By making the pillbox sufficiently thin (l → 0), any contribution from integrating over the curved sides of the cylinder will be negligible compared to that from integrating over either end. The contribution from the right-end of the cylindrical pillbox
nt
is
∫ Aright end
Et2 Et2 cos θt ⋅ = ⋅ = ⋅ = Af n θ da n θt A f n nt A f 2µ0 v2 2µ0 v2
θt n mat'l #1
where we have substituted qt from (165), replacing the subscript “j” by “t”. The contribution from the left-end is slightly more involved simply because we have two contributions to the flux (one from the incident wave and another from the reflected wave): n ⋅ θ da =n ⋅ ( θθ i + r ) Af
∫ Aleft end
=
(
nr
)
1 Ei2 n ⋅ ni + Er2 n ⋅ n r A f 2µ0 v1
π − θi
ni
θr
n
1 2 2 Ei cos ( π − θi ) + Er cos θr A f = ( ( 2µ0 v1 cos θi − cos θi
mat'l #2
θi
ni
Adding the contributions from the two ends and requiring the sum to vanish, leads to
mat'l #1
mat'l #
Ei2 Et2 Er2 cos θi cos θi + cos θt = 2µ0 v1 2µ0 v1 2µ0 v2 Dividing through by the left-hand side:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
85
Fall, 2014 2
2
E v cos θt E = 1 r + t 1 Ei Ei v2 cos θi R
c n1 c n2
T
Now the ratio of the electric field amplitudes are R and T — the two Fresnel coefficients for the electric field. In addition, recalling (32)
c n
(32)
n cos θt 2 = 1 R2 + 2 T n1 cos θi R
(166)
n≡
c v
v=
or
Thus conservation of energy requires
c
where
R ≡ R2
and
n cos θt 2 c ≡ 2 T n1 cos θi
(167)
The factor of T2 is accounting for the different electric field amplitudes in the transmitted and incident waves while the ratio n2/n1 accounts for the different velocities of the two waves (which affects the fluxes). On the other hand, the ratio cosθt /cosθi accounts for the different areas of the two beams. To see this, consider the figure at right showing the reflection and transmission of two parallel rays having an angle of incidence of θi which define the width of the incident beam having a cross-sectional area Ai. The projection of the incident beam onto the interface has an area of Af while the crossection area of the reflected and transmitted beams are Ar and At. The red angles are all θi while the blue angles are θt. From the three right triangles we see Ai Ar cos θi = = Af Af
Thus
and
At = cos θt Af
cos θt At = cos θi Ai
is an area ratio. And power is energy flux times area. Thus the Fresnel ratios R and T defined by (167) are for power — not flux. Since the area of the reflected beam equals the area of the incident beam, the area ratio Ar/Ai = 1 and no area correction appears in the formula for R .
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
86
Fall, 2014
Recall our previous results for Fresnel’s coefficients for the electric field in TE mode:
and
R cos θ − n21 cos θt R⊥ ≡ 0 = i I 0 cos θi + n21 cos θt
(37)
T 2 cos θi T⊥ ≡ 0 = I 0 cos θi + n21 cos θt
(38)
or in TM mode:
R ≡
cos θt − n21 cos θi cos θt + n21 cos θi
(39)
and
T =
2 cos θi cos θt + n21 cos θi
(40)
Using either set of R and T we can evaluate R and T from (167). The calculated results for n21 = 1.5 are shown below.
Notice that R + T = 1 as required by conservation of energy (166).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
87
Fall, 2014
Brewster Angle Microscopy As with the Fresnel coefficient for the electric field amplitude (recall figure on page 16), the Fresnel coefficient for power of the reflected wave (when E is parallel to the POI, or TM mode) also vanishes at the Brewster angle. Brewster angle — angle at which R =0 for an external reflection in TM mode.♣ Setting R =0 in Fresnel’s equation (39) leads to
cos = θt n21 cos θi
(168)
In addition, these two angles are related by Snell’s law (33): sin θi = n21 sin θt
(33)
Eliminating θt between these two equations leads to (see Hwk) n2 sin 2 θi = 21 2 1 + n21 Taking the square root of both sides and dropping the negative root (since θi can’t be negative): sin θi =
also
cos θi =
n21 = 2 1 + n21 1 − sin 2 θi =
n2 n12 + n22 n1 n12 + n22
(169)
(170)
Taking the ratio of sin/cos = tan gives a simpler result: tan θiBrew = n21
which yields θiBrew = 56.3º when n21 = 1.5. The Brewster angle is also called the “polarizing angle” because when unpolarized light (i.e. an arbitrary mixture of TE and TM modes) is incident upon the interface at this angle, the reflected light contains ♣
Sir David Brewster (1781-1868), Scottish physicist, mathematician Figure and inventor, known for 2 developing the laws of polarization by reflection. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
88
Fall, 2014
only the electric field component which is normal to the plane of incidence (TE mode). Thus upon reflection at this angle, unpolarized light becomes linearly polarized. Finally we point out that — at the Brewster angle — θi and θt are trigonometric complements of each other (see Hwk). From the geometry of the triangle in Figure 3: sin θi= cos θt=
n2 n12
(171)
+ n22
Figure 3 cos θi = sin θt =
n1 n12
(172)
+ n22
Moreover, the parts of (171) and (172) dealing with θi satisfy (169) and (170). The parts dealing with θt also satisfy (168) and (33) above. In other words, their sum θi + θt = 90º so the reflected ray and the transmitted ray are orthogonal (see Figure 2). The effect of polarization is illustrated visually in the two side-by-side photos copied below:♦
without polarizer
with polarizer
The center photo shows the view (without the polarizer) taken at the Brewster angle, measured in a horizontal plane of incidence, of a store window having three metallic mannequins. This center photo also contains a reflection (off the store window) of another building having three windows on the second floor and even more windows on the first floor. The photo at far right was taken at the same Brewster angle but after the light passes through the polarizer, which is designed to block light having a vertical electric field. Given that the plane of incidence is ♦
This is Fig. 5.6 in Lipson, 4th ed.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
89
Fall, 2014
horizontal, the polarizer is blocking TE or s-polarized light. Since the act of reflection at the Brewster angle is eliminating the TM or p-polarized light, none of the reflection should appear. The reflection of the other building has almost disappeared. Light coming from the mannequins is not being reflected and has both s- and p-polarized light. The p-polarized light passes through the filter, but the reflection of the other building does not. One fairly recent♠ innovation which exploits this polarization is Brewster-angle microscopy (BAM) which can be used to observe dynamics of microstructure of adsorbed monolayers at the air-water interface. A commercial instrument is now available, whose schematic is shown below:
The horizontal monolayer is illuminated with a laser oriented at the Brewster angle with respect to the vertical to the interface. Water has a refractive index of n2 = 1.33 while air is essentially n1 = 1.00. This yields a Brewster angle of tan–1(1.33/1.00) = 53.0º. The reflection is photographed. The three figures below are sample images of a steric acid monolayer, which changes as the monolayer is squeezed by the Langmuir trough:
♠
developed independently by Honig & Mobius, J. Phys. Chem. 95, 4590-4592 (1991) and by Henon and Meunier, Rev. Sci. Instrum. 62 936 (1991). See also Cohen Stuart et al., Langmuir 12 2863-2865 (1996). Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
90
Fall, 2014
These images (perhaps 100 µm by 100 µm) are viewed through a microscope. The black areas indicate no reflection and no adsorbed layers. Reflectometry for the analysis of adsorbed surface layers is performed at angles near the Brewster angle because even a thin surface layer will cause the reflection coefficient to become nonzero. The magnitude of the reflection coefficient can be used to infer (for example) the refractive index of the adsorbed layer, which can be related to the mass of adsorbate per unit area. We will delay further discussion of this until we have covered interference effects below. Lecture #13 begins here Internal Reflection In the examples we looked at above, the incident and reflected waves are on the side of the interface having the smaller refractive index (n1 < n2). This situation is described as external reflection because it usually arises from reflecting a wave propagating in air off the exterior of a glass prism or other condensed matter. external reflection — light is incident from the less dense side of the interface; i.e. from “outside” the solid (n21>1) When n21 ≡ n2/n1 > 1, Snell’s law (33) predicts
sin= θt
sin θi < sin θi n21
so that θt < θi which means that the transmitted ray is bent towards the normal. If instead the incident light comes from the more dense side of the interface we have an internal reflection — light is incident from the more dense side of the interface; i.e. from “inside” the solid (n21 θi, which means that the transmitted ray is bent away from the normal.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
91
Fall, 2014
Under conditions of internal reflection, there exists one particular angle of incidence at which the refracted ray is parallel to the interface (as shown at right). This is known as the critical angle. So at the critical angle, the angle of the transmitted ray θt = 90º or
n1 sin= θi n2 sin θt 1
From Snell’s law (33), the corresponding angle of incidence is given by
sin θcrit ≡ n21
(173)
For glass-air (i.e. n21 = 0.67), this corresponds to θcrit = 41.8°. For glass-water (n21 = 1.33/1.5), this corresponds to θcrit = 62.5º. Fresnel’s formulas (37)-(40) still apply. In the plot below we show the Fresnel’s coefficients vs angle of incidence, under conditions of internal reflection for the particular case in which n21 = 0.67 (glass/air) for which the critical angle is 41.8º. First we examine the response for angles of incidence below the critical angle. In the next section, we will look at the response above the critical angle.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
92
Fall, 2014
Notice that the electric-field coefficients (T’s) for the transmitted wave are all larger than one for this internal reflection, suggesting that the transmitted wave is more intense than the incident one. Indeed T approaches 3 times the incident wave as we tend toward the critical angle. By contrast the electric-field amplitudes were always less than one for external reflections (both R’s and T’s). Below are the Fresnel coefficients for power, calculated from (167). Of course, the total power of the reflected and transmitted waves still equals that of the incident wave, as can be seen in the lower graph above.
To see how T can be greater than unity and still conserve power, recall (167) which gives the ratio of power of the transmitted wave to the power of the incident wave:
= lim c θi →θcrit
n2 cos θt 2 T → 0 n1 cos θi 2 3 = 9
0.667
0
The numbers beneath each term give the asymptotic values of the quantities as we approach the critical angle. Although the intensity of the electric field is much greater than for the incident wave, the cross-sectional area of the transmitted beam is shrinking to zero (at the critical angle θt →90º and cos θt →0). Above the critical angle of 41.8º for the glass-air interface, the coefficients become complex numbers (see next section). We have only displayed their values in the graphs above for angles of incidence below the critical angle, for which they are purely real. Notice that when Ei is parallel to the plane of incidence (i.e. for TM mode), the reflection coefficient still vanishes at a
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
93
particular angle below the critical. reflections.
Fall, 2014
Thus the Brewster angle still exists even for internal
Total Internal Reflection For angles of incidence greater than the critical angle, Snell’s law (33) predicts
sin θi sin θcrit sin θi = >1 n21 n21 sin θcrit
sin= θt
for θi>θcrit:
(174)
>1
1
thus θt becomes complex and we no longer have a LPPW in material #2. This situation is called total internal reflection. To see why this name is used, let’s examine this situation more closely. Fresnel’s formulas for the reflection and transmission coefficients are still valid for θi > θcrit, except that the cosθt will be purely imaginary: 2
2
sin θi sin θi cos θt = 1 − sin 2 θt = 1 − = ±i − 1 = −i β sin θcrit (( sin θcrit >1 ((((
(β > 0)
(175)
θcrit:
for θi > θcrit:
cos θi + in21β 2 cos θi = T⊥ cos θi − in21β cos θi − in21β
n cos θi + iβ R ≡ − 21 n21 cos θi − iβ
T =
2 cos θi n21 cos θi − iβ
(176)
(177)
Properties of the Reflected Wave for TIR Both reflection coefficients have the form:
R =
u − iv = e −i α u + iv
(178)
where u and v are real (but not necessarily positive)
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
94
Fall, 2014
= a arctan ( 2= uv, u 2 − v 2 ) 2 arctan ( v, u ) ♥
(179)
The polar form of the complex coefficient is more useful for our application because the exponent contains phase information only whereas the amplification is given by the preexponential factor, as we will now illustrate. Let’s focus on s-polarized light (TE mode) and suppose the plane of incidence is the xyplane. Then the electric field of all three waves (incident, reflected and transmitted) have only a z-component:[see (21) and (22) on page 10]
Eiz = I 0 e (
i ωt − k i .r )
(180)
i ωt − k i .r ) Erz = R0 e (
(181)
i ωt − k i .r ) Etz = T0 e (
(182)
Let the incident light be a LPPW, then I0 in (180) is purely real and the real part of the incident electric field is simply a cosine:
{
}
i ωt − k i .r ) Re {= Eiz } Re I 0 e ( = I 0 cos ( ωt − k i .r )
For supercritical angles of incidence, the reflection coefficient is given by (176) is complex. Writing the coefficient R0 in the form of (178), the reflected wave (181) becomes : = Erz
) r = R0 e ( i ωt − .r
( e−iα= ) I0ei (ωt − .r ) r
i ωt − r .r −α ) I0 e (
R⊥ I 0
or
Re (= Erz ) I 0 cos ( ωt − k r .r − α )
Thus the reflected wave has the same amplitude I0 as the incident wave, which results because the pre-exponential factor of R in (178) represents the magnitude R of the complex number, which is unity; but the oscillations are shifted in phase by α radians. Since R = 1, the power of the reflected wave contains all of the power of the incident wave, leaving no power in the transmitted wave. This why this situation is called total internal reflection: all of the power is being reflected.
♥
Even when the angle is restricted to a particular full period (say –π < α ≤ +π), arctan is a multivalued function of the ratio of its arguments (i.e. of the tangent). To get a single-valued function, we need to specify not only the ratio of the two sides of the right triangle, we also need to specify which quadrant the angle is in. This is usually accomplished by specifying the length of the two sides (including sign) as separate arguments, as we have done in (179).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
95
Fall, 2014
The phase angle α depends on the polarization of the incidence wave (i.e. on whether Ei is parallel or perpendicular to the plane of incidence). For perpendicular (s) polarization, the reflection coefficient is given by (176). Comparing this with (178), we have u = cosθi Then (179) gives
and
v = –n21β
n β a ⊥ =2 arctan ( −n21β, cos θi ) =−2 tan −1 21 cos θi
♦
(183)
For parallel (p) polarization, the reflection coefficient is given by (177). Comparing with (178), we have u = n21 cosθi or
and
v = –β
n cos θi a = π + 2 arctan ( −β, n21 cos θi ) = 2 tan −1 21 β
♣
(184)
The absolute magnitude R and phase angle α [calculated from either (183) or (184)] of the reflection coefficients for either polarization are shown in the plots below for the glass-air interface (n21 = 2/3): θBrew θcrit
θBrew θcrit
[taken from Fig. 5.7 on p141 of Lipson, 3rd ed.] Notice that R⊥ vanishes at θi = θBrew = 33.7º. This corresponds to the Brewster angle for internal reflection. The phase angle α changes discontinuously from zero to π at this same ♦
Since –n21β < 0 and cosθi > 0, the resulting α lies in the 4th quadrant (–π/2 < α < 0), so we can simplify the result by using the single-argument tan–1 instead of the dual-argument arctan.
♣ The π added to the arctan accounts for the –1 in front of the remainder of (177). eiπ. In other words, multiplying by –1 adds (or subtracts) π to the phase angle.
Copyright© 2014 by Dennis C. Prieve
Note that –1 =
PDF generated November 30, 2014
06-712
96
Fall, 2014
angle. This change in phase angle occurs because R⊥ changes sign from positive for smaller angles to negative at larger angles. Above the critical angle of incidence (θcrit = 41.8º), the phase angles for either polarization begin to decrease continuously and at nearly the same rate. Indeed the difference between the two phase angles appears almost constant at π radians. The figure below shows the difference between the phase angles over the entire range of incidence angles.
π − ( α − α ⊥ )
[taken from Fig. 5.8 on p143 of Lipson, 3rd ed.] A Circular Polarizer One important application of this phase difference is in the design of a circular polarizer. Recall from (133) that circularly polarized light is just the superposition of two LPPW, of equal intensity, but out of phase by π/2. This phase shift can be accomplished using total internal reflection, except the maximum phase difference in the figure above is about π/4 which occurs at an angle of incidence of 52º. Fresnel designed an optical component — now called the Fresnel rhomb — which produces two internal reflections at 52º, thus generating a total phase difference of π/2. The device is shown in the figure above at right: the overall shape is a rhombus having an angle of 52º. A LPPW is input to rhomb at normal incidence. The electric field of the LPPW is oriented at 45º to the plane of incidence, thus corresponding to equal components of spolarized and p-polarized LPPW. After two reflections, the two components leave π/2 out of phase and their sum is a circularized polarized plane wave. Goos-Hänchen Effect Reference: Jackson, p308 and problem 7.7. In the next section, we will show that the transmitted evanescent wave only penetrates a distance (denoted as δ) into the second media before being totally reflected. For incident beams of significant width (e.g. our He-Ne laser is about 1 mm wide which is significant compared with the wavelength), the reflected rays appears to have translated laterally along the Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
97
Fall, 2014
interface some distance before being reflected. From the geometry of the figure above, this lateral distance appears to be 2δ sinθi. A more careful analysis reveals that the lateral distance depends on the polarization of the incident light. The first-order expressions for D for the two states of linear polarization are D⊥ =
sin θi λ π sin 2 θ − sin 2 θ crit i
sin 2 θ
D = D⊥
and
sin 2 θi − cos 2 θi sin 2 θcrit
This lateral shift in the reflected rays is called the Goos-Hänchen effect. Properties of the Transmitted Wave for TIR The transmitted wave is even more interesting, because — for angles of incidence above the critical angle — it has an altered spatial dependence, which can be seen if we write its mathematical description out in full. (182) becomes . Etz = T0 eiωt e−ik2 nt r
but
so
. n= t r
( e x cos θt + e y sin θt ).( xe x + ye y + ze z )
= x cos θt + y sin θt
Etz = T0 eiωt e−ik2 x cos θt e−ik2 y sin θt
(185)
Now recall (174) and (175):
sin= θt
sin θi sin θcrit sin θi = >1 n21 n21 sin θcrit
(174)
>1
1
cos θt = 1 − sin 2 θt = −i β
(β > 0)
(175)
We can also write sinθt in terms of β:
sin θ= t
1 − cos θt2= 2
where
= β
sin θi n= −1 21
1 − ( −i β ) = 2
1 + β2
(186)
2
sin θi sin θ − 1 crit
(187)
With these substitutions, (185) becomes
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
98
Fall, 2014
−ik2 x ( −iβ ) −ik2 y 1+β2 Etz = T0 eiωt e( ( e e− k2βx
Notice that the exponent with the x-dependence no longer contains an i. This means that the x-dependence is now purely exponential, rather than sinusoidal. Of course, the y-dependence is still sinusoidal as is the t-dependence. To avoid Etz blowing up as x →∞, we needed to chose the negative square-root in (175). 2 Etz = T0 eiωt e− k2 βx e−ik2 y 1+β
(188)
Because the amplitude of the transmitted wave decays to zero as we move away from the interface, we call it an evanescent wave.* Below is a surface plot of electric field intensity for the particular case of internal reflection of an s-polarized LPPW for n21 = 0.667 and θi = 60º.
incident wave
reflected wave
total
Internal Reflection at 60º (above Critical Angle) Compare this to the plot that on page 17 for external reflection of an s-polarized LPPW at the same angle of incidence:
* “Evanescent” means “vanishing” or “tending to vanish.” Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
99
incident wave
Fall, 2014
reflected wave
total
External Reflection at 60º The most noticible difference is in the top-right square which depicts the evanescent wave. This square is mostly green, which corresponds to zero electric field amplitude (red is positive and blue is negative). The blue and red checker-board squares (in lower right square) march to the right at a constant speed with time t. Lecture #14 begins here Penetration Depth Since Etz decays exponentially with x, of interest is how far into the medium #2 does this wave penetrate. Based on (188), the characteristic decay length (or penetration depth) for the electric field amplitude is k2 =
λ pE =
1 = k2 β
2p λ2
λ2 2pβ
(189)
The electric field amplitude drops by a factor of e = 2.71 for any increase in x by this amount. The second equation above was obtained by substituting (18). Example: compare the penetration depth to the wavelength for an angle of incidence which is 1º larger than the critical angle for reflection off a glass-air interface (n21 = 2/3). Solution: The critical angle of incidence is first calculated from (173): sin θcrit ≡ n21 = 2 = 0.6667 3
(173)
which yields θcrit = 41.81º. So the desired angle of incidence is 42.81º and sinθi = 0.6796. Finally β is calculated from (187): Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
100 = β
Fall, 2014
2
0.6796 = − 1 0.20 0.6667
The ratio of the penetration depth to the wavelength can be calculated from (189): λ pE
1 = 0.80 = λpβ 2 2
Thus the penetration depth is similar to the wavelength of normal propagating light in the same media.
The figure above shows how the decay length depends on the angle of incidence for an HeNe laser (l0 = 546 nm in vacuum, l = 364 nm in glass) reflecting off a glass-air interface. The decay length is infinite at the critical angle and approaches an asymptotic constant at glacing incidence (θi = 90°). The Poynting Vector To calculate the Poynting vector Π = E×H for the transmitted wave, we need both the electric field and the magnetic field. For s-polarized light, the electric field is given by (188) as 2 Etz = T0 eiωt e− 2 βx e−i2 y 1+β
and
Etx = Ety = 0
(188)
T⊥ I 0
where T⊥ is given by (176) as a complex number. Using vector notation, we can write the electric field as i ωt − 2 βx −i2 y 1+β i φ − 2 β x = Et T= e ez ⊥ I 0 e e e z T⊥ I 0 e e 2
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
101
Fall, 2014
= Re ( Et ) T⊥ I 0 e− k2 βx cos φ e z φ ≡ ωt − k 2 y 1 + β 2
where
The magnetic field associated with this is given by (20): = Ht
1 k t × Et µ0 ω
where the wave vector was deduced in (185) = nt θt + e y sin θt t 2= 2 e x cos − iβ 1+β2
Substituting (186) and (175) for sinθt and cosθt:
k t = −iβk2 e x + k2 1 + β2 e y
(190)
Now this wave vector must be crossed with Et which has only a z-component. Crossing with ez: k t × e z = −iβ k2 ( e x × e z ) + k2 1 + β2 ( e y × e z ) = k2 (( (( −e y
(
1 + β2 e x + iβ e y
)
ex
Multiplying the rest of the scalars in Et: = Ht
k2T⊥ I 0 iφ − k2 βx 1 k= e e t × Et µ0 ω µ0 ω
(
1 + β2 e x + i β e y
)
where complex conjugate is
= Ht *
k2T⊥ I 0 −iφ − k2 βx e e µ0 ω
(
1 + β2 e x − i β e y
)
Averaging over one cycle of time can be calculated as
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
102
Fall, 2014
1 Et × Ht * 2 1 k T I = T⊥ I 0 eiφ e− k2 βx e z × 2 ⊥ 0 e−iφ e− k2 βx 2 µ0 ω
Π = t
(
=
)
k2 (T⊥ I 0 ) 2µ0 ω
2
(
e−2k2 βx i βe x + 1 + β2 e y
(
)
1 + β2 e x − i β e y
)
When the wave vector k is real, then this cross-product is also real [see (164)]. In this case, k in (190) was complex and so is this cross-product. The real part of this result represents the actual time-averaged energy flux:
{
}
= Re Π t
k2 (T⊥ I 0 ) 2µ0 ω
2
e−2k2 βx 1 + β2 e y
(191)
The real, time-average Poynting vector is parallel to the interface (y-direction) and its amplitude also decays exponentially with distance from the interface, but its decay length is half that of the electric field. In other words, the energy flux decays twice as fast with distance from the interface as the electric field itself. There is no energy flux normal to the interface (x-direction) (unlike for subcritical refracted waves). This is another reason why this situation is referred to as total internal reflection: all of the energy is reflected, not transmitted across the interface.
Applications Total Internal Reflection has been exploited in a number of applications, some of which we will now describe. Wave Guides
Suppose we cause total internal reflection inside a two-sided glass plate, as shown above. The ray will undergo multiple reflections off the two surfaces each surface at the same angle of incidence (see sketch above) until the end of the plate is reached. Except for the entrance to and exit from the plate, the all remaining reflections are total internal reflections, meaning there is no loss of energy, regardless of the number of reflections. Instead of a glass plate, we could also use a long cylinder, as shown in the video frame below. If the cylinder is gently bent, the light will follow the curved path. The bend in the cylinder changes slightly the angle of incidence of adjacent reflections, but as long as the angle Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
103
Fall, 2014
of incidence remains above the critical angle, the light stays inside. The cylinder below or the plate above is acting as a wave guide: to guide the path of the light. One important commercial application is fiber-optic communication of TV signals.
(taken from http://youtu.be/rlo2XeB2qt4) Attenuated Total Reflection Spectroscopy (ATR) Ideally reflections inside a wave guide are total, but in real experiments losses often occur. Once source of loss is absorption of light energy by any material coating the outer surface of the wave guide. This loss is exploited in ATR spectroscopy, were the loss is carefully measured as function of the wavelength or frequency of the incident light. Fourier-Transform Infrared (FT-IR) spectroscopy is a widely used analytical tool for identifying functional groups such as C=O, C-H or N-H which have characteristic adsorption peaks at certain wavelengths. Before ATR became popular, transmission methods were used in which the spectrum of light transmitted through the sample was analyzed. Transmission methods are difficult to apply to opaque materials or powders, since an extremely thin sample has to be used. In ATR, the spectrum of the reflected light is measured; the thickness of the sample seen by the reflected wave is on the order of the penetration depth of the evanescent wave. The figure at right shows a commercial ATR unit.♠ A drop of a liquid sample is being placed on the ATR crystal — in this case a crystal of diamond. The cone-shaped device above the crystal can apply pressure to solid samples, like powders or fibers in order to force a significant amount of material close to the crystal where they will be exposed to the evanescent wave. The figure below shows the ATR spectra obtained by placing a single polyamide granule on either a germanium crystal (red curve) or ♠ Taken
from http://www.azom.com/article.aspx?ArticleID=5958
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
104
Fall, 2014
a diamond crystal (blue curve).
The two curves have absorption peaks at the same wave numbers but the diamond crystal (blue curve) is giving a stronger signal. That’s because diamond has a lower refractive index (n1 = 2.40) than germanium (n1 = 4.01). A lower refractive index gives a higher penetration depth for the evanescent wave — according to (189) and (187) — and a better ATR signal. The photo at right shows a black piece of rubber with a white powder on its surface. Suppose we wanted to identify this white powder. We can scrape some of this white powder off the sample and place it on the ATR crystal for analysis. The figure below shows the ATR spectra for the clean rubber (blue) and the powder (red).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
105
Fall, 2014
Clearly the powder has a different composition from the rubber substrate. The usual procedure to identify an unknown sample like the white powder is to compare its measured spectrum with the spectra taken from a library of known samples. Below we see a good match:
Total Internal Reflection Fluorescence (TIRF) This is another technique exploiting the evanescent wave for illuminating the sample of interest. Here the wavelength of the evanescent wave is chosen to generate fluorescence in the molecules of interest (e.g. proteins). The intensity of fluorescence is then measured. One application is to measure the adsorption of proteins at the interface where the evanescent wave is generated. The main advantage of TIRF is that only molecules at the Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
106
Fall, 2014
interface experience the evanescent excitation. This allows determining the concentration of adsorbed protein in the presence of a vast excess of unadsorbed protein, since the unadsorbed protein is not being excited by the evanescent wave. Fluorescence Recovery After Photobleaching (FRAP) In addition to determining the concentration of adsorbed protein, TIRF can also be used to measure the mobility of proteins on the surface; in particular to measure their diffusion coefficient. In the sequence of images below,♦ “pre” denotes a region of adsorbed, fluorescently tagged protein. It appears bright owing to its fluorescence excited by the evanescent wave. At time “0s” an intense laser beam is focused on this same spot, rendering the proteins nonfluorescent (or photo-bleaching them). The result is a dark spot at the center of this region, where the bleaching is complete. The bleaching laser beam is then turned off while the evanescent wave continues to excite fluorescence in unbleached protein. In the frames marked “10 s” and “60 s”, you can the recovery of fluorescence owing to the diffusion of unbleached protein in the region.
The rate of recovery of fluorescence can be used to infer the diffusion coefficient of the proteins on the surface. This technique is called fluorescence recovery after photobleaching (FRAP). TIRF Microscopy Another variation of TIRF is TIRF microscopy which is used to image the distribution of fluorescent entities on the surface. Below is a schematic of the device which employs an inverted microscope (objective lens 1 is below the sample and the illumination from the evanescent wave is from the top of the sample).
♦
http://malone.bioquant.uni-heidelberg.de/methods/mobility/mobility.html
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
107
Fall, 2014
Below are two images♥ of the same fluorescently tagged structure: one taken using TIRF and the other used epi-illumination to excite the fluorescence rather than the evanescent wave. In epi-illumination, the light is sent through the objective lens 1 (in front of the sample) rather than coming from behind the sample. Epi-illumination penetrates much deeper into the sample than the evanescent wave, causing more entities to fluoresce, thus blurring the fine structures in the fore-ground.
TIRF
epi-illumination
Frustrated TIR When a second dielectric medium (e.g. a glass plate) is brought sufficiently close to the surface at which an evanescent wave is generated (see figure at right), a new propagating wave is generated in the second medium which transmits some of the energy in the incident wave into the second medium. Since we no longer have total ♥http://lightmicroscopy.ucdenver.edu/doc/AMCF/
seminars/20091120.pdf Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
108
Fall, 2014
reflection, this situation is called frustrated total internal reflection. Lipson (p144) provides the following equation for the electric field amplitude in the second medium T=
1 β n cos θi − cosh k βd + 1 2 n cos θi β
sinh k βd
where d is the spacing between two identical media. As their arguments become large, both sinh(s) and cosh(s) behave like (1/2)es, leading to the following asymptotic behavior for the transmission coefficient Ae− k βd T (d ) = 1
where
A=
as d → ∞ for d → 0
(192)
4 n cos θi β 2+ − β n cos θi
Thus the second medium needs to be within a few penetration depths (kβ)–1 of the first medium in order for any transmission of energy to be accomplished. In the other extreme of d=0, T becomes unity; thus the transmitted wave is just a continuation of the incident wave — as if the air gap never existed. Frustrated total internal reflection can be seen in sketch (b) below, which is an image of the reflected light bouncing off the triangular glass prism in (a) when a spherical glass lens is placed in contact with the top of the prism. Most of the reflected image (see in b) is white owing to TIR. The black spot seen in the middle of (b) results when the incident light passes out the lens on top (as a result of frustrated TIR) rather than reflects
Image (b) was recorded for an angle of incidence greater than the critical angle so that TIR occurs. For comparison, image (c) shows the same view when the angle of incidence is subCopyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
109
Fall, 2014
critical. A black spot also occurs but is surrounded by many rings. These are called “Newton rings” and they (as well as the black spot in (c)) arise from interference rather than frustrated TIR. Total Internal Reflection Microscopy FRAP and TIRF-M use the evanescent wave to excite fluorescence in proteins or cells located very near the interface where the evanescent wave is generated. This fluorescence can then be viewed through a microscope. Rather than using the evanescent wave to excite fluorescence, total internal reflection microscopy (TIRM)♣ exploits the scattering of the evanescent wave to monitor the elevation of microscopic Brownian particles levitated very close to the interface. The schematic on the left below shows a single microscopic polystyrene sphere (10 µm diameter), immersed in water and levitated a distance h (on the order of 100-300 nm) above a glass microscope slide. The bent rays below the thick horizontal line (denoting the glass-water interface) represents the incident light undergoing total internal reflection. The horizontal rays of varying length above the thick horizontal line represent the evanescent wave propagating along the interface.
The red rays coming out of the top of the sphere represent the scattered light, whose intensity I(h) is exquisitely sensitive to the elevation h of the particle:
h = I ( h ) I 0 exp − dp
(193)
where this exponential is the same as the exponential appearing in the Poynting vector (191) for the intensity of the evanescent wave; in other words, dp is the penetration depth for the power, which is half the decay length for the electric field: ♣
D.C. Prieve, “Measurement of Colloidal Forces with TIRM,” Adv. Colloid & Interf. Sci. 82, 93-125 (1999).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
110
Fall, 2014 k2 =
= dp
λ pE 1 = = 2 2k2 β
2p λ2
λ2 4pβ
(194)
Under typical optical conditions for TIRM, dp is on the order of 100 nm. (193) then implies that a change in h of just 1 nm will cause a 1% change in scattering intensity; this is easily measured with a photomultiplier tube. Thus TIRM can resolve Brownian changes in elevation on the order of 1 nm. This “scattering” can also be thought of as a consequence of frustrated TIR. Indeed (193) is consistent with (192) in the limit d → ∞. The penetration depth (kβ)–1 in (192) is for the electric field; whereas dp in (193) and (194) is the penetration depth for power, which is half as large (2kβ)–1. The intensity of a single 10 µm polystyrene latex sphere is monitored every 10 ms for 10 min. The raw scattering data is shown at right. Remember that low scattering intensities I correspond to high elevations h according to (193).
A histogram of the scattering intensity values is shown at right. The number of observations of a particular intensity (or of a particular elevation) is proportional to the probability p(h)dh of finding the particle in that interval (width is dh) of elevations which, in turn, is related to the potential energy φ(h) of the sphere at that elevations by Boltzmann’s equation: φ (h) = p (h) A exp − kT
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
111
Thus the histogram of elevations can be converted into potential energy profile for the sphere, which is shown at right. At large elevations, the only force felt by the sphere is gravity so the slope of 0.29 pN represents the net weight of the sphere. The upturn in energy as the elevation becomes small represents electrostatic repulsion between the negatively charged sphere and the negatively charged glass plate. This repulsion is levitating the sphere.
Fall, 2014 8
Potential Energy (kT)
06-712
7 6 5
0.29 pN 4 3
10 µm 0.5 mM
2 1 0 80
100
120
140
160
180
200
220
240
Separation Distance (nm)
Lecture #15 begins here Light Propagation in Wave Guides Reference: §10.1.2 in Lipson Let’s now go back and look at the behavior of light inside a wave guide such as that used in ATR. Let the direction of propagation along the wave guide be denoted by z and let the direction along which the refractive index changes be denoted by x (as in the figure at right). We will eventually solve Maxwell’s equations for a 2-D slab (which is mathematically simpler than a cylindrical fiber) so there are no changes along the y-direction. For simplicity, let’s consider only TE polarized light. In the coordinate system of our figure above, this means the electric field has only a y-component denoted as Ey(x,z,t). The wave equation (141) requires ∂2 E y
∂2 E y
2
n2 ∂ E y + = ∂x 2 ∂z 2 c 2 ∂t 2
(195)
where the refractive index n takes on different values inside the wave guide (i.e. x < a ) and outside the wave guide (i.e. x > a is called the cladding). We will look for a particular solution having the form = E y ( x, z , t ) E ( x ) exp i ( k z z − ωt )
(196)
which kz is an unspecified constant. The above equation has the form of a LPPW in the zdirection, but leaves the x-dependence arbitrary. Indeed the fact that the refractive index depends on x means that the usual LPPW cannot occur in the x-direction. Substituting our particular solution into the wave equation (195) [ ∂/∂z = ikz and ∂/∂t = –iω]:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
112
Fall, 2014
d2E
n2 2 2 + ( ik z ) E =( iω) E 2 2 dx c
after the common exponential factor is divided out. Rearranging:
d2E dx
2
− k z2 E = −n2
ω2 c
2
E
or
( k z2 − n2 k02 ) E
d2E = dx 2
(197)
k0 ≡ ω/c
where
would be the wave number for a LPPW propagating through unbounded vacuum. Owing to two different values of n, the coefficient on the right-hand side takes on two different values, depending on x. While the constant kz is currently unspecified, we will shortly see that the guided-wave solutions of this problem correspond to kz being bounded by the two values of nk0:
n1k0 < k z < n2 k0 where the larger refractive index n2 occurs inside the wave guide and the smaller value n1 occurs outside the wave guide (inside the cladding). Thus the coefficient on the right-hand side of (197) takes on opposite signs inside and outside the wave guide. −a 2 E for x < a = 2 dx 2 +β E for x > a
d2E
where
2 −α= k z2 − n22 k02 < 0
and
2 +β= k z2 − n12 k02 > 0
(198)
Thus inside the wave guide, E(x) satisfies
d2E
for x < a :
dx whose general solution is
2
+ α2 E =0
( x ) E2a sin ax + E2 s cos ax E=
(199)
where α is strictly positive. The subscripts on the integration constants are “a” for antisymmetric part (i.e. odd function of x) and “s” for symmetric part (i.e. even function of x). Outside the wave guide, E(x) satisfies for x > a :
d2E dx
2
− β2 E = 0
whose bounded general solution is
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
113
Fall, 2014
E eβx for x < − a 1l E ( x) = E1r e−βx for x > + a
(200)
where β is strictly positive. The subscripts on the integration constants are “l” for left side (i.e. x+a). Thus E(x) given by (199) is oscillatory inside the wave guide while E(x) given by (200) is evanescent outside the wave guide. What remains is to match the inner and outer solutions. Continuity of Ey at the boundary x = +a requires: −βa E1r e= E2a sin aa + E2 s cos aa
(201)
Continuity of Hz requires continuity of ∂Ey/∂x at the boundary x = +a. Differentiating (199) and (200), evaluating at x = +a: −βE1r e−βa = aa E2a cos a − aa E2 s sin a
(202)
Similar matching at x = –a leads to
and
E1l eβa =− E2a sin aa + E2 s cos aa
(203)
E2a cos a + aa E2 s sin a βE1l e−βa = aa
(204)
Equations (201), (202), (203) and (204) constitute four equations in five unknowns: E1l, E1r, E2a, E2s and kz [recall that α and β are related to kz by (198)]. The missing fifth constraint arises from specifying the total energy carried by the wave guide which would arise from an additional matching condition at the inlet of the wave guide (located at z=0). Without getting further bogged down in a detailed solution, some important general features can be deduced by looking for solutions for E(x) which are either even or odd functions of x. Case #1: Let’s assume that E(x) is an even function. Then for E(x) even:
E1l = E1r ≡ E1
and
E2a = 0
−βa E= E2 s cos aa 1e
(205)
βE1e−βa = aa E2 s sin a
(206)
(201) and (202) become:
while (203) and (204) duplicate these. Dividing (206) by (205) eliminates the unknown E1 and E2s. Multiplying the remainder by a gives: βa =aa a tan a
Copyright© 2014 by Dennis C. Prieve
(207)
PDF generated November 30, 2014
06-712
114
Fall, 2014
which represents one constraint between values of βa and αa. A second independent constraint between these values is provided by (198), which can be rearranged to give: 2 α 2 + β=
( n22 − n12 ) k02 ≡ ( ∆k )2
(208)
(207) and (208) represent two transcendental algebraic equations. These algebraic equations can be solved graphically. The solid curves labelled “s” in the figure at right represent the relationship β (α) imposed by (207) for positive α and β. The solid quarter circles in this same figure represent β (α) imposed by (208) for three different values of ∆k, which is known. For small ∆k, there is only one intersection of (207) and (208), which we will denote by “mode 0”. But as ∆k increases, there are two other intersections, which we will denoted by “mode 2” and “mode 4”. Even numbered modes represent even solutions for E(x). Case #2: Let’s assume that E(x) is an odd function. Then for E(x) odd:
E1l = –E1r ≡ –E1
(201) and (202) lead to
and
βa = −aa a cot a
E2s = 0 (209)
which now replaces (207). (209) is shown in the figure above as the dotted curves labelled “a”. Intersections of these dotted curves with the solid quarter circles give rise to the odd E(x) which we denote by “mode 1” and “mode 3.” E(x) for all five modes are shown in the figure below:
Each of the modes corresponds to a particular discrete value of kz which satisfies all the conditions. While the solution to the overall problem presented above is incomplete, it points out some of the complexities of wave guides. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
115
Fall, 2014
Lecture #16 begins here Graded Index Wave Guides Commercial optical fibers often have a continuously variable refractive index denoted n(x) so there is no discrete surface at which TIR occurs. These are called graded index wave guides. The position dependence in n(x) leads to an interesting departure from the wave equation, which we will now point out. An intermediate step in the derivation of the wave equation is given by (12):
∇ × ∇ × E = −µε
∂2E
(12)
∂ 2t
We then applied a mathematical identity E.5: ∇ × ∇ × E = ∇ ( ∇.E ) − ∇ 2 E
(210)
To obtain the wave equation given by (13) or (126), we dropped the first term on the right-hand side by arguing that there are no free charges (i.e. ρ = 0) in a vacuum. While this term can be dropped when ρ = 0 and when n and ε are spatially uniform, any position dependence in n(x) leads to a nonzero ∇.E even when ρ = 0. To see this, recall the first of Maxwell’s equation (68): ∇.D = ρ
(68)
Now let’s substitute the constitutive equation D = εΕ, but we will assume ρ = 0.
∇.( εE ) = 0
(211)
In the absence of absorption, the refractive index is related to the permittivity by (83): n2 =
ε ε0
Thus a graded index wave guide has a position dependence in the permittivity, which we will denote by ε(x). Applying mathematical identity C.1 ∇.( εE ) = ∇ε.E + ε ( ∇.E )
Substituting (211) and solving for the divergence:
1 1 dε Ex ∇.E = − ∇ε.E = − ε ε dx Substituting this result into (210) and the result for the curl-squared into (12): Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
116
µε
Fall, 2014
∂2E
1 dε = ∇2 E + ∇ Ex 2 ε dx ∂ t
which reduces to the usual wave equation if ε = const. Propagation of Light in Conductors Reference: Born & Wolf, p735f and Lipson §5.6 Our LPPW is a particular solution to the wave equation, which was derived from Maxwell’s equations for dielectric by taking the current J to vanish. While this works well for dielectric materials (nonconductors) and even for moderate conductors like aqueous solutions, it is not appropriate for metal. Using the usual constitutive equations from page 3 (assuming the materials are isotropic and the properties do not vary with position), Maxwell’s equations (1) through (4) become: ∇.E = ∇×H = ε
ρ ε
(212)
∂E + KE ∂t
(213)
∂H ∂t
(214)
∇ × E = −µ
∇.H = 0
(215)
It turns out that any space charge ρ which might accumulate inside the good conductor, like a metal, has a very short lifetime (owing to the high conductivity). To show this, we take the divergence of (213):
∂E ∇.( ∇ × H ) = ε∇. + K ∇.E (( ∂ t
(216)
0
Recall that the divergence of the curl of any vector vanishes (a vector identity). Taking the partial time derivative of (212):
∂ . 1 ∂ρ (∇ E) = ∂t ε ∂t Interchanging the order of the derivatives: ∂E 1 ∂ρ ∇. = ∂t ε ∂t
Copyright© 2014 by Dennis C. Prieve
(217)
PDF generated November 30, 2014
06-712
117
Fall, 2014
After substituting the divergence of the time derivative from (217) and the divergence itself from (212), (216) becomes −
∂ρ K = ρ ∂t ε
ρ = ρ0 e − t t
which integrates to
τ=
where
ε K
According to the CRC Handbook,♠ the resistivity of copper at 25ºC is 1.712×10–8 ohm-m. The conductivity is just the reciprocal of this = K
1 1.712 × 10
-8
S = 5.84 × 107 m Ω-m
Using the electric permittivity of vacuum:♦ C2 8.854 × 10-12 e N-m 2 =1.5 × 10-19 sec τ= 0 = S K 5.84 × 107 m
Even if the electric field oscillates at 1016 per second, this relaxation time is infinitesimal compared to the period of oscillation. Thus any charge which might arise, decays to zero virtually instantaneously. This is called charge relaxation. So we can still take ρ = 0 and replace (212) by:
∇.E = 0
(218)
Taking the curl of (214) and substituting (213):
♠ http://www.hbcpnetbase.com/ ♦In
case you are wondering about the cancellation of units in the above expression, here are the details. The SI unit of conductivity is the siemens (namesake identified on page 59), which is defined as
amp coul sec C2 -s S≡ = = volt joul coul N-m The SI unit “siemens” replaces the older unit “mho” which had the same magnitude. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
118
Fall, 2014
∂ ∂ ∂E ∂2E ∂E ∇ × ( ∇ × E ) = −µ ( ∇ × H ) = −µ ε + KE = −µε − µK (( ∂t ∂t ∂t ∂t ∂t 2 2
∇ ( ∇.E ) −∇ E
Since ∇.E = 0 , the left-hand side reduces to the laplacian and we have
∇ 2 E = µε
∂2E ∂t 2
+ µK
∂E ∂t
(219)
which is no longer the wave equation owing to the new term in red. The LPPW is still a particular solution of (219), except that the wave number must have a complex magnitude, as we will now show. Align the x-axis in the direction of propagation of the LPPW; then the LPPW solution can be described as♠ ˆ − ωt ) = E ( x, t ) E0 exp i ( kx
(220)
Recall (see page 9) that one of the advantages of complex exponentials was ease of differentiation. In particular ∇ × E= ik × E
(221)
∂E → −iωE ∂t
and
The right-hand sides of the above two equations are the complex conjugates of the corresponding expressions on page 9 since (220) is the complex conjugate of (19). Then the right-hand side of (219) becomes: µε
∂2E ∂t 2
+ µK
(
)
∂E 2 = µε ( −iω) E + µK ( −iω) E = − µεω2 + iµK ω E ∂t
(222)
−ω2
Applying (221) to the left hand side of (219) becomes
∇ 2 E = −∇ × ( ∇ × E ) = −i 2 k × ( k × E ) = kˆ 2 e x × ( e x × e y ) E = − kˆ 2 E (( ez ( (((
(223)
−e y
♠
This solution is the complex conjugate of our earlier solution (130). Either (130) or (220) are solutions of the wave equation and of PDE (219), but (220) is preferred because it leads to complex properties which have positive imaginary parts (more conventional in the literature) whereas (130) leads to negative imaginary parts. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
119
Fall, 2014
Notice that the same result would have been obtained of we taken E to be in the z-direction instead of the y-direction. Substituting (222) and (223) into (219):
(
)
−kˆ 2 E = − µεω2 + iµK ω E After cancelling out the minus sign and E, we have♣ K kˆ 2 ≡ µεω2 + iµK ω = µω2 ε + i ω
(224)
εˆ
The analogy with nonconducting media becomes closer if — in addition to a complex wave number — we also allow the permittivity, velocity and refractive index to be complex numbers — but keep the same relationship among them which apply for nonconducting media. For example, for nonconducting media, the wave number is related to the permittivity by (23) and (24):
k=
ω = ω µε v
or
k 2 = µεω2
(224) has the same form if we define a complex permittivity as:
εˆ ≡ ε + i
K , ω
(225)
Similarly, the relationship (24) between velocity and permittivity suggest the following definition for a complex velocity: vˆ ≡
1
(226)
µεˆ
Finally, for nonconducting media, the relationship between refractive index and velocity is given by (32). The same relationship between the complex quantities is
nˆ =
c vˆ
(227)
Optical properties of metals are usually reported in terms of the real and imaginary parts of the complex refractive index [for example, see Optical Props of Elements (CRC 12_23_93).pdf]. The complex refractive index is frequently expressed in terms of two real constants (n and κ):
nˆ ≡ n (1 + i κ )
♣ We will
(228)
put a “hat” over symbols in order to remind us that they are complex.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
120
Fall, 2014
where n is the real part of nˆ and nκ is the imaginary part, where κ is called the attenuation index or the extinction coefficient. The reason for this name will become apparent below. These new real quantities (n and κ) can now be related back to the original (real) properties of the media: µ, ε and K. Eliminating vˆ between (226) and (227): c2 =
1 µε0
nˆ 2= c 2 µεˆ →
εˆ ε0
εˆ ε K ≡ +i ε0 ε0 ωε0
Dividing both sides of (225) by ε0:
Substituting this into the equation above: K ε +i ε0 ε0 ω
2 nˆ=
(229)
So we need to square a complex number. In general, if a and b are real then the square of an arbitrary complex number (a + ib) squared is
( a + ib )2 = a 2 + 2iab + ( ib )2 = a 2 − b 2 + i 2ab Substituting (228) [i.e. taking a = n and b = nκ] into the real part of (229) eˆ Re ( nˆ 2 ) = Re e0 e 2 ( )2 −( n( n κ = e0 2 2 n (1− κ
(230)
)
Similarly, equating the imaginary parts of (229) requires: ε K ( n( ) κ= ≡s n 2 ( ε0 ω ε0 2
(231)
2n κ
s≡
where
K εω
(230) and (231) can be thought of two algebraic equations relating s and ε/ε0 to the tabulated values of n and κ. Solving (230) and (231) simultaneously: 2 n=
Copyright© 2014 by Dennis C. Prieve
ε 2 1 + 1 + s 2ε0
PDF generated November 30, 2014
06-712
121
Fall, 2014
s2 κ2 = s 2 + 2 1 + 1 + s 2
Let’s express the complex wave vector in terms of our new complex refractive index. Substituting (229) into (224): ω2 nˆ 2 nˆ 2 = kˆ 2 = µω2 εˆ = µ ω2 c2 µ c2 Taking the square root of both sides and substituting (228): ωnˆ ωn (1 + i ) = (1 + i ) ˆ = = c c
(232)
Our particular solution (220) becomes = E ( x, t ) E0 e−kkx exp i ( kx − ωt )
(233)
−x Re {E} E0 e cos ( x − ωt ) whose real part is =
(234)
attenuation
which is still oscillating in both x and t, but the oscillations in x have an amplitude which decays exponentially with x (see figure below). The rate of decay is determined by the imaginary part κ. 1
l
0.5 E exp ( − κ ⋅ x)
0 0
− exp ( − κ ⋅ x) 0.5
1
0
1
2
3
4
5
x
kx
The exponential decay length is called the skin depth:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
122
Fall, 2014
l0 1 c = = kk nωk 2πn k
lskin =
and the wavelength for the position oscillations with x is
2π 2πc λ 0 = = k nω n
= λ
Example: calculate the exponential decay length and wavelength for l0 = 546 nm (wavelength in vacuum) light incident on copper. This wavelength is green light. Solution: Using a frequency of
2πc = 3.45 × 1015 s −1 λ0
= ω
The corresponding energy is♥
ω = 2.271 eV
Based on tabulated data for copper, we have
n = 1.04 lskin = and
and
nκ =2.59
c = 34 nm ωn k
λ λ = 0 = 525 nm n
Because of the much shorter decay length, the wave completely decays to zero in much less than one wavelength. Lecture #17 begins here Reflection from a Conductor Reflection from a conductor can be treated just like reflection from a dielectric, except that the complex refractive index nˆ2 must be used in place of the real n2. Lipson (Fig. 5.12) gives some results for the coefficient of “power” or R
♥
2
shown below:
Since our frequency is in radians rather than cycles, we should use = h 2π rather than h.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
123
Fall, 2014
The figure on the left above is calculated for metallic silver at long wavelengths for which the real part of the refractive index is dwarfed by the imaginary part. The figure on the right is for a wavelength of 600 nm, close to our ruby-red He-Ne laser. At optical frequencies, the real and imaginary parts of nˆ are not much different. Notice that the reflection coefficient for power is nearly equal to unity for all angles. This is why metallic silver is used as the backing for mirrors. Absorption of Light Energy by a Conductor Reference: Sect. 1.1.4 of Born & Wolf (p7-10) Although light does not penetrate far into a metal conductor, the energy is not all reflected. Notice that the reflection coefficient for power is less than unity in the the figures above. What happens to the the remainder of the incident energy? It is not propagated through the metal. We will now show that the remainder of the energy is absorbed as joule heat associated with the conductivity of the metal. Let’s revisit our energy balance performed on page 80: dU dt
dW + ∫ n.( E × H ) da = − dt A (((( accumulation rate of of internal energy
net rate of energy transported out of system
(235)
wor done on system
Recall that this energy balance was derived by calculating the work done by magnetic and electric fields on a point charge q moving at velocity v through the magnetic and electric fields [see (147)]. For a set of such charges inside a region V, the corresponding expression is
dW = ∑ F.v = ∑ q E.v = ∫ E. vρ dV= ∫ E.J dV dt V V ρdV V V J
Copyright© 2014 by Dennis C. Prieve
(236)
PDF generated November 30, 2014
06-712
124
= U
and
Fall, 2014
1 ( E.D + B.H ) dV 2∫ V
In our derivation leading to (235), the current was given by: dq v = J dV
J=
or
(148)
dq v = ρv dV ρ
We might consider this “convection” of charge. For conductors, there is a second contribution to current J =ρ KE v + Jv
(237)
Jc
where KE might be considered the transport of charge by conduction. Substituting (237) into (236)
) dV ∫ E.( J v + J c= V
∫ E.JvdV
V
v dq (( dqE.v ((
+ ∫ E.J c dV V
rate of work done by system
Now the first integral (representing force times velocity) can thought of as the rate at which the system does work on its surroundings. What does the second integral represent? Substituting Ohm’s law
( KE ) dV E.J c dV ∫ E.= ∫= V
V
2 K= ∫ E.E dV K ∫ E dV ≥ 0 V
V
This is Joule heating — much like what occurs when we send electric current through a long wire. The voltage drop between the ends of the wire multiplied by the current (in amps) flowing through the wire gives watts of electrical power dissipated as heat associated with conduction through the wire. This heat goes into raising the temperature and internal energy of the wire — or is conducted outside the system. Combining (235) through (237), our energy balance can be written as
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
125
Fall, 2014
dU dt
+ ∫ n.( E × H ) da = − ∫ E.J v dV − ∫ E.J c dV A V V rv (((( (( (( accumulation net rate of energy transported out of system
of internal energy
(238)
irreversible loss of energy
rate of wor done on system
In the case of propagation of a LPPW through a conductor, the accumulation term vanishes (after time-averaging over one cycle, dU/dt = 0) and rate-of-work-by-convective-current term also vanishes (recall ρ = 0), leaving − K ∫ E.E dV ∫ n.( E × H )da = A
V
Recognizing that E×H = Π and applying the divergence theorem, the above leads to ∇.( E × H ) = − K ( E.E )
But we need to time-average both sides over one cycle: ∇. E × H = − K E.E
(239)
Since we have products of fields [plus the wave vector whose magnitude kˆ appears in , let’s be on the safe side and resort to real parts. The real part of the electric field is given by (234): Re {E} =
E0 e−x cos ( x − ωt )
(234)
E0 e y
We will use this real part as E in (239). Then we will calculate the associated magnetic field from one of Maxwell’s equations (214): ∇ × E = −µ
∂H ∂t
(214)
Substituting the curl of (234) and substituting it into (214): kE 1 ∂H = − ∇ × E = 0 e−kkx [ k cos ( kx − ωt ) + sin ( kx − ωt )] e z ∂t µ µ
Integrating both sides with respect to t and dropping the integration constant: = H
kE0 −kkx [ −k sin ( kx − ωt ) + cos ( kx − ωt )] e z e µω
Crossing this result with (234):
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
126
Fall, 2014
kE02 −2 kkx 2 = E× H e −k sin ( kx − ωt ) cos ( kx − ωt ) + cos ( kx − ωt ) e x µω Time averaging over one cycle: 2π
E× H ≡ 2π Π
ω
kE 2 E × H dt =0 e−2 kkx e x 2µω ω 0
∫
1
(240)
Πx
This expression for the Poynting vector of an attenuated wave is identical with (160) for a LPPW provided we lump the exponential attenuation factor e–κkx in with the amplitude E0 evaluated at x=0. Now we return to evaluating (239). Since the Poynting vector above has only an xcomponent, the divergence is
dΠ ∇. E × H = x dx
(241)
The right-hand side of (239) requires us to evaluate E.E. Dotting (234) with itself: = E.E E02 e−2 kkx cos 2 ( kx − ωt )
Time averaging over one cycle: 2π
E.E ≡
1 2π
ω
∫
ω 0
E02 −2 kkx µω e Πx E.E dt == 2 k
(242)
where in the last equation, we have substituted (240). Substituting (241) and (242) into (239):
dΠx µω = −K Πx dx k Solving (231) for K and substituting the result above:
(
)
dΠx µω = − 2n 2 kε0 ω Πx dx k Next, we substitute ω = vk, µε0 = 1/c2 and n2v2 = c2. After cancellations, we get
dΠx =−2 kk Π x dx
Copyright© 2014 by Dennis C. Prieve
(243)
PDF generated November 30, 2014
06-712
127
Fall, 2014
which when integrated yields the same exponential decay for Πx as we obtained using the wave equation. So why is the power of the wave decreasing with x? Where is this power going? Answer: the right-hand side of (243) came from the right-hand side of (238): it represents the irreversible loss of energy due to Joule heating. The light energy is being converted into heat as a consequence of the conductivity of the metal. Surface Plasmons Reference: Lipson §13.7 In our analysis of wave guides [recall Light Propagation in Wave Guides on page 111], we saw that it is possible to trap light energy in a layer (or in an optical fiber) having an index of refraction which is higher than the surrounding media. In the slab geometry we analyzed, there were two interfaces which confined the light energy. The figure at right shows the mode 0 solution, which (like all modes) is sinusoidal inside the wave guide and evanescent outside. It turns out to be possible to trap light energy at a single interface provided one of the two materials has a negative permittivity (as many metals do). The figure at right shows the electric field at the interface between a dielectric (on top) and a metal (on bottom). Before analyzing this behavior, let’s show that the permittivity is negative for many metals. Example: Evaluate the conductivity and permittivity of metallic copper at a frequency corresponding to green light (l0 = 546 nm). Solution: In the earlier example (see page 122), we evaluated the real and imaginary parts of refractive index from the CRC handbook tables:
n = 1.04 = κ
So the attenuation coefficient is
From (231):
and
nκ =2.59
nκ = 2.49 n
= K 2n 2 κε0= ω 1.65 × 105
S m
This is about 0.3% of the static (i.e. ω = 0) conductivity used to calculate the relaxation time. From (230):
ε = n 2 (1 − κ 2 ) = −5.63 ε0
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
128
Fall, 2014
Thus we see that any material having an attenuation constant larger than unity will have a negative permittivity. A surface plamon is a wave which propagates along the interface x = 0 separating the metal from the dielectric. So let’s look for a particular solution which oscillates with t and z but has some possibly different functionality (i.e. exponential) with respect to x: = E ( x, z , t ) E0 ( x ) exp i ( k z z − ωt ) e
(244)
where e is a unit vector denoting the direction of the electric field. For x 0
ε 2 k xm = k z2 − k02 m ε0
(250)
(251)
PDF generated November 30, 2014
06-712
130
Fall, 2014
ε 2 k xd = k z2 − k02 d ε0
and
(252)
Boundary conditions at the interface are given by (6). When the magnetic permeability µ is the same for both phases, all components of H must be continuous across the interface. In particular continuity of the y-component requires
Hy + = Hy = x 0= x 0− When (249) and (250) substituted, the common exponential cancels out leaving: A=B or
Aek xm x ( ) H x = Ae− k xd x
for x < 0
(253)
for x > 0
We need an additional matching condition which must involve the electric field. Recall that the electric field is related to the magnetic field by Maxwell’s equation (110). After substituting our linear constitutive equations (D = εE and J = KE), (110) becomes ∇ × H = KE + ε
∂E K = ( K − iωε ) E = −iω ε + i E ∂t ω (( εˆ
where the last two equations arise when the time-dependence of E is e–iωt as in (244). For TM mode, the electric field E will have an x and a z component but no y-component. The zcomponent yields
(∇ × H )z =
∂H y
K = −iω ε + i E z ∂x ω (( εˆ
Now recall that boundary condition (6) requires the tangential components of E (including Ez) to be continuous across the interface. Then the above equation requires 1 ∂H y 1 ∂H y = + εˆ ∂x x 0= εˆ ∂x x 0− =
Substituiting (253) and replacing εˆ by either εm or εd: k xm A k A = − xd εm εd
Copyright© 2014 by Dennis C. Prieve
(254)
PDF generated November 30, 2014
06-712
131
Fall, 2014
Now (251), (252) and (254) represent three algebraic equations in three unknowns (kxm, kxd and kz). The solution is kz = k0
εd εm , ( ε d + ε m ) ε0
k xd = k0
εd
− ( ε d + ε m ) ε0
and
k xm = k0
−ε m
− ( ε d + ε m ) ε0
(255)
For surface plasmons to exist, kxd and kxm must have positive real parts. Otherwise (253) does not yield an evanescent wave on both sides of the interface. Assuming the ε’s are both real, we need εd + εm < 0
or
–εm > εd
to avoid taking the square-root of a negative numbers. In addition, we need εm < 0 so that both numerator and denominator in kz are negative; thus leading to a positive quotient. Otherwise we would obtain purely imaginary values for one or more the k’s and would not obtain a surface plamon. To summarize, necessary conditions for a surface plasmon to occur are 1) εm < 0 2) –εm > εd There are two methods to generate surface plasmons. In figure (a) at right a thin metal film is deposited onto a glass prism. Total internal reflection of a laser occurs at the glass-metal interface producing an evanescent wave inside the metal which, in turn, excite surface plasmons at the metal-solution interface below. This is called the Kretschmann configuration. Clearly the metal film thickness must be comparable to or less thin that the skin depth of the metal. Figure (b) above shows the Otto configuration in which the glass prism is separated from the metal slab by a thin layer of solution. Total internal reflection again occurs inside the prism, creating an evanescent wave in the solution which, in turn, excites surface plasmons on the solution-metal interface below. Let’s plug some physical property data into the formulas above and see what values are obtained for the wave numbers etc. Example: Calculate the k’s and associated decay lengths (or wavelength) for surface plasmons formed at the gold-water interface using incident light having a wavelength l0 of 632 nm (red HeNe laser). Solution: The wavelength is first converted into frequency:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
132 = ω
Fall, 2014
2πc rad = 2.98 × 1015 s λ0 ω =1.962 eV
and then into a photon energy:
Let’s round this up to 2 eV, for which the CRC handbook tables for gold give:
n = 0.13 = κ
So the attenuation coefficient is
From (230):
and
nκ =3.16
nκ = 24.3 n
εm = n 2 (1 − κ 2 ) = −9.97 ε0
Water has a refractive index of about n = 1.33 for visible light gives
εd 2 = n= 1.77 ε0 Notice that both of the necessary conditions listed above are met. The wave numbers given by (255) are kz = k0 k xd = k0 k xm = k0
εd εm = 1.466 ( ε d + ε m ) ε0 εd = 0.618 − ( ε d + ε m ) ε0 −ε m = 3.48 − ( ε d + ε m ) ε0
where the wave number in vacuum is k0=
ω 2π = = 9.94 mm −1 c λ0
Note that all of the k’s are positive real numbers. According to our particular solution given by (249) and (250), the wave number kz is associated with the sinusoidal variation of the electric field along the z-axis (i.e. along the interface). The associated wavelength for those sinusoidal oscillations is
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
133 λ z=
Fall, 2014
2π = 431 nm kz
which is somewhat smaller than the wavelength of the incident light (632 nm). The other two k’s are associated with exponential decays. The associated penetration depths of the evanescent wave in the metal and the water are 1 k xm
= 29 nm
and
1 k xd
= 163 nm
Below is a surface plot of the profile of Hy(x,z,0) for t = 0 where the colors represents values of Hy. The dimensions of the green square are 1000 nm × 1000 nm. The graph on the right is the profile of Hy(0,z,0) along the z-axis (x = 0) which is sinusoidal in shape. The graph at the bottom is profile of Hy(x,0,0) along the x-axis (z = 0) which is evanescent on either side of the interface.
As time progresses, the colors march to the right at a constant speed which can be determined by setting to zero the argument of the exponential in (249) and solving for z:
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
134 k z z − ωt = 0
Fall, 2014
or
z=
ω t z vz
Substituting our values from the example, the speed of propagation of our plamon is vz c = 0.682 . Recall that the refractive index n = 0.13 is less than unity. This small value of the refractive index suggests that light should travel faster in metal than in vacuum — but our plasmon is travelling slower than the speed of light in vacuum. This is because the wave number kz given by (255) depends on the properties of the dielectric as well as of the metal. Use of Complex Permittivity In our development of the surface plasmon above, we argued that the imaginary part of the metal’s permittivity εˆ could be dropped, leaving εm purely real. Actually, this was not necessary. Example: repeat the example above including an imaginary part for εˆ m . Solution: The complex expression for εˆ was introduced in (224): εˆ ≡ ε + i
K ω
(256)
so we will need to evaluate the gold conductivity K
= K 2n 2 κε0= ω 2.17 × 104
from (231):
S m
εˆ = −9.97 + 0.822i ε0
Then (256) gives
Substituting this complex value in place of εm in (255) leads to: kz = k0 k xd = k0 k xm = k0
εd εm = 1.465 + 0.013i ( ε d + ε m ) ε0 εd
= 0.615 + 0.031i − ( ε d + ε m ) ε0 −ε m
= 3.48 − 0.113i − ( ε d + ε m ) ε0
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
135
Fall, 2014
The real parts of these wave numbers are virtually (but not identically) equal to the values obtained in the previous example. So what do the imaginary parts of these wave numbers mean? Inside the metal, the functional form given by (249) and (253) is for x 0
Ey is tangent to the interface. Continuity of tangential components across the interface require A = B or Aek xm x ( ) E0 x = Ae− k xd x
for x < 0 for x > 0
(258)
We need an additional matching condition which must involve the magnetic field. The magnetic field related to the electric field by (107) another of Maxwell’s equation: ∇×E = −
The z-component yields
(∇ × E)z = −
∂Bz ∂t
∂B ∂t
or
∂E y ∂x
= i ωB z
Since i and ω are same on either side of the interface, continuity of Bz across the interface requires continuity of ∂Ey/∂x: ∂E y ∂E y = + ∂x x 0= ∂x x 0− =
Substituting (257) and (258) and cancelling out common factors leaves −k xd = k xm But both of these wave numbers must have positive real parts if we are to have an evanescent wave on both sides of the interface. So the above condition is incompatible with a surface plasmon. Lecture #19 begins here Surface Plasmon Resonance Reference: Lipson §13.7.1 Over certain ranges of frequencies (mainly infrared), electrons in metals are known to behave like gaseous plasmas in that the real part of their frequency-dependent dielectric constant can be fit to
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
137
Fall, 2014
2 ε m ( ω) Ω = 1− ε0 ω
(259)
where the parameter Ω is known as the plasma frequency. Substituting (259) for εm into (255), the wave number along the interface is given by:
εd ε0
ω 2 − 1 Ω
ck z ( ω) ω = Ω Ω ω 2 ε d + 1 − 1 Ω ε0
(260)
Now this function yields a positive real value for the wave number for many frequencies, Figure 4: Graphical representation of equation (260). but not all (see plot at right, which assumes εd/ε0 = 1.52 = 2.25). To understand the figure, notice that for low frequencies (ω≪Ω) both numerator and denominator inside the squareroot of (260) are negative, so their quotient is positive and a real value of kz is obtained. As frequency increases, the sign of the denominator changes when 2 ω ω εd or + 1 − 1 = 0 = Ω ε0 Ω
1 = 0.555 εd +1 ε0
For 0.555 < ω/Ω < 1, the numerator and denominator have opposite signs, leading to a purely imaginary kz (not a plasmon). Finally the numerator changes sign at ω/Ω = 1 and we obtain the second branch (shown in figure) of real positive values for kz. Note that this second branch corresponds to εm > 0 which does not lead to surface plasmons (recall criteria on page 131). Of particular interest is the asympt at zero frequencies: c z ( ω) = Ω ω→ 0 lim
εd ω 3 + O ( ω Ω ) ε0 Ω nd
which can also be written as
ω lim k z (= ω) nd= nd k0 ( ω) c ω→ 0 ω
Copyright© 2014 by Dennis C. Prieve
(261)
c
PDF generated November 30, 2014
06-712
138
Fall, 2014
Surface plasmon resonance occurs when an electromagnetic wave is incident upon the interface with both kz (the component along the interface) and ω equal to those of the plasmon, as specified by (261) (as least for low frequency). Let’s suppose we are trying to excite resonance at a gold-water interface using a LPPW in air (see figure at right). The value of kz we are creating in the water is related to the wavelength l0 of the incident light at the angle θi of incidence. From the geometry of the right triangle:
λ0 2π 2π = = k z sin θi k0 sin θi = k z k0 sin θi
or
Regardless of θi in air we will always generate kz < k0 but (261) requires kz = ndk0 > k0 (recall nd = 1.33 for water). Thus we cannot excite plasmons with this configuration. In the Otto configuration (see schematic on page 131), we replace air with a glass prism (ng = 1.5). Then we can generate = z ng 0 sin θi ω
(262)
c
which for a particular angle of incidence matches kz given by (261). (261) only applies at very low frequencies. The more general relationship for kz(ω) is (260) and is shown as the two solid curves in the figure at right. These two solid curves are identical to those in Figure 4, except the axes have been reversed. The straight dashed line is (261). Notice that the “surface plasmon branch” of (260) becomes tangent to the dashed line at low frequencies. The solid straight line represents (262) when sinθi = 1 (θi = 0). We can generate any kz(ω) to the right of this solid line (i.e. smaller kz) by adjusting the angle of incidence. In other words, the shaded region of the graph shows the possible kz(ω) for these particular geometry using a glass prism. Some of these points in the shaded region also lie on the “surface plasmon branch.” These common points represent the “region where resonance is possible.”
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
139
Fall, 2014
Using the reflection coefficient for power is measured as a function of the angle of incidence. The figure at right shows the resulting reflection coefficient and phase angle measured in the Kretschmann configuration (see schematic on page 131). In this arrangement, the incident light generates an evanescent wave in the metal at the glass-metal interface. The evanescent wave penetrates a short distance through the thin metal film and excites plasmon resonance (over a narrow range of angles of incidence) at the metal-solution interface. Notice that the reflection coefficient increases with angle of incidence to reach near unity at the critical angle labelled θc, where TIR would be expected to occur if the metal film were thick. The measured R are not quite unity because of tunneling of energy across the metal (even without any plasmon resonance): recall Frustrated TIR on page 107. As the angle of incidence approaches the angle at which resonance occurs — labelled θp — the reflection coefficient plunges to near zero as nearly all of the incident energy is absorbed by the surface plasmons. Surface plasmon resonance (SPR) is used to detect very small changes in refractive index of the dielectric phase (i.e. the water) arising from adsorption of biomolecules (for example). The figure at right shows very tiny changes (10–5) in the refractive index of the aqueous solution next to the Au-water interface which occur when biomolecules adsorb (sub-monolayer) at the interface at t = 20 sec. At t = 80 sec, you can see the desorption of these molecules as wash with solute-free water. Reflection from a Uniform Film: Multiple Reflections Reference: Sect. 7.6.1 of Born & Wolf (p360f)
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
140
Fall, 2014
Radiation Pressure Reference: Sect. 5.8 in Lipson, Consider a LPPW with normal incidence at an air-metal interface. According to the figure on page 123 (Fig. 5.12b in Lipson), virtually all (perhaps 96%) of the power of the incident light is reflected. If we assume the reflection coefficient for the electric field is –1, then the sum of the incident and reflected LPPW’s gives a standing wave.♥ The figure at right shows a standing wave Ex(z,t): in air:
E x= E x 0 cos ( ωt − kz ) + − E x 0 cos ( ωt= + kz ) 2 E x 0 sin ( kz ) sin ( ωt ) ( z, t ) (((( (( (((( Eiy
Ery
This in turn produces a standing wave in Hy(z,t): in air:
2E0 cos ( z ) cos ( ωt ) µω
= H y ( z, t )
(263)
2 E0 ≡ H0 µc
We expect the magnetic field will propagate a short distance (the skin depth) into the metal before decaying to zero (as shown in the figure above). In accordance with the Ampere-Maxwell equation (110), this magnetic field will induce an electric current inside the conductor: J+
∂D ∂t
= ∇×H = −
∂H y ∂z
ex
(110)
0 because E = 0 inside the metal
Jx = −
or
∂H y
(264)
∂z
This induced current interacting with the magnetic field exerts a force on the metal according to Lorentz force equation (146): F= q ( E + v × B )
(146)
Replacing qv → JdV = JAdz and setting E = 0 inside the metal, (146) can be written as
( J × B ) Adz = dF = J x µH y ( e x × e y ) Adz (( ez
♥
recall graph of a standing wave on page 67.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
141
Fall, 2014
where we have made the force a differential quantity because we are using a differential volume dV = A dz. Here A is the area of the metal which is illuminated by the LPPW. Clearly, the force acts in the z-direction:
dFz= J x µH y Adz Substituting (264) and integrating over the metal plate (from z = 0 to z = +∞) gives the total force on the metal: Fz = −µ A
z = +∞
∫
z =0
∂H y
= −
H y dz ∂z ( (
H z ( z = +∞ ) = 0 µ µ H z2 = H 2 ( 0) ( ) H z z= 0= H 0 2 2
( )
(265)
1 ∂H z 1 dz = dH z2 2 ∂z 2 2
where this force per unit area is called the radiation pressure. Continuity of the magnetic field across the air-metal interface requires that H(0) must also be the amplitude of the oscillations in air, which is given by (263) as = H ( 0)
2 E0 cos ωt µc
Fz 2 E02 = cos 2 ( ωt ) Substituting this result into (265): A µc 2
Time averaging:
Fz A
=
E02
(266)
µc 2
Recall that the magnitude of the Poynting vector (energy flux) is given by (165) as:
E2 Π z =0 2µc
(267)
after we have substituting v = c (in air). Eliminating E02 between (266) and (267): Fz A
=
2 Πz c
(268)
Example: calculate the radiation pressure exerted on the surface of a mirror if the incident light has the same energy density as sunlight on the surface of the earth (about 103 W/m2) Solution: Substituting Π z = 103 W/m2 and c = 3×108 m/s into (268):
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
142 Fz A
=
Fall, 2014
2 P z = 6.7 × 10−6 Pa = 6.7 × 10−11 atm c
This explains why we have never experienced radiation pressure in everyday life: the radiation force exerted by natural sunlight is too weak to detect. Another way of expressing the same result is 2 nN = 6.7 c W
and this assumes that all of the momentum of the incident light is reflected (as in a mirror). If it is simply absorbed, the force is half of the above, or 1 nN = 3.4 c W
(269)
which represents the amount of momentum carried by the incident wave. Lecture #20 begins here Historically, radiation pressure is as old as the astronomer Kepler (1571-1630) who suggested that comets’ tails point away from the sun because of the “solar wind.” The phrase “solar wind” suggests an analogy between the effect of sunlight on comets’ tails and the effect of an ocean wind on the sails of ships. During the 18th century, believers in the particle-theory of light thought that any force exerted by light would be direct proof that light is a particle not a wave. Owing to the weakness of the force, it wasn’t until 1899 that radiation pressure was measured (after a search lasting more than two centuries). However, by this time Maxwell had shown that EM waves can also exert a force. Owing to the weakness of the force, no significant applications of radiation pressure were made until the 1960’s when lasers were invented. In 1970 Arthur Ashkin (Phys. Rev. Lett. 24:152) used a focused laser to maneuver organelles inside single cells. Today this technique is called optical tweezers. Another way to think about this force is to consider that light is a stream of photons — particles of light which possess energy and momentum. If the mass of a photon♦ is m and its speed is c, then the energy and momentum are given by energy =
so that
1 2 mc 2
and
momentum = mc
mc momentum 2 = = 1 2 c energy mc 2
♦
In Mass of the Photon on page 20, we saw that the mass of the photon was over 20 orders of magnitude smaller than that of an electron. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
143
Fall, 2014
This differs by a factor of 2 from what we calculated in (269). Momentum Balance Reference: Jackson p260f In Energy Flux on page 75, we derived a general equation representing an energy balance starting with the Lorentz force equation (146): F= q ( E + v × B )
(146)
which represents the force acting on a particle having charge q, moving at velocity v through space having electric and magnetic fields E and B. In this section, we use a very similar approach to derive a momentum balance. Any force felt by the particle will cause it to accelerate at a rate given by Newton’s second law: dp = F dt
m= a
If p = mv is the momentum of the particle, and acceleration a = dv/dt, then mass times acceleration can be thought of as the time-rate of change in momentum dp/dt as we have written in the first equation above. Next we sum over all particles inside some volume V: dPmech dpi = ∑ = ∑ qi ( E + vi × B ) = ∫ ( ρE + J × B ) dV dt dt i in V i in V V
(270)
In writing the sum of the forces as a volume integral we have replaced q → ρdV and qv → JdV just as we did in the energy balance. After eliminating ρ using Coulomb’s law (68) and eliminating J using Maxwell-Ampere’s law (110),♣ then applying some math identities, the above momentum balance eventually becomes:
d = + Pfield ) ( Pmech dt (( (((( accumulation of momentum
where
net force . = da n T exerted ∫ on system A((
(271)
transport of momentum into system
= Pfield
1
∫ E × H dV
c2 V
♣
The relationships used by Jackson employ the constitutive equations for vacuum: i.e. ε0 and µ0 rather than ε and µ. This might limit the applicability of the final result. Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712 and
144
Fall, 2014
(
1 T= ε0 EE + c 2 BB − E.E + c 2 B.B 2
) I
(272)
is called the Maxwell stress tensor and I is the identity tensor. We encountered the Maxwell stress tensor in 06-607 when we calculated the double-layer force between interacting charged particles. In 06-607, we were dealing with static electric fields. In the above definition of T, we have additional contributions from the magnetic field as well as from the electric field. The surface integral of the Maxwell stress yields the net force felt by the system from the electric and magnetic fields outside. In the derivation of (270), we saw that Pmech represents the net momentum of all charged particles (e.g. ions in solution or electrons bound to the nucleus of atoms) inside the volume V which serves as the system on which the momentum balance is being performed. Pfield represents additional momentum the system has by virtue of electric and magnetic fields acting internal to it. Optical Tweezers In principle, the net force acting on a microscopic colloidal particle by light can be calculated from (271) by integrating the normal component of the Maxwell stress given by (272) over the surface of the particle. Indeed, this is the most general and most rigorous approach. However, it is hard to anticipate the direction of the force from this integral. Another approach is to use ray optics or geometric optics. Consider the situation shown at right of a colloidal particle (represented by the circle) with a laser beam impinging on it from above. In geometric optics, a laser beam is treated like a bundle of parallel rays of light, one of these rays is shown as the vertical line near point A which is impinging on the top of the particle. Refraction causes the direction of the ray to change as it passes through the transparent particle (perhaps a polystyrene latex bead). The direction of the ray exiting the particle differs from the direction of the incident ray: the ray at B has a component acting to the right. Since the total momentum must be conserved, the particle feels a net force to the left as indicated by the left-pointing arrow inside the particle. This illustrates the basic origin of optical forces acting on particles. In the figure above, we have only shown the transmitted portion of the ray as it encounters two interfaces. Each encounter with a phase boundary also generates a reflected ray, whose momentum must also be considered. There are also multiple reflections occurring within the sphere. Let’s assume that all the reflection coefficients are small so that reflections can be neglected, leaving just the transmitted portion of the ray as shown in the figure. If the incident laser beam is uniform in its intensity across the beam, then a second ray will be incident on the sphere an equal Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
145
Fall, 2014
distance on the other side of the north pole, as shown at right. Owing to symmetry, this ray will be refracted in the opposite direction to that of the first ray. They net effect is that the change in horizontal momentum of the two equal rays is zero: a uniform beam would exert no horizontal force on the particle. However, even a uniform beam would exert a downward force on the particle owing to the diversion of some of the downward momentum of the rays into horizontal momentum. Laser beams usually have a Gaussian intensity profile across the beam, with the center of the beam being the brightest. Suppose that leads to the vertical incident beam on the left being more intense than the vertical beam on the right. Then the nonuniformity of the laser leads to a net force on the particle pulling it to the left — in the direction of more intense axis of the laser beam. Using a lens to focus the laser beam causes the beam to have its maximum intensity at the focal point. Particles are attracted to this point. The potential energy U of the particle in the nonuniform beam can be estimated from:♠ 1 U = − v (ε p − ε f ) E2 2
where εp and εf are the permittivities of the particle and the surrounding fluid and E is the local amplitude of the oscillating electric field. The force felt by the particle is the gradient in this energy: F = −∇U =
1 v (ε p − ε f ) ∇ ( E2 ) 2
Provided εp > εf, the particle will be pulled to the most intense region. Of course, an air bubble in the same fluid has εp < εf and will be pushed away from the intense region.
A highly focused (but low divergent) laser beam is incident on the bottom of the particle and exerts an upward force on the 10 µm polystyrene sphere in 0.15 mM NaCl solution. Recalling that the
8
0.88 mW 7 6
Potential, kT
The component of the force acting along the direction of the laser beam is evident in the potential energy profiles shown at right. These were measured using TIRM.♥
♠ see
5 4 3 2
2.00 mW
1 0
(5.97) in Lipson. ♥ Walz & Prieve, Langmuir 8:3043 (1992). Copyright© 2014 by Dennis C. Prieve
1.34 mW
100
150
200
250
Separation Distance, nm
PDF generated November 30, 2014
06-712
146
Fall, 2014
slope of the linear portion of the profile represents the apparent net weight of the sphere, increasing the power of the laser appears to make the particle lighter — suggesting the laser is exerting a upward force on the particle. 0.8
Apparent Weight (pN)
By regressing the data in PE profiles like those above, we can determine the apparent weight of the particle as a function of the beam intensity. The results are summarized in the figure at right.♦ Focussing the beam on the bottom of the particle makes it lighter while focusing the beam on the top makes it heavier. The true net weight of the particle is about 0.22 pN.
0.6
9 µm PS latex 10 µm beams top
0.4
slope = 0.34 0.2
nN watt
bottom
The plot above the slope as the 0.0 net weight of the PS bead for 0.0 0.5 1.0 1.5 2.0 different laser powers, both for experiments where we push up and Net Power (mW) make the sphere lighter, and for other experiments where we push down and make the sphere heavier. Although the slopes of the two curves have opposite sign, their magnitude is the same 0.34 nN/watt. However, this is only 1/10 of our proportionality constant 1/c in (269). Clearly, we need to take a closer look to understand this quantitatively. A LPPW can be thought of as a stream of photons which possess momentum. The amount of momentum flux can be calculated from the energy flux:
P 1 N ] P = =εE02 εk [= 2 v m2 where E0 is the amplitude of the electric field and ek is a unit vector pointing in the direction of propagation of the LPPW (i.e. the direction of the wave vector k).
♦
Rebecca Liebert, PhD thesis, CMU, 1995.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
147
Fall, 2014
When this wave reflects off an interface, it produces two daughter waves: the reflected wave and the transmitted wave. Each carries momentum. In order to conserve momentum upon reflection, a force is exerted on the interface. The amount of the force per unit area P can be calculated from
Pr
Pt
P Pi = Pr + Pt + P The units of this equation are those of pressure. Complicating the calculation of force on the sphere is the infinite number of reflections which occur inside the sphere:
Pi
These have been summed analytically by Roosen & Imbert (1976). The axial component of the force (acting in the direction of the incident ray) and the radial component are expressed in terms of dimensionless efficiencies Q:
dFaxial =
n.qdA Qaxial v
dFradial =
n.qdA Qradial v
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712 where
and
148 Qaxial = 1 + R cos 2θi −
= R sin 2θi − Qradial
Copyright© 2014 by Dennis C. Prieve
Fall, 2014
T 2 cos ( 2θi − 2θr ) + R cos 2θi 1 + R 2 + 2 R cos 2θr
T 2 sin ( 2θi − 2θr ) + R sin 2θi 1 + R 2 + 2 R cos 2θr
PDF generated November 30, 2014
06-712
149
Fall, 2014
Lecture #21 begins here
Light Scattering Reference: Lipson, 4th ed., Sect. 5.3.1 In the section titled Polarization of Dielectrics (starting on page 27), we saw that electric fields can induce a dipole moment in all material — even if the atoms or molecules do not possess a permanent dipole of their own. The dipole in turn exerts an electrostatic force on other charges in the universe. For distances that are large compared to the displacement between the changes of the dipole, the electric field is given by* far field:
E (r ) =
)− 1 3n ( n.ππ 4πε0 r3
r
θ
test charge
p point dipole
(273)
where p is the dipole moment, r is the magnitude of the position vector r. And
n≡
r r
is a unit vector pointing in the direction of the position vector. To emphasize that (273) is valid when r is large, we sometimes refer to (273) as being the electric field emanating from a point dipole. If we adopt spherical coordinates, with the z-axis pointing in the direction of the dipole moment, the dipole moment is expressed as: p = pez and (273) yields:
Er =
2 p cos θ r3
, Eθ =
p sin θ r3
, and Eφ = 0
The direction of this electric field is neither aligned with the dipole p, nor with the position vector r or n. Radiation from an Oscillating Point Dipole Next, consider that externally applied electric field and the resulting point dipole oscillate with a frequency ω. at r = 0:
= p p0 cos ( ωt )
(274)
* Homework.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
150
Fall, 2014
This oscillating dipole can be thought of as an oscillatory electric current. Recall that the dipole moment can be written as p(t) ≡ qx(t)
(57)
where q>0 is the magnitude of the plus and minus charges which are separated and x is a displacement vector drawn from the minus charge to the positive charge. To see that the oscillating dipole presents an electric current, let’s fix the negative charge portion of the dipole at the origin so that the positive charge is located at r = x and the motion of the positive charge represents a current. Then
dp dx = q= qv dt dt
(275)
We have previously encountered this product of charge and velocity and related it to an electric current in a wire (recall Laws of Ampere, Biot and Savart on page 44): I = qv J= dV ( A= dr ) I dr A Recall from (88) that any current I in a wire segment of length dr generates a magnetic field at position x (relative to the location of the wire segment) given by µ I dr × x dB ( x ) = 0 4π x 3
(88)
Normally we would integrate along the entire length of the wire to get B(x) from this equation. In our case, the current arises from a fluctuating point dipole located at x = 0; no integration is needed. The fluctuating position in the positive charge of the dipole represents a current. We just replace I dr in (88) by qv: µ q v (t ) × r B ( r, t ) = 0 3 4π r
Any changes in the velocity of the charge q will create changes in the magnetic field B: ∂B µ0 q v ( t ) × r = 3 4π ∂t r
where v ( t ) denotes the acceleration of the charge q. This fluctuating magnetic field generates an electric field according to Faraday’s law
∇×E +
Copyright© 2014 by Dennis C. Prieve
∂B =0 ∂t
(107)
PDF generated November 30, 2014
06-712
151 ∇×E = −
or
Fall, 2014
µ q v ( t ) × r µ0 q r × v ( t ) ∂B =− 0 = 3 3 ∂t 4π 4π r r
Integrating this PDE gives the electric field:♣ µ q r × ( r × v ) E ( r, t ) = 0 3 4π r
In spherical coordinates, we can write the two vectors r and v ( t ) as:
r = re r
and
= v ve= z v ( e r cos θ − eθ sin θ )
ez θ
er
Then
unit circle on φ = const surface
r × v = rv er × er cos θ − er × eθ sin θ = − rv ( sin θ ) eφ 0 eφ and
leaving
eθ
r 2 v ( sin θ ) eθ − rv ( sin θ ) r er × eφ = r × ( r × v ) = −eθ
= E ( r , q, φ, t )
µ0 qv sin q eq ( q, φ ) 4π r
While this derivation is very suggestive of the origin and functional form of E induced by the oscillating dipole, the result is not quite correct. The starting equation (88) is for magnetostatic (or slowly varying currents); more precisely, (88) does not include relativistic effects arising from the finite speed of light. We might try to “patch things up” by accounting for the finite speed of light as follows. It takes time r/c for light to propagate from the oscillating dipole at r=0 to some position r=r where the electric field is being evaluated. Instead of evaluating v at the current time t, we should evaluate it at an earlier time, say t – r/c. Then r µ0 qv t − c sin q e q, φ = E ( r , q, φ, t ) ) q( 4π r
(276)
Let’s assume the acceleration v arises from the oscillating dipole given by (274), where p0 = p0ez. The velocity can then be calculated from (275)
♣ Hwk.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
152
Fall, 2014
qv = ωp0 sin ( ωt ) e z
and the acceleration is
(276) becomes
qv = ω2 p0 cos ( ωt ) e z
µ0 ω2 p0 sin θ E ( r , θ, φ, t ) eθ ( θ, φ ) = cos ( ωt − kr ) r 4p
ω c
k=
where
(277)
Spherical Waves Instead of using (88) from magnetostatics, we should have used the Ampere-Maxwell equation (110) from electrodynamics.
∇×= H
∂D +J ∂t
(110)
The current density J is only nonzero at the origin where our oscillating dipole resides. Everywhere else J can be taken as zero. If we also use our usual constitutive equations (B = µH and D = εE), then we can once again [recall Special Case: No Free Charges (e.g. in a Vacuum) on page 5] eliminate B and H between (110) and Faraday’s law (107) to obtain a single PDE in the electric field: 2
∇ E=
µε 1
∂2E
(13)
∂t 2
c2
which we called the wave equation. The exact solution of the wave equation (adding a point source of oscillating current at the origin) for an oscillating point dipole p = p0e-iωtez at the origin is♦ E =
( ) p0 ei kr −ωt k 2 1 ik ( n × e z ) × n + 3 − 2 3n ( n.e z ) − e z 4pe0 r r (((( r direction of E caused by static dipole
While the direction of the first term is transverse (orthogonal) to n, the second term (which is the electric field generated by a quasi-static dipole [see (273)] has a component of E in the direction of n (longitudinal). But for kr ≫ 1, the second term becomes negligible, leaving ♦
taken from equation (9.18) on p411 of Jackson, 3rd ed.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
153 = lim E kr →∞
Fall, 2014
( ) p0 k 2 ei kr −ωt n × ez ) × n (( ( 4pe0 r
(278)
− ( sin θ ) eθ
which is equivalent to (277). The corresponding magnetic field is lim H = kr →∞
( ) p0 k 2 c ei kr −ωt n × ez ) ( (( 4p r
− ( sin θ ) eφ
(278) is a spherical wave. Now n and ez both lie a φ=const plane, so that n×ez must be normal to this plane; thus H is parallel to eφ. On the other hand, the electric field E has the direction of n×H, which lies in the direction of eθ. Of particular interest to us is the energy flux vector
Π Re {E} × Re {H} = =
k 4 p02 c sin 2 θ cos
2
( kr − ωt )
2 16pe 0
r2
(279)
er
Energy is propagating radially outward in all directions from the oscillating dipole. This is called radiation. Note that the flux is proportional to r-2. If we integrate the radial component of flux over a spherical surface of radius r, the result is the power emanating from the oscillating dipole: p
= P
(
)
2pr 2= sin θ d θ ∫ ( εr .P ) (( (( 0
k 4 p02 c cos 2 ( kr − ωt ) 6pε0
da
which fluctuates with time and position. Averaging this over one period of oscillation (in time) yields an average power which is independent of r: 4 p02 c 4p3 p02 c = P = ∝ λ −4 4 12pε0 = 2 p 3ε0 λ
(280)
λ
Note the sensitivity to the wavelength of the incident light, whose frequency and speed of propagation are the same as the scattering light.
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
154
Fall, 2014
Visible light varies in wavelength from about violet or blue:
l = 400 nm
red:
l = 700 nm
Red light has a longer wavelength and scatters much less light than does blue:
P red 700 −4 = = 0.11 P blue 400 This is why the daytime sky appears blue and sunsets and sunrises appear red. During daytime, we don’t look directly at the sun, but at white light which is scattered off of molecules of air and water in the atmosphere. Since the blue portion of the spectrum is scattered more intensely, the sky appears blue.
noon sun white blue blue red
white
setting sun
earth
On the other hand, at sunset we are looking almost directly at the sun. So we see the light which has not been scattered away. Thus the setting sun appears red. Lecture #22 begins here Intensity of Scattering The radial component of the averaged energy flux for the scattered light is obtained by integrating (279) over one period in t, which yields
k 4 a 2 E02 c sin 2 θ sca Πr = 32π2 ε0 r 2
Copyright© 2014 by Dennis C. Prieve
(281)
PDF generated November 30, 2014
06-712
155
Fall, 2014
after substituting p0 = αE0 from (58). The corresponding expression for the energy flux of the incident plane wave is:
E02 Π inc = x 2µ0 c
(282)
Dividing (281) by (282) eliminates the dependence on E0.
Π rsca 4 a 2 c 2 µ0 sin 2 θ = = 2 2 π ε r 16 2 Π inc 0 x c =1 Recall that c 2 =
1
µ0ε0
4 a 2 sin 2 θ µ0ε0
16π2 ε02
r2
was used in deriving the wave equation. To see that this relationship is
dimensionless, we could substitute α e = 4πe0 R3 [see equation after (58) and recall Prob. 2 of Hwk #2] and k = 2 π λ :
Π rsca
4R
4
2
R = 16π sin 2 θ l). Thus a very small fraction of the incident energy flux is scattered in the far field (at least for a single oscillating dipole). To eliminate the dependence on r, consider the power scattered per unit solid angle rather than per unit area. Just like we compute the arc length ds subtended by a angle dθ as ds = r dθ
arc length
we can calculate the area subtended by a solid angle dΩ as da = r2 dΩ
area
Recall that there are 4π steradians in a sphere (just like there are 2π radians in a circle). Thus 4πr2 is the surface area of a sphere. Recognizing that energy flux is power per unit area, we can express the energy flux in terms of the power per unit of solid angle: P rsca =
dP 1 dP = da r 2 d Ω
(284)
Multiplying the numerator of (283) by r2 and the denominator by πR2, we obtain
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
156
Fall, 2014
r 2 P rsca
dP 4 3R Ω d = = 16π sin 2 θ inc inc 2 2 λ πR P x πR P x
(285)
which represents the power of the scattered radiation per unit solid angle per unit power incident upon the atom being polarized (πR2 is the area of the incident plane wave which actually hits the spherical atom). Integrating the numerator over all solid angles♦ in a sphere yields the total (time-averaged) scattered power, which is given by (280):
∫
dΩ dΩ P 128p4 sphere = = 3 R 2 inc R 2 inc pPpP x x dP
4
R −8 ≈ 10 λ
where the second equation was obtained either by integrating (285) or by substituting (280). The above expression represents the ratio of the total scatter power to the power of the LPPW incident upon the spherical scattering atom having radius R. Substituting some typical values (R = 1 nm and l = 1 µm), this ratio is seen to be very tiny. Effect of Scattering Angle Let’s try to depict the scattering intensity as a function of direction of scattering [since (285) clearly does not depend on r]. In spherical coordinates, direction corresponds to a particular pair of values for (θ,φ). So we will draw a surface r(θ,φ) in which the distance r of the surface from the origin (chosen to coincide with the location of the oscillating dipole) is proportional to the intensity or sin2θ. Because (285) also does not depend on φ, this surface will be axisymmetric around the z-axis. At right is that surface which resembles a donut or bagel (with no hole). Color in this plot represents the z-coordinate of the surface. Light scattering experiments today typically use a laser as a light source. The sample is held in ♦
The general formula for integrating and axisymmetric function f over solid angle Ω in spherical coordinates is
∫ sphere
p
f d= Ω
∫ f ( θ ) 2p sin θ d θ .
0
Notice that if f = 1, then
∫ dΩ =
4π .
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
157
Fall, 2014
a 1 cm diameter glass cylinder a few cm long (the small black square in the schematic at right). The detector is usually a photomultiplier tube (PMT) which is held by a goniometer, a device which allows the detector to be rotated around a vertical axis through the center of the sample to detect light scattered in various angles relative to the direction of the incident light. Usually the goniometer rotates in a horizontal plane. This plane contains both the wave vector kinc for the incident beam as well as the wave vector ksca for the scattered rays. The plane containing both of these wave vectors is called the scattering plane. The angle between these two vectors is the scattering angle θsca. The scattering angle θsca is not the same as the polar angle in spherical coordinates θ which is the angle measured from the axis aligned with the electric field for the incident light (see sketch at right). The angular dependence of scattering depends on the polarization of the incident light. If the incident light is polarized so that Einc also lies in the horizontal plane, then the scattering angle θsca is related to the spherical coordinate θ used in (279): θsca = π/2 - θ
Einc horizontal:
sin2θ = cos2 θsca
(286)
This “cut” through the bagel produces the scattering pattern resembling a figure-8, as in the figure below.
On the other hand, if the incident light is vertically polarized (i.e. Einc is vertical), then scattering angle is the spherical coordinate φ: Einc vertical:
θsca = φ and θ = π/2
Since the scattering intensity does not depend on φ, we measure the same scattering intensity regardless of the position of the detector, or r(φ) = const. In effect, we are looking at the slice of the bagel through the x-y plane. The color picture below shows the top view of the bagel, but we are only measuring the instensity in the z=0 plane (green in this figure).
Copyright© 2014 by Dennis C. Prieve
PDF generated November 30, 2014
06-712
158
Fall, 2014
Sunlight or the light from a Hg vapor lamp is unpolarized, which means that there is a uniform distribution of Einc with all polarizations between horizontal and vertical. In this case the scattering profile is the arithmetic mean of the two profiles above: unpolarized light:
Πr ∝
1 + cos 2 θsca 2
Scattering from Small Particles (279) represents the energy flux radiating from a single H atom or other fluctuating dipole. Suppose instead of one, we have two H atoms close enough together that they see the same incident electic field so that the two induced dipole oscillate in unison. Generally, this requires that the distance separating them is very small compared to the wavelength of the incident light.
incident wave
λ
p1 p2
∆r
∆r
