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Are optical forces derived from a scalar potential?

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Abstract

The expression of optical forces provoked by an incident light illuminating particles can be deduced from the Lorentz law. It is shown that these forces derive from a scalar potential in the 2D problem and s-polarization, with light propagating in the cross-section plane of the particles, a fact which shows that the separation between gradient and scattering forces could be questioned. This property does not extend to the p-polarization and 3D problem. In the general case, it is shown that one of the components of the optical force is intimately linked with the reactive energy inside the particle. A possible application is given.

©2007 Optical Society of America

1. Introduction

Radiation pressure in Optics has been the subject of an increasing number of articles in the recent years [127]. The interested reader can find a description of the pioneering work by Ashkin and a review by the same author of the subsequent achievements (until 1997) in [1,2]. Some of the studies are devoted to the use of optical forces for trapping a set of particles. In this frame, the phenomenon of “optical binding” has been reported [37,12,13].

The present work follows two recent papers by the same authors [20,21]. The first one describes a rigorous numerical analysis of the trapping and binding phenomena observed with 2D particles in 2D interference fringes. The second one provides a quantitative phenomenological interpretation of the binding phenomena observed when 2D particles are trapped in order to make a structure close to a photonic crystal. This phenomenological theory is able to predict with a precision better than 3% the period of the trapped photonic crystal (significantly different from that of the system of interference, due to binding) using only a numerical tool giving dispersion curves of photonic crystals.

It is acknowledged that particles in a system of interference fringes are attracted by light spots. However, this remark is rather vague: are the light spots the maximal values of the electric or magnetic power density? Moreover, it is interesting to find a precise mathematical explanation of this behavior and to answer the following question: is the optical force derived from a potential linked with light intensity?

In this paper, it is shown that in 2D problems and s-polarization, the motion of the particles can be explained in a very simple manner: the optical forces derive from a scalar potential which is proportional to the opposite of the electric power density, thus the particles are attracted by the spots of electric power. On the other hand, for p-polarization, the potential is the sum of two terms. The first one has a magnetic origin and derives from a potential which is proportional to the magnetic power density. The second one is a surface term and has an electric origin. For arbitrary polarization, it emerges that a component of the optical force parallel to the particles exists.

We show that the magnetic force can be expressed in terms of the imaginary part of the Poynting vector, which allows us to show the vital role played by the difference between magnetic and electric power densities, the so-called ‘reactive energy’. A possible practical application of the expression of the optical force in 2D problems is envisaged.

2-Expression of the optical force.

Throughout the paper we consider harmonic fields with time dependence in exp(-iωt). The 3D non-magnetic particles of index νr are placed in vacuum. These particles are illuminated by an incident light of wavelength λ0=2π/k0. We call εr0ν2r the permittivity of the particles, µ0 being the permeability of both media.

The optical force on a given particle Ω of regular external surface Ω, with a normal vector n oriented towards the exterior is the consequence of the existence of a polarization vector P inside the dielectric, where no free current or free charge can exist. However, it is equivalent to represent the effects of the polarization vector in terms of bound currents and charges [2326]. First, in order to get rid of the difficulties linked with the discontinuities of the permittivity, we will write Maxwell equations in the more general case where the permittivity is a continuous function ε of space coordinates(x,y,z), with ε(x,y,z)=ε0 outside Ω. One can imagine for example that Ω is divided in two regions: an homogeneous internal region where the permittivity is equal to εr and a transition region of width η, close to the boundary ∂Ω, where the permittivity goes continuously from ε r to ε0. The case of homogeneous particles in vacuum is obtained by imposing η to tend to zero. The optical force can be deduced from Maxwell stress tensor [2830]. Here, we achieve a direct and rigorous derivation from the Lorentz law.

2.1 The optical force in a continuous medium.

Let us write Maxwell equations in harmonic regime:

×Eiωμ0H=0
×H+iωεE=0

Equation (2) can be written in the equivalent form:

×H+iωε0E=iω(ε0ε)E

Bearing in mind that the field generated at any point of space by a volume current density j placed in vacuum satisfies the equation:

×H+iωε0E=j

it emerges that the bound volume current density is given by:

j=iω(ε0ε)E

Using the conservation of charge:

.j=iωρ

and bearing in mind that ∇.(ε E)=∇.D=0 it comes out that the bound volume charge density is given by:

ρ=ε0.E

In the following, we will distinguish the electric force f E, viz. the force generated by the electric field on the bound charges, from the magnetic forces f M, viz. those generated by the magnetic field on the bound currents.

From the Lorentz law, the volume magnetic force density is given by:

fMV=μ02Re{j×H*}
=ω(εε0)μ02Im{E×H*}

while the electric force density can be written:

fEV=12Re{ρE*}=ε02Re{E*.E}

2.2 The optical force on homogeneous particles in vacuum.

In that case, the total magnetic force f M can be deduced from Eq. (8):

fM=ωμ0(εrε0)2ΩIm{E×H*}dV

It is worth noting that, by contrast with the magnetic force, the electric force vanishes in the regions where the permittivity is constant since ∇.E=0. As a consequence, the electric force vanishes outside the transition region and tends to a distribution when η tends to zero [2326]. It is worth noting that in this limit case the expression of this force given by Eq. (9) is ambiguous. Indeed, the right-hand member is the product of the distribution ∇.E whose support lies on the boundary of the particle by the vector E* which has a normal component which is discontinuous on the same boundary. In order to remove this ambiguity, we separate the tangential and normal fields in the transition region:

E=Et+nEn

by defining the normal at any point of the transition region as the unit vector collinear with the gradient of the permittivity, assuming that the variation of permittivity in the orthogonal directions is smooth, in such a way that the normal in the entire transition region tends to the normal to Ω as η tends to zero. Inserting the right-hand member of Eq. (11) in Eq. (9) yields:

fE,t*=ε02Re{Et*.E}

By contrast with Eq. (9), Eq. (12) is not ambiguous since the tangential component of the field is continuous us on ∂Ω as η tends to zero. Using the theory of distributions, we can write that at the limit:

.E=n.[E]δΩ

with ∂Ω defined (incorrectly from a mathematical point of view but conveniently for the physicist) by:

allspaceuδΩdV=Ωu(M)dS

with u being an integrable function defined on ∂Ω, M a point of ∂Ω, dV a volume element of space, dS a surface element around M, and [E] the jump of the electric field on ∂Ω in the direction of the normal. From Eqs. (13) and (14), we deduce that the total contribution of the electric tangential forces on ∂Ω is given by:

fE,t=allspaceε02Re{Et*(n.[E])}δΩdV=ε02ΩRe{[En]Et*}ds

Bearing in mind that the normal components of the electric field on both sides of the normal are linked by the equation:

ε0Enout=εrEnin

it emerges that [En]=EnoutEnin=(εrε01)Enin and thus:

fE,t=(εrε0)2ΩRe{EninEt*}dS

The normal volume electric force density is given by:

fE,nV=ε02Re{nEn*.(Et+nEn)}

Since the tangential component of the electric field is continuous at the limit when η tends to zero, ∇.(E t) remains a function and since this function is summed on a volume which tends to zero, its contribution to the force tends to zero as well, which entails:

fE,nV=ε02Re{nEn*.(nEn)}

In order to remove the ambiguity in Eq. (19) when η tends to zero, let us re-write this equation in the form:

fE,nV=ε02Re{n(εEn*)1ε.(nεEnε)}

Using the vector formula ∇.(aV)=a∇.V+∇a.V and bearing in mind that εEn remains continuous as η tends to zero, thus that its divergence tends to a function, we can write:

fE,nV=ε02Re{n(εEn*)[1ε(1ε).nεEn]}
=ε04Re{n(εEn*)[(1ε2).nεEn]}

As η tends to zero, εEn and its conjugate remain continuous and the gradient tends to a distribution, in such a way that the right hand member of Eq. (21) is no more ambiguous, and since at the limit:

(1ε2)(1ε021εr2)nδΩ

we can finally express the normal force in the form:

fE,nV=ε0εr24(1ε021εr2)Enin2.nδΩ
=ε04(εr2ε021)Enin2nδΩ

Thus, from Eq. (14) the total contribution of the normal electric force to the optical force on the particle can be expressed by:

fE,n=ε04(εr2ε021)ΩEnin2nds

It is interesting to notice that the above result, which gives the total force on the particle from the limit value of the electric field inside the particle, can take a more classical form by writing:

fE,n=ε04Ω(εrε0+1)nEn*in(εrε01)Eninds

and, remembering that Enout=εrε0Enin,

fE,n=12ΩnEn*ε0(EnoutEnin)ds

where 〈En〉 stands for the mean value of the normals of the electric fields on both sides of the surface. Remarking that ε0(Eoutn-Einn) is the surface charge density, it comes out that the force is given by:

fE,n=12ΩnEn*σds

a result which can be understood in the following intuitive way: the normal electric force at a point of the surface is half of the product of the surface charge density by the conjugate of local normal field on this surface current, this local normal field being obtained by removing from the total normal field that produced by the surface current, a result already mentioned in [2326].

In conclusion, the total force on the particle can be written in the form:

f=fM+fE,t+fE,n=ωμ0(εrε0)2ΩIm{E×H*}dV
+(εrε0)2ΩRe{EninEt*}ds+(εr2ε024ε0)ΩEnin2nds

3. Expression of the optical force in the 2D problem

3.1 Calculation of the optical force in the 2D problem.

Now we deal with the 2D problem (Fig. 1): the particles and the incident wave are invariant by translation on the z axis of a cartesian coordinates system xyz. In order to keep the same notations as in the last section, Ω and ∂Ω will denote the cross section of a cylinder and its boundary respectively, dV and dS becoming a small surface element in Ω and a small curvilinear element on ∂Ω respectively. In these conditions, Eq. (28) holds for the force exerted by the incident field on a unit length of the cylinder. It must be noticed that the force exerted along the z axis is contained in the first and second terms of the right-hand member of Eq. (28).

It is well known that in that case, all the components of the fields can be expressed from the z components of the electric and magnetic fields. Thus we separate the xy (transverse) components E xy and H xy from the z (longitudinal) components E zẑ and Hzẑ of the fields. It can be shown easily from Maxwell equations and vector differential relationships that inside Ω :

Exy=iωεrẑ×Hz
Hxy=iωμ0ẑ×Ez

Inserting these representations of the fields in Eq. (28) yields:

f=fxy+ẑfz
fxy=ΩUdV+(εrε0)2ΩRe{EninEt,xy*}ds
+(εrsε024ε0)ΩEnin2ndS
U=14(1ε0εr)(μ0Hz2εrEz2)
fz=12ω(εrε0εr)ΩIm[ẑ.(Hz×Ez*)]dV
+(εrε0)2ΩRe{EninEz*}dS

It is interesting to notice from Eq.(32) that, using Gauss’ theorem, f xy can also be written in the form:

fxy=ΩUndS+(εrε0)2ΩRe{EninEt,xy*}dS
+(εr2ε024ε0)ΩEnin2ndS

Eq. (32) shows that the magnetic transverse component of the optical force derives from a potential. This potential is proportional to the difference between the magnetic and electric power densities. This component of the force can be interpreted as the sum of elementary forces ces dfxy=-∇U dV applied inside Ω.

It is worth noting that for s or p-polarization, Hz or Ez vanishes, thus the first term in the right-hand member of Eq. (34) vanishes. This conclusion holds for the second term since in Einn and E*z are equal to zero for s and p-polarization respectively, thus fz=0. In these two fundamental cases of polarization, the total optical force reduces to f xy. Moreover, for s-polarization, the normal component of the electric field Einn=0 and thus the total optical force derives from the potential U=14(εrε0)Ez2. As a consequence, the elementary optical forces inside Ω are directed towards the smallest values of the potential, i.e. the spots of electric power density. Since ∇U is continuous across ∂Ω (Ez and its normal derivative are continuous), one can state that a particle moves towards the spots of electric power density outside Ω. For p-polarization, the magnetic component of the force (first term in the right-hand member of Eq. (32)) derives from the potential U=μ04(1εεr)Hz2 and the corresponding elementary optical forces are oriented towards the regions of small magnetic power density. However, ∇U is not continuous across Ω since the normal derivative of Hz is not continuous but the normal derivatives inside and outside the particle are proportional and thus one can state again that this component of the force attracts a particle towards regions of small magnetic power density. Due to the existence of two additive surface integrals (electric force) in Eq. (32), it is not possible to state a simple rule for the motion of particles for that polarization.

In the general case of polarization, the z component of the force is the sum of two terms, a volume term of magnetic origin and a surface term of electric origin. Remembering the symmetry with respect to the xy plane of the structure, this result seems surprising. In fact, it should be noticed that the incident field is not symmetrical with respect to the xy plane. A similar transverse effect can be found in [31]: the Imbert-Fedorov effect. This effect is an optical phenomenon in which circularly or elliptically polarized light undergoes a small shift, transverse to the direction of propagation, when totally internally reflected on a plane. This effect is analogous to the Goos-Hänchen effect, but for transverse direction, and arises for elliptically or circularly polarized light. Of course, this remark does not prove that the z component of the force does not vanish in our problem, but shows that this result is not contradictory with the laws of Physics. We have not been able to show theoretically that the two terms representing this force cancel each other, nor the opposite. Moreover, we have not at hand numerical tools for computing both terms. Thus, we have conjectured a priori that fz≠0.

3.2 About gradient and scattering optical forces.

Another important conclusion to be drawn from the existence of a potential in the 2D problem for s-polarization is that the separation between gradient and scattering forces is questionable. In general, it is acknowledged that two kinds of forces can be distinguished [27,5]: gradient forces, which move particles towards the spots of light intensity, and scattering forces, such as the force acting on a particle placed in the field generated by a single plane wave, this second kind of force being easily explained by the transfer of momentum between the electromagnetic waves scattered by the particle and the particle itself. In fact, we have shown in this section that even the scattering forces reduce to gradient forces for s-polarization. It is worth noting that the difficulty to distinguish gradient and scattering forces in some cases was already pointed out in [5]. It is interesting to notice that in the present paper, we assume the permittivity to be real. In [32], it is stated that the existence of a “radiative reaction” (and thus scattering forces) for elementary dipoles requires the polarizability of an elementary dipole to include an imaginary part, at least from a formal point of view. This remark does not entail that scattering forces vanish for real permittivities since the same author showed that a real permittivity does not imply a real polarizability.

The fact that the scattering forces can reduce to a gradient force in some cases is not straightforward from an intuitive point of view. Indeed, it means that the transfer of momentum between the electromagnetic waves scattered by the particle and the particle itself generates a gradient of the square modulus of the field inside the particle, and that this gradient is at the origin of the scattering force. This remark was previously pointed out in the case of Mie particles in [26]. In order to validate this remark, let us give an unexpected consequence of Eq. (35). We consider a particular case of 2D particle, which is in fact a 1D particle: a dielectric slab parallel to the xz axis and illuminated in normal incidence, the direction of incidence being thus the y axis (Fig. 2). Due to the transfer of momentum the vertical optical force on the slab must be oriented in the same direction as the incident wavevector, i.e. towards y=-∞, except in the particular case where all the light is transmitted by the slab. Thus from Eq. (35), the value of the potential at the bottom of the slab (shadow side of the slab) must be smaller than that on the front of the slab (illuminated side). Since the problem is invariant by rotation of polarization, the value of E.E* should be greater at the bottom of the slab than at the front. This elementary property can be verified by using the analytic formulae giving the amplitudes of the plane waves in the three different regions of space [33], but we give here a much simpler demonstration.

We consider in Fig. 2 a dielectric slab of width w and index νr placed in vacuum and illuminated in normal incidence by a plane wave of unit amplitude, with the electric field parallel to the z axis. The system of cartesian coordinates is chosen in such a way that the xz plane coincides with the bottom of the slab.

 figure: Fig.2.

Fig.2. he dielectric slab

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We give in Fig. 2 the amplitudes (electric field) of the plane waves propagating in the three regions of space, the amplitude of the incident wave being unity. It must be noticed that at the bottom of the slab, the interface is illuminated by a single plane wave of amplitude a and thus the amplitude b of the reflected plane waves is given by

b=ρa

with ρ being the reflection coefficient given by Fresnel formula:

ρ=vr1vr+1

Thus, if we assume that the incident electric field is parallel to the z axis, E=Ez and the value of Ez inside the slab is given by:

Ez=a[exp(ik0vry)+ρexp(+ik0vry)]

The square modulus of the electric field is given by:

Ez2=E2=a2exp(ik0vry)+ρexp(+ik0vry)2
=a21+ρexp(+2ik0vry)2

Since νr>1, the coefficient ρ=vr1vr+1 is a real positive number and it is easy to see that the maximum value of |E|2 is reached when exp(+2ik0νry)=1, or equivalently:

y=pλ02vr,p=0,1,2,

It can be concluded that the square modulus of the electric field reaches its maximum value at the bottom of the slab and that its value at the top of the slab takes a smaller value, except when w=pλ02vr, p=0,1,2,…. In that case, the square modulus at the top and at the bottom of the slab are equal, and it can be shown that the reflected field vanishes, which explains why the optical forces generated by scattering vanish too.

As regards the magnetic field, which is parallel to the x axis, the Fresnel formulae show that the reflection coefficient is just the opposite to that given by Eq. (37) and thus, from Eq. (39), it emerges that the minimum value of |H|2 is reached at the bottom of the slab. This is not surprising since in that case, Einn=0, thus the optical force reduces to a gradient. Another way to understand this result is to bear in mind that in a system of interference generated by two identically polarized plane waves propagating in opposite directions, the maximum values of the square modulus of the electric field correspond to the minimum values of the square modulus of the magnetic field. This conclusion can be compared with the results reported in [23], where the author states the link between the force density and currents on a mirror.

The validation obtained analytically for the dielectric slab can be extended to the case of a circular cylinder from numerical results. We have represented in Fig. 3 the square modulus of the electric (resp. magnetic) field at the boundary of a circular cylinder of index νr=3 and radius R=0.3µm illuminated in the external medium of index ν=1.3 by a s (resp. p) polarized plane wave of unit amplitude and wavelength λ0=0.546µm in vacuum. In that case, the index of the external medium is not equal to unity but the conclusions of the present study remain identical.

 figure: Fig. 3.

Fig. 3. Square modulus of the field (SI) at the boundary of a circular cylinder illuminated by a plane wave

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It can be predicted from considerations about transfer of momentum from the electromagnetic fields and the cylinder that the optical force must be oriented towards y=-∞. Thus, from Eqs. (32) and (33), the integral J given by:

J=ΩUndS

should be oriented towards y=-∞ for s-polarization. Taking into account the symmetry of the problem, J is parallel to the y axis, thus, after projecting J on the y axis and simplification, it emerges that the projection J given by:

J=εrε04ΩRcos(θ)Ez2dθ

should be positive. We have found J=3.9 10-19 in the International System of Units (SI). This property is explained by the fact that the largest values of the electric field are obtained in the shadow side (θ close to 0) while those of the magnetic field are reached on the illuminated side (θ close to 180°). For large values of the radius, this result seems to be linked to that obtained for the parallel slab, but it is worth to notice that it holds in the resonance domain. Even though the optical force does not derive from a potential for p-polarization, we have found J=-4.3 10-19 (SI) in that case by replacing Ez by Hz. However, this result does not allow one to believe that the potential term is the predominant one in the expression of the force. It would be interesting to evaluate numerically the importance of the gradient and surface terms in that case.

The same question arises for 3D particles like spheres, where the separation between gradient and scattering forces is generally invoked. Even though the heuristic argument about the separation between these two kinds of forces is basically the same in both cases, it must be acknowledged that the second one is more complicated in nature. Here too, numerical results would be necessary to enlighten the adequacy of this separation.

3.3 About the optical force in the 3D problem.

The conclusions of the last section show that optical force does not derive from a scalar potential for 3D problem. However, let us notice that the expression of the magnetic part of the optical force in terms of the imaginary part of the complex Poynting vector shows that, even in the 3D problem, this part of the force is related to the ‘reactive energy’, i.e. the difference between the electric and magnetic power densities. Indeed, let us consider the flux Φ of the Poynting vector through ∂Ω :

Φ=12Ωn.(E×H*)dS

Using the divergence theorem, it can be shown that:

Φ=12Ω.(E×H*)dV
=12Ω(H*.×EE.×H*)dV

thus, using Maxwell equations:

Φ=2iωΩ(μ04H2εr4E2)dV=2iωΩ(WMWE)dV

Thus in harmonic regime, the flux through the closed surface ∂Ω of the imaginary part of the Poynting vector P is linked to the time average of the difference between the total magnetic and electric powers inside Ω, generally denoted as reactive energy:

This conclusion must be compared to Eq. (28), which shows that the magnetic force is proportional to the flux of the imaginary part of the Poynting vector, thus to the reactive energy inside the particle. However, we have not been able to find a physical interpretation to this result.

3.4 A possible application of optical forces on 2D particles.

It has been shown that in the 2D problem, there exists a longitudinal component Ez of the optical force, which vanishes for s and p-polarizations. Let us suppose that a particle obtained by truncating a circular cylinder is allowed to move along the axis of the cylinder only and that this particle is illuminated by an unpolarized incident light propagating in the cross-section of the cylinder and converging on the cylinder. We can imagine for example that the particle is maintained on the same axis by a mechanical means (insertion in a transparent micropipe). If a polarizer is placed on the incident beam, the longitudinal optical force will vanish for s or p polarized light. Now, rotating the polarizer from this equilibrium position will create a longitudinal optical force which can move the particle in the longitudinal direction on both sides, according to the sense of rotation. Bearing in mind that the direction of polarization of some kinds of polarizers can be changed very rapidly using for example quartz oscillations in MOEMS, such a device could be used in order to generate rapid liquid flows in microfluidics or rapid arbitrary translations of cylinders in micromechanics. It is worth noting that when the particle is placed in a background medium, a mechanical contribution to the force should be added [3439].

4. Conclusion

It has been shown that in 2D problems, the magnetic part of the transverse optical force exerted on a particle derives from a scalar potential linked to the reactive energy. For s-polarization, this conclusion applies to the total optical force, a result which is rather surprising in a domain where it is generally assumed that this force separates into a gradient and a scattering part. For 3D problem, we have shown the strong link between the optical force and the reactive energy. An application to microfluidics or micromechanics has been envisaged.

Acknowledgements

This work has been carried out in the framework of the Nanosciences and Nanotechnology French program. The support of the EC-funded project PHOREMOST (FP6/2003/IST/2-511616) is gratefully acknowledged. The content of this work is the sole responsibility of the authors. The authors would like to thank Dr Jean-Marc Fournier (EPFL, Lausanne) for helpful discussions.

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Figures (3)

Fig.2.
Fig.2. he dielectric slab
Fig. 3.
Fig. 3. Square modulus of the field (SI) at the boundary of a circular cylinder illuminated by a plane wave

Equations (54)

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× E i ω μ 0 H = 0
× H + i ω ε E = 0
× H + i ω ε 0 E = i ω ( ε 0 ε ) E
× H + i ω ε 0 E = j
j = i ω ( ε 0 ε ) E
. j = i ω ρ
ρ = ε 0 . E
f M V = μ 0 2 Re { j × H * }
= ω ( ε ε 0 ) μ 0 2 Im { E × H * }
f E V = 1 2 Re { ρ E * } = ε 0 2 Re { E * . E }
f M = ω μ 0 ( ε r ε 0 ) 2 Ω Im { E × H * } d V
E = E t + n E n
f E , t * = ε 0 2 Re { E t * . E }
. E = n . [ E ] δ Ω
all space u δ Ω d V = Ω u ( M ) d S
f E , t = all space ε 0 2 Re { E t * ( n . [ E ] ) } δ Ω d V = ε 0 2 Ω Re { [ E n ] E t * } d s
ε 0 E n out = ε r E n in
f E , t = ( ε r ε 0 ) 2 Ω Re { E n in E t * } d S
f E , n V = ε 0 2 Re { n E n * . ( E t + n E n ) }
f E , n V = ε 0 2 Re { n E n * . ( n E n ) }
f E , n V = ε 0 2 Re { n ( ε E n * ) 1 ε . ( n ε E n ε ) }
f E , n V = ε 0 2 Re { n ( ε E n * ) [ 1 ε ( 1 ε ) . n ε E n ] }
= ε 0 4 Re { n ( ε E n * ) [ ( 1 ε 2 ) . n ε E n ] }
( 1 ε 2 ) ( 1 ε 0 2 1 ε r 2 ) n δ Ω
f E , n V = ε 0 ε r 2 4 ( 1 ε 0 2 1 ε r 2 ) E n in 2 . n δ Ω
= ε 0 4 ( ε r 2 ε 0 2 1 ) E n in 2 n δ Ω
f E , n = ε 0 4 ( ε r 2 ε 0 2 1 ) Ω E n in 2 n d s
f E , n = ε 0 4 Ω ( ε r ε 0 + 1 ) n E n * in ( ε r ε 0 1 ) E n in d s
f E , n = 1 2 Ω n E n * ε 0 ( E n out E n in ) d s
f E , n = 1 2 Ω n E n * σ d s
f = f M + f E , t + f E , n = ω μ 0 ( ε r ε 0 ) 2 Ω Im { E × H * } d V
+ ( ε r ε 0 ) 2 Ω Re { E n in E t * } d s + ( ε r 2 ε 0 2 4 ε 0 ) Ω E n in 2 n d s
E xy = i ω ε r z ̂ × H z
H xy = i ω μ 0 z ̂ × E z
f = f xy + z ̂ f z
f xy = Ω UdV + ( ε r ε 0 ) 2 Ω Re { E n in E t , xy * } d s
+ ( ε r s ε 0 2 4 ε 0 ) Ω E n in 2 n d S
U = 1 4 ( 1 ε 0 ε r ) ( μ 0 H z 2 ε r E z 2 )
f z = 1 2 ω ( ε r ε 0 ε r ) Ω Im [ z ̂ . ( H z × E z * ) ] d V
+ ( ε r ε 0 ) 2 Ω Re { E n in E z * } d S
f xy = Ω U n d S + ( ε r ε 0 ) 2 Ω Re { E n in E t , xy * } d S
+ ( ε r 2 ε 0 2 4 ε 0 ) Ω E n in 2 n d S
b = ρ a
ρ = v r 1 v r + 1
E z = a [ exp ( i k 0 v r y ) + ρ exp ( + i k 0 v r y ) ]
E z 2 = E 2 = a 2 exp ( i k 0 v r y ) + ρ exp ( + i k 0 v r y ) 2
= a 2 1 + ρ exp ( + 2 i k 0 v r y ) 2
y = p λ 0 2 v r , p = 0 , 1 , 2 ,
J = Ω U n d S
J = ε r ε 0 4 Ω R cos ( θ ) E z 2 d θ
Φ = 1 2 Ω n . ( E × H * ) d S
Φ = 1 2 Ω . ( E × H * ) d V
= 1 2 Ω ( H * . × E E . × H * ) d V
Φ = 2 i ω Ω ( μ 0 4 H 2 ε r 4 E 2 ) d V = 2 i ω Ω ( W M W E ) d V
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