Physical Meaning of the Dipole Radiation Resistance in Lossless and Lossy Media M

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Physical meaning of the dipole radiation resistance in lossless and lossy media M. S. Mirmoosa, Member, IEEE, S. Nordebo, and S. A. Tretyakov, Fellow, IEEE

Abstract—In this tutorial, we discuss the radiation from a tradictory discussions on the definition and physical meaning Hertzian dipole into uniform isotropic lossy media of infinite of key parameters of antennas in absorbing media, such as extent. If the medium is lossless, the radiated power propagates to the input impedance and radiation resistance [4], [8]–[12]. infinity, and the apparent dissipation is measured by the radiation resistance of the dipole. If the medium is lossy, the power expo- In this tutorial paper we carefully examine the notion of nentially decays, and the physical meaning of radiation resistance radiation resistance of a Hertzian dipole in infinite isotropic needs clarification. Here, we present explicit calculations of the homogeneous media and discuss the physical meaning of this power absorbed in the infinite lossy host space and discuss the model in both lossless and lossy cases. We explicitly calculate limit of zero losses. We show that the input impedance of dipole the absorbed power in the infinite space and show that the antennas contains a radiation-resistance contribution which does not depend on the imaginary part of the refractive index. This radiation resistance as a model of power transported to infinity fact means that the power delivered by dipole antennas to is also nonzero for dipoles in lossy media. Subsequently, we surrounding space always contains a contribution from far fields study the limit of zero loss. Finally, we discuss implications of unless the real part of the refractive index is zero. Based on this this theory for understanding and engineering power transfer understanding, we discuss the fundamental limitations of power between antennas in media. coupling between two antennas and possibilities of removing the limit imposed by radiation damping. From the applications point of view, antennas embedded in lossy media have received attention in geophysics, marine Index Terms—Absorbed power, Hertzian dipole, lossy media, technology [3], [13], [14], medical engineering [15]–[17], etc. radiation resistance, radiated power. Hence, it is worthy to develop a thorough understanding of radiation from the Hertzian dipole which is placed in a I.INTRODUCTION lossy medium. These results will also help to understand The radiation resistance of a dipole antenna is a simple sub-wavelength emitters or nanoantennas immersed in dis- classical concept which is used to model the power radiated sipative media, since we can make an analogy between the into surrounding infinite space, as explained in many antenna Hertzian dipole and the sub-wavelength emitter/nanoantenna textbooks, e.g. [1]. However, its physical meaning is not al- that operates in the optical range. The investigation of light ways easy to grasp. Indeed, if the surrounding space is lossless, interaction with plasmonic or all-dielectric nanoantennas is the energy cannot be dissipated. Yet, if the space around the mainly conducted in the assumption that the host medium is antenna is infinite, the radiation resistance is nonzero, which is not dissipative [18]–[20]. However, from the practical point tantamount to absorption of power. One can perhaps say that of view, in most applications nanoantennas (nanoparticles) are the radiated energy is transported all the way to infinity. On the placed in absorptive environments (e.g. see Refs. [21], [22]). other hand, if the surrounding medium is lossy, the radiated This review is also relevant to the studies of absorption and power is exponentially decaying, and it is sometime assumed scattering by small particles in lossy background [23]–[30]. that the usual definition of the radiation resistance does not If a point source dipole is embedded in a lossy host, there apply, because at the infinite distance from the antenna the is also a theoretical problem of singularity of power absorbed radiated fields are zero. One can perhaps say that all the power in the medium, which is due to singularity of the dipole fields is dissipated in the antenna vicinity. Within this interpretation, at the source point. This issue has been discussed e.g. in [7]. we have a confusing “discontinuity”: If the medium is lossless, The problem of source singularity should be addressed taking all the radiated power is transported to infinity, but if the into account the final size and shape of the antenna. Here, we arXiv:1901.02777v2 [physics.class-ph] 18 Dec 2019 loss factor is nonzero (even arbitrarily small), no power is focus our discussion on radiation phenomena, studying power transported to infinity. absorbed outside of a small sphere centred at the source point. Although there is extensive classical literature on antennas in absorbing media (see the monograph [2] and e.g. Refs. [3]– II.RADIATION FROM HERTZIANDIPOLESINISOTROPIC [7]), we have found only limited and sometimes even con- HOMOGENEOUSLOSSYMEDIA This work has been partly supported by the Swedish Foundation for Let us consider a Hertzian dipole antenna located in an Strategic Research (SSF) under the programme Applied Mathematics and the project Complex Analysis and Convex Optimization for EM Design. infinite homogeneous space, as illustrated in Fig. 1. The M. S. Mirmoosa and S. A. Tretyakov are with the Department of Electronics current amplitude in the dipole is fixed and denoted by I and Nanoengineering, Aalto University, P.O. Box 15500, FI-00076 Aalto, Fin- (we consider the time-harmonic regime, assuming ejωt time land (email: mohammad.mirmoosa@aalto.fi and sergei.tretyakov@aalto.fi). S. Nordebo is with the Department of Physics and Electrical Engineering, dependence). The dipole length is l. For simplicity, we assume Linnæus University, 351 95 Vaxj¨ o,¨ Sweden (email: [email protected]). that the medium is non-magnetic (µ = µ0), which does not 2

A. Time-averaged power and energy conservation If the host medium is lossy and absorbs power, the calculated “radiated” power depends on the radius of the fictitious spherical surface, and careful considerations are necessary. Since the fields of the point source are singular at the source location, we consider the outward power flux through the surface of a small sphere of the radius R , with the dipole 푅0 0 at its center, as shown in Fig. 1 using yellow color. The law of energy conservation tells that all the outward flowing power must be absorbed by the exterior infinite volume (r > R0 in Fig. 1). In other words,

Prad = Pabs, (3)

Fig. 1. The Hertzian dipole is radiating in a lossy medium. where I 1 ∗ Prad = Re E × H · dS (4) 2 S0 compromise the generality of our discussion. We characterize and Z the background medium by its complex relative permittivity 1 2 0 00 0 Pabs = σ|E| dV. (5) = − j = − jσ/(ω0), where σ is the effective 2 V conductivity. We will also use the complex refractive index √ Here, S0 denotes the surface of radius R0 and V represents n = = n0 − jn00 (the square root branch is defined along 00 the volume of the spherical shell extending from r = R0 to the positive real axis so that n ≥ 0). r → ∞. Furthermore, σ is the finite conductivity of the lossy The power delivered from the ideal current source I is medium. Since Eq. (3) must always hold for a lossy medium, usually written in terms of the equivalent resistance Rin, as it should be also true in the limiting case when the dipole is located in a lossless medium. It may be difficult to perceive 1 2 Prad = Rin|I| . (1) because the conductivity would be zero (σ = 0) in this case 2 and the right-hand side of Eq. (3) seems to vanish (Pabs = 0). In fact, the resistance Rin is the real part of the ratio of However, the left-hand side is apparently not zero (Prad 6= 0). the voltage to electric current at the input terminals of the Thus, care should be taken in considering the limit of zero dipole [1]. Hence, Rin has the meaning of the real part of the losses. Let us first evaluate the power radiated from the sphere dipole input impedance. We remind that the Hertzian dipole of radius R0, given by Eq. (4) for the general case of lossy model assumes that the antenna current is fixed, that is, it is an media when σ 6= 0, and then study the case when σ tends to ideal current source without any dissipation inside the antenna zero. This surface integral was calculated in Ref. [7]. If we structure. If the infinite medium surrounding the dipole is substitute the known expressions of the electric and magnetic lossless, such as free space, the input resistance is identical to fields [1] the radiation resistance. In this case, the usual calculation of Il 1 k the Poynting vector flux through a fictitious spherical surface E = −j cos θ + j exp(−jkr), r 2πω r3 r2 (which is easy to evaluate in the far zone, see, e.g. [1]) results 0 2 in Il 1 k k Eθ = −j sin θ + j − exp(−jkr), (6) 2 4πω r3 r2 r (k0l) 0 0 Rin(lossless) = Rrad = η0 n , (2) Il 1 k 6π H = sin θ + j exp(−jkr), φ 4π r2 r in which Rrad represents the radiation resistance. Also, η0 and k0 are the free-space intrinsic impedance and wave number, generated by the Hertzian dipole into Eq. (4) considering that respectively. Note that the refraction index n = n0 is a real, k = k0n, and calculate the integral, the result reads non-negative number1. However, if the medium is dissipative, Prad = the situation may remarkably change. The input resistance R in (k l)2|I|2 2n00 4n002 may differ extremely from the radiation resistance since it must 0 0 η0 n 02 002 2 3 3 + 02 002 2 2 2 also model the dissipation in the near zone. In the following, 12π (n + n ) k0R0 (n + n ) k0R0 00 2n 00 this difference will be discussed in detail. −2k0n R0 + 02 002 + 1 e . (n + n )k0R0 1 In isotropic non-magnetic media characterized√ by relative permittivity , (7) the choice of the square root branch for n = according to the passivity condition n00 ≥ 0 ensures that n0 ≥ 0. Thus, the resistance (2) is always Derivation of the above equation is straightforward (see Sup- non-negative. The real part of the refraction index n0 is negative in double- plementary Information). However, calculating Pabs does not negative media, where the real parts of both permittivity and permeability are negative. However, also in that case the resistance value in (2) is non-negative, appear to be that simple. Paper [7] contains a statement that 2 (k0l) equality (3) holds, but calculation of the volume integral (5) because for lossless media we get Rin = η0 6π µn, where both µ and n are negative. is not given, and the limit of σ → 0 is not discussed. 3

Z ∞ −κr 00 2 002 e k0n 2k n To calculate the integral (5), we firstly write the square of 2 : 2k n00 dr = − 0 e−κR0 0 r3 R2 R the absolute value of the electric field components, which gives R0 0 0 (16) Z ∞ e−κr 2 2 2 + 4k3n003 dr, 2 |I| l cos θ 0 |E | = · R r r 4π2ω22(n02 + n002)2 0 0 and 00 2 02 002 (8) 1 2k0n k (n + n ) 00 ∞ 0 −2k0n r k2(n02 + 5n002) Z e−κr k2(n02 + 5n002) 6 + 5 + 4 e , 0 0 −κR0 r r r 3 : 2 dr = e 3 R0 r 3R0 and 2k3n00(n02 + 5n002) Z ∞ e−κr − 0 dr, |I|2l2 sin2 θ 3 r |E |2 = · R0 θ 2 2 2 02 002 2 (17) 16π ω 0(n + n ) 1 2k n00 k2(3n002 − n02) where we have introduced the notation κ = 2k n00. We can + 0 + 0 (9) 0 r6 r5 r4 combine the above results (15–17) and write 3 00 02 002 4 02 002 2 Z ∞ 2k n (n + n ) k (n + n ) 00 00 0 0 −2k0n r f (r)e−2k0n rdr = + 3 + 2 e . 1 r r R0 1 2k n00 k2(n02 + n002) Using the above expressions, we find the square of the absolute 0 0 −κR0 3 + 2 + e (18) value of the electric field, and next we can evaluate the volume 3R0 3R0 3R0 integral in Eq. (5) to calculate the power P absorbed in 2k3n00(n02 + n002) Z ∞ e−κr abs − 0 dr. the exterior infinite spherical shell. Upon integration over the 3 R0 r angles, we find that (see Supplementary Information) Comparing Eq. (18) with Eq. (11) we find that the last term 2 2 0 00 ∞ |I| l n n Z 00 in Eq. (18) (the coefficient before the integral) is the same as −2k0n r Pabs = f(r)e dr, (10) 02 002 2 function f2(r) but with the opposite sign. Therefore, only the 2πω0 (n + n ) R 0 first three terms inside the square brackets in Eq. (18) remain where f(r) = f1(r) + f2(r) + f3(r), with and, interestingly, the last term vanishes. Thus, 00 2 02 002 Z ∞ 1 2k0n k0(n + 5n ) −κr f (r) = + + , f1(r) + f2(r) e dr = 1 r4 r3 3r2 R0 (19) 2k3n00(n02 + n002) 00 2 02 002 0 1 2k0n k0(n + n ) f2(r) = , (11) −κR0 3 + 2 + e . 3r 3R 3R 3R0 4 02 002 2 0 0 k0(n + n ) f (r) = . The last step in this long derivation of Pabs in Eq. (10) is the 3 3 evaluation of the integral for function f3(r), which gives Here, we have used the relation between the imaginary part Z ∞ k3(n02 + n002)2 of the permittivity and the corresponding conductivity f (r)e−κrdr = 0 e−κR0 . (20) 3 6n00 σ R0 00 = 2n0n00 = , (12) η /k = 1/(ω ) ω Using Eqs. (19) and (20) and the identity 0 0 0 , we 0 finally arrive to the following formula for the power absorbed which gives in the exterior environment (in the spherical shell extending 0 00 σ = 2ω0n n . (13) from R0 to infinity):

Note that the term f3, which does not depend on the distance, Pabs = 2 2 0 00 comes from the 1/r terms in the expression for the electric (k0l) |I| 2n n η0 02 002 2 3 ﬁeld. To simplify Eq. (10), we use partial integration, and 12π (n + n ) (k0R0) derive the following identity: 4n0n002 2n0n00 + + + n0 e−κR0 . Z ∞ 1 (n02 + n002)2(k R )2 (n02 + n002)(k R ) e−κr = 0 0 0 0 m (21) R0 r m−1 i−1 i−1 −κR m−1 m−1 ∞ −κr Now, if we compare Eq. (21) with Eq. (7), we see that they are X (−1) κ e 0 (−1) κ Z e + dr, identical, meaning that Eq. (3) holds for any non-zero value P (m − 1, i)Rm−i (m − 1)! r i=1 0 R0 of the conductivity (or the imaginary part of the permittivity), (14) including the limiting case as σ tends to zero. where P (x, y) refers to the permutation formula. The function Considering the expression (21) for the absorbed power, we note that there are several terms which are proportional to n00 f1(r) consists of three terms and for each of them we can use the expression given by Eq. (14). Hence, and singular at R0 → 0. They can be interpreted as the power absorbed in the near vicinity of the dipole. The singularity is Z ∞ −κr 00 2 002 e 1 k0n 2k n 1 : dr = − + 0 e−κR0 due to the fact that the ﬁelds of a point dipole are singular at 4 3 2 00 R0 r 3R0 3R0 3R0 the dipole position. Thus, if the medium is lossy (n 6= 0), the 3 003 Z ∞ −κr (15) 4k0n e absorbed power diverges. Naturally, these terms tend to zero − dr, if n00 → 0. 3 R0 r 4

B. Radiation resistance surrounding medium between spherical radii R0 and R, and 2 2 00 (k0l) |I| 0 −2k0n R0 the second term is the power radiated away beyond radius R. The last term in Eq. (21), η0 12π n e , depends 00 Now we can accurately calculate the limit for R → ∞ and on n and R0 only in the exponential factor, in contrast to 00 the singular near-field terms. The exponential factor tends to vanishing σ (and n ). Calculating the integral (10) with a finite 00 upper limit R and considering vanishing n00 and infinitely unity when either R0 or n tends to zero. Because this last growing R, the power balance relation (23) takes the form term is not singular, we can let R0 → 0 and interpret this term as the dipole radiation resistance, which measures the “power 2 2 (k l) |I| 00 delivered to infinity”. Therefore, 0 0 −2k0n R Prad = η0 n 1 − lim e 2 2 12π n00→0,R→∞ h (k0l) |I| 0 −2k n00R i lim η n e 0 0 2 2 2 R0→0 0 (k0l) |I| 00 12π (k0l) 0 0 −2k0n R Rrad = 2 = η0 n . + η0 n lim e . (24) |I|2 6π 12π n00→0,R→∞ (22) Importantly, this term results from integration of 1/r (“wave”) Here, the first line gives the power absorbed in the infinite terms in the expression for the dipole fields, and it is exactly space, and the second line is the power crossing the spherical the same as the power radiated into infinite lossless media surface of the infinite radius. Clearly, the value of the double- (see Eq. (2)). Thus, the radiation resistance of electric dipoles limit depends on the order in which the two limits are taken. in lossy media does not depend on the imaginary part of the Thus, there are two possible interpretations of the absorption refractive index, and it is expressed by the same formula as for in the lossless infinite space. lossless background. This formula for the radiation resistance 1. Taking first the limit R → ∞ and then the limit of was given in paper [9] (the first term in Eq. (10)), derived vanishing loss factor n00 → 0, in the spirit of the principle of on the basis of re-normalizing the power flow density, which vanishing absorption [31], we can say that in this interpretation is not offering clear physical insight. It also agrees with the all the radiated power is dissipated in the vacuum and there is result presented in paper [10], Eq. (2)2, which is basically no radiation to infinity (because the expression in the second equivalent to substituting complex-valued material parameters line tends to zero). into the formula derived for the free-space background. On 2. Taking first the limit n00 → 0 and then extending the the other hand, contradictory expressions can be also found in volume integration over the whole space letting R → ∞, the the literature, for example, in Refs. [11], [12]. expression in the first line (absorption in the medium) tends to zero. This is a very common interpretation of the lossless III.DISCUSSION infinite space. There are no losses in the “vacuum”, and all the As we see from Eq. (21), the absorbed power calculated radiated power is radiated away throughout the infinite space, as the volume integral (5) is not zero even in the limit of beyond any finite radius R (in this sense “transported all the zero conductivity (lossless background). This result seemingly way to infinity”). contradicts to the fact that for σ = 0 the integrand of (5) is However, we stress that in the expression for the total identically zero. However, we cannot conclude that in this “absorbed power” (24) the two double-limit expressions cancel case the integral is zero, because for lossless background out, meaning that whatever is the interpretation, the radiation R 2 resistance defined as 2P /|I|2 is a well-behaving, continuous the integral V |E| dV diverges. Care should be taken in rad considering the limit of vanishing absorption if the absorption function of σ or n00, and based on the first interpretation above volume is infinite. we can use the expression (21) for both lossy and lossless Let us discuss this limit. The expression for the power ab- backgrounds3. sorbed in the medium has a form of a product of conductivity, Another, perhaps more important, implication of this consid- which tends to zero in the lossless limit, and the integral of the eration is that the input resistance given by Eq. (2) for lossy energy density, which diverges when we extend the integration background contains exactly the same term as the radiation volume to infinity. Thus, we should write the power balance resistance of Hertzian dipole antenna in lossless media or vac- relation considering absorption in a sphere of a finite radius R uum. Thus, it is misleading to assume that if the background and then properly calculate the limit. For this consideration, medium is lossy, there is no radiation resistance as such, as the energy conservation relation (3) is written as the fields exponentially decay. We see that the total input Z resistance is the sum of the radiation resistance, proportional 1 2 0 Prad = σ |E| dV to n , and the near- and intermediate-zone loss resistance, 2 00 VR\VR0 proportional to n . Importantly, as already discussed, the term 1 I proportional to n0 does not vanish if n00 becomes zero, and it + Re {E × H∗}· dS, (23) 2 SR has exactly the same form in both lossy and lossless cases. 00 0 00 The only scenario where there is no radiation resistance at where σ = ω0 = 2ω0n n , and VR and VR denote 0 all is when the real part of the refraction index n0 vanishes. spheres of radii R and R0, respectively. The first term on the right-hand side of (23) is the power dissipated in the 3Actually, there can be infinitely many intermediate “intepretations” since 2Note the misprint in Eq. (2) of [10]: the right-hand side should read this double-limit expression can take any value from zero to unity depending Z(nω, 0)/n. on the way of taking the limit. 5

Perhaps counter-intuitively, this case corresponds to lossless which is always possible. The ideal scenario where the de- background media. Indeed, calculating livered power can be arbitrarily high corresponds to Rin = 0 (then we have P = |E|2l2/R which diverges for R → = n2 = n02 − n002 − 2jn0n00, (25) load load 0). Now we should look at the expression for the effective we see that if n0 = 0, then = −n002 is purely real, meaning input resistance and find under what conditions this expression that the medium is lossless. We also note that the permittivity gives the smallest value. Apparently, conventional lossless is negative that is the case when wave propagation is not background (n00 = 0) is better than the lossy one, but even in possible. that case the received power is limited, because the radiation Formula (21) can be written in terms of the contribution resistance (2) is not zero. This consideration brings us to the to the real part of the input impedance of the antenna due well-known limit of the effective absorption cross section for to dissipation in the surrounding medium (upon dividing by any dipole antenna/scatterer in lossless background [32]. But |I|2/2): notice an important special case of n0 = 0. In this case the input resistance is zero, and the delivered power has no upper R = in bound. 2 2 (k0l) 0 (k0l) 0 00 2 Accordingly, we reach an enlightening conclusion. If two η0 n + η0 n n 02 002 2 3 + 6π 6π (n + n ) (k0R0) (26) antennas are in an environment which does not allow wave 4n00 2 propagation, the power delivered from one antenna to the other 02 002 2 2 + 02 002 . (n + n ) (k0R0) (n + n )(k0R0) can be arbitrarily high. In contrast, if wave propagation is allowed, the delivered power is fundamentally limited by the Here, we have assumed that k R is very small so that 0 0 ultimate absorption cross section of a dipole antenna, which, in e−κR0 ≈ 1. As mentioned before, the first term is the same turn, is determined by its non-zero radiation/input resistance. as the radiation resistance in lossless media, and can be Another case when the power delivered to a load from an interpreted as the radiation resistance in lossy media. The ideal current source is limited only by the parasitic resistance second term, singular at R → 0, is due to dissipation in 0 of conductors, is the case of zero frequency (DC) circuits. the near and intermediate zones. This second term naturally Interestingly, the reason why there is no fundamental limit is vanishes if n00 = 0. Thus, the same: There is no radiation loss at DC. Rin = Rrad + Rnrad, (27) The scenario of such unlimited-capacity power delivery channel corresponds, for instance, to the case of two dipole where the non-radiative resistance is given by antennas inside a lossless-wall waveguide below cut-off. The (k l)2 2 two antennas are coupled only by reactive fields, and at the R = η 0 n0n00 + nrad 0 6π (n02 + n002)2(k R )3 resonance of the antenna pair inside the waveguide, all the 0 0 (28) 4n00 2 effective reactances are compensated. At this frequency, the 02 002 2 2 + 02 002 . receiving antenna is effectively connected to the ideal current (n + n ) (k0R0) (n + n )(k0R0) source feeding the transmitting antenna by a zero-impedance This consideration clarifies the physical meaning of radiation link. Here, the power delivered from one antenna to the other resistance in the general case of isotropic background media: is limited only by the power available from the source, and It vanishes only if propagation is not possible. In other words, by parasitic losses in the waveguide walls and in the antenna the radiation resistance is not zero even in lossy media, wires. Another example is coupling between two antennas in because the propagation constant is not equal to zero and the lossless plasma. outward power flux is not zero. In terms of applications, this conclusion is important for understanding of ultimate limits for power transported from IV. CONCLUSIONS one antenna to another. Let us position a receiving Hertzian In this tutorial review we have addressed the question dipole antenna of length l in the field created by our radiating of radiation resistance of electric dipole antennas in lossy dipole antenna (in an infinite isotropic medium characterized background. We have shown that the power absorbed in the by the refractive index n = n0 − jn00). The current induced in background medium (excluding a small sphere of radius R0 the receiving antenna I is proportional to the electric field rec around the source) contains terms which diverge at R0 → 0 E created by the transmitting antenna at the receiver position and measure the power delivered by the antenna near fields, and inversely proportional to the impedance: and, importantly, the radiation resistance term, which measures El the power delivered to the lossy host by the wave fields Irec = . (29) Rload + jX + Rin (decaying as 1/r). An important conclusion is that this radiation resistance does not depend on the imaginary part of the Let us assume that the receiving antenna is loaded by a resistor refractive index: It differs from the well-known formula for the R at its center. The power delivered to the load reads load radiation resistance of a dipole in free space by multiplication 1 2 by the real part of the refractive index. We have discussed P = Rload|Irec| . (30) 2 in detail the limit of zero loss factor, comparing two possible Obviously, to maximize the delivered power we should bring interpretation of power “loss” in lossless vacuum. Finally, we the antenna to resonance, making the total reactance X = 0, presented some considerations on fundamental limitations on 6

power coupling between two antennas and possible scenarios [27] Q. Fu and W. Sun, “Mie theory for light scattering by a spherical particle of removing the limit imposed by radiation damping. in an absorbing medium,” Appl. Opt., vol. 40, no. 9, pp. 1354–1361, 2001. [28] P. Yang, et al, “Inherent and apparent scattering properties of coated or uncoated spheres embedded in an absorbing host medium,” Appl. Opt., REFERENCES vol. 41, no. 15, pp. 2740–2759, 2002. [29] S. Nordebo, G. Kristensson, M. Mirmoosa and S. Tretyakov, “Op- [1] C. A. Balanis, Antenna Theory: Analysis and Design, New Jersey: John timal plasmonic multipole resonances of a sphere in lossy media,” Wiley & Sons, 2005. Phys. Rev. B, vol. 99, p. 054301, 2019. [2] R. W. P. King and G. S. Smith, with M. Owens and T. T. Wu, Antennas [30] S. Nordebo, M. Mirmoosa and S. Tretyakov, “On the quasistatic op- in Matter: Fundamentals, Theory, and Applications, Cambridge, MA: timal plasmonic resonances in lossy media,” J. Appl. Phys., vol. 125, The MIT Press, 1981. p. 103105, 2019. [3] R. W. P. King and C. W. Harrison, Antennas and Waves: A Modern [31] B. R. Vainberg, “Principles of radiation, limit absorption and limit Approach, MA: The MIT Press, 1969. amplitude in the general theory of partial differential equations,” [4] R. K. Moore, “Effects of a surrounding conducting medium on antenna Russ. Math. Surv., vol. 21, pp. 115–193, 1966. analysis,” IEEE Trans. Antenn. Propag., vol. 11, no. 3, pp. 216–225, [32] S. Tretyakov, “Maximizing absorption and scattering by dipole parti- May 1963. cles,” Plasmonics, vol. 9, pp. 935–944, 2014. [5] A. Karlsson, “Physical limitations of antennas in a lossy medium,” IEEE Trans. Antenn. Propag., vol. 52, no. 8, pp. 2027–2033, Aug. 2004. [6] J. R. Wait, “The magnetic dipole antenna immersed in a conducting medium,” Proc. IRE, vol. 40, no. 10, pp. 1244–1245, Oct. 1952. [7] C. T. Tai and R. E. Collin, “Radiation of a Hertzian dipole immersed in a dissipative medium,” IEEE Trans. Antenn. Propag., vol. 48, no. 10, pp. 1501–1506, 2000. [8] L. A. Ames, J. T. DeBettencourt, J. W. Frazier and A. S. Orange, “Radio communications via rock strata,” IEEE Trans. Commun. Syst., vol. 11, no. 2, pp. 159–169, June 1963. [9] C. K. H. Tsao, “Radiation resistance of antennas in lossy media,” IEEE Trans. Antenn. Propag., vol. 19, no. 3, pp. 443–444, 1971. [10] G. Deschamps, “Impedance of an antenna in a conductive medium,” IRE Trans. Antenn. Propag., vol. 10, no. 5, pp. 648–650, 1962. [11] C. Ancona, “On small antenna impedance in weakly dissipative media,” IEEE Trans. Antenn. Propag., vol. 26, no. 2, pp. 341–343, 1978. [12] Y. Liao, T. H. Hubing and D. Su, “Equivalent circuit for dipole antennas in a lossy medium,” IEEE Trans. Antenn. Propag., vol. 60, no. 8, pp. 3950–3953, 2012. [13] R. W. P. King, “Antennas in material media near boundaries with application to communication and geophysical exploration, part I: The bare metal dipole,” IEEE Trans. Antenn. Propag., vol. 34, no. 4, pp. 483– 489, Apr. 1986. [14] H. A. Wheeler, “Fundamental limitations of a small VLF antenna for submarines,” IRE Trans. Antenn. Propag., vol. 6, no. 1, pp. 123–125, Jan. 1958. [15] J. Kim and Y. Rahmat-Samii, “Implanted antennas inside a human body: Simulations, designs, and characterizations,” IEEE Trans. Mi- crowave Theory Tech., vol. 52, no. 8, pp. 1934–1943, Aug. 2004. [16] D. Nikolayev, M. Zhadobov, L. Le Coq, P. Karban and R. Sauleau, “Robust ultraminiature capsule antenna for ingestible and implantable applications,” IEEE Tran. Antenn. Propag., vol. 65, no. 11, pp. 6107– 6119, Nov. 2017. [17] A. K. Skrivervik, M. Bosiljevac and Z. Sipus, “Fundamental limits for implanted antennas: Maximum power density reaching free space,” IEEE Tran. Antenn. Propag., vol. 67, no. 8, pp. 4978–4988, Aug. 2019. [18] C. F. Bohren and D. R. Huffman, Absorption and Scattering of Light by Small Particles, New York: John Wiley & Sons, 1983. [19] S. A. Maier, Plasmonics: Fundamentals and Applications, Berlin: Springer-Verlag, 2007. [20] P. Bharadwaj, B. Deutsch and L. Novotny, “Optical antennas,” Adv. Opt. Photon., vol. 1, no. 3, pp. 438–483, 2009. [21] F. A. Miller and C. H. Wilkins, “Infrared spectra and characteristic frequencies of inorganic ions,” Anal. Chem., vol. 24, no. 8, pp. 1253– 1294, 1952. [22] D. B. Tanner, A. J. Sievers and R. A. Buhrman, “Far-infrared absorption in small metallic particles,” Phys. Rev. B, vol. 11, no. 4, pp. 1330–1341, 1975. [23] W. C. Mundy, J. A. Roux and A. M. Smith, “Mie scattering by spheres in an absorbing medium,” J. Opt. Soc. Am., vol. 64, no. 12, pp. 1593– 1597, 1974. [24] P. Chylek,´ “Light scattering by small particles in an absorbing medium,” J. Opt. Soc. Am., vol. 67, no. 4, pp. 561–563, 1977. [25] C. F. Bohren and D. P. Gilra, “Extinction by a spherical particle in an absorbing medium,” Journal of Colloid and Interface Science, vol. 72, no. 2, pp. 215–221, 1979. [26] I. W. Sudiarta and P. Chylek, “Mie-scattering formalism for spherical particles embedded in an absorbing medium,” J. Opt. Soc. Am. A, vol. 18, no. 6, pp. 1275–1278, 2001.