Electromagnetic Resonances of a Straight Wire

With an interest in finding wires and distinguishing them from other electrically conducting objects, we have looked for an electromagnetic “fingerprint” in terms of resonances of a straight wire of length 2h and radius a. The resonances of the wire are formulated using the theory of the linear antenna, leading to an integral equation for the current on the wire. Complex-valued resonant frequencies are defined as those for which the homogeneous integral equation for the current on the wire has non-zero solutions. By applying a variational technique we obtain approximate numerical solutions for the resonant frequencies and their widths. A table of the first five resonances is given for several ratios of wire half-length h to wire radius a. In a subsequent paper we propose to extend the method described here to deal with wires on an earth-air interface, for example as used to command the detonation of improvised explosive devices.

for the antisymmetric cases, which, so far as we know, have not previously been found.The important advantage of the variational technique is its relative insensitivity to errors in the current, allowing a rough approximation to the resonant currents to be used to obtain a close approximation to resonant frequencies and the widths of the resonances.

II. FORMULATION
We represent a straight wire subject to electromagnetic illumination by a thin, perfectly conducting cylinder of radius a and length 2h, embedded in an infinite uniform lossless medium with dielectric constant ≤ and magnetic permeability µ 0 .The axis of the wire coincides with the x-axis of a coordinate system and we consider an incident field at a single angular frequency ω, which induces a spatially varying current I(x) along the wire.This is defined by: along with the boundary condition that I(±h) = 0.In (1), E(x) is the x-component of the incident electric field, k = ω/c (with c the speed of light) is the propagation constant, and the kernel K is defined in Appendix A as a function of x and k by If there is no incident field, that is, if E x (x) = 0 on −h ≤ x ≤ h, it follows that (1) specializes to At first glance, one might expect the only solution to be I(x) = 0.However, the key to defining resonance is to note that the kernel K depends not only on position along the wire, but also on the frequency ω = ck.Therefore the solution to the integral equation (3) depends on ω and for certain discrete special values of ω, ω 1 = ck 1 , ω 2 = ck 2 , . . ., the equation has non-zero solutions.These values are expected to be complex, so that, listed in increasing order of their real parts, we have Re ω n = the n-th resonant frequency, −Im ω n = the half-width at half-maximum of the n-th resonance. (4) We assume that the resonant currents are proportional to 1/(ω − ω n ).The issue is how to find the ω n , or equivalently the k n , at which (3) has non-zero solutions.
III. APPROXIMATE SOLUTION FOR RESONANT FREQUENCIES Symbolically, we let A(k) denote the linear operator in (3), so that the equation is abbreviated as A(k) * I n (k) = 0, where * denotes an integral over the spatial variable, and we have omitted writing the spatial variables x and x 0 while we have made explicit the dependence of A and I n on the propagation constant k.For a thin wire, it is known that the resonant frequencies correspond to k n near nπ/2h.Our problem of resonance is to determine for n = 1, . . ., 5 the k n near nπ/2h such that A(k n ) * I n (k n ) = 0 has a solution for non-zero I n (k n ).What we require is only k n ; we do not seek an accurate solution to the current I n (k) ≡ I n (k, x).
To find k n , we start by considering the functional Suppose that for some value k n there is a non-zero solution I n (k n , x) to (3).As shown in Appendix B, the variation with respect to I (as a function of x) around this But since the first variation of the left-hand side is zero, replacing I n (k n ) by an approximation I ap n makes no first-order error in the expression I ap n * A(k n ) * I ap n .Thus we will determine k n as the solution, for a suitable approximating current I ap n , to For computational convenience we carry an x-derivative under the integral, note that dK(x − x 0 )/dx = −dK(x − x 0 )/dx 0 , and integrate by parts to obtain for the equation for The dependence on k n , the sought value of k, is now limited to the dependence of the kernel K(x − x 0 ) on k as expressed in (2).Now we choose an approximating current.As noted above, for sufficiently small a one expects resonant frequencies at values of k near for n = 1, 2, . . ., with the currents corresponding to odd values of n symmetric in x while the currents corresponding to even values of n are antisymmetric in x [2].
For the symmetric case, there are good reasons to believe that the resonance current is given roughly by the shifted-cosine form cos kx − cos kh [1].One might expect to get a good approximation to the complex resonant frequency by choosing the current to be the shifted cosine; however, use of the shifted cosine can result in possibly spurious values of the resonant frequency.Examination of how these values arise shows that they are indeed spurious artifacts of the shifted-cosine form; hence we need to attend more carefully to the choice of the approximating current.The shifted-cosine form leads to an approximating current dependent on the frequency ω ∼ k.Question: should the approximating current depend on the frequency ω or not?Having encountered spurious values from the shifted-cosine form, we choose instead an approximating current that is independent of the frequency ω.This independence essentially determines the approximation.To begin with, it can depend only on the geometric parameters h and a.Although we could force the approximation to depend on a, we cannot see how to do this in a physically sensible way.Thus we take the approximation to the resonant current to depend only on h.Then we can hardly avoid choosing I ap n ∼ cos κ n x = cos(nπx/2h) for the symmetric case where n is odd.Correspondingly, for the cases of resonance in which the current is antisymmetric about the center point of the wire, for which n is even, our approximation to the resonant current is I ap n ∼ sin κ n x.We expect these approximate currents to be adequate for use in (7) for the first five resonant frequencies, but not for much higher resonances.

A. Resonances
With the chosen approximating currents, (7) for determining k n becomes Each of this pair of expressions involves the same two integrals In terms of these integrals, ( 9) and (10) can be put in a form convenient for calculating k n : We now make a zero-th order check.(Given any positive numbers ≤ and η and a wire of half-length h, there is some a 0 , depending on ≤ and η, such that for a < a 0 the kernel in the integral in (3) acts essentially as a delta-function.More precisely the integral of the absolute value of the kernel over the integration range 2h > |x − x 0 | > ≤ can be made less than η times the integral of the absolute value of the kernel over the small integration range |x − x 0 | ≤ ≤.) Upon replacing K(x − x 0 ) by a delta function δ(x − x 0 ), carrying out the integrals in (9), and using the definition of κ n = nπ/2h, the result for n odd is Carrying out the same procedure on (10) yields this relation for n even-confirming our claim that for sufficiently small radius, the resonant propagation constants should be near κ n .
Returning to the use of K(x − x 0 ) and not just the delta function, we want to hold fixed the geometrical and material parameters h, a, ≤, and µ 0 and vary only the propagation constant k = ω/c, which amounts to varying the frequency ω in order to find solutions to (13) for the symmetric resonances and (14) for the antisymmetric resonances.The first step, derived in Appendix C, is to reduce the double integrals to single integrals to obtain: When the wire radius a is much smaller than all the other dimensions in this scattering problem, we can approximate the kernel K defined in (2).For x ¿ a, the kernel is very close to and furthermore the integral over the logarithmic singularity in K is closely matched by the integral over the approximation defined in (17).Note that k is complex.With this approximation one obtains from ( 16) and the definition of κ n in (8) where the last approximation makes negligible error when a is much smaller than the other dimensions.
It will turn out that the resonant frequencies correspond to Im k < 0, so that the integrand in ( 18) is an increasing function of y.As evaluated in Appendix D, we obtain Putting all this together, we need to solve numerically for complex k the equation Numerical analysis then yields the examples shown in Table I.

IV. DISCUSSION
Naively, one pictures a resonance as a large response to a small incident field.Thought of this way, the calculation of resonance seems to demand choosing one or more incident fields, and the choice of these fields defies any simple physical basis.For example, one might consider plane waves or one might consider an incident field generated by one or another transmitting antenna located at some distance and orientation from the scattering wire.Except for the case of an incident field transverse to the wire, for which very little scattering occurs, one expects the resonant frequencies to be largely insensitive to the incident field.Here we have taken advantage of the near-independence of resonances from the choice of incident field to define resonances in terms of non-zero solutions to the homogeneous Pocklington equation.
The other noteworthy feature is to make use of the variational technique to show that the complex resonant frequencies are insensitive to small errors in the resonant current, which justifies replacing the resonant current, which is exceedingly difficult to determine to high accuracy, by an approximate current, as discussed above.

APPENDIX A KERNEL
The kernel K in ( 1) is defined as a function of x and k by the Green's function as follows.Consider cylindrical coordinates (ρ, φ, x) oriented around the x-axis, so that the distance to a point from the x-axis is given by ρ = √ y 2 + z 2 ; correspondingly we have the cartesian components y = ρ cos φ, z = ρ sin φ.The retarded Green's function defines the vector potential on the wire surface at a point (a, φ, x) arising from a current density J(a, φ 0 , x 0 ) on the wire by We now average over φ to obtain where we define The kernel K(x) enters both the Hallén integral equation [1] and the Pocklington equation.For a wire with ka ø 1, we assume that only the x-components of the current density and of the vector potential are relevant, and (1) follows.

APPENDIX B VARIATIONAL FORM OF INTEGRAL EQUATION
We study the variational of S[I] defined in (5) with respect to I(k, x), with k fixed.Computing the variation one finds Denote the second term by T 2 and invoke the boundary condition I(k, x) → 0 as x → ±h to obtain, dropping the explicit mention of k, Thus this term has been put in the form of a part of the first line of (B1).Similarly but more simply, the third term of (B1) is equal to the corresponding part of the first line, so that one has altogether so that (3) implies that δS[I] = 0.

APPENDIX C INTEGRATIONS FOR I c AND I s
Because K(x − x 0 ) depends only on the difference between x and x 0 , it must be possible to rewrite the integrals (11) and (12) as single integrals.We do this without using the fact that K is an even function of its argument.Define Changing integration variables x → −x 0 and x 0 → −x produces the relation With this relation one expresses I s and I c as (C6) For the reduction to single integrals we compute The case α 0 = −α is worked out directly to show Substitution of (C7) and (C8) into (C5) and (C6) yields ) In the special case, which we have here, in which K(y) = K(−y) these simplify slightly to which are the single integrals that we wanted to obtain.

APPENDIX D EVALUATION OF I s + I c
Now we evaluate the sum I s + I c , starting by writing where Of note is the appearance (in the denominator of the second term) of the difference k − κ n , which is small (and complex).We are interested in the case that |k − κ n | ø κ n but 2h|k − κ n | may or may not be small.Very roughly, we expect |k − κ n |/κ n to be of the order of 1/2 ln(2h/a), which is perhaps 0.05.For the fifth resonance, κ 5 = 5π/(2h), leading to 2h|k − κ n | = (0.05)5π ∼ 0.8.

B. Evaluation of the integral I 1
We want to evaluate

TABLE I COMPLEX
VALUES OF kh AT RESONANCE n (WHERE κn = nπ/2h)