# Exact solution to Maxwell’s equations for the infinite ideal solenoid with a
time-dependent surface current

Edward Parker<EMAIL_ADDRESS>
###### Abstract

Very little previous literature has considered the _exact_ solution to
Maxwell’s equations for an infinite ideal cylindrical solenoid with an
arbitrary time-dependent azimuthal surface current $K(t)\hat{\bm{\phi}}$. Most
of the previous literature has focused on special cases and has approached the
problem by calculating the magnetic vector potential $\bm{A}$, which requires
performing some very complicated surface integrals over the cylinder. In this
article, we take a simpler approach and directly tackle Maxwell’s equations
without ever invoking a vector potential. The high symmetry of the geometry
allows us to reduce Maxwell’s equations to just two coupled partial
differential equations for two functions of two real variables, which can be
readily solved numerically. We find the general analytic solution to these
PDEs and derive the Green’s functions for the electromagnetic fields, which
allow us to calculate the fields directly from the surface current $K(t)$. We
also briefly discuss a family of exact formal solutions that (the author
believes) has not appeared in the previous literature because it corresponds
to a current $K(t)$ that does not have a Fourier transform.

##  I Introduction

Two of the workhorse examples in the study of classical electromagnetism (EM)
are the parallel-plate capacitor and the cylindrical solenoid of azimuthal
surface current. In the static context, these produce the simplest possible
nontrivial electromagnetic fields far away from their edges: a uniform
electric or magnetic field, respectively. Their highly symmetric geometries
allow many quantities to be calculated exactly. In the dynamic context, the
solenoid is arguably the “nicer” of these two examples, because we can
consider a reasonably realistic idealization in which the surface current
changes over time. By contrast, the continuity equation prevents us from
moving charge between the plates of a parallel-plate capacitor without
introducing a current that typically reduces the geometric symmetry.

Moreover, the solenoid with a slowly time-varying surface current is an
excellent illustration of Faraday’s law of induction that is discussed in most
EM textbooks [1, 2]. The high symmetry of the geometry allows us to calculate
the EM fields in the quasistatic approximation. But textbook discussion do not
always make completely clear that this solution to Maxwell’s equations is only
approximate [3].

The _exact_ solution to Maxwell’s equations for a solenoid with an arbitrary
time-dependent azimuthal surface current $K(t)\hat{\bm{\phi}}$ is quite
complicated, but it can be solved exactly. The high symmetry of the geometry
provides a nice example of how symmetries can greatly reduce the number of
degrees of freedom in a problem. Surprisingly, there has been relatively
little literature exploring this problem, and to the author’s knowledge, no
previous literature has directly solved Maxwell’s equations in full generality
without using the vector potential (which requires performing some very
complicated surface integrals over the solenoid). This article does so.

The article proceeds as follows: In Section II, we set up the mathematical
problem of the infinite ideal thin cylindrical solenoid with a time-dependent
azimuthal surface current. In Section III, we review the prior literature for
this problem. In Section IV, we use the symmetry of the problem to reduce
Maxwell’s equations (ordinary four partial differential equations for six
unknown functions of four real variables) to just two coupled PDEs for two
unknown functions of two real variables. The author was unable to find these
simple PDEs in any of the published literature. In Section V, we consider the
limiting case of slowly-varying (i.e. constant or quasistatic) currents
$K(t)$. In Section VI, we calculate the general solution for arbitrary
currents. In Subsection VI.2, we discuss a class of solutions that (to the
author’s knowledge) has never appeared in the previous literature, because it
corresponds to physically questionable (but mathematically well-defined)
currents $K(t)$ that do not have a Fourier transform. In Subsection VI.3, we
restrict ourselves to currents that do have a Fourier transform and work out
the Green’s functions for the electric and magnetic fields (_not_ the more
common Green’s function for the vector _potential_) in both the frequency and
the time domains, which (to the author’s knowledge) have not been published in
the literature before.

Most of the material in this article should be accessible to someone who has
taken a graduate-level course in classical electromagnetism. Some of the
material in Subsection VI.3 uses some somewhat advanced concepts in harmonic
analysis; in Appendix B we give a self-contained and pedagogical explanation
of the necessarily mathematical concepts. Appendix C contains asymptotic forms
of certain Bessel functions that are used in the main text. The main text can
be read independently of the appendices if the reader is willing to skip
certain proofs.

##  II Setup and assumptions

Clearly, cylindrical coordinates $(r,\phi,z)$ are the natural coordinate
system to use for this cylindrical geometry.

In this article, we define an “infinite ideal solenoid” be be a current
distribution with the following properties:

  1. 1.

It is an infinitely long, infinitely thin cylindrical surface current with
radius $R$ and surface current density $\bm{K}$.

  2. 2.

The current flow is purely in the azimuthal (i.e. transverse) direction:
$\bm{K}\propto\hat{\bm{\phi}}$.

  3. 3.

the magnitude of the current at any given time is independent of the position
on the cylinder: the magnitude $K$ is a function only of $t$ and not of $z$ or
$\phi$.

  4. 4.

There is no net electric (or magnetic!) charge anywhere: $\rho\equiv 0$.

Together, these assumptions imply that $\bm{K}(\phi,z,t)\equiv
K(t)\,\hat{\bm{\phi}}$. The volumetric current density
$\bm{J}(r,\phi,z,t)=K(t)\,\delta(r-R)\,\hat{\bm{\phi}}$, where $\delta$
represents the Dirac delta function.

Each of these assumptions can be translated into an idealized approximation
for a physical solenoid, e.g. one made of a tightly coiled current-carrying
wire. For example, the “inifinitely thin” part of assumption #1 holds if the
wire is very thin compared to $R$ and is coiled so tightly that we can neglect
the spacing between coils. The “infinitely long” part holds if (a) the
physical solenoid is much longer than its radius $R$, and (b) we restrict our
attention to the region near the center of the cylinder, where we can neglect
the fringing effects at the ends of the solenoid.111Note that once we start
talking about the emitted electromagnetic fields, a new length scale will come
into play: the wavelength of the electromagnetic radiation, or (closely
related) the time scale over which the surface current $K(t)$ changes. The
solutions that we derive below will only hold in the regime where this new
length scale is far from the length scales that we are implicitly dropping in
making these approximation. For example, our results below will stop being
valid for wavelengths nearly as short as the diameter of the wires in the
physical solenoid, or nearly as long as the length of the physical solenoid.
Assumption #2 holds if the coils are angled nearly perpendicular to the axis
of the cylinder (or if they double-wrap the cylinder with oppositely slanted
layers), so that there is a negligible longitudinal current flow in the
$z$-direction.

Of course, each of these simplifying assumptions can be – and has been –
eliminated to get a more realistic model for a physical solenoid. Many, many
articles consider the magnetic field for a finite-length solenoid and a steady
(i.e. time-independent) current; Ref. 4 reviews the literature and gives an
exact analytic solution for thin solenoids, and Ref. 5 compares several
numerical models for both thin and thick solenoids. As Refs. 3 and 6 discuss,
assumptions #2 and #3 are very difficult to satisfy in practice: the natural
way to create the current in a real solenoid is by driving one or both ends
with a (potentially time-varying) voltage difference, but the signals from
these driven ends can only propagate down the solenoid at the speed of light,
so the instantaneous current will be different at different points along the
solenoid. Ref. 6 addresses this issue by considering a much more complicated
model that captures the finite propagation speed of signals down the length of
the solenoid and the dependence of $K$ on the $z$-coordinate. In this article,
we will neglect this practical consideration for simplicity, noting that
assumptions #2 and #3 above are perfectly consistent with special relativity,
even though they are practically difficult to satisfy experimentally.

##  III Prior literature

The previous literature has extensively considered the infinite ideal solenoid
with a time-varying current (and many variations). The first exact solution
that the author has found was published in Ref. 7, which considers a
sinusoidal time-varying current $K(t)\propto\sin(\omega t)$. The authors
perform some truly heroic feats of double integration over the cylinder to
calculate the exact magnetic vector potential $\bm{A}$ (including all
retardation effects) outside the solenoid (only), and then differentiate
$\bm{A}$ to obtain exact analytic expressions for the outside electromagnetic
fields in terms of Bessel functions of the first and second kind.222Ref. 7
also considers azimuthal current densities with time dependence $K(t)$
proportional to the Heaviside step function $K(t)\propto\Theta(t)$,
$K(t)\propto t\,\Theta(t)$, and $K(t)\propto\delta(t)$, but only in the far-
field limit $r\gg R$. It also considers the case of longitudinal rather than
azimuthal current $\bm{K}\parallel\hat{z}$. (Ref. 7 also proves a theorem that
is not particularly relevant to this article, but which is so remarkable that
we state it in Appendix A without proof.)

Ref. 3 finds the same results by directly solving Maxwell’s equations for a
sinusoidal current (only). Refs. 3, 8, 9 contain an interesting back-and-forth
conceptual discussion about whether the electric field _outside_ the solenoid
is best thought of as being induced by the changing magnetic field that is
located _inside_ or _outside_ of the solenoid.

We can use the Fourier transform to decompose an arbitrary current $K(t)$ into
a superposition of its Fourier modes $\tilde{K}(\omega)$. Since we know the
exact electromagnetic fields produced by a sinusoidal current with frequency
$\omega$, we can superpose the contributions from each Fourier mode
$\tilde{K}(\omega)$ to find formal expressions for the fields generated by an
arbitrary current $K(t)$; these expressions for the fields are rather
complicated integral transforms of $\tilde{K}(\omega)$. Ref. 10 expresses the
fields both inside and outside the solenoid in terms of Fourier transforms.

Remarkably, the author was unable to find any published literature that takes
the simplest approach of directly solving Maxwell’s equations for a general
current $K(t)$. (Ref. 3 comes close, but only considers the sinusoidal case.)
In this paper, we do so. Unlike in most of the literature discussed above, we
do not use the vector potential $\bm{A}$; nor do we ever need to perform any
geometric integrals over the solenoid surface. One of our main results
simplifies the problem down to a simple pair of coupled PDEs for two functions
of two variables (equations (6)) that can be readily solved numerically.
Another main result solves those PDEs analytically and directly expresses each
electromagnetic field as a single convolution of the current $K(t)$ with an
appropriate Green’s function $g(r,t)$ (equations (17) and (19)). Both results
are exact and do not make any approximations. The intermediate steps used to
derive them use various tools from harmonic analysis, but the final results do
not require any Fourier transforms or even complex numbers – only a single
convolution integral for each electromagnetic field.

##  IV Maxwell’s equations for the ideal solenoid

Both inside and outside the solenoid, the electromagnetic fields satisfy the
vacuum Maxwell’s equations

$\displaystyle\bm{\nabla}\cdot\bm{E}$ $\displaystyle=0,$ (1)
$\displaystyle\bm{\nabla}\cdot\bm{B}$ $\displaystyle=0,$
$\displaystyle\bm{\nabla}\times\bm{E}$ $\displaystyle=-\frac{\partial
B}{\partial t},$ $\displaystyle\bm{\nabla}\times\bm{B}$
$\displaystyle=\frac{\partial E}{\partial t},$

where we work in Heaviside-Lorentz units where $\epsilon_{0}=\mu_{0}=c=1$. At
the solenoid, we have the interface boundary conditions that $\bm{E}$ is
continuous,

$\bm{E}_{\text{in}}(r=R)=\bm{E}_{\text{out}}(r=R),$ (2)

and

$\displaystyle B_{\text{in}}^{r}(r=R)$
$\displaystyle=B_{\text{out}}^{r}(r=R),$ (3)
$\displaystyle\bm{B}_{\text{in}}^{\parallel}(r=R)-\bm{B}_{\text{out}}^{\parallel}(r=R)$
$\displaystyle=-\bm{K}\times\hat{\bm{r}}=K\hat{\bm{z}},$

where $\bm{B}^{\parallel}$ refers to the component of $B$ that lies in the
$\phi$-$z$ plane tangent to the solenoid.

The boundary conditions at spatial infinity (or more precisely, far from the
solenoid as $r\to\infty$) are somewhat more complicated. If $K(t)$ vanishes
identically before some starting time $t_{0}$, then the $\bm{E}$ and $\bm{B}$
fields vanish identically outside of the future light cone $|r-R|=t-t_{0}$.
But if the support of $K(t)$ extends infinitely far back in the past, then we
need to instead impose the condition that the electromagnetic waves propagate
purely outward rather than inward in order to respect causality. There are
several different ways to formalize this physical requirement mathematically,
but the most common one is the _Sommerfeld radiation condition_ [11]

$\lim_{r\to\infty}r\mathopen{}\mathclose{{}\left(\frac{\partial}{\partial
r}-ik}\right)\tilde{\bm{A}}(\bm{x},\omega)=0,$ (4)

where $\tilde{\bm{A}}(\bm{x},\omega)$ is the Fourier transform of the vector
potential $\bm{A}(\bm{x},t)$ with respect to time.

A fully general electromagnetic field on flat Minkowski spacetime is
represented by a two-form field $F:\mathbb{R}^{4}\to\mathbb{R}^{6}$, and
Maxwell’s equations represent eight independent equations.333In $D$ spacetime
dimensions,
$F:\mathbb{R}^{D}\to\mathbb{R}^{\binom{D}{2}}=\mathbb{R}^{\frac{1}{2}D(D-1)}$,
and Maxwell’s equations represent $D(D^{2}-3D+8)/6$ independent equations. But
in this case, we can use the symmetry of the source distribution to enormously
simplify Maxwell’s equations:

  1. 1.

The time-varying current density $K(t)$ explicitly breaks time-translation and
time-reversal symmetry, so the relevant symmetries are the cylindrical
symmetries of the solenoid. The coordinates $\phi$ and $z$ are cyclic, so the
electromagnetic fields must be independent of $\phi$ and $z$ when expressed in
cylindrical coordinates, which reduces the domain from four dimensions to just
two: $r$ and $t$.

  2. 2.

Since the source current $\bm{K}$ is purely azimuthal and $\rho=0$, the source
four-vector $J$ is symmetric under the reflection $z\to-z$ about the $x$-$y$
plane. The $\bm{E}$ field is a true vector, so each coordinate component
changes sign under an inversion of that coordinate, and so $E_{z}$ is odd
under the reflection $z\to-z$: $E_{z}(z)=-E_{z}(-z)$ for all $z$. But using
the translational invariance in $z$, $E_{z}(-z)=E_{z}(z)$, so
$E_{z}(z)=-E_{z}(z)$, and so $E_{z}\equiv 0$ by reflectional and translational
symmetry in $z$.

  3. 3.

Similarly, the $\bm{B}$ field is a pseudovector: under the reflection
$z\to-z$, the component $B_{z}$ is unchanged but the other components
$\bm{B}_{\parallel}$ parallel to the $x$-$y$ plane change sign, and so
$\bm{B}_{\parallel}(z)=-\bm{B}_{\parallel}(-z)$. But then by translational
invariance in $z$, we have $\bm{B}_{\parallel}(-z)=\bm{B}_{\parallel}(z)$ and
so $\bm{B}_{\parallel}(z)=-\bm{B}_{\parallel}(z)$, so
$\bm{B}_{\parallel}(z)\equiv\bm{0}$ and $\bm{B}\parallel\hat{\bm{z}}$.

Solely from cyclindrical and reflection symmetry, we have reduced the initial
six components of the electromagnetic field down to just three ($E_{r}$,
$E_{\phi}$, and $B_{z}$) and simplified the electromagnetic field to a much
simpler map $\mathbb{R}^{2}\to\mathbb{R}^{3}$. Since $\hat{\bm{r}}$ and
$\hat{\bm{\phi}}$ are singular at the $z$-axis, the requirement that $\bm{E}$
be continuinuous there requires that $E_{r}$ and $E_{\phi}$ both approach $0$
as $r\to 0^{+}$. We will now drop the subscript $z$ and note that for the rest
of this article, $B$ will denote the signed quantity $B_{z}$, not the
nonnegative magnitude $|\bm{B}|$.

Using all these symmetries, Maxwell’s equations (1) reduce to

$\displaystyle\frac{1}{r}\frac{\partial}{\partial r}(rE_{r})$
$\displaystyle=0$ (5) $\displaystyle\frac{1}{r}\frac{\partial}{\partial
r}(rE_{\phi})$ $\displaystyle=-\frac{\partial B}{\partial t}$
$\displaystyle-\frac{\partial B}{\partial r}$ $\displaystyle=\frac{\partial
E_{\phi}}{\partial t},$

with Gauss’s law for magnetism satisfied automatically.

Gauss’s law for electricity (5) is easy to solve: it implies that $rE_{r}$ is
constant both inside and outside the solenoid, and so $E_{r}\propto 1/r$. The
requirement of continuity at the $z$-axis requires that $E_{r}$ vanish inside
the solenoid. If there were a net surface charge on the solenoid, then the
proportionality constant would be different inside and outside the solenoid –
but since we are assuming no net charge on the solenoid, the boundary
condition (2) implies that the proportionality constant is the same in both
regions. So $E_{r}\equiv 0$ everywhere, and the $\bm{E}$ field must be purely
azimuthal: $\bm{E}\parallel\hat{\bm{\phi}}$. Therefore, we have reduced the
electromagnetic field to an even simpler map
$\mathbb{R}^{2}\to\mathbb{R}^{2}$. As before, we will drop the subscript on
$E_{\phi}$ and let $E$ denote the signed quantity $E_{\phi}$, not the
magnitude $|\bm{E}|$.

We are finally left with the quite simple coupled partial differential
equations

$\displaystyle\frac{1}{r}\frac{\partial}{\partial
r}(rE)=\frac{E}{r}+\frac{\partial E}{\partial r}$
$\displaystyle=-\frac{\partial B}{\partial t}$ (6a)
$\displaystyle-\frac{\partial B}{\partial r}$ $\displaystyle=\frac{\partial
E}{\partial t}.$ (6b)

Equation (6a) is Faraday’s law, and equation (6b) is Ampère’s law. The real-
valued fields $E(r,t)$ and $B(r,t)$ are defined on the half-plane $(r\geq
0,t\in\mathbb{R})$. The interface boundary conditions (2) and (3) and the
requirement of continuity reduce to the requirements that

  1. 1.

$E$ be continuous on the whole half-plane $r\geq 0$,

  2. 2.

$E\to 0$ as $r\to 0^{+}$,

  3. 3.

$B$ have a jump discontinuity

$B(r\to R^{-},t)-B(r\to R^{+},t)=K(t)$

along the coordinate line $r=R$, and

  4. 4.

$E$ and $B$ respect the boundary conditions discussed below equations (3) as
$r\to\infty$.

The equations (6) are the first main result of this article. To the author’s
knowledge, they have not been explicitly published in any previous textbooks
or scientific journals (although they can be derived from the results in Ref.
3). This system of equations, which is exactly equivalent to Maxwell’s
equations for the ideal solenoid, is probably the best formulation of the
problem to use for purely numerical methods. As a pair of coupled linear
homogeneous first-order PDEs in only two variables, they are relatively
straightforward to solve numerically. Note that nothing up to this point has
required any complications such as gauge potential fields, complex numbers, or
even any integrals.

##  V Slowly-varying currents

Let us check the familiar textbook situations in which the currents are static
or quasistatic.

###  V.1 Constant current

Suppose that $K(t)\equiv K$ is constant. In this case, we can solve equations
(6) by inspection. Since the boundary conditions are time-independent, it is
natural to guess that $E$ and $B$ are constant inside and outside the
solenoid, so that Ampère’s law (6b) is trivially satisfied. Faraday’s law (6a)
implies that $E\propto 1/r$, and then boundary condition #2 requires that
$E\equiv 0$. Boundary condition #4 requires that $B_{\text{out}}\equiv 0$
outside the solenoid, and then boundary condition #3 requires that
$B_{\text{in}}\equiv K$ inside. This is the familiar exact magnetostatic
solution for an ideal solenoid with constant current.

###  V.2 Quasistatic current

When first introducing Faraday’s law, EM courses often discuss the example of
a quasistatic solenoid – although they do not always make fully clear that
this is an approximate rather than an exact solution to Maxwell’s equations
[3]. Textbooks often explain that this quasistatic solution holds when the
current is “slowly varying”, although they do not always explain exactly what
this means (i.e., slow compared to what?).

In this approximation, the magnetic field is assumed to be spatially uniform
inside the solenoid with a magnitude that instantaneously matches the solenoid
current ($B_{\text{in}}(r,t)\equiv K(t)$) and to vanish outside
($B_{\text{out}}\equiv 0$). The electric field is assumed to be
$E_{\text{in}}=-\frac{1}{2}\frac{dK}{dt}r$ inside the solenoid and
$E_{\text{out}}=-\frac{1}{2}\frac{dK}{dt}\frac{R^{2}}{r}$ outside, in
accordance with the integral form of Faraday’s law. Faraday’s law (6a) is
exactly satisfied, as are the interface boundary conditions #1-3,444There are
some conceptual difficulties around the meaning of the Sommerfeld radiation
condition in this case, since Maxwell’s equations are not satisfied. but
Ampère’s law (6b) is not: the LHS $-\frac{\partial B}{\partial r}$ vanishes,
but the RHS

$\frac{\partial E}{\partial t}=\frac{1}{2}\frac{d^{2}K}{dt^{2}}r\neq 0$

inside the solenoid and

$\frac{\partial E}{\partial
t}=\frac{1}{2}\frac{d^{2}K}{dt^{2}}\frac{R^{2}}{r}\neq 0$

outside. In this case the quasistatic approximation is “non-perturbative”,
since we are not dropping a subleading correction term but instead the _only_
nonzero term. Noting that $\frac{d^{2}K}{dt^{2}}$ is the key parameter that
set the scale in Ampère’s law (6b), we must resort to dimensional analysis and
conclude that the quasistatic approximation holds if the dimensionless ratio

$\frac{R^{2}}{K(t)}\frac{d^{2}K}{dt^{2}}\ll 1$

at all times.

As noted in Refs. 7, 8, if $K(t)$ depends linearly on time, then
$\frac{d^{2}K}{dt^{2}}\equiv 0$, the quasistatic approximation becomes exact,
and the magnetic field vanishes identically outside of the solenoid. It may
seem somewhat counterintuitive that in this case, Maxwell’s equations are
_exactly_ solved by using the Biot-Savart law with the time-varying current
source evaluated at the _present_ time rather than the retarded time, but
nevertheless it is true [12]. The situation is somewhat reminiscent of the
fact that the electric field created by a relativistic electric point charge
moving with constant velocity points exactly at the charge’s _present_
position, rather than at its retarded position as we might expect. In any
case, a surface current that depents linearly on $t$ for all time is somewhat
artificial, since it becomes unboundedly large in the far past and far future.

##  VI The general solution

The partial differential equations (6) are in the most useful form for
numerical solutions, because coupled first-order differential equations are
usually more convenient to tackle numerically than uncoupled second-order
differential equations. Nevertheless, we can decouple the equations (6) by
taking the partial derivative of (6a) with respect to $r$, taking the partial
derivative of (6b) with respect to $t$, and equating them to get the decoupled
PDE for $E$

$\frac{\partial}{\partial
r}\mathopen{}\mathclose{{}\left(\frac{1}{r}\frac{\partial}{\partial
r}(rE)}\right)=\frac{\partial^{2}E}{\partial t^{2}}.$ (7)

Similarly, we can take the partial derivative of (6a) with respect to $t$,
apply the differential operator

$\frac{1}{r}\frac{\partial}{\partial
r}(r\underline{\phantom{x}})=\frac{1}{r}+\frac{\partial}{\partial r}$

to (6b), and equate them to get the decoupled PDE for $B$

$\frac{1}{r}\frac{\partial}{\partial
r}\mathopen{}\mathclose{{}\left(r\frac{\partial B}{\partial
r}}\right)=\frac{\partial^{2}B}{\partial t^{2}}.$ (8)

Equations (7) and (8) are somewhat similar – but not identical – to the usual
1D wave equation.

Before we dive into the messy details of solving these PDEs, we will outline
our plan of attack. The typical textbook approach for dealing with an
uncoupled linear PDE is to first make a separation-of-variables ansatz that
breaks the PDE into an equality between decoupled ODEs of different dependent
variables, which must equal a common unknown constant. The boundary conditions
typically restrict the values that the constant can take on, and the full
solution is a linear combination over the allowed constants, with coefficients
determined by the boundary conditions.

Plugging in the ansatzes $E(r,t)=\mathcal{R}_{E}(r)T_{E}(t),\
B(r,t)=\mathcal{R}_{B}(r)T_{B}(t)$ and separating variables gives the
eigenvalue equations

$\displaystyle\frac{1}{\mathcal{R}_{E}}\frac{d}{dr}\mathopen{}\mathclose{{}\left(\frac{1}{r}\frac{d}{dr}(r\mathcal{R}_{E})}\right)$
$\displaystyle=\frac{1}{T_{E}}\frac{d^{2}T_{E}}{dt^{2}}=\lambda,$ (9)
$\displaystyle\frac{1}{r\mathcal{R}_{B}}\frac{d}{dr}\mathopen{}\mathclose{{}\left(r\frac{d\mathcal{R}_{B}}{dr}}\right)$
$\displaystyle=\frac{1}{T_{B}}\frac{d^{2}T_{B}}{dt^{2}}=\lambda.$

$\mathcal{R}_{E}(r)$ must be continuous everywhere by boundary condition #1,
but it can have a kink (i.e. fail to be differentiable) at $r=R$.555The letter
$\lambda$ denotes an eigenvalue, _not_ a wavelength. The eigenfunctions
$\mathcal{R}$ and $T$ implicitly depend on the eigenvalue $\lambda$, but we
will omit this dependence in order to simplify the notation. A priori, the
eigenvalue $\lambda$ could take on different values $\lambda_{E}$ and
$\lambda_{B}$ across the two equations (9). But as we will see below, the
coupled equations (6) require that
$\lambda_{E,\text{in}}=\lambda_{B,\text{in}}$ and
$\lambda_{E,\text{out}}=\lambda_{B,\text{out}}$ in order for the time
dependence to have the same frequency. And the requirement that $E$ be
continuous at the solenoid $r=R$ for all $t$ requires that
$\lambda_{E,\text{in}}=\lambda_{E,\text{out}}$, so
$\lambda_{E}=\lambda_{B}\equiv\lambda$ everywhere.

In general, the eigenvalues $\lambda$ can range over a continuous set of real
values. Therefore, we will need to solve equations (9) for each possible
eigenvalue $\lambda$ – which is already quite nontrivial – and then superpose
them not through a discrete linear combination, but through an integral over
the allowed eigenvalues $\lambda$.

###  VI.1 Case 1: $\lambda=0$.

This case is the simplest. It turns out to yield exactly the quasistatic
solution

$\displaystyle K(t)=K_{0}+\dot{K}t,$ (10) $\displaystyle
B_{\text{in}}(r,t)\equiv K(t),$ $\displaystyle B_{\text{out}}(r,t)\equiv 0,$
$\displaystyle E_{\text{in}}(r,t)=-\frac{1}{2}\dot{K}r,$ $\displaystyle
E_{\text{out}}(r,t)=-\frac{1}{2}\dot{K}\frac{R^{2}}{r}$

discussed in the previous section, where $\dot{K}$ is a _constant_. This is
the contribution to the electromagnetic fields from any _linear_ component of
$K(t)$ – although as we mentioned above, it is not clear how physically
realistic is a surface current that gets unboundedly large in the far past and
far future.

###  VI.2 Case 2: $\lambda>0$.

This case is even more physically questionable. It leads to mathematically
valid solutions to Maxwell’s equations (whose radial functions
$\mathcal{R}_{E}(r)$ and $\mathcal{R}_{B}(r)$ are modified Bessel functions)
that technically satisfy all four boundary conditions listed above. However,
these solutions correspond to surface currents $K(t)$ that grow exponentially
with time in the far past and/or far future. As such, the function $K(t)$ is
not a tempered distribution and does not have a well-defined Fourier transform
– not even in the distributional sense – and so much of our analytical
machinery breaks down. The author is not aware of any previous literature that
mentions the existence of these formal solutions to Maxwell’s equations for
the ideal solenoid, since most previous literature has implicitly assumed that
$K(t)$ is a tempered distribution with a well-defined Fourier transform. We
will not work out these physically unrealistic solutions to Maxwell’s
equations in detail, but it is interesting to note their existence.

Moreover, it is interesting to consider the example of a surface current
$K(t)$ that starts from $0$ in the infinite past and then grows exponentially
(with some growth constant $\tau$) up until some time $t_{\text{change}}$,
when it then changes to a different time dependence that stays bounded over
future times. This current profile is reasonably physically realistic. Of
course, we could not experimentally arrange for it told hold exactly out to
the infinite past – but it is negligibly small in the far past, so we could
closely approximate it by turning on a very small current and then letting the
current grow exponentially for several $e$-folding times $\tau$ before we
taper off its growth.

In this example, the spacetime region $|r-R|>t-t_{\text{change}}$ lies outside
the future light cone of the “new” current. Therefore, the past light cones of
every point in this spacetime region only intersect with the exponentially
growing current. Therefore, this region has no way to “know” that the current
will eventually change from exponential growth, and so the electromagnetic
fields in this region are the same as those produced by a hypothetical (and
physically unrealistic) current that grows exponentially for _all_ time. In
this spacetime region (only), the radial functions $\mathcal{R}_{E}$ and
$\mathcal{R}_{B}$ are given _exactly_ by _modified_ Bessel functions that take
the dimensionless argument $r/\tau$.

###  VI.3 Case 3: $\lambda<0$.

This is the most physically relevant case. It will be convenient to change
variables to $\lambda=-\omega^{2}$, where
$\omega\in\mathbb{R}\setminus\\{0\\}$. We see that $T(t)$ is sinusoidal with
frequency $|\omega|$. We can introduce the nondimensionalized variable666Since
we are working in vacuum and in units where $c=1$, the wave number $k$ equals
the frequency $\omega$. Note that this substitution fails if $\omega=0$, so
taking the limit $\omega\to 0$ of the expressions derived in this subsection
does _not_ yield the quasistatic solution (10), but only the static solution
discussed in subsection V.1. $x:=\omega r$ and rearrange the radial equations
to

$\displaystyle\frac{d^{2}\mathcal{R}_{E}}{dx^{2}}+\frac{1}{x}\frac{d\mathcal{R}_{E}}{dx}+\mathopen{}\mathclose{{}\left(1-\frac{1}{x^{2}}}\right)\mathcal{R}_{E}$
$\displaystyle=0,$
$\displaystyle\frac{d^{2}\mathcal{R}_{B}}{dx^{2}}+\frac{1}{x}\frac{d\mathcal{R}_{B}}{dx}+\mathcal{R}_{B}$
$\displaystyle=0$

which are Bessel’s equations of order 1 and 0 respectively.

There is no single Bessel function that satisfies both the boundary conditions
#2 and #4, so $\mathcal{R}_{E}(x)$ must have a kink (or higher-derivative
discontinuity) at $x=\omega R$, and $\mathcal{R}_{B}(x)$ must have a jump
discontinuity there by boundary condition #3. Boundary condition #2 requires
that $\lim\limits_{x\to 0^{+}}\mathcal{R}_{E}(x)=0$, and continuity requires
that $\lim\limits_{x\to 0^{+}}\mathcal{R}_{B}(x)$ must converge to a finite
value. These requirement are satisfied by the Bessel functions of the first
kind $\mathcal{R}_{E}(x)\propto J_{1}(x)$ and $\mathcal{R}_{B}(x)\propto
J_{0}(x)$ inside the solenoid.

Boundary condition #4 requires that the waves outside the solenoid be purely
outgoing. If $\omega>0$, then this means that they must have the form

$\displaystyle E_{\text{out}}$
$\displaystyle\propto\operatorname{Re}\mathopen{}\mathclose{{}\left[e^{i(-\omega
t+\delta_{E})}H^{(1)}_{1}(\omega r)}\right],$ $\displaystyle B_{\text{out}}$
$\displaystyle\propto\operatorname{Re}\mathopen{}\mathclose{{}\left[e^{i(-\omega
t+\delta_{B})}H^{(1)}_{0}(\omega r)}\right],$

where $H^{(1)}_{\alpha}(x)$ are the Hankel functions of the first kind and the
implicit proportionality constants can depend on $\omega$ (but not on $r$ or
$t$).777These are the standard Bessel functions, not the modified Bessel
functions mentioned above for the unphysical $\lambda>0$ case. If $\omega<0$,
then the outward-propagating waves instead correspond to the Hankel functions
of the second kind $H^{(2)}_{\alpha}(x)$.

We can now play the usual game of assigning coefficients to each of the four
functions $E_{\text{in}}$, $E_{\text{out}}$, $B_{\text{in}}$, and
$B_{\text{out}}$ and then using equations (6) and the boundary conditions to
fix their values. The process is straightforward but requires some algebra and
some obscure Bessel function identities.

Ref. 3 worked it out for the special case of a sinusoidal surface current
$K_{\omega}(t)$ oscillating at a single positive frequency $\omega$. If we
characterize the sinusoidal current $K_{\omega}(t)$ by a _complex_ constant
$\tilde{K}_{\omega}$ via888Ref. 3 only considered the case of positive $K$ and
$K(t)\propto\cos(\omega t)$, but it easy to generalize the author’s results to
the case where $K$ is complex and $K(t)$ has a phase offset.

$K_{\omega}(t)=\operatorname{Re}\big{[}\tilde{K}_{\omega}\,e^{-i\omega
t}\big{]}=|\tilde{K}_{\omega}|\cos\big{(}\omega
t-\arg{\tilde{K}_{\omega}}\big{)},$ (11)

then the solution works out to be

$\displaystyle E_{\text{in}}^{\omega}(r,t)$
$\displaystyle=-\frac{1}{2}\pi\omega R\,J_{1}(\omega
r)\operatorname{Re}\mathopen{}\mathclose{{}\left[\phantom{i}\tilde{K}_{\omega}H^{(1)}_{1}(\omega
R)\,e^{-i\omega t}}\right],$ $\displaystyle r$ $\displaystyle\in[0,R]$ (12)
$\displaystyle E_{\text{out}}^{\omega}(r,t)$
$\displaystyle=-\frac{1}{2}\pi\omega RJ_{1}(\omega
R)\operatorname{Re}\mathopen{}\mathclose{{}\left[\phantom{i}\tilde{K}_{\omega}H^{(1)}_{1}(\omega
r\,)\,e^{-i\omega t}}\right],$ $\displaystyle r$ $\displaystyle\geq R$
$\displaystyle B_{\text{in}}^{\omega}(r,t)$
$\displaystyle=\phantom{-}\frac{1}{2}\pi\omega RJ_{0}(\omega
r\,)\operatorname{Re}\mathopen{}\mathclose{{}\left[i\tilde{K}_{\omega}H^{(1)}_{1}(\omega
R)\,e^{-i\omega t}}\right],$ $\displaystyle r$ $\displaystyle\in[0,R]$
$\displaystyle B_{\text{out}}^{\omega}(r,t)$
$\displaystyle=\phantom{-}\frac{1}{2}\pi\omega RJ_{1}(\omega
R)\operatorname{Re}\mathopen{}\mathclose{{}\left[i\tilde{K}_{\omega}H^{(1)}_{0}(\omega
r\,)\,e^{-i\omega t}}\right],$ $\displaystyle r$ $\displaystyle\geq R.$

For negative $\omega$, all the Hankel functions of the first kind
$H^{(1)}_{\alpha}$ in eqs. (12) are replaced by Hankel functions of the second
kind $H^{(2)}_{\alpha}$.

For a general (tempered-distribution) surface current $K(t)$, we can superpose
the fields (12) sourced by a single Fourier mode (11) over all the Fourier
modes with amplitude (and phase) specified by $\tilde{K}(\omega)$ at frequency
$\omega$. But in this case, doing so turns out to be somewhat subtle; Ref. 10
is the only source the author could find that actually discusses how to do
so.999Griffiths’ textbook [1] only says “Any wave can be written as a linear
combination of sinusoidal waves, and therefore you know how sinusoidal waves
behave, you know in principle how _any_ wave behaves.” Similarly, Jackson’s
textbook [2] only says “Assum[e] solutions with harmonic time dependence
$e^{-i\omega t}$, from which we can build an arbitrary solution by Fourier
superposition.”

The difficulty is that, as mentioned above, for positive $\omega$ the outward-
propagating waves correspond to the Hankel functions of the first kind
$H^{(1)}_{\alpha}(x)$, but for negative $\omega$ they correspond to the Hankel
functions of the second kind $H^{(2)}_{\alpha}(x)$. There are two possible
ways to deal with this difficulty.

The first approach is to decompose $K(t)$ via the usual Fourier transform

$K(t)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\tilde{K}(\omega)e^{-i\omega
t}\,d\omega,$

where $\omega$ ranges over all of $\mathbb{R}$. Under this approach, we can
find the general fields $E_{\text{in}}$, $E_{\text{out}}$, $B_{\text{in}}$,
and $B_{\text{out}}$ by promoting the constant $\tilde{K}_{\omega}$ in (12) to
the function $\tilde{K}(\omega)$ and then integrating over all
$\omega\in\mathbb{R}$. This is the approach implicitly taken by Ref. 10. The
advantage of this first option is that we can use all the familiar machinery
of Fourier analysis; the disadvantage is that the integrand is complicated and
is discontinuous at $\omega=0$, because we need to define it piecewise using
expressions (12) for $\omega>0$ and the corresponding expressions with
$H^{(2)}_{\alpha}$ for $\omega<0$.

The second approach is to instead decompose $K(t)$ using the modified inverse
Fourier transform

$K(t)=\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int\limits_{0}^{\infty}\tilde{K}(\omega)e^{-i\omega
t}\,d\omega}\right],$ (13)

where now we only integrate over positive $\omega$. It turns out that since
$K(t)$ is real, this modified Fourier decomposition holds if we use the
Fourier coefficients $\tilde{K}(\omega)$ given by the usual forward Fourier
transform

$\tilde{K}(\omega)=\int\limits_{-\infty}^{\infty}K(t)\,e^{i\omega t}\,dt.$

Loosely speaking, we need to use a prefactor $\frac{1}{\pi}$ in eq. (13)
instead of the usual $\frac{1}{2\pi}$ because there is an implicit factor of
$2$ that corrects for the fact that we are only integrating over positive
frequencies. See Appendix B for a more rigorous proof of eq. (13) and a
conceptual explanation of why it works. The advantage of this second option is
that the integrands are simpler and continuous; the disadvantage is that we
need to use slightly more complicated Fourier machinery to generalize results
like the convolution theorem.

Since it is already much more convenient to express the integrands (12) as the
real parts of complex functions, we will choose the second option. We can find
the general fields $E_{\text{in}}$, $E_{\text{out}}$, $B_{\text{in}}$, and
$B_{\text{out}}$ by promoting the constant $\tilde{K}_{\omega}$ in (12) to the
function $\tilde{K}(\omega)$ and then integrating over only positive $\omega$:

$\displaystyle E_{\text{in}}(r,t)$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}E_{\text{in}}(r,t;\omega)\,d\omega$
(14) $\displaystyle E_{\text{out}}(r,t)$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}E_{\text{out}}(r,t;\omega)\,d\omega$
$\displaystyle B_{\text{in}}(r,t)$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}B_{\text{in}}(r,t;\omega)\,d\omega$
$\displaystyle B_{\text{out}}(r,t)$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}B_{\text{in}}(r,t;\omega)\,d\omega.$

If we plug the expressions (12) into equations (14), then we see that

$E_{\text{in}}(r,t)=\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{g}_{E_{\text{in}}}(r,\omega)\,\tilde{K}(\omega)\,e^{-i\omega
t}\,d\omega}\right]$ (15)

and similarly for the other three fields, where

$\displaystyle\tilde{g}_{E_{\text{in}}}(r,\omega)$
$\displaystyle:=\frac{-1}{2}\pi\,\omega R\,J_{1}(\omega
r\,)\,H^{(1)}_{1}(\omega R),$ (16)
$\displaystyle\tilde{g}_{E_{\text{out}}}(r,\omega)$
$\displaystyle:=\frac{-1}{2}\pi\,\omega R\,J_{1}(\omega R)\,H^{(1)}_{1}(\omega
r),$ $\displaystyle\tilde{g}_{B_{\text{in}}}(r,\omega)$
$\displaystyle:=\frac{+i}{2}\hskip 1.0pt\pi\,\omega R\,J_{0}(\omega
r\,)\,H^{(1)}_{1}(\omega R),$
$\displaystyle\tilde{g}_{B_{\text{out}}}(r,\omega)$
$\displaystyle:=\frac{+i}{2}\hskip 1.0pt\pi\,\omega R\,J_{1}(\omega
R)\,H^{(1)}_{0}(\omega r).$

Ref. 10 derives a modified version of equations (15) and (16), but does not
consider the time-domain duals of the frequency-domain Green’s functions (16).
We will derive the time-domain Green’s functions in the rest of this section,
which will allow us to directly solve the problem in the time domain without
every needing to go into the frequency domain (or needing to use any complex
numbers).

Equation (15) is similar to (the real part of) an inverse Fourier transform,
but not identical because we only integrate over positive frequencies. If the
integral in eq. (15) extended over all $\omega\in\mathbb{R}$ (and the
prefactor were $\frac{1}{2\pi}$), then we would be able to apply the
convolution theorem, and we would find that
$E_{\text{in}}(r,t)=\operatorname{Re}[g_{E_{\text{in}}}(r,t)\ast
K(t)]=\operatorname{Re}[g_{E_{\text{in}}}(t)]\ast K(t)$ because $K(t)$ is
real, where $g_{E_{\text{in}}}(r,t)$ would be the inverse Fourier transform of
$\tilde{g}_{E_{\text{in}}}(r,\omega)$. So
$\operatorname{Re}[g_{E_{\text{in}}}(r,t)]$ would be the time-domain Green’s
function for the electric field inside the solenoid.

The fact that the integral in equation (15) only runs over positive
frequencies means that we cannot apply the standard convolution theorem.
Nevertheless, as we prove in Corollary 6 of Appendix B, the argument goes
through with only minimal changes, and

$\displaystyle E_{\text{in}}(r,t)$ $\displaystyle=g_{E_{\text{in}}}(r,t)\ast
K(t)$ (17) $\displaystyle E_{\text{out}}(r,t)$
$\displaystyle=g_{E_{\text{out}}}(r,t)\ast K(t)$ $\displaystyle
B_{\text{in}}(r,t)$ $\displaystyle=g_{B_{\text{in}}}(r,t)\ast K(t)$
$\displaystyle B_{\text{out}}(r,t)$ $\displaystyle=g_{B_{\text{out}}}(r,t)\ast
K(t),$

where all convolutions are taken with respect to time. Equations (17) are the
second main result of this article. Equation (34) gives that since $K(t)$ is
real, each of the four Green’s functions has the form

$g(r,t)=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{g}(r,\omega)\,e^{-i\omega
t}\,d\omega}\right]$ (18)

where $\tilde{g}(r,\omega)$ denotes the corresponding frequency-domain
expression (16). Explicitly, we get the time-domain Green’s
functions101010Note that $\tilde{g}_{E_{\text{in}}}(r=0,\omega)\equiv
g_{E_{\text{in}}}(r=0,t)\equiv 0$, so $E_{\text{in}}(r=0,t)\equiv 0$ in
accordance with boundary condition #2.

$\displaystyle g_{E_{\text{in}}}(r,t)$
$\displaystyle=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{-1}{2}\int_{0}^{\infty}\omega
R\,J_{1}(\omega r\,)\,H^{(1)}_{1}(\omega R)\,e^{-i\omega t}\,d\omega}\right]$
(19) $\displaystyle g_{E_{\text{out}}}(r,t)$
$\displaystyle=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{-1}{2}\int_{0}^{\infty}\omega
R\,J_{1}(\omega R)\,H^{(1)}_{1}(\omega r)\,e^{-i\omega t}\,d\omega}\right]$
$\displaystyle g_{B_{\text{in}}}(r,t)$
$\displaystyle=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{+i}{2}\int_{0}^{\infty}\omega
R\,J_{0}(\omega r\,)\,H^{(1)}_{1}(\omega R)\,e^{-i\omega t}\,d\omega}\right]$
$\displaystyle g_{B_{\text{out}}}(r,t)$
$\displaystyle=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{+i}{2}\int_{0}^{\infty}\omega
R\,J_{1}(\omega R)\,H^{(1)}_{0}(\omega r)\,e^{-i\omega t}\,d\omega}\right].$

Note that in the special case where $K(t)\equiv K$ is constant, equations (17)
and (18) give that

$\displaystyle E(r,t)$ $\displaystyle\equiv
K\int_{-\infty}^{\infty}g(r,\tau)\,d\tau$
$\displaystyle=K\,\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}\tilde{g}(r,\omega)\,e^{-i\omega\tau}\,d\omega\,d\tau}\right]$
$\displaystyle=K\,\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{g}(r,\omega)\mathopen{}\mathclose{{}\left(\int_{-\infty}^{\infty}e^{-i\omega\tau}\,d\tau}\right)d\omega}\right]$
$\displaystyle=K\,\mathrm{Re}\mathopen{}\mathclose{{}\left[2\int_{0}^{\infty}\tilde{g}(r,\omega)\,\delta(\omega)\,d\omega}\right]$
$\displaystyle=K\,\mathrm{Re}\mathopen{}\mathclose{{}\left[\tilde{g}(r,\omega=0)}\right],$

where we interpret $\tilde{g}(r,\omega=0)$ to mean $\lim\limits_{\omega\to
0^{+}}\tilde{g}(r,\omega)$ since the integral only runs over nonnegative
values of $\omega$. We prove in equations (37) of Appendix C that

$\displaystyle\lim_{\omega\to
0^{+}}\tilde{g}_{E_{\text{in}}}(r,\omega)=\lim_{\omega\to
0^{+}}\tilde{g}_{E_{\text{out}}}(r,\omega)=\lim_{\omega\to
0^{+}}\tilde{g}_{B_{\text{out}}}(r,\omega)$ $\displaystyle=0,$
$\displaystyle\lim_{\omega\to 0^{+}}\tilde{g}_{B_{\text{in}}}(r,\omega)$
$\displaystyle=1.$

We therefore get the familiar magnetostatic solution $E(r,t)\equiv 0$,
$B(r,t)=K\,\theta(R-r)$.

Since all four frequency-domain Green’s functions (16) remain bounded in the
low-frequency limit $\omega\to 0^{+}$, all four integrals in equations (19)
converge at the lower ends of the intervals of integration. But as we prove in
Appendix C, in the high-frequency limit all four frequency-domain Green’s
functions (16) asymptotically approach

$\tilde{g}(r,\omega)\sim\tilde{g}_{\text{HF}}(r,\omega):=\frac{1}{2}\sqrt{\frac{R}{r}}\mathopen{}\mathclose{{}\left[\pm
e^{i\,|r-R|\,\omega}+e^{i\mathopen{}\mathclose{{}\left((R+r)\omega-\frac{1}{2}\pi}\right)}}\right]$
(20)

as $\omega\to+\infty$, where we choose the $+$ branch of the $\pm$ symbol for
$\tilde{g}_{B_{\text{in}}}$ and the $-$ branch for
$\tilde{g}_{E_{\text{in}}}$, $\tilde{g}_{E_{\text{out}}}$, and
$\tilde{g}_{B_{\text{out}}}$.111111Just as the magnetic field $B_{\text{in}}$
behaves differently from the other three fields in the static case $\omega=0$
– namely, by being nonzero – it also behaves differently in the high-frequency
limit $\omega r\gg 1$. Along the $r=0$ axis,
$\tilde{g}_{E_{\text{in}}}(r=0,\omega)\equiv 0$ and
$\tilde{g}_{B_{\text{in}}}(r=0,\omega)\sim\sqrt{\frac{\pi}{2}\omega
R}\,e^{i\mathopen{}\mathclose{{}\left(\omega R-\frac{1}{4}\pi}\right)}$ as
$\omega\to+\infty$. We neglect this measure-zero case for the rest of this
article; we can handle it using a similar approach as in the $r>0$ case, but
doing so complicates the results by adding more special cases that need to be
handled separately. Since the frequency-domain Green’s functions do not go to
zero in the high-frequency limit $\omega\to+\infty$, they are not square-
integrable, and so the integrals (19) do not converge to proper functions
$g(r,t)$ but to distributions that contain Dirac delta functions.

To separate out the “well-behaved” contributions to the Green’s functions
(19), we decompose each of the four frequency-domain Green’s functions (16)
into the sum of the asymptotic exponential $\tilde{g}_{\text{HF}}(r,\omega)$
given in equation (20) and a remainder term

$\tilde{g}_{\text{SI}}(r,\omega):=\tilde{g}(r,\omega)-\tilde{g}_{\text{HF}}(r,\omega)$
(21)

that will turn out to be square-integrable:

$\tilde{g}(r,\omega)=\tilde{g}_{\text{SI}}(r,\omega)+\tilde{g}_{\text{HF}}(r,\omega).$
(22)

It turns out that the $\tilde{g}_{B,\text{HF}}(r,\omega)$ terms (20) for the
$B$ field contain the entire jump discontinuity of $\tilde{g}_{B}(r,\omega)$
at $r=R$, so the remaining square-integrable term
$\tilde{g}_{B,\text{SI}}(r,\omega)$ is continuous at $r=R$. We can distribute
the modified inverse Fourier transform (18) across the two terms of the
decomposition (22) and define

$\displaystyle g_{\text{SI}}(r,t)$
$\displaystyle:=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{g}_{\text{SI}}(r,\omega)\,e^{-i\omega
t}\,d\omega}\right],$ (23) $\displaystyle g_{\text{HF}}(r,t)$
$\displaystyle:=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{g}_{\text{HF}}(r,\omega)\,e^{-i\omega
t}\,d\omega}\right],$ (24)

with $g(r,t)=g_{\text{SI}}(r,t)+g_{\text{HF}}(r,t)$.121212The subscript “HF”
(for “high-frequency”) in $g_{\text{HF}}(r,t)$ should not be interpreted too
literally. It is simply a mnemonic for the inverse Fourier transform of the
particular choice of high-frequency limiting expression
$\tilde{g}_{\text{HF}}(r,\omega)$ of $\tilde{g}(r,\omega)$ that is given in
equation (20) – but $g_{\text{HF}}(r,t)$ includes the low-frequency modes in
$\tilde{g}_{\text{HF}}(r,\omega)$. The particular choice of function (20) is
not the unique asymptotic function for $\tilde{g}(r,\omega)$, but it is the
most convenient choice for our purposes. Therefore, the particular choice of
decomposition (22) is not a unique decomposition of $\tilde{g}(r,\omega)$ into
the sum of a “square-integrable part” and a “high-frequency part”.

We can calculate the second modified inverse Fourier transform (24) to express
$g_{\text{HF}}(r,t)$ in closed form. We prove in equation (31) of Appendix B
that if $A\in\mathbb{C}$ and $t_{0}\in\mathbb{R}$, then

$\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}A\,e^{i\omega
t_{0}}\,e^{-i\omega
t}\,d\omega}\right]=\operatorname{Re}[A]\,\delta(t-t_{0})+\frac{\operatorname{Im}[A]}{\pi(t-t_{0})}.$
(25)

Both terms of $\tilde{g}_{\text{HF}}(r,\omega)$ in equation (20) take the form
$A\,e^{i\omega t_{0}}$ with $A=\pm\frac{1}{2}\sqrt{\frac{R}{r}},\ t_{0}=|r-R|$
and $A=-\frac{1}{2}\sqrt{\frac{R}{r}}\,i,t_{0}=R+r$, respectively. Therefore,
equations (20), (24), and (25) give that

$g_{\text{HF}}(r,t)=\frac{1}{2}\sqrt{\frac{R}{r}}\mathopen{}\mathclose{{}\left[\pm\delta(t-|r-R|)-\frac{1}{\pi(t-(R+r))}}\right].$
(26)

It turns out that $g_{E}(r,t)$ is continuous across the solenoid $r=R$, and
$g_{B}(r,t)$ is continuous except for the sign of the delta-function term,
which switches across the solenoid.

Returning to equations (17), we see that the first term in equation (26) (the
delta function) corresponds to the contribution to the electromagnetic fields
from the wavefront that is just arriving at the point $(r,t)$ after having
been emitted at the solenoid at time $t-|r-R|$ and traveled radially inward or
outward at the speed of light. This delta-function term represents the very
first influence that a given point “feels” as the initial wavefront arrives.
The second term is inversely proportional to $t-(R+r)$, which is the retarded
time at which the _back_ end of the solenoid (antipodal to the observation
point) emitted the EM waves that reach the observation point at time $t$.

Causality requires that the time-domain Green’s functions $g(r,t)$ all vanish
outside the future light cone $t=|r-R|$.131313For this geometry, the radius
$r$ represents a cylinder rather than a point, so the future light cone
$t=|r-R|$ is a three-dimensional null hypersurface. It starts out aligned with
the solenoid (as a cylinder of radius $R$) and then splits into two concentric
cylinders. One cylinder shrinks in radius at the speed of light before
vanishing at time $t=R$ when it degenerates at the solenoid axis, and the
other increases in radius without bound at the speed of light. The “outside”
of this light cone corresponds to the region $|R-r|>t$ that is spacelike
separated from the entire solenoid. For $t<R$, this region consists of two
disconnected components, one bounded in radius; for $t>R$, it consists of a
single connected component unbounded in radius. Therefore, outside the light
cone we must have that $g_{\text{SI}}(r,t)\equiv-g_{\text{HF}}(r,t)$ for all
four Green’s functions. From equations (23) and (26), we see that causality
implies the purely mathematical fact that

$\mathrm{Re}\mathopen{}\mathclose{{}\left[\int_{0}^{\infty}\tilde{g}_{\text{SI}}(r,\omega)\,e^{-i\omega
t}\,d\omega}\right]=\frac{\sqrt{R/r}}{2(t-(R+r))}$ (27)

if $0<t<|R-r|$ for all four functions (16). This mathematical result is easy
to check numerically, but the author is not aware of any way to prove it
directly without using this causality argument.

The modified inverse Fourier transform (23) for $g_{\text{SI}}(r,t)$ cannot
(to the author’s knowledge) be computed in closed form for any of the four
electromagnetic fields described in equations (19). But since the frequency-
domain functions $\tilde{g}_{\text{SI}}(r,\omega)$ defined in equation (21)
are all square-integrable, we can calculate the Fourier transforms numerically
(although the fact that the integrands are oscillating means that we need to
be careful with numerical stability).

In Figure 1, we plot the time-domain Green’s functions
$g(r,t)=g_{\text{SI}}(r,t)+g_{\text{HF}}(r,t)$ for the electric and magnetic
fields inside and outside the solenoid. As expected, the Green’s functions
vanish identically outside of the solenoid’s future light cone. This fact
serves as a check of the correctness of our calculations.141414We did not
manually truncate the Green’s functions in Figure 1. We calculated the
functions numerically for all times within the plots’ domains; outside of the
solenoid’s future light cone, the two terms cancel identically because of
equation (27).

Figure 1:  Plots of the time-domain Green’s functions $g_{E}(r,t)$ for the
electric field (in blue) and $g_{B}(r,t)$ for the magnetic field (in orange)
(a) inside the solenoid, at $r=0.5R$, (b) at the solenoid, at $r=R$, and (c)
outside the solenoid, at $r=2R$. The Green’s functions vanish identically if
$t<|R-r|$ (outside the solenoid’s future light cone). Each Green’s function
has a $-1/(t-(R+r))$ singularity at $t=r+R$. The vertical rays at $t=|r-R|$
represent Dirac delta functions. Both Green’s functions are non-differentiable
at the solenoid $r=R$, but $g_{E}(r,t)$ is continuous there, while
$g_{B}(r,t)$ is continuous except for the delta-function term, whose
coefficient jumps discontinuously from being positive inside the solenoid to
being negative outside of it. Upward-pointing rays indicate a delta function
with a positive coefficient, downward-pointing rays indicate one with a
negative coefficient, and black rays indicate a blue and orange ray overlaid
at the same point in time. Panel (b) displays orange rays pointing both upward
and downward to indicate the sign change across the solenoid. We calculated
the Green’s functions $g(r,t)=g_{\text{SI}}(r,t)+g_{\text{HF}}(r,t)$ from
equations (16), (20), (21), (23), and (26). We discretized the independent
variable $t$ into time steps $\Delta t=0.05R$ and evaluated the improper
integral (23) numerically. The integral appears to converge, but its integrand
oscillates in sign, so for the purpose of numerical stability we truncated the
upper bound of integration to $\omega_{\text{max}}R=2000$.

##  Appendix A Vanishing electromagnetic fields outside of an ideal solenoid

As mentioned in the main text, Ref. 7 proves a truly remarkable theorem about
infinite ideal solenoids. This theorem is not entirely simple to state, but it
is worth the effort.

######  Theorem.

Suppose that $\kappa(t)$ is _any_ continuous real-valued function defined over
a time interval of finite duration $T$. Then there exists an extension $K(t)$
of $\kappa(t)$ to the entire real line such that the following propositions
hold:

  * •

$K(t)$ is defined over all of $\mathbb{R}$.

  * •

$K(t)\equiv\kappa(t)$ on the original domain of $\kappa(t)$.

  * •

If $K(t)$ is the current density for an infinite ideal solenoid of diameter
$cT$ (where $c$ is the speed of light), then the _exact_ electromagnetic
fields vanish _identically_ outside the solenoid at all times.∎

(In fact, the original function $\kappa(t)$ does not even have to be
continuous; it can have a finite number of jump discontinuities.) The extended
function $K(t)$ is _not_ periodic in general.

##  Appendix B Modified inverse Fourier transforms evaluated over positive
frequencies

Let

$\mathcal{F}[f(t)]:=\int_{-\infty}^{\infty}f(t)\,e^{i\omega t}\,dt$

and

$\mathcal{F}^{-1}\big{[}\tilde{f}(\omega)\big{]}:=\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{f}(\omega)\,e^{-i\omega
t}\,d\omega$

denote the forward and inverse Fourier transforms and let $\theta(x)$ denote
the Heaviside step function. Let the _Hilbert transform_ of a function $f(t)$
be another function $\hat{f}(t)$ or $H[f](t)$ defined by

$\hat{f}(t)=H[f](t):=\frac{1}{\pi}\,\mathrm{p.v.}\int_{-\infty}^{\infty}\frac{f(\tau)}{t-\tau}d\tau,$
(28)

where $\mathrm{p.v.}\int$ denotes the Cauchy principal value of an improper
integral [13].151515Strictly speaking, the Hilbert transform is only defined
for functions in $L^{p}$ with $p\in(1,\infty)$ by the Titchmarsh theorem [13].
But in this appendix, we will ignore questions of functional domains and
assume that all functions are sufficiently well-behaved. It will sometimes be
clearer for us to write $H[f(t)]$ instead of $H[f](t)$ or $\hat{f}(t)$; these
expressions all refer to the same Hilbert transform. Note that unlike with the
Fourier transform, the arguments of $f(t)$ and $\hat{f}(t)$ lie in the same
domain.

All the results in this Appendix follow from the follow theorem:

######  Theorem.

If $\tilde{f}(\omega)=\mathcal{F}[f(t)]$ and
$f(t)=\mathcal{F}^{-1}\big{[}\tilde{f}(\omega)\big{]}$ are a Fourier pair of
functions $\mathbb{R}\to\mathbb{C}$, then

$\mathcal{F}^{-1}\big{[}2\theta(\omega)\tilde{f}(\omega)\big{]}=f(t)-i\hat{f}(t),$
(29)

where $\hat{f}(t)$ denotes the Hilbert transform $H[f](t)$.161616Under the
standard sign convention for a _spatial_ Fourier transform, where the
_forward_ Fourier transform contains $e^{-ikx}$ and the _inverse_ transform
contains $e^{ikx}$, the RHS becomes $f(t)+i\hat{f}(t)$.

###### Proof.

$\displaystyle\mathcal{F}^{-1}\big{[}2\theta(\omega)\tilde{f}(\omega)\big{]}$
$\displaystyle=\mathcal{F}^{-1}\big{[}\mathopen{}\mathclose{{}\left(1+\operatorname{sgn}(\omega)}\right)\tilde{f}(\omega)\big{]}$
$\displaystyle=f(t)+\mathcal{F}^{-1}\big{[}\operatorname{sgn}(\omega)\tilde{f}(\omega)\big{]}.$

By the convolution theorem, the final inverse Fourier transform

$\displaystyle\mathcal{F}^{-1}\big{[}\operatorname{sgn}(\omega)\tilde{f}(\omega)\big{]}$
$\displaystyle=\mathcal{F}^{-1}\mathopen{}\mathclose{{}\left[\operatorname{sgn}(\omega)}\right]\ast
f(t)$ $\displaystyle=-\frac{i}{\pi t}\ast f(t)$ $\displaystyle=-i\,H[f](t),$

where $\ast$ denotes convolution with respect to time.171717Strictly speaking,
the inverse Fourier transform of $\operatorname{sgn}(\omega)$ is actually the
(tempered) _distribution_ $\mathrm{p.v.}\frac{-i}{\pi t}$ that maps a
compactly supported smooth test function $f:\mathbb{R}\to\mathbb{C}$ to
$\mathrm{p.v.}\int_{-\infty}^{\infty}\frac{-if(t)}{\pi t}\,dt$ [13]. ∎

######  Corollary 1.

If $f:\mathbb{R}\to\mathbb{C}$ is complex-valued, then

$\displaystyle\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,e^{-i\omega
t}\,d\omega}\right]$
$\displaystyle=\operatorname{Re}[f(t)]+\operatorname{Im}\big{[}\hat{f}(t)\big{]}$
$\displaystyle=\operatorname{Re}[f(t)]+H[\operatorname{Im}[f(t)]].$

###### Proof.

$\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,e^{-i\omega
t}\,d\omega=\mathcal{F}^{-1}\big{[}2\theta(k)\tilde{f}(\omega)\big{]}=f(t)-i\hat{f}(t).$
(30)

The Hilbert transform is an integral transform, so it is linear and we can
apply it separately to the real and imaginary parts of $f(t)$. Its integral
kernel is real-valued, so the Hilbert transform of the real part of $f(t)$ is
real, and the Hilbert transform of the imaginary part of $f(t)$ is imaginary.
∎

######  Corollary 2.

If $A\in\mathbb{C}$ and $t_{0}\in\mathbb{R}$, then

$\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}A\,e^{-i\omega(t-t_{0})}\,d\omega}\right]=\operatorname{Re}[A]\,\delta(t-t_{0})+\frac{\operatorname{Im}[A]}{\pi(t-t_{0})}.$
(31)

###### Proof.

Let $f(t)=A\,\delta(t-t_{0})$ and $\tilde{f}(\omega)=A\,e^{i\omega t_{0}}$.
Then

$\hat{f}(t)=\frac{A}{\pi(t-t_{0})}$

from definition (28), and

$\frac{1}{\pi}\int_{0}^{\infty}A\,e^{-i\omega(t-t_{0})}\,d\omega=A\,\delta(t-t_{0})-i\frac{A}{\pi(t-t_{0})}.$

Equation (31) follows from decomposing $A$ into its real and imaginary parts
and then taking the real part of the above equation. ∎

We use equation (31) in equation (25) of the main text.

######  Corollary 3.

If $f:\mathbb{R}\to\mathbb{R}$ is real-valued, then

$f(t)\equiv\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,e^{-i\omega
t}\,d\omega}\right].$

Letting $f(t)$ be the surface current $K(t)$ discussed in the main text proves
eq. (13).

The proof of the main theorem is rather formal, but there is a way to
understand it more intuitively. Because we take the real part in eq. (11), the
sign of $\omega$ is ambiguous; the complex constants $\tilde{K}_{\omega}$ and
$\tilde{K}^{*}_{-\omega}$ correspond to the same physical (i.e. real-valued)
current $K(t)$. This corresponds to the fact that the Fourier transform of a
real function $f(t)$ is Hermitian (i.e.
$\tilde{f}(-\omega)\equiv\tilde{f}^{*}(\omega)$), while the Fourier transform
of a pure imaginary function $g(t)$ is anti-Hermitian (i.e.
$\tilde{g}(-\omega)\equiv-\tilde{g}^{*}(\omega)$). Therefore, conversely only
the Hermitian part of $\tilde{K}(\omega)$ will affect the physical (i.e. real)
part of $K(t)$. Loosely speaking, we have a sort of “gauge freedom” to add any
imaginary part we want to $K(t)$ (or equivalently, to add any anti-Hermitian
part to $\tilde{K}(\omega)$) without changing any physical observables.

We can think of this “gauge freedom” as allowing us to, for each pair of
opposite-sign frequencies $\pm\omega_{0}$, independently redistribute Fourier
weights between $\tilde{K}(\omega_{0})$ and $\tilde{K}^{*}(-\omega_{0})$,
since only their sum is physically meaningful. The most natural “gauge
fixing”, corresponding to the usual Fourier transform of a pure real function
$K(t)$, is to distribute the amplitude (and phase) of each real mode with
frequency $\omega_{0}$ symmetrically between $\tilde{K}(\omega_{0})$ and
$\tilde{K}^{*}(-\omega_{0})$. This choice leads to a Hermitian function
$\tilde{K}(\omega)$. Adding a pure imaginary component to $K(t)$ before taking
the Fourier transform redistributes the Fourier weights non-symmetrically
between the pairs $\tilde{K}(\omega_{0})$ and $\tilde{K}^{*}(-\omega_{0})$ by
adding an anti-Hermitian part to $\tilde{K}(\omega)$.

The Hermitian “gauge choice” is _usually_ the most convenient, but in some
situations other choices are more convenient. The most common alternative
choice is to add a pure imaginary component to the original real signal that
corresponds to placing all of the Fourier weights on the nonnegative
frequencies only. A complex-valued function $f(t)$ whose Fourier transform is
nonzero only for $\omega\geq 0$ (i.e. $\tilde{f}(\omega)\equiv 0$ if
$\omega<0$) is referred to as an _analytic signal_ [14].

If $f:\mathbb{R}\to\mathbb{R}$ is a _real_ -valued function, then its
_analytic representation_ $f_{a}:\mathbb{R}\to\mathbb{C}$ is defined to be the
complex-valued function $f_{a}(t):=f(t)-i\hat{f}(t)$. By theorem 29,
$f_{a}(t)=\mathcal{F}^{-1}\big{[}2\theta(\omega)\tilde{f}(\omega)\big{]}$.
Every analytic representation of a real-valued function is a (complex-valued)
analytic signal, and vice versa:

######  Corollary 4.

$f_{a}:\mathbb{R}\to\mathbb{C}$ is an analytic signal iff
$f_{a}(t)=f(t)-i\hat{f}(t)$ for some $f:\mathbb{R}\to\mathbb{C}$.

###### Proof.

For the forward direction: if $f_{a}$ is an analytic signal then
$\tilde{f}_{a}(\omega)\equiv\theta(\omega)\tilde{f}_{a}(\omega)$. Taking the
inverse Fourier transform of both sides and applying theorem (29) (but without
the factor of $2$), we find that

$f_{a}(t)\equiv-i\,\hat{f}_{a}(t).$

If we write
$f_{a}(t)=\operatorname{Re}[f_{a}(t)]+i\operatorname{Im}[f_{a}(t)]$ and equate
the imaginary parts of the equation above, then we get that
$\operatorname{Im}[f_{a}(t)]=-\operatorname{Re}\big{[}\hat{f}_{a}(t)\big{]}=-H[\operatorname{Re}[f_{a}(t)]]$
because the Hilbert transform’s integral kernel is real. Let
$f(t)=\operatorname{Re}[f_{a}(t)]$.

For the reverse direction: Theorem (29) gives that
$f_{a}(t)=\mathcal{F}^{-1}\big{[}2\theta(\omega)\tilde{f}(\omega)\big{]}$.
Taking the forward Fourier transform of both sides gives that
$\tilde{f}_{a}(\omega)=2\theta(\omega)\tilde{f}(\omega)$. Therefore,
$f_{a}(t)$ is an analytic signal. ∎

This corollary is essentially the Fourier dual of the Kramers-Kronig
relations. The Kramers-Kronig relations describe the _forward_ Fourier
transform of a causal response function that vanishes for negative _time
displacement_ , while this corollary describes the _inverse_ Fourier transform
of a function that vanishes for negative _frequency_. For square-integrable
functions $f_{a}$, the corollary can be extended to the statement that the
analytic continuation of $f_{a}$ to the complex plane is analytic on the
closed upper half-plane and approaches zero on the upper half-plane as
$|z|\to\infty$; this is another version of Titchmarsh’s theorem [15].

We can now understand what is going on behind the scenes in eq. (13) in the
main text. We can think of the “one-sided inverse Fourier transform”

$\frac{1}{\pi}\int\limits_{0}^{\infty}\tilde{K}(\omega)\,e^{-i\omega
t}\,d\omega$

inside the brackets in eq. (13) as
$\mathcal{F}^{-1}\big{[}2\theta(\omega)\tilde{f}(\omega)\big{]}$. By theorem
(29), this is not the real-valued physical function $K(t)$, but instead the
analytic representation $K_{a}(t)=K(t)-i\hat{K}(t)$ with only positive-
frequency components. Taking the real part drops the imaginary Hilbert
transform term, restores the negative-frequency components, and leaves the
physical function $K(t)$.

We can prove (at a physicist’s level of rigor) a “one-sided” variant of the
convolution theorem:

######  Corollary 5.

If $\big{(}f(t),\tilde{f}(\omega)\big{)}$ and
$\mathopen{}\mathclose{{}\left(g(t),\tilde{g}(\omega)}\right)$ are each
Fourier pairs of functions $\mathbb{R}\to\mathbb{C}$, then

$\displaystyle\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,\tilde{g}(\omega)\,e^{-i\omega
t}\,d\omega$ $\displaystyle=\big{(}f(t)-i\,\hat{f}(t)\big{)}\ast g(t)$
$\displaystyle=f(t)\ast\mathopen{}\mathclose{{}\left(g(t)-i\,\hat{g}(t)}\right).$

###### Proof.

$\displaystyle\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,\tilde{g}(\omega)\,e^{-i\omega
t}\,d\omega$
$\displaystyle=\mathcal{F}^{-1}\big{[}\tilde{f}(\omega)\,2\theta(\omega)\,\tilde{g}(\omega)\big{]}$
$\displaystyle=\mathcal{F}^{-1}\big{[}\tilde{f}(\omega)\,2\theta(\omega)\big{]}\ast
g(t)$ (32)
$\displaystyle=f(t)\ast\mathcal{F}^{-1}\big{[}2\theta(\omega)\,\tilde{g}(\omega)\big{]}.$

∎

More generally, we have that

$\frac{1}{\pi}\int_{0}^{\infty}\prod_{k=1}^{n}\tilde{f}_{k}(\omega)\,e^{-i\omega
t}\,d\omega$ (33)

equals $1/2^{m-1}$ times the $n$-fold convolution of the $\\{f_{k}(t)\\}$,
$k=1,...n$, but with any $m$ of the convolved functions $f_{k}(t)$ replaced by
$f_{k}(t)-i\,\hat{f}_{k}(t)$, where $m$ is any natural number $m=1,\dots,n$.
This is because
$2\theta(\omega)=2\theta(\omega)^{m}=\frac{1}{2^{m-1}}(2\theta(\omega))^{m}$,
and we can multiply each factor $\tilde{f}_{k}(\omega)$ in the integrand of
the integral (33) by a different factor $2\theta(\omega)$.

Finally, we have

######  Corollary 6.

If $\mathopen{}\mathclose{{}\left(g(t),\tilde{g}(\omega)}\right)$ is a Fourier
pair and $g:\mathbb{R}\to\mathbb{R}$ is real-valued, then

$\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,\tilde{g}(\omega)\,e^{-i\omega
t}\,d\omega}\right]\\\
=\mathrm{Re}\mathopen{}\mathclose{{}\left[\frac{1}{\pi}\int_{0}^{\infty}\tilde{f}(\omega)\,e^{-i\omega
t}\,d\omega}\right]\ast g(t).$ (34)

The proof follows from equations (32) and (30) and from the fact that $g(t)$
is real.

Letting $g(t)$ be the surface current $K(t)$ discussed in the main text and
using equation (15) proves eqs. (17).

##  Appendix C Asymptotics of the frequency-domain Green’s functions for the
electromagnetic fields

The Bessel functions of the first kind $J_{0}(x)$ and $J_{1}(x)$ and the
Hankel functions of the first kind $H^{(1)}_{0}(x)$ and $H^{(1)}_{1}(x)$ have
the asymptotic forms

$\displaystyle J_{0}(x)$ $\displaystyle\sim 1,$ $\displaystyle J_{1}(x)$
$\displaystyle\sim\frac{1}{2}x$ $\displaystyle H^{(1)}_{0}(x)$
$\displaystyle\sim\frac{2i}{\pi}\ln x$ $\displaystyle H^{(1)}_{1}(x)$
$\displaystyle\sim-\frac{2i}{\pi x}$

as $x\to 0^{+}$ and

$\displaystyle J_{0}(x)$ $\displaystyle\sim\sqrt{\frac{2}{\pi
x}}\cos\mathopen{}\mathclose{{}\left(x-\frac{1}{4}\pi}\right)$ (35a)
$\displaystyle J_{1}(x)$ $\displaystyle\sim\sqrt{\frac{2}{\pi
x}}\cos\mathopen{}\mathclose{{}\left(x-\frac{3}{4}\pi}\right)$ (35b)
$\displaystyle H^{(1)}_{0}(x)$ $\displaystyle\sim\sqrt{\frac{2}{\pi
x}}e^{i\mathopen{}\mathclose{{}\left(x-\frac{1}{4}\pi}\right)}\hskip 44.0pt$
(36a) $\displaystyle H^{(1)}_{1}(x)$ $\displaystyle\sim\sqrt{\frac{2}{\pi
x}}e^{i\mathopen{}\mathclose{{}\left(x-\frac{3}{4}\pi}\right)}$ (36b)

as $x\to+\infty$ [2].181818Strictly speaking, the asymptotic relations (35)
for the Bessel functions $J_{n}(x)$ are not quite correct in the standard
sense, because the zeros of the Bessel functions come arbitrary close to – but
never exactly equal to – the zeros of the cosine approximation at
$\mathopen{}\mathclose{{}\left(m+\frac{1}{2}\mathopen{}\mathclose{{}\left(n-\frac{1}{2}}\right)}\right)\pi$,
$m\in\mathbb{Z}$. Those relations should be interpreted as saying that the
ratio of the two sides comes arbitrarily close to 1 as $x\to+\infty$, except
on an infinite sequence of real intervals where the ratio diverges – but those
intervals (where the ratio deviates from $1$ by more than $\epsilon$ for any
fixed $\epsilon$) become arbitrarily short as $x\to+\infty$. The asymptotic
relations (36) for the Hankel functions $H^{(1)}_{n}(x)$ are correct in the
standard sense.

Therefore, the frequency-domain Green’s functions (16) have the asymptotic
forms

$\displaystyle\tilde{g}_{E_{\text{in}}}(r,\omega)$
$\displaystyle\sim\frac{ir}{2}\omega$ (37)
$\displaystyle\tilde{g}_{E_{\text{out}}}(r,\omega)$
$\displaystyle\sim\frac{iR^{2}}{2r}\omega$
$\displaystyle\tilde{g}_{B_{\text{in}}}(r,\omega)$ $\displaystyle\sim 1$
$\displaystyle\tilde{g}_{B_{\text{out}}}(r,\omega)$
$\displaystyle\sim-\frac{1}{2}R^{2}\omega^{2}\ln(\omega r)$

as $\omega\to 0^{+}$ and

$\displaystyle\tilde{g}_{E_{\text{in}}}(r,\omega)$
$\displaystyle\sim\sqrt{\frac{R}{r}}\cos\mathopen{}\mathclose{{}\left(\omega
r-\frac{3}{4}\pi}\right)e^{i\mathopen{}\mathclose{{}\left(\omega
R+\frac{1}{4}\pi}\right)}$ (38)
$\displaystyle\tilde{g}_{B_{\text{in}}}(r,\omega)$
$\displaystyle\sim\begin{cases}\sqrt{\frac{R}{r}}\cos\mathopen{}\mathclose{{}\left(\omega
r-\frac{1}{4}\pi}\right)e^{i\mathopen{}\mathclose{{}\left(\omega
R-\frac{1}{4}\pi}\right)}\qquad&\text{if }r>0\\\ \sqrt{\frac{\pi\omega
R}{2}}e^{i\mathopen{}\mathclose{{}\left(\omega
R-\frac{1}{4}\pi}\right)}\qquad&\text{if }r=0\end{cases}$
$\displaystyle\tilde{g}_{E_{\text{out}}}(r,\omega)$
$\displaystyle\sim\tilde{g}_{B_{\text{out}}}(r,\omega)\sim\sqrt{\frac{R}{r}}\cos\mathopen{}\mathclose{{}\left(\omega
R-\frac{3}{4}\pi}\right)e^{i\mathopen{}\mathclose{{}\left(\omega
r+\frac{1}{4}\pi}\right)}$

as $\omega\to+\infty$.191919The high-frequency asymptotic relations (38) as
$\omega\to+\infty$ hold in the same sense as the asymptotic relations (35)
discussed in the previous footnote. Recall that
$\tilde{g}_{E_{\text{in}}}(r=0,\omega)\equiv 0$. By expanding out the cosines
as sums of complex exponentials, we can rewrite the high-frequency asymptotics
(38) as equation (20) in the main text. This equation shows that
$\tilde{g}_{E_{\text{in}}}(r,\omega)\sim\tilde{g}_{E_{\text{out}}}(r,\omega)\sim\tilde{g}_{B_{\text{out}}}(r,\omega)$
as $\omega\to+\infty$, which is not immediately obvious from equations (38).

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