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The solution proposed by Matzner and outlined in the other answer is only approximate, not exact. The problem is, how much of an approximate solution it really is.

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and \begin{align} A_x(\theta, \phi) & = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) & = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] \end{align} But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k\ll1$.

1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

   But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact \begin{align} A_r(\theta, \phi) & = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta \\ & = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \\ & \neq 0 \end{align} Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

   If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr\gg1$ and/or $k \rightarrow 0$.

So what can be done about the desired exact monochromatic solution with a spherically symmetric intensity?

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209 (eprint). However, finding an exact solution is still a tricky task.

Formally, it amounts to solving for the electric field as a divergenceless solution to a Helmholtz equation, $$ \left(\Delta + k^2 \right) \vec{E} = 0 \;, \;\;\; \nabla \cdot \vec{E} =0 $$ under the additional requirement that the field intensity $\vec{E}^2$ be spherically symmetric.

One standard strategy is to consider a multipole expansion with scalar radial factors given by spherical Bessel functions and vectorial directional factors as linear superpositions of spherical harmonics. But then imposing the spherical intensity condition brings up a hornet's nest of Clebsch-Gordon coefficients. Or, potentially, some really neat irrep argument.

Anyone up to it?

The solution proposed by Matzner and outlined in the other answer is only approximate, not exact. The problem is, how much of an approximate solution it really is.

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and \begin{align} A_x(\theta, \phi) & = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) & = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] \end{align} But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k\ll1$.

1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

   But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact \begin{align} A_r(\theta, \phi) & = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta \\ & = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \\ & \neq 0 \end{align} Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

   If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr\gg1$ and/or $k \rightarrow 0$.

So what can be done about the desired exact monochromatic solution with a spherically symmetric intensity?

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209 (eprint). However, finding an exact solution is still a tricky task.

Formally, it amounts to solving for the electric field as a divergenceless solution to a Helmholtz equation, $$ \left(\Delta + k^2 \right) \vec{E} = 0 \;, \;\;\; \nabla \cdot \vec{E} =0 $$ under the additional requirement that the field intensity $\vec{E}^2$ be spherically symmetric.

One standard strategy is to consider a multipole expansion with scalar radial factors given by spherical Bessel functions and vectorial directional factors as linear superpositions of spherical harmonics. But then imposing the spherical intensity condition brings up a hornet's nest of Clebsch-Gordon coefficients. Or, potentially, some really neat irrep argument.

Anyone up to it?

The solution proposed by Matzner and outlined in the other answer is only approximate, not exact. The problem is, how much of an approximate solution it really is.

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and \begin{align} A_x(\theta, \phi) & = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) & = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] \end{align} But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k\ll1$.

1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

   But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact \begin{align} A_r(\theta, \phi) & = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta \\ & = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \\ & \neq 0 \end{align} Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

   If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr\gg1$ and/or $k \rightarrow 0$.

So what can be done about the desired exact monochromatic solution with a spherically symmetric intensity?

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209 (eprint). However, finding an exact solution is still a tricky task.

Formally, it amounts to solving for the electric field as a divergenceless solution to a Helmholtz equation, $$ \left(\Delta + k^2 \right) \vec{E} = 0 \;, \;\;\; \nabla \cdot \vec{E} =0 $$ under the additional requirement that the field intensity $\vec{E}^2$ be spherically symmetric. But since the Laplacian separates into a radial and angular part, this amounts to requiring $$ {\hat L}^2 \left(\vec{E}^2\right) = 0 $$ One standard strategy is to consider a multipole expansion with scalar radial factors given by spherical Bessel functions and vectorial directional factors as linear superpositions of spherical harmonics. But then imposing the spherical intensity condition brings up a hornet's nest of Clebsch-Gordon coefficients. Or, potentially, some really neat irrep argument.

Anyone up to it?

The solution proposed by Matzner and outlined in the other answer is only approximate, not exact. The problem is, how much of an approximate solution it really is.

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and \begin{align} A_x(\theta, \phi) & = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) & = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] \end{align} But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k\ll1$.

1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

   But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact \begin{align} A_r(\theta, \phi) & = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta \\ & = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \\ & \neq 0 \end{align} Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

   If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr\gg1$ and/or $k \rightarrow 0$.

So what can be done about the desired exact monochromatic solution with a spherically symmetric intensity?

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209 (eprint). However, finding an exact solution is still a tricky task.

Formally, it amounts to solving for the electric field as a divergenceless solution to a Helmholtz equation, $$ \left(\Delta + k^2 \right) \vec{E} = 0 \;, \;\;\; \nabla \cdot \vec{E} =0 $$ under the additional requirement that the field intensity $\vec{E}^2$ be spherically symmetric.

One standard strategy is to consider a multipole expansion with scalar radial factors given by spherical Bessel functions and vectorial directional factors as linear superpositions of spherical harmonics. But then imposing the spherical intensity condition brings up a hornet's nest of Clebsch-Gordon coefficients. Or, potentially, some really neat irrep argument.

Anyone up to it?

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and $$ A_x(\theta, \phi) = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] $$ But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k<<1$.

1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact $$ A_r(\theta, \phi) = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \neq 0 $$ Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr>>1$ and/or $k \rightarrow 0$.

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209. But finding an exact solution is still a tricky task.

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and \begin{align} A_x(\theta, \phi) & = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) & = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] \end{align} But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k\ll1$.

1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

   But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact \begin{align} A_r(\theta, \phi) & = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta \\ & = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \\ & \neq 0 \end{align} Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

   If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr\gg1$ and/or $k \rightarrow 0$.

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209 (eprint). However, finding an exact solution is still a tricky task.