## Show Solution

The strategy to find the solution for this problem is very general and used throughout the natural sciences. We will express the **general solution** of the guiding differential equation and solve the **boundary conditions**. Here, it is suitable to use the electrostatic potential \(\phi\), for which the Laplace equation holds if no charges are present: \[\Delta\phi\left(\mathbf{r}\right) = 0\ .\]Since the sphere is rotationally symmetric, it is convenient to treat the problem in **spherical coordinates** \(r\), \(\theta\) and \(\varphi\).

## A General Solution for the Potential

In spherical coordinates, the solution be can expressed in terms of **Legendre polynomials** \(P_{l}\left(\cos\theta\right)\) - the Laplace equation is fullfilled both by \(r^{l}P_{l}\left(\cos\theta\right)\) and \(P_{l}\left(\cos\theta\right)/r^{l+1}\) for all \(l=0\dots\infty\). We have left out any dependency in \(\varphi\) since the exciting field is invariant in this angular direction. If this would not have been the case, we should have use spherical harmonics \(Y_{lm}\left(\theta,\varphi\right)\). The index \(m\) then denotes an angular variation of \(\exp\left(\mathrm{i}m\varphi\right)\). A constant function in \(\varphi\), as in our case, is thus described by \(m=0\) and the \(Y_{l0}\) should be related to the \(P_{l}\) by a constant factor which is indeed the case.

The overall potential will be a **superposition** of the exciting potential and the potential of the sphere. Furthermore, inside the sphere, the field should be **finite** allowing only solutions in \(r^{l}\) whereas the same condition leads to the solutions with \(r^{-l-1}\) outside of it. Hence we make the following ansatz for the potentials:\[\begin{eqnarray*}\phi_{\mathrm{in}}\left(r,\theta\right) & = & \sum_{l=0}^{\infty}A_{l}r^{l}P_{l}\left(\cos\theta\right)\\ \phi_{\mathrm{out}}\left(r,\theta\right) & = & \underbrace{-E_{0}rP_{1}\left(\cos\theta\right)}_{\mathrm{exciting\ field}}\\ & & +\sum_{l=0}^{\infty}C_{l}P_{l}\left(\cos\theta\right)/r^{l+1}\ .\end{eqnarray*}\]

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## Employing the Boundary Conditions

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The \(A_{l}\) and \(C_{l}\) are **constants** we have to find employing the boundary conditions. Also, \(z=r\cos\theta\equiv rP_{1}\left(\cos\theta\right)\) and since \(\mathbf{E}_{\mathrm{exc}}\left(\mathbf{r}\right)=E_{0}\mathbf{e}_{z}=-\nabla\phi_{\mathrm{exc}}\left(\mathbf{r}\right)\) lead us to the given form of the exciting potential. The regions in- and outside of the sphere have been denoted by the subscripts “in” and “out”, respectively, which will hold also for the rest of the solution.

Now we have to use the **boundary conditions**. From \(\nabla\times\mathbf{E}\left(\mathbf{r}\right)=0\) we know that the **tangential** part of the electric field is continuous whereas \(\nabla\cdot\mathbf{D}\left(\mathbf{r}\right)=0\) implies that the **normal** part of \(\mathbf{D}\left(\mathbf{r}\right)\) is continuous, see also the figure on the right. Along with the **constitutive relation** \(\mathbf{D}\left(\mathbf{r}\right)=\varepsilon\left(\mathbf{r}\right)\mathbf{E}\left(\mathbf{r}\right)\) results in the continuity of \(\varepsilon\left(\mathbf{r}\right)\mathbf{E}\left(\mathbf{r}\right)\). Using \(\mathbf{E}\left(\mathbf{r}\right)=-\nabla\phi\left(\mathbf{r}\right)\), we find\[\begin{eqnarray}-\frac{1}{r}\partial_{\theta}\phi_{\mathrm{in}}\left(r=r_{s},\theta\right) & = & -\frac{1}{r}\partial_{\theta}\phi_{\mathrm{out}}\left(r=r_{s},\theta\right)\nonumber \\-\varepsilon_{\mathrm{i}}\partial_{r}\phi_{\mathrm{in}}\left(r=r_{s},\theta\right) & = & -\varepsilon_{\mathrm{o}}\partial_{r} \phi_{\mathrm{out}}\left(r=r_{s},\theta\right)\ .\label{eq:Boundary_equations}\end{eqnarray}\]Then, using\[P_{l}^{1}\left(\cos\theta\right) = \frac{dP_{l}\left(\cos\theta\right)}{d\theta}\ ,\]we get\[\begin{eqnarray*}\partial_{\theta}\phi_{\mathrm{in}}\left(r=r_{s},\theta\right) & = & \sum_{l=1}^{\infty}A_{l}r_{s}^{l}P_{l}^{1}\left(\cos\theta\right)\\& \overset{!}{=} & -E_{0}r_{s}P_{1}^{1}\left(\cos\theta\right)\\& & +\sum_{l=1}^{\infty}C_{l}P_{l}^{1}\left(\cos\theta\right)/r_{s}^{l+1}\ .\end{eqnarray*}\]The summation can start at \(l=1\) since \(P_{0}^{1}\left(x\right)=0\). Comparing the different series and using the **orthogonality** of the associated Legendre polynomials \(P_{l}^{m}\), we derive the relations\[\begin{eqnarray*}A_{1}r_{s} & = & -E_{0}r_{s}+\frac{C_{1}}{r_{s}^{2}}\\A_{n}r_{s}^{2n+1} & = & C_{n}\ ,\ n\geq2\ .\end{eqnarray*}\]Employing the second boundary condition\[\begin{eqnarray*}\varepsilon_{\mathrm{in}}\partial_{r}\phi_{\mathrm{in}}\left(r=r_{s},\theta\right) & = & \varepsilon_{\mathrm{in}}\sum_{l=1}^{\infty}l\cdot A_{l}r_{s}^{l-1}P_{l}\left(\cos\theta\right)\\& \overset{!}{=} & -\varepsilon_{\mathrm{out}}E_{0}P_{1}\left(\cos\theta\right)\\ & & -\varepsilon_{\mathrm{out}}\sum_{l=0}^{\infty}\left(l+1\right)\cdot C_{l}P_{l}\left(\cos\theta\right)/r_{s}^{l+2}\ ,\end{eqnarray*}\]we can further conclude that\[\begin{eqnarray*}0 & = & C_{0}\ ,\\ \varepsilon_{\mathrm{in}}\cdot A_{1} & = & -\varepsilon_{\mathrm{out}}\cdot\left(E_{0}+2C_{1}/r_{s}^{3}\right)\ \mathrm{and}\\ \varepsilon_{\mathrm{in}}nA_{n}r_{s}^{2n+1} & = & -\varepsilon_{\mathrm{out}}\left(n+1\right)C_{n}\ \mathrm{for}\ n\geq2\ .\end{eqnarray*}\]This system of equations can only hold if \(A_{n}=C_{n}=0\) for all \(n\neq1\).

The remaining part can be formulated as \[\left(\begin{array}{cc} r_{s}^{3} & -1\\ \frac{\varepsilon_{\mathrm{in}}}{\varepsilon_{\mathrm{out}}}r_{s}^{3} & 2 \end{array}\right)\left(\begin{array}{c} A_{1}\\ C_{1} \end{array}\right) = \left(\begin{array}{c} -E_{0}r_{s}^{3}\\ -E_{0}r_{s}^{3} \end{array}\right)\ .\] This matrix-like representation seems a bit over the top for our task. This form, however, can be very useful if we want to **generalize** our approach to **fields of arbitrary form**. We can solve the latter equation to find\[\begin{eqnarray*}A_{1} & = & -\frac{3E_{0}}{2+\varepsilon_{\mathrm{in}}/\varepsilon_{\mathrm{out}}}\\C_{1} & = & r_{s}^{3}E_{0}\frac{\varepsilon_{\mathrm{in}}-\varepsilon_{\mathrm{out}}}{\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}}\ .\end{eqnarray*}\]

## The Resulting Potential and Polarizability

We can put our findings now into the expression for the **potential**: \[\begin{eqnarray*}\phi_{\mathrm{in}}\left(r,\theta\right) & = & -\frac{3E_{0}}{2+\varepsilon_{\mathrm{in}}/\varepsilon_{\mathrm{out}}}\cdot rP_{1}\left(\cos\theta\right)\\ \phi_{\mathrm{out}}\left(r,\theta\right) & = & -E_{0}rP_{1}\left(\cos\theta\right)\\ & & +E_{0}\frac{\varepsilon_{\mathrm{in}}-\varepsilon_{\mathrm{out}}}{\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}}\cdot\frac{r_{s}^{3}}{r^{2}}P_{1}\left(\cos\theta\right)\ . \end{eqnarray*}\]Now, what is the **polarizability** \(\alpha\) of the sphere? Usually, this quantity is related to the dipole moment of an object subject to a constant electric field via \(\mathbf{p}=\alpha\mathbf{D}\). So, the only thing left to do is to relate our result do a **dipole moment** \(\mathbf{p}\) in the right units. The electrostatic potential of a **dipole** is given by \[\phi_{\mathrm{dipole}}\left(\mathbf{r}\right) = \frac{1}{4\pi\varepsilon_{0}\varepsilon_{out}}\cdot\frac{\mathbf{p}\cdot\mathbf{r}}{r^{3}}\ .\]We have to compare this dipole potential to the part of the sphere in \(\phi_{\mathrm{out}}\left(r,\theta\right)\). Since \(\mathbf{p}\cdot\mathbf{r}=prP_{1}\left(\cos\theta^{\prime}\right)\) with \(\theta^{\prime}\) being the angle between \(\mathbf{p}\) and \(\mathbf{r}\),\[\begin{eqnarray*}\phi_{\mathrm{out}}^{\mathrm{sphere}}\left(r,\theta\right) & = & E_{0}\frac{\varepsilon_{\mathrm{in}}-\varepsilon_{\mathrm{out}}}{\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}}\cdot\frac{r_{s}^{3}}{r^{2}}P_{1}\left(\cos\theta\right)\\& \overset{!}{=} & \frac{1}{4\pi\varepsilon_{0}\varepsilon_{\mathrm{out}}}\cdot\frac{\mathbf{p}\cdot\mathbf{r}=prP_{1}\left(\cos\theta^{\prime}\right)}{r^{3}}\ .\end{eqnarray*}\]So, we can conclude that \(\theta=\theta^{\prime}\) and also \(\mathbf{p}=p\mathbf{e}_{z}\) for our problem. In general, the polarizability of a sphere should not depend on the direction of the exciting field and we can write the result directly as\[\begin{eqnarray*}\mathbf{p} & = & 4\pi r_{s}^{3}\cdot\varepsilon_{0}\varepsilon_{\mathrm{out}}\cdot\frac{\varepsilon_{\mathrm{in}}-\varepsilon_{\mathrm{out}}}{\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}}\mathbf{E}_{0}\\ & \equiv & \varepsilon_{0}\varepsilon_{\mathrm{out}}\cdot\alpha\mathbf{E}_{0}\ \mathrm{with}\\ \alpha & = & 4\pi r_{s}^{3}\cdot\frac{\varepsilon_{\mathrm{in}}-\varepsilon_{\mathrm{out}}}{\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}}\ . \end{eqnarray*} \]

## The Fröhlich Condition and Cloaking Spheres

We have found a very **important result** that is at the heart of nanophotonics. If the denominator of \(\alpha\) vanishes, the system is said to have a **dipole resonance**. Then, the induced field of the sphere goes to infinity. For real media, however, one has to take into account also that the permittivity of the sphere can be a complex number implying losses and limiting the achievable induced field. Then, one defines the resonances by the minimum of \(\left|\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}\right|\) which becomes for low \(\mathfrak{I}\left(\varepsilon_{\mathrm{in}}\right)\) approximately \[\mathfrak{Re}\left(\varepsilon_{\mathrm{in}}+2\varepsilon_{\mathrm{out}}\right) = 0\ .\]The latter equation is called the **Fröhlich condition**.

Using this result we can also explain the different polarization directions from the publication we discussed earlier in the background. If the sphere is a **metal** with \(-2\varepsilon_{\mathrm{out}}<\varepsilon_{\mathrm{in}}<-\varepsilon_{\mathrm{out}}\), the polarizability has a different sign than if it was a **dielectric** one with \(\varepsilon_{\mathrm{in}}>\varepsilon_{\mathrm{out}}\), for example for some glass in air. That means that the polarization fields of metallic and dielectric spheres, if arranged accordingly, may cancel each other - the dielectric sphere is **cloaked** as in “Cloaking dielectric spherical objects by a shell of metallic nanoparticles”. Naturally, losses in real metals and higher-order couplings will limit the effectivity of the cloak.