The Green's function can be used to find the electric field or electrostatic potential for arbitrary **charge distributions**. You will learn that the **polarization** of a **dipole** in front of a flat metallic surface enormously affects the overall field.

## Problem Statement

Use the **Green's function** to calculate the electrostatic potential 𝜙*( r)* for a dipole

*at*

**p***in front of a flat metallic surface terminated at*

**r**_{D}*z = 0*.

Determine the main contributions to the potential for |* r_{D}*| ≪ |

*|.*

**r**Which contribution has a stronger fall-of: x- or z-**polarization**? Can you explain this in simple words?

## Hints

What is the **Green's function** *G( r,r')* of a flat metallic surface? Remember the mirror charge technique (grounded metallic corner problem) and that the Green's function is basically just the field of a point charge at

*. Then,*

**r**'\[\phi\left(\mathbf{r}\right)=\int G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}\]

The charge distribution of a mathematical dipole is formally given by

\[\rho_{D}\left(\mathbf{r}\right)=-\mathbf{p}\cdot\nabla\delta\left(\mathbf{r}-\mathbf{r}_{D}\right)\]

A **Taylor expansion** can be a quite useful tool for approximations and also outlined in the dipole calculations.

Let us find the electric field for a dipole in front of a flat metallic surface!

## The Electrostatic Potential

Let us write down the **Green's function** for the given system. Of course we know that for a single charge at

\[\mathbf{r}^{\prime} = \left(\begin{array}{c}x^{\prime}\\y^{\prime}\\z^{\prime}\end{array}\right)\ ,\]

the **mirror charge** would have to be placed at

\[\mathbf{r}_{m}^{\prime} = \left(\begin{array}{c}x^{\prime}\\y^{\prime}\\{\color{red}-}z^{\prime}\end{array}\right)\ .\]

Then, the Green's function is given by

\[G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = \frac{1}{4\pi\varepsilon_{0}}\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}-\frac{1}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|}\right)\ .\]

To find the **electrostatic potential** for the dipole, we have to use the charge distribution of it:

\[\rho_{D}\left(\mathbf{r}\right) = -\mathbf{p}\cdot\nabla\delta\left(\mathbf{r}-\mathbf{r}_{D}\right)\ .\]

We obtained this formula in "The Electric Field of a Dipole" , if you want to check the expression. So, we use the solution of the **Poisson equation** to formally find

\[\begin{eqnarray*} \phi\left(\mathbf{r}\right)&=&\int G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}\\&=&-\mathbf{p}\cdot\int G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\nabla_{\mathbf{r}^{\prime}}\delta\left(\mathbf{r}^{\prime}-\mathbf{r}_{D}\right)dV^{\prime}\\&=&\mathbf{p}\cdot\int\left[\nabla_{\mathbf{r}^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\right]\delta\left(\mathbf{r}^{\prime}-\mathbf{r}_{D}\right)dV^{\prime} \end{eqnarray*}\]

by partial integration. We now have to determine the **gradient** of the Green's function with respect to its second variable! First for the *x* component:

\[\begin{eqnarray*} \frac{\partial}{\partial x^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right)&=&\frac{1}{4\pi\varepsilon_{0}}\frac{\partial}{\partial x^{\prime}}\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}-\frac{1}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|}\right)\\&=&\frac{1}{4\pi\varepsilon_{0}}\left(\frac{x-x^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}-\frac{x-x^{\prime}}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|^{3}}\right) \end{eqnarray*}\]

and likewise for the *y* component. Note that we have used

\[\left|\mathbf{r}-\mathbf{r}^{\prime}\right|=\sqrt{\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}+\left(z-z^{\prime}\right)^{2}}\]

and that we have **differentiated** with respect to the **second variable** of the Green's function! Now for the *z* component:

\[\frac{\partial}{\partial z^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = \frac{1}{4\pi\varepsilon_{0}}\left(\frac{z-z^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}{\color{red}+}\frac{z-z^{\prime}}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|^{3}}\right)\ .\]

We can already see the **difference** between both components of the Green's function's gradient: a **sign change** (denoted red). Now we put these components of the gradient into the solution of the Poisson equation and perform the integration:

\[\begin{eqnarray*} \phi\left(\mathbf{r}\right)&=&\mathbf{p}\cdot\int\left[\nabla_{\mathbf{r}^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\right]\delta\left(\mathbf{r}^{\prime}-\mathbf{r}_{D}\right)dV^{\prime}\\&=&\frac{1}{4\pi\varepsilon_{0}}\mathbf{p}\cdot\left(\begin{array}{c}\left(x-x_{D}\right)\left[\left|\mathbf{r}-\mathbf{r}_{D}\right|^{-3}-\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{-3}\right]\\\left(y-y_{D}\right)\left[\left|\mathbf{r}-\mathbf{r}_{D}\right|^{-3}-\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{-3}\right]\\\left(z-z_{D}^{\prime}\right)\left[\left|\mathbf{r}-\mathbf{r}_{D}\right|^{-3}+\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{-3}\right]\end{array}\right)\ \text{with}\\\mathbf{r}_{mD}&=&\left(\begin{array}{c}x_{D}\\y_{D}\\{\color{red}-}z_{D}\end{array}\right)\ . \end{eqnarray*}\]

This is the **solution** of the electrostatic potential of a dipole in front of a flat metallic surface. We can directly see that it corresponds to a two-dipole-solution: a mirror-dipole appears at * r_{mD}* (see figure).

Now let us try to **understand the potential** a little further

## Taylor Expansion of the Potential

From a heuristic point of view it is clear that the dipole in *z* direction causes a mirror dipole pointing in the same direction. The *x*-polarization, parallel to the surface, corresponds to dipoles with opposite direction. Hence, the z-polarization should correspond to an overall dipolar field and the *x* polarization to a quadrupolar one. This is what we should find out.

Far away from the sources for |* r_{D}*| ≪ |

*|, we may*

**r****Taylor-expand**the denominators in the potential. The terms look rather similar, so we just have to calculate:

\[\frac{1}{\left|\mathbf{r}-\mathbf{r}_{D}\right|^{3}} \approx \frac{1}{\left|\mathbf{r}\right|^{3}}+\frac{3\mathbf{r}\cdot\mathbf{r}_{D}}{\left|\mathbf{r}\right|^{5}}\]

and analog for * r_{mD}*, where the

*z*component switches the sign. This is the

**important difference**and we find

\[\begin{eqnarray*} \frac{1}{\left|\mathbf{r}-\mathbf{r}_{D}\right|^{3}}-\frac{1}{\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{3}}&\approx&\frac{6z\, z_{D}}{\left|\mathbf{r}\right|^{5}}\ \text{and}\\\frac{1}{\left|\mathbf{r}-\mathbf{r}_{D}\right|^{3}}+\frac{1}{\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{3}}&\approx&\frac{2}{\left|\mathbf{r}\right|^{3}}+\frac{6\left(x\, x_{D}+y\, y_{D}\right)}{\left|\mathbf{r}\right|^{5}}\ .\end{eqnarray*}\]

In this far-field approximation, the potential reads as

\[\phi\left(\mathbf{r}\right) \approx \frac{1}{4\pi\varepsilon_{0}}\mathbf{p}\cdot\left(\begin{array}{c}6x\, z\, z_{D}/\left|\mathbf{r}\right|^{5}\\6y\, z\, z_{D}/\left|\mathbf{r}\right|^{5}\\2z/\left|\mathbf{r}\right|^{3}+6z\left(x\, x_{D}+y\, y_{D}\right)/\left|\mathbf{r}\right|^{5}\end{array}\right)\]

and we have shown that the potential has a stronger fall-off behaviour for the *x* and *y* polarizations which goes effectively like \(\propto Q_{ij}x^{i}x^{j}/r^{5}\).

On the other hand the *z* polarization behaves as \(\propto p_{z}z/r^{3}\).

The figure below illustrates both cases:

These potentials correspond to the quadrupolar and dipolar fields. We will find out later that oscillating dipoles radiate much faster than quadrupoles.

This is the reason why a dipole polarized parallel to the surface will basically heat the metal whereas the perpendicula polarization has a chance to escape the system. The polarization has thus a **dramatic effect** on the effectivity of emitter/metal-plate-systems.

Check again the multipole moment section for more information on multipole moments.

## Background: Dipole Emitters in Front of Flat Metallic Surfaces

In this problem, we discuss the electrostatic dipole in the vicinity of a flat metal which is of huge interest both in **applied** and **fundamental physics**: think of a dipole **antenna** with a metal backplate or an emitting **molecule**. Of course such systems demand an understanding of full electrodynamics. The point is, however, that the **main features** will remain; *x*- and *z*-polarization will radiate differently.

It gets even more interesting if we do **not** have a **perfect metal**. Then, the electric field can penetrate the conductor and a closely placed dipole can excite surface waves as predicted by **Sommerfeld** more than one hundred years ago.

Even though this effect was not the explanation for long-range radio transmissions, the theory of Sommerfeld is still at the core of **plasmonics**, the field that deals with these so-called surface plasmon polaritons. For much **more information** on dipoles in front of flat metallic surfaces please have a look at Lukas Novotny's freely available book chapter on the topic.