Dipole in a spherical cavity in an infinite dielectric

Question

In this example the assumption is made that inside a spherical cavity (with an infinite dielectric) with a dipole placed at its centre, the potential takes the following form: $$\phi_{in}= \frac{p cos(\theta)}{r^2}-E_{in}r \cos(\theta)$$ and $$\phi_{out}=\frac{p' cos(\theta)}{r^2}-E_{\infty}r cos(\theta)$$ I cannot see the justification for these forms of the potentials, so please can someone explain it to me?

I am specifically uncertain on:

1. As $r\rightarrow 0$ , why $\phi_{in} \rightarrow \frac{p cos(\theta)}{r^2}$ when you should have from polarization charges on the surface of the cavity and also from $E_\infty$.

2. Why can we assume the form of the first term in $\phi_{out}$?

Edit

I found this solution to Griffiths problem 3.34, which answers this question expect for the point about when $r \rightarrow 0$ the potential is only that of the dipole. I still do not understand why this has to be, you are inducing charges that are going to produce a field in the sphere and thus will not necessarily give you 0 potential at the centre.

Answer 1

Let me try to offer a different way to see it based on linearity. Note that your two links are reversed in the situation they consider; your second link has the dipole embedded in the dielectric embedded in vacuum, whereas your first has the dipole in a vacuum embedded in a dielectric. So, equation (3) contains an $\epsilon$ where it should have $\epsilon_0.$ The process I'm asking you to visualize looks like this:

1. Freeze the polarization field.
2. Remove the dipole charge and its vacuum field.
3. Examine what the remaining field must be doing at $r = 0$.
4. Re-insert the dipole charge using the principle of superposition.

During step (3) we know that within the cavity we are solving Laplace's equation due to the bound charges on the sphere -- but the solution needs to be continuous at $r=0$ because there is nothing there! We know that the general solution is: $$ V(r, \theta) = \sum_{\ell=0}^\infty \left[A_\ell~ r^\ell + B_\ell ~ r^{-\ell - 1}\right]~P_\ell(\cos\theta).$$ Continuity is obviously broken if any $B_\ell \ne 0$ as a singularity is not continuous.

Therefore during step 4 when we use the principle of superposition to re-insert the dipole, the voltage inside the cavity due to the surface charge must be simply: $$ V(r, \theta) = \frac {p \cos\theta}{r^2} + \sum_{\ell=0}^\infty A_\ell~ r^\ell ~P_\ell(\cos\theta).$$

Furthermore, as your second link discovers, the fact that $V$ and $\vec D \cdot \hat r$ are continuous across the boundary means that essentially the voltage fields induced by the contained moments must have the same $\theta$-dependence as the moments themselves: quadrupole moments will induce surface charges characteristic of quadrupole moment fields just the same as a dipole moment inducing a surface charge characteristic of the dipole moment field.

Answer 2

In general, no justification needs to be made - we can assume whatever we want. If a solution which satisfies both the equations and the boundary conditions is found, it is the only solution. By focusing on a subset of all possible solutions, we can make solving the problem much easier.

Now, as we have "guessed" the form of solution (so that the equations are satisfied), we need to find constants so that the boundary terms are also satisfied.

1. As $r\to 0$, the second term, resulting from the homogeneous field, is zero, since this is how the homogeneous field potential is defined. This leaves us with the dipole potential, which is exactly the term you wrote, $\frac{pcos(\theta)}{r^2}$. Polarization charges on the surface of the cavity are still present, but they are caused by this form of potential, and are in agreement with it.

2. Say we guessed the solution inside the cavity. Now, for the solution outside, $\phi_{out}$, the first term must be there to provide a solution to the boundary conditions at $r=R$. To satisfy both conditions, you must have a term proportional to $cos(\theta)/r^{2}$, or only one (or none) of the boundary conditions can be satisfied.

Answer 3

So I am not aware whether you have regarded the previous answer properly or you know about this, but there is a uniqueness theorem for the solution of the Poisson equation. Thus this solution is already correct because you have solved the boundary problem:
You have found a solution for the Laplace equation in the area where no free charges exist ($\mathbb{R}^3\backslash\{0\}$), namely seperately for the cavity $\Omega$ and its outside $\mathbb{R}^3\backslash\{\Omega\}$. You can superpose the dipole potential upon this, since the Poisson/Laplace equation are linear. The boundary here is the border of the cavity $\partial\Omega$ and you impose Neumann boundary conditions according to the behaviour of the electric (displacement) field on the boundaries of dielectrics. You can then formally find the coefficients by using the orthonormality of the Legendre polynomials, which is the function that is the general solution to the radially symmetric Laplace equation (as already stated in the solution). With the uniqueness theorem, the problem is solved.

So your objection appears to boil down to the fact that you doubt that the external field and polarisation are not being taken into account at the center of the cavity. Thus I might add 2 remarks:

1. When you're at the origin, the external potential doesn't matter since the potential diverges anyway. In a way the origin is an irregular, singular point. Solving the Poisson equation rigorously yields an additional delta at the origin, and being an irregular distribution this really makes the solution of this equation miss out the origin, strictly speaking.
2. Trying to explain the field component by arguing over collective polarisation components seems unreasonable to me. The dipole-dipole interaction is not really trivial, every induced dipole can and will contribute to other dipoles and finally the central dipole also influences the induced charges. Also mind the form of a dipole field which is not radially symmetric but also has a polar angular dependency.

Due to these points, I do not think that it is helpful to depart from just solving the boundary condition problem. Also note the gauge freedom of potential, the electric field is only determined by the spatial variation of the potential (static case).

Answer 4

TL;DR: The external electric field through the dielectric causes small displacements on the bound charges of the dielectric. These accumulate as an induced surface charge on the dielectric/vacuum interface. The total electric potential is then given as a sum of the external potential, the induced potential of the induced point-like dipole, and the induced potential of the induced surface charge at the dielectric/vacuum interface as given by the multipole rules governing the Poisson equation.

The longer explanation:

I think I can answer the question where the forms of these potentials come from, but be slightly cautious (especially on nomenclature) as by background is in Density Functional Theory. We solve electrostatic equations like this constantly (Hartree-potential) in a polarizable medium (electrons), but in our world, everything is microscopic and nicely continuous.

As the example PDF assumes, there are no charges inside the cavity $0 < r < R$, where the Eq. (1) is fulfilled. And it also is true that there are no free charges outside the cavity, r>R and the Eq. (1) is also fulfilled. What particular detail it omits (which might lead to confusion) is that there will be induced charge shell at $r=R$. The differential equation can be solved with mathematical constraints, but I think it is crucial for the physical picture to understand the induced charges.

When there are extra charges in the system, the potential can be solved using following equation, so called Poisson equation

$$ \nabla^2 \phi = n,$$

where n is the total charge density. This can be divided into external, and induced charge densities

$$ \nabla^2 \phi = n_{ext} + n_{ind}. $$

The induced charges are due to the bound charges in the dielectric medium polarizing (i.e. moving away from their neutralizing background charge), or due to the dipole moment of the point-like object. In the case of the surface charge, the width of this charge will be infinitesimal, but the charge density on this charged shell diverges too to keep everything in control. First, let's verify that this is indeed the case. Let's take the left hand side of the Poisson equation (the Laplace operator) and plug in the already solved potential. I will omit LaTeX, and do this in Maple, as it gets rather tedious.

First, let's define the potential they solved

> F:=piecewise(r<R, p*cos(theta)/r^2 - 1/(1+2*eps)*(2*(eps-1)*p/R^3+3*eps*Einf)*r*cos(theta), r>=R, 1/(1+2*eps)*(3*p-(eps-1)*R^3*Einf)*cos(theta)/r^2-Einf*r*cos(theta));

  {                /2 (eps - 1) p             \
  {                |------------- + 3 eps Einf| r cos(theta)
  {                |      3                   |
  { p cos(theta)   \     R                    /
  { ------------ - -----------------------------------------        r < R
  {       2                        1 + 2 eps
  {      r
  {
  {                   3
  { (3 p - (eps - 1) R  Einf) cos(theta)
  { ------------------------------------ - Einf r cos(theta)        R <= r
  {                         2
  {            (1 + 2 eps) r

We can verify that the potential is continuous at R.

> simplify(limit(F,r=R,right)-limit(F,r=R,left));
                                                                 0

Now we can apply the Laplacian operator (in spherical coordinates) to the potential.

> simplify(convert(1/r*diff(r*F,r$2)+1/(r^2*sin(theta))*diff(sin(theta)*diff(phi,theta), theta), piecewise,r));
                                                  {     0            r~ < R~
                                                  {
                                                  { undefined        r~ = R~
                                                  {
                                                  {     0            R~ < r~

We indeed find, that the induced density is zero in the regions described by the example, and Eq. (1) can be used. However, Maple cannot immediately do the $r=R$ case. This is because already the gradient of this potential is discontinuous. Let's see what the discontinuity in the gradient looks like. There is no $\phi$ dependence, so gradient discontinuity to $\phi$ direction will be 0. There is $\theta$ dependence, but we get:

> limit(diff(phi,theta)/r,r=R,right)-limit(diff(F,theta)/r,r=R,left);
                                                              0

which was one of the boundary conditions given by the example, and for gradient discontinuity in $r$ direction we get

> simplify(limit(diff(F,r),r=R,right)-limit(diff(F,r),r=R,left));
                                        3
                 3 cos(theta~) (Einf~ R~  + 2 p~) (eps~ - 1)
                 -------------------------------------------
                                             3
                              (1 + 2 eps~) R~

Note, that this also has the $\cos(\theta)$ symmetry. Indeed, there is a discontinuity in the gradient of the potential i.e. the electric field. Discontinuity in the radial component of the electric field means that there is surface charge at $r=R$.

Now, that we understand the form of induced charge in the system, understanding the forms of chosen potentials is easier. If you look at this link, you will see how you can write the electrostatic potential as an integral over the Coulomb kernel $\frac{1}{|r-r'|}$. There exists a particular expansion for this potential in spherical symmetry, given with r< and r> terms, where $r\leq \min(r,r')$ and $r\geq \max(r,r')$. $Y_lm$ are the spherical harmonic functions, and all we need to know that $A\cdot\cos(\theta)$ is one of the $l=1, m=0$ spherical harmonic, and these functions are orthogonal. Therefore the angular integrals all vanish except the $l=1,m=0$ term, where it yields a constant (omitted, and so is the angular part of the integral for brevity).

Now, finally we can get to the forms of these potentials. First, let's look at the case r>R and r'<=R. This means, let's look at the potential outside the sphere, induced by charge inside the sphere (and the surface).

We can write

$$ \phi_{out,due~in}(r) = \frac1{r^2} \cos(\theta) \int_{0}^{R+} dr' r' n_{ind}(r') $$.

The integral integrates the 'dipole moment' of the density inside the sphere. And since we multiply this integrated constant by 1/r^2 cos(theta), it shows outside the sphere exactly the same way as a point like dipole. The integral of dipole moment, $p_{out}$ will be sum of the point dipole dipole moment at the center, $p$, plus the induced dipole moment on the dielectric surface $p_{ind}$:

$$ \int_{0}^{R+} dr' r' n_{ind}(r') = p_{out} = p + p_{ind}. $$

Let's then analyse the case where r>R, r'>R, but r' is very large. We can write

$$ \phi_{out,due~far~away} = r \cos(\theta) \int_{large}^{\infty} dr' n(r') / r'^2. $$

We see that the linear electric field in the problem, E_inf, could be pictured to be cause by an external charge density with $cos(\theta)$ symmetry far away, integrating to $E_{\infty} = \int_{large}^{\infty} dr' n(r') / r'^2$.

Let's then analyse the case where $r'=0$, and $0<r<R$. We have the point like dipole at $r'=0$, which can be described if we allow distribution like induced density.

$$ \phi_{in,due~dipole} = \frac{1}{r^2} \cos(\theta) p $$,

thus we get what we expect. The dipole moment integral over a dipole moment yields naturally the dipole moment.

And then finally, let's analyse the case where $0<r<R$, and $r'>R$.

$$ \phi_{in,due~out} = r \cos(\theta) \int_{R}^{\infty} dr' n_{ind} / r^2 $$.

Here the integral $\int dr' n_{ind} / r^2$, gives $E_{in}$.

The Poisson equation is additive, so we can sum the in due dipole and in due out, to get

$$ \phi_{in} = \frac{1}{r^2} \cos(\theta) p + r \cos(\theta) E_{in} $$

and

$$ \phi_{out} = r \cos(\theta) E_{\infty} + \frac{1}{r^2} \cos(\theta) (p + p_{ind}) $$.

This concludes, why the potentials in and out, look the way they look.

What we actually have come up against, is the powerful theory of multipoles in a Laplacian equation. Given a Laplace equation, we notice, that if we know certain integrals of charge density shells with domain outside of our own shell, and certain integrals of the inside spherical shells of our location (called multipole moments), we can solve the potential at our location.

And one more remainder, that the integrals I present are only sketched to give the idea, they miss factors and part of angular integrals etc.