Date fre 02 februari 2018

We are given a task to characterize the solution to maxwell's equations that enables a non radiative source term.

We will base our analysis of subset of solution to maxwells equations in free space of the following form and for that we will use the linear mapping:

$$[F(\hat z)]_{lm} = \int_S F(\hat n) exp(i k \hat r \cdot \hat n)Y_{l,m}(\hat n) \exp(i w t) dS(\hat n)$$

Worth noting is that we have the plane wave expanded in spherical harmoics as:

$$\exp(i k \hat r \cdot \hat n) = 4 \pi \sum_{l=0}^\infty \sum_{m=-l}^{l} i^l j_l(k r) Y_{l,m}^*(\hat n) Y_{l,m}(\hat r)w$$

This and the orthogonality of the spherical hormonics means that

$$[1]_{lm} = 4 \pi i^l j_{l}(k r)Y_{l,m}(r)\exp(iwt)$$

Also because every multiplicatoin of spherical harmonics can be represented as

$$Y_{l',m'}Y_{l,m} = \sum a_{l,m l',m',l'',m''} Y_{l'',m''}$$

we can integrate all those combinations, more specifilly if $$l'' = l_0$$ then we have

$$[Y_{l',m'}]_{l,m} = 4 \pi i^{l_0}j_{l_0}(k r) Y_{l',m'}Y_{l,m} \exp(i w t)$$

More specifically if $$F(\hat n)$$ can be decomposed in a set of linear combinations of spherical harmoics $$Y_{l',m'}$$ so that for all those we have the same $$l_0$$ then

$$[F(\hat n)]_{l,m} = 4 \pi i^{l_0}j_{l_0}(k r) F(\hat r / r) Y_{l,m} \exp(i w t)$$

With some analysis one can show that $$\hat \varphi_{\hat n}$$ and $$\hat \theta_{\hat n}sin(\theta)$$ has coordinates that can be decomposed in spherical harmonics with $$l' = 1$$ and that if we take $$l=0$$ we obviously get

$$[\hat \varphi_{\hat n} \sin(\theta)]_{0,0} = 4 \pi i^{l_0}j_{l_0}(k r) \hat \varphi_{\hat r} \sin(\theta) \exp(i w t)$$

and

$$[\hat \theta_{\hat n}]_{0,0} = 4 \pi i^{l_0}j_{l_0}(k r) \hat \theta_{\hat r}\exp(i w t).$$

Set

$$f_1 = \hat \varphi \sin(\theta) = \hat n \times \hat z$$

and

$$f_2 = \hat \theta \sin(\theta)$$

and

$$\nabla \times [f_1 + i f_2]_{l,m} = i k [f_2 - if_1]_{l,m} = k[f_1 + i f_2]_{k,l}$$

e.g. a force free field.

we now have

$$[f_1]_{l,m} = 4 \pi j_1(k r) f_1 \exp(iwt) = 4 \pi j_1(k r) \hat \varphi \sin(\theta) \exp(iwt)$$

and

$$i k [f_2]_{l,m} = \nabla \times [f_1]_{l,m} = 4 \pi \nabla \times j_1(k r) \varphi \sin(\theta) \exp(iwt)$$

Now

$$\nabla \times j_1(k r) \varphi \sin(\theta) = - \frac{k}{r}(r j_1'(kr) + j_1(kr))\hat \theta \sin(\theta) + \frac{j_1(kr)}{r} \cos(\theta) \hat n$$

So,

$$i k [f_2]_{l,m} = 4 \pi (- \frac{k}{r}(r j_1'(kr) + j_1(kr))\hat \theta \sin(\theta) + \frac{j_1(kr)}{r} \cos(\theta) \hat n) \exp(iwt)$$

A final rule to note is that

$$\frac{\partial}{\partial_i} [F]_{l,m} = i k [n_i F]_{l,m}$$

if $$F$$ is independent of $$\hat r$$,

with this we see that

$$\nabla \times [f_1]_{l,m} = ik [f_2]_{l,m}$$

and

$$\nabla \times [f_2]_{l,m} = -ik [f_2]_{l,m}$$

and

$$\nabla \times \nabla [f_1]_{l,m} = k^2 [f_1]_{l,m}$$

and

$$\nabla \times \nabla [f_2]_{l,m} = k^2 [f_2]_{l,m}.$$

Let's define the scalar potential $$\phi$$, with $$\rhoi$$ constant as

$$\phi = [\rho]_{1,0}$$

and the vector potential, with $$a,b$$ constant as

$$A = [af_1 + bf_2]_{0,0} + [d \hat n]_{1,0}$$

Now assume the Lorenz gauge and the we must fullfill the continuity equation:

$$\nabla \cdot A = -\frac{1}{c^2}\frac{\partial \phi}{\partial t}$$

Which translates to

$$i k [\hat n \cdot (a f_1 + b f_2)]_{0,0} + i k [\hat n \cdot d \hat n]_{1,0} = - \frac{i w}{c^2}[\rho]_{1,0}.$$

$$f_1$$ and $$f_2$$ is orthogonal to $$\hat n$$ and $$\hat n \cdot \hat n = 1$$ so we have:

$$i k d = - \frac{i w}{c^2}$$

Set $$d -> \rho d$$ and we have

$$w = - d k c^2.$$

Furthermore we have that $$\phi$$ shall satisfy the wave equation with the speed $$c$$ so that means:

$$w^2 = c^2 k^2$$

and hence

$$d^2 = 1 / c^2$$

With this we have a solution to maxwells equations in free space. The next step is to introduce source terms at a spherical shell at a certain distance. This is a bit tricky. First we must patch solutions together so that the postential is continuous as we move radially. How to do this? well we note from the discusion about $$f_1,f_2$$ that

$$\phi = j_1(kr) p$$

and

$$A = j_1(kr) P + j_0(k|r|)Q(b f_2)$$

and hence let assume at a certain distance $$r_0$$,

$$j_1(k r_0) = 0$$

This means that according to theory, not only do we know that Maxwell's equations will not radiate, we can take the scalar potentials equal to zero and patch with a new solution for $$r > r_0$$ where wi replace $$j_0(kr)$$ with $$y_0(kr)$$ and assume $$j_1'(k r_0)=h y_1'(k r_0)$$ which imply

$$h = j_1'(k r_0)/y_1'(k r_0)$$

Because $$j_i,y_i$$ both follows the same recurrence equation we coclude that the resulting solution indeed solves Maxwells equations e.g we use

$$\frac{d j_i}{dr} = a j_{i+1} + b j_{i}$$

and

$$\frac{d y_i}{dr} = a y_{i+1} + b y_{i}.$$

We have $$E = - \nabla \phi - \frac{\partial}{\partial t}A$$ that is

$$E = - \nabla E - i w A$$

Now for a change in the interface for $$E$$ over a phase we know that A is continuous and the difference comes from $$\phi$$. Therefore we calculate

$$\nabla \phi = \nabla [\rho]_{1,0} = \nabla 4 \pi i \rho j_1(k r) Y_{1,0}\exp(i w t).$$

The derivative with respect to the spherical harmonics can be neglected because that the bessel function is untouched and hence at $$r_0$$ zero. So:

$$\nabla j_1(k r) = k j_1'(k r) \hat n$$

So over the interface 1.

$$0 = \hat n (E_2 - E_1) = - \hat n \times E_1 = 0$$
$$\sigma_s = \epsilon \hat n \cdot (E_2 - E_1) = - \epsilon \hat n \cdot E_1 = \hat n \cdot \nabla \phi = 4 \pi i \epsilon \rho k j_1'(k r_0) Y_{1,0}\exp(i w t).$$

Turning the focus to the magnetic part we have $$B = \nabla \times A$$ which we can write

$$B = i k [a f_2 - b f_1]_{0,0}$$

Turning over to the boundary condition we see that if we take

$$B_2 = - i h k b [f_1]'_{0,0},$$

where $$[F]_{i,j}'$$ is $$[F]_{i,j}$$ with $$j_i$$ replaced by $$y_i$$. e.g. we assume that the maxwell solution

$$\phi'=0$$

and

$$A'= h n[f_2]'_{0,0}$$

for $$r > r_0$$.  0 = \hat n \cdot (B_2 - B_1) = \hat n \cdot (-ikb[f_1]'_{0,0} - B_1) = C(\hat n \cdot ([f_1]'_{0,0} - [a f_2 - b f_1]_{0,0}) = 0 $$because the radial part has $$j_1(kr_0) = 0$$ as a factor. Furthermore:$$ J = \frac{1}{\mu_0} \hat n_r \times (B2 - B1) $$As said the radial part of $$B1$$ dissapears and left is a form of type$$ C(r_0,t)\hat n \times \hat \theta f(\theta) = - C(r_0,t)\phi f(\theta) $$But$$ - i b k h \hat n \times [f_1]'_{0,0} = b k h \hat n y_1(kr_0) \hat \phi sin(\theta) = b k h y_1(k r_0) sin(theta) \hat \theta $$The divergence of this is$$ \frac{1}{r_0 sin(\theta)}\frac{\partial A_{\theta}}{\partial \theta} = b k h y_1(k r_0) \frac{2}{\mu_0 r_0} cos(\theta)exp(i w t) = b D Y_{1,0} exp(i w t) $$Furthermore for the source terms we have$$ \frac{\partial \sigma_s}{\partial t} + \nabla \cdot J = 0 = i w + \nabla \cdot J 

for a flow on the sphere. Now we can match $$b$$ so that the continuity equation for the source terms are satisfied and we are done. note that a is arbritary in this evaluation and can be taken 0 or we could have design it with the help of complex numbers so that both $$B$$ satisfy $$\nabla \times B = k B$$ and $$\nabla \times B' = k B'$$ e.g. force free.