Determining H from E (or vice-versa)

Now that we have both $\nabla \times$ and $\times$ operations defined on our fields we can go from one to the other (e.g. from E to H) using those operations. Given the stationary Maxwell Equations previously mentioned

$\begin{eqnarray} \label{eq:E} \nabla \times E &=& \frac{-i\omega}{c}\mu \cdot H \\ \nabla \times H &=&\frac{i\omega}{c}\epsilon \cdot E \end{eqnarray}$
It follows that

$\begin{eqnarray} \label{eq:curlE} \frac{ic}{\omega} \mu^{-1} \cdot ( \nabla \times E) &=& H \\ \frac{-ic}{\omega} \epsilon^{-1} \cdot ( \nabla \times H) &=& E \end{eqnarray}$

Supposing we had the E field solution to our vector Helmholtz equation. We could determine our H field by computing curl of E and applying some other multiplicative operations on E. However, the derivation of one field from another is complicated by the $\mu^{-1}$ and $\epsilon^{-1}$ on the left hand side of (3) and (4). These quantities are tensors that can vary from element to element. Fortunately, Curly3d has the ability to track these properties for each element and is able to calculate the required tensor product.
It is also important to point out a relationship that arises in the above equations. By making the following transformation

$\begin{eqnarray} (\epsilon,E) &\Leftrightarrow& (\mu^{*},H^{*}) \\ (\mu, H) &\Leftrightarrow& (\epsilon^{*},E^{*}) \end{eqnarray}$

(1) transforms into (2), also conversely (2) becomes (1). Furthermore, the transformation affects (3) and (4) identically.
Curly3d has modules that treat $\nabla$ 's as $\nabla + ik$ .

Poynting Vectors

With the functionality of computing the H field from the E field and vice-versa we can now determine the Poynting vectors through the unit cell. If we have solved the vector Helmholtz equation for E and we want to find the Poynting vectors, given

$\begin{equation} S \propto E^{*} \times H \end{equation}$

and using (3) we get

$\begin{equation} S \propto E^{*} \times [i \mu^{-1} \cdot ( \nabla \times E)] \end{equation}$

Using the transformations in (5) and (6) on (8) S can be calculated the same way when we have the H field.

Ryan McClarren

Last modified: Thu Aug 15 14:30:35 EDT 2002