How is the physical angle between two complex electromagnetic fields defined? If I have two electromagnetic field vectors, which are in general complex, how do I sensibly define an angle between these two fields?
Some books refer to the angle between electromagnetic fields, especially in crystal optics. How is this angle defined for in general complex fields like
$$
\vec{E}=E\hat{e}\exp(i(\vec{k}\vec{r}-\omega t))
$$
$$
\vec{D}=D\hat{d}\exp(i(\vec{k}\vec{r}-\omega t))
$$
where $\hat{d}, \hat{e}\in\mathbb{C^3}$ for elliptical polarization. I encountered this problem when studying normal modes in crystals.
 A: While the question seems simple it deserves some discussion:
In dealing with electrical fields one often uses complex extensions $\vec{E}, \vec{H} \in \mathbb{C}^3$  of the real physical fields $\vec{E}_r=\Re(\vec{E}), \vec{H}_r =\Re(\vec{H}) \in \mathbb{R}^3$.
The physically sensible angle is between the two real field vectors, not between the two complex fields.
$$
\angle(\vec{E_r},\vec{H_r})=\arccos\left(\frac{\vec{E_r}\cdot\vec{H_r}}{|\vec{E_r}||\vec{H_r}|}\right)
$$
One must always be careful when taking the product between two generalized complex field: The extension of the field only makes sense as long everything is linear for the fields, such that the real physical field $\Re(\vec{E})$ and the auxialliary fields $\Im(\vec{E})$ don't mix. If one has produtcts the two fields start to mix:
$$
\vec{E}_r\cdot\vec{H}_r\neq\Re(\vec{E}\cdot\vec{H}) \quad \quad \vec{E}_r\times\vec{H}_r\neq\Re(\vec{E}\times\vec{H})
$$
As the angle between two fields is defined by a scalar product, finding it requires caution!
The mathematical definition of an angle between two complex field vectors
$$
\angle(\vec{E_r},\vec{H_r})\neq\angle(\vec{E},\vec{H})=\arccos\left(\frac{\Re(\vec{E}\cdot\vec{H})}{|\vec{E}||\vec{H}|}\right)
$$
does not coincide with the angle between the two real field vectors. In conflict to the mathematical defintion many authors in the optics literature use $\angle(\vec{E},\vec{H})$ and $\angle(\vec{E_r},\vec{H_r})$ interchangably.
This might come up as a problem when deriving that for normal modes in a uniaxial crystal $\angle(\vec{S},\vec{k})=\angle(\vec{E},\vec{D})$ and is seldomly discussed.
A: The definition for $\vec{H} = \frac{\vec{B}}{\mu_0}-\vec{M}$, where $\vec{M}$ is the magnitization vector.
Since the angle between any two vectors $\vec{a}$ and $\vec{b}$ is given by:
$$
\cos(\theta)= \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}
$$
It follows that the angle between $\vec{B}$ and $\vec{H}$ is:
$$
\theta=\cos^{-1}\left( \frac{\vec{B}\cdot\vec{H}}{|\vec{B}||\vec{H}|}
 \right)
$$
$$
\vec{B}\cdot\vec{H}=\frac{|\vec{B}|^2}{\mu_0}+\vec{B}\cdot\vec{M}
$$
In a vacuum, since $\vec{M}=\vec{0}$, $\theta=0$.
Hope this helps!
A: $\vec{B}$ and $\vec{H}$ are both vectors. As such, they have magnitude and direction. One way to find the angle between them is to calculate $\vec{H}\cdot \vec{B}/(|\vec{H}||\vec{B}|)$ then take the inverse cosine. That gives you the angle between them.
There might be other ways to find it with more details.
