2
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ Welcome to Physics! Could you provide a bit more detail on how you got to this calculation? While this might initially seem superfluous, knowing a bit more about the context might help us understand what you are trying to do and what is going wrong. For example, what do you mean by "two electromagnetic fields"? What do you mean with the symbols $\vec{B}$ and $\vec{H}$?Are these just the standard magnetic induction and magnetic field? $\endgroup$ Jan 25, 2022 at 21:33
  • $\begingroup$ @AnonPhoton what do you mean a little Q&A mark? Questions should be clear and well defined, if people are asking for clarification it may mean it's unclear what you mean. You should also show any work or research you made in helping to answer your own question (just searching Google for angle between vectors will get you the answer you need right away for instance). $\endgroup$
    – Triatticus
    Jan 25, 2022 at 21:57
  • $\begingroup$ Triatticus, googeling doesn't give you the right answer. The Q&A is about the ambiguity between the angle between the complex and the real physical fields. $\endgroup$
    – user326366
    Jan 25, 2022 at 22:00
  • $\begingroup$ I edited your title since it seemed to be confusing a lot of people. Doing weird stuff with complex valued electromagnetic fields is more of an electrical engineering thing. In physics we don't do it so much, precisely because of the issues you raised. $\endgroup$
    – knzhou
    Jan 25, 2022 at 22:29
  • $\begingroup$ @AnonPhoton while leaving a Q&A is certainly very much welcomed in this website, if the question is not clear it will hardly be useful for other users. From your answer, I can see you were thinking of complex electromagnetic fields, and that information is certainly essential to understand the problem (otherwise, people understand you were asking about real vectors, which seems weird, bu possible). Clarifying this would make yours a better question and allows other users to also provide answers to it, and add even more information to your Q&A $\endgroup$ Jan 25, 2022 at 22:35

3 Answers 3

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$\vec{B}$ 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.

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.