1
$\begingroup$

enter image description here

I'm learning graduate level E&M. Textbook is a famous Jackson book. What I would talk now is about pp.295-298 in 3rd ed. I attached the photo of p.298.

It says (paragraph above eq.(7.15) and footnote in the photo) that $\vec{n}\cdot \vec{n}=1$ doesn't mean n is unit vector if n is complex vector. And it discusses about the form of n satisfying above relation.

But it looks weird to me. When I learned linear algebra/mathematical physics, I learned that in complex domain it is more natural to define inner product as $\vec{a}\cdot\vec{b}=\Sigma a_i^\ast b_i$. If we use this definition there would be no problem of being not unit vector. Why did Jackson stick to definition of dot product in real domain?

Improve this question
$\endgroup$
1
  • $\begingroup$ Please, do not post pictures. Type the question instead (and using MathJax for formulae). There are many reasons, including facilitation for algorithms and helping users whose device doesn't display them well. Thank you $\endgroup$ Commented Feb 1, 2019 at 0:07

2 Answers 2

Reset to default
1
$\begingroup$

The vector $\hat{n}$ is meant to have real components, so the definitions are equivalent.

The conjugation is applied to calculate the Poynting's vector only because we like to work with complex exponentials, but you should only care about the real part. Consequently, you can either use $S\propto\Re e(\vec{E})\times\Re e(\vec{H})$ or use the cross product normally but conjugating the second one.

Improve this answer
$\endgroup$
1
$\begingroup$

The reason is quite simple: the use of the “standard” (i.e. real) dot product is the more familiar notation to physicists. Of course, you could absolutely rewrite everything in terms of some Hermitian inner product $\langle\cdot,\cdot\rangle$ and the equations might become a bit cleaner, but this would come at the expense of being slightly out-of-touch with your audience, who have been using one notation for a long time. And ultimately, a textbook needs to be understandable by a wide audience.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.