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(Before marking this question as a duplicate, please consider I've read this post but it I didn't find the answers to it quite satisfactory regarding my doubt).

I'm trying to derive the expression of Poynting's vector (averaged over time) for plane harmonic waves. I start from the well-known expression for harmonic waves and then substitute the fields by the ones corresponding to plane harmonic waves: $$\langle \vec{S}\rangle = \frac{1}{2}Re(\vec{E}_0\times \vec{H}_0^\star) \\ =\frac{1}{2\mu_0\omega}Re(\vec{E}_0\times (\vec{k}^\star\times \vec{E}_0^\star)) \\ =\frac{1}{2\mu_0\omega}Re(\vec{k}^\star(\vec{E}_0\cdot \vec{E}_0^\star))$$ where $\cdot$ is the dot product defined in $\mathbb{C}^n$. The next step would be to admit $\vec{E}_0\cdot \vec{E}_0^\star=|\vec{E}_0|^2$, but I don't find that correct since a vector's norm is given by the dot product of said vector with itself. That is to say, $|\vec{E}_0|^2\equiv \langle\vec{E}_0,\vec{E}_0\rangle = \vec{E}_0\cdot \vec{E}_0\neq \vec{E}_0\cdot \vec{E}_0^\star$.

Of course, I know my issue has everything to do with the fact that I'm not understanding how to use the dot product correctly in this operation, and I'm also aware that, for a complex vector, the dot product with any other complex vector requires to transpose and conjugate the first one, such that: $\langle\vec{u},\vec{v}\rangle=\vec{u}^\star\cdot\vec{v}$ where $\cdot$ is now just notation to represent the product of a row vector with a column vector. It would all make sense if the dot product in my calculations were just the product of these row and column vectors, but I don't think it's that way since the identity I've used to get from the second to the third line is: $$\vec a\times(\vec b\times \vec c) = \vec b(\vec a\cdot \vec c)-\vec c(\vec a\cdot \vec b)$$

EDIT:

I think my problem arises from the identity:

$$\vec a\times(\vec b\times \vec c) = \vec b(\vec a\cdot \vec c)-\vec c(\vec a\cdot \vec b)$$

Is this dot an actual dot product defined in $\mathbb{C}^n$ or is it just notation for the product of a row and a column vector such that: $\vec{E}_0^\star\cdot\vec{E}_0=<\vec{E}_0,\vec{E}_0>=|\vec{E}_0|^2$?

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  • $\begingroup$ hint: $\vec k$ is a unit vector in the direction of propagation and is not a "complex" vector as $\vec E$ or $\vec H$, so $\vec H^* \propto \vec k \times \vec E^*$. $\endgroup$ Commented Apr 7, 2024 at 18:42
  • $\begingroup$ But $\vec{k}$ is not a unit vector, is it? It's $\vec{k} = k\vec{e}_k$ and it can be complex if the wave propagates through a material medium. I don't quite get your point $\endgroup$ Commented Apr 8, 2024 at 10:26
  • $\begingroup$ are you talking about a lossy absorptive medium? $\endgroup$ Commented Apr 8, 2024 at 12:25
  • $\begingroup$ Seems to be all about definitions. Why not write the Poynting vector as $\vec{E} \times \vec{H}$ (which is what it is). Then if, $\vec{H} = \vec{k}\times\vec{E}/\mu_0\omega$, the Poynting vector is $(E^2/\mu_0\omega) \vec{k}$ and you can time-average it as you wish. $\endgroup$ Commented Apr 8, 2024 at 12:54

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When we use complex vectors, we work with bivectors. In general, the algebra of these bivectors is a natural and simple extension of the algebra of real vectors so that we do not insist on the questions that may arise.

But, in some cases, these complex vectors must be handled with great care. For example, we naturally define a scalar product which is an extension of the scalar product for real vectors but which in the end is no longer a true scalar product. for example, the vector $\underline{\vec{E_0}}=\vec{e_x}+j \vec{e_y}$ has a scalar square which is zero while the vector is non-zero!

A simple discussion of this topic can be found in Gibbs's very famous article on vector analysis : Gibbs. Scientific papers. Vector analysis. p 84

By following this link, we find that the definition of the module of a bivector (complex vector) should be specified :

In biscalar analysis, the product of a biscalar and its conjugate is a positive scalar. The positive square root of this scalar is called the modulus of the biscalar. In bivector analysis, the direct product of a bivector and its conjugate is, as seen above, a positive scalar. The positive square root of this scalar may be called the modulus of the bivector. When this modulus vanishes, the bivector vanishes, and only in this case.

So your question simply becomes a definition problem.

Hope it can help and sorry for my poor english.

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  • $\begingroup$ Thanks for your answer, your English is perfect! So you're saying we are using the dot product defined for real vectors but using complex vectors? $\endgroup$ Commented Apr 8, 2024 at 13:51
  • $\begingroup$ @AlanFox86 In physics classes, this is what is usually done. We generalize the complex representation for vectors without much comment. In general, this does not pose a problem provided you do not believe that a complex vector has a fixed real direction. In your case, the real part is in general a vector rotating on an ellipse and its real norm changes over time. $\endgroup$ Commented Apr 8, 2024 at 14:38
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Note that the scalar product on $\Bbb{C}^n$ must be hermitian instead of symmetric in order to ensure positive-definiteness. In consequence, one has indeed $\vec{E}_0 \cdot \vec{E}_0^* = \langle \vec{E}_0,\vec{E}_0 \rangle_{\Bbb{C}^n} = |\vec{E}_0|^2$. Moreover, the variable $\vec{k} = k\hat{n}$, where $k$ is the wavenumber and $\hat{n}$ a unit vector, is a real quantity, because it corresponds to the Fourier variable associated to position $\vec{x}$. These considerations imply that $$ \langle S \rangle = \frac{\Re\left((\vec{E}_0 \cdot \vec{E}_0^*)\vec{k}^*\right)}{2\mu_0\omega} = \frac{k}{2\mu_0\omega}|\vec{E}_0|^2\hat{n} = \frac{1}{2} \sqrt{\frac{\varepsilon_0}{\mu_0}} |\vec{E}_0|^2\hat{n}, $$ as stated in the post you linked, since $\frac{1}{c} = \frac{k}{\omega} = \sqrt{\mu_0\varepsilon_0}$.

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  • $\begingroup$ Thanks for your answer! Nonetheless, then I don't understand why this holds: $\vec{a}\times(\vec{b}\times\vec{c})=\vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b}) \neq \vec{b}(\langle \vec{a},\vec{c} \rangle_{\mathbb{C}^n})-\vec{c}(\langle \vec{a},\vec{b} \rangle_{\mathbb{C}^n})$. If the last equality did hold, then $\vec{E}_0\cdot\vec{E}_0^\star\neq|\vec{E}_0|^2$ $\endgroup$ Commented Apr 8, 2024 at 13:55

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