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Note I am not sure if this post is of relevance for this platform, but I already asked the question in Physics Stack Exchange and in Mathematics Stack Exchange without success.

Setup Let's suppose a homogeneous dielectric with a (spatially) local dielectric response function $\underline{\underline{\epsilon}} (\omega)$ (in general a tensor), such that we have the linear response relation $$\mathbf{D}(\mathbf{x}, \omega) = \underline{\underline{\epsilon}} \left(\omega \right) \mathbf{E}(\mathbf{x}, \omega) \, ,$$ for the displacement field $\mathbf{D}$ in the dielectric. We can now write down a Lagrangian in Fourier space, describing the EM-field coupling to the dielectric body $$\mathcal{L}=\frac{1}{2}\left[\mathbf{E}^{*}\left(x,\omega\right) \cdot (\underline{\underline{\epsilon}} \left(\omega \right)-1) \mathbf{E}\left(x,\omega\right)+|\mathbf{E}|^{2}\left(x,\omega\right)-|\mathbf{B}|^{2}\left(x,\omega\right) \right] \, .$$ If we choose a gauge \begin{align} \mathbf{E} &= \frac{i \omega}{c} \mathbf{A} \\ \mathbf{B} &= \nabla \times \mathbf{A} \, , \end{align} such that we can write the Lagrangian (suppressing arguments) in terms of the vector potential $\mathbf{A}$ as $$\mathcal{L} =\frac{1}{2}\left[\frac{\omega^2}{c^2} \mathbf{A}^{*} \cdot (\underline{\underline{\epsilon}}- \mathbb{1}) \mathbf{A}+ \frac{\omega^2}{c^2} |\mathbf{A}|^{2}-|\nabla \times \mathbf{A}|^{2}\right] \, . $$ And consequently we have the physical action $$ S[\mathbf{A}] = \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \mathcal{L} \left(\mathbf{A}\right) \, . $$

Goal My goal is to derive the EM-wave equation for the electric field in the dielectric media.

Idea So my ansatz is the following: If we use Hamilton's principle, we want the first variation of the action to be zero \begin{align} 0 = \delta S[\mathbf{A}] &= \left.\frac{\mathrm{d}}{\mathrm{d} \varepsilon} S[\mathbf{A} + \varepsilon \mathbf{h}] \right|_{\varepsilon=0} \\ &= \left.\frac{\mathrm{d}}{\mathrm{d} \varepsilon} \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \mathcal{L} (\mathbf{A} + \varepsilon \mathbf{h}) \right|_{\varepsilon=0} \\ &= \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \frac{1}{2} \Bigg( \frac{\omega^2}{c^2} \mathbf{A}^* \cdot ({\underline{\underline{\epsilon}}}-\mathbb{1}) \mathbf{h} + \frac{\omega^2}{c^2} \mathbf{h}^* \cdot ({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} + \frac{\omega^2}{c^2} \mathbf{A}^* \cdot \mathbf{h} + \frac{\omega^2}{c^2} \mathbf{h}^* \cdot \mathbf{A} \\ &\quad \quad \quad \quad \quad \quad \quad \quad- (\nabla \times \mathbf{A}^* ) \cdot ( \nabla \times \mathbf{h}) - (\nabla \times \mathbf{h}^* ) \cdot ( \nabla \times \mathbf{A}) \Bigg) \\ &= \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \frac{1}{2} \Bigg( \frac{\omega^2}{c^2} \left[ ({\underline{\underline{\epsilon}}}^{\dagger}-\mathbb{1}) \mathbf{A} \right]^* \cdot \mathbf{h} + \frac{\omega^2}{c^2} \left[({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} \right] \cdot \mathbf{h}^* + \frac{\omega^2}{c^2} \mathbf{A}^* \cdot \mathbf{h} + \frac{\omega^2}{c^2} \mathbf{A} \cdot \mathbf{h}^* \\ &\quad \quad \quad \quad \quad \quad \quad \quad- (\nabla \times \nabla \times \mathbf{A}^* ) \cdot \mathbf{h} - (\nabla \times \nabla \times \mathbf{A} ) \cdot \mathbf{h}^* \Bigg) \\ &= \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \frac{1}{2} \Bigg( \underbrace{\left[ \frac{\omega^2}{c^2} \left[ ({\underline{\underline{\epsilon}}}^{\dagger}-\mathbb{1}) \mathbf{A} \right]^* + \frac{\omega^2}{c^2} \mathbf{A}^* - \nabla \times \nabla \times \mathbf{A}^* \right]}_{\stackrel{!}{=} 0} \cdot \mathbf{h} \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + \underbrace{\left[ \frac{\omega^2}{c^2} \left[({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} \right] + \frac{\omega^2}{c^2} \mathbf{A} - \nabla \times \nabla \times \mathbf{A} \right]}_{\stackrel{!}{=} 0} \cdot \mathbf{h}^* \Bigg) \, , \end{align} for all $\mathbf{h}(\mathbf{x}, \omega)$. And consequently we get the equations \begin{align} \frac{\omega^2}{c^2} \left[ ({\underline{\underline{\epsilon}}}^{\dagger}-\mathbb{1}) \mathbf{A} \right]^* + \frac{\omega^2}{c^2} \mathbf{A}^* - \nabla \times \nabla \times \mathbf{A}^* &= 0 \\ \frac{\omega^2}{c^2} \left[({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} \right] + \frac{\omega^2}{c^2} \mathbf{A} - \nabla \times \nabla \times \mathbf{A} &= 0 \, . \end{align} If we suppose a lossy dielectric body, such that $\underline{\underline{\epsilon}}^{\dagger} \neq \underline{\underline{\epsilon}}$, the equations are in contradiction.

Time-domain An analogues derivation in the time-domain (which I can post here on request), yields the wave equation (in Fourier space) $$\frac{\omega^2}{2 c^2} (\underline{\underline{\epsilon}}-1) \mathbf{A} + \frac{\omega^2}{2 c^2} (\underline{\underline{\epsilon}}^{\dagger} -1) \mathbf{A} + \frac{\omega^2}{c^2} \mathbf{A} - \left( \nabla \times \nabla \times \mathbf{A} \right) = 0 \, .$$ This result is also not resembling the expected result, for $\underline{\underline{\epsilon}}^{\dagger} \neq \underline{\underline{\epsilon}}$.

Question What went wrong in the calculation?

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    $\begingroup$ if you have losses, the dynamics of the polarization field $P$ that enters in $D=\epsilon_0 E+P$ has to be explicitly considered when you wish to derive the wave equation; so you will need some model for the dielectric, for example the Drude model; you cannot work exclusively with the electromagnetic fields. $\endgroup$ Sep 20, 2021 at 14:55

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Expanding on Carlo Beenakker's comment, one can't just expect to substitute in a complex dielectric function to properly describe absorption. Rather, the relevant question is how to structure a Lagrangean such as to generate the desired damping term in the equation of motion. For example, to generate linear damping in a harmonic oscillator, one can use $$ L=e^{\gamma t} \left( \frac{m\dot{q}^2 }{2} -\frac{kq^2 }{2} \right) . $$ Since the electromagnetic field is a collection of oscillators, one might be able to use a generalization of this.

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A simple observation which proves that the Lagrangian does not represent the reality of physical phynomemes, we can divide the Lagrangian into two parts, the part which contains the free field (without the presence of charges) and the part which contains the interaction part between the EM field and the material: A simple observation that proves that the Lagrangian does not represent the reality of physical phenomena, we can divide the Lagrangian into two parts, the part that contains the free field (without the presence of charges) and the part that contains the part of the interaction between the EM field and matter :

$ \mathcal{L}=\left[\textbf{E*}(\textbf{r},\omega).(\underline{\underline{\epsilon}}-1)\textbf{E}(\textbf{r},\omega)+|\textbf{E}|^{2}(\textbf{r},\omega)-|\textbf{B}|^{2}(\textbf{r},\omega)\right]$

$\mathcal{L}=\mathcal{L}_{field}+\mathcal{L}_{interaction}$

where :

$\begin{cases}\mathcal{L}_{f}=|\textbf{E}|^{2}(\textbf{r},\omega)-|\textbf{B}|^{2}(\textbf{r},\omega)\\ \mathcal{L}_{i}=\textbf{E*}(\textbf{r},\omega).(\underline{\underline{\epsilon}}-1).\textbf{E}(\textbf{r},\omega)\end{cases}$

it is sufficient that $\underline{\underline{\epsilon}}=1$ so that it is not interaction ( $\mathcal{L}_{i}=0)$

This means that vacuum is the only medium that does not absorb or disperses EM waves for all values of the pulse, which is physically false, there are transparent media such as glass that allow light to pass for a certain pulse band.

I suggest to replace the term $(\underline{\underline{\epsilon}}-1)$ by $(\underline{\underline{\epsilon}}- \underline{\underline{\epsilon}}^{\dagger})$ in the interaction Lagrangian to at least contain the case of transparent media, i.e. in the case where $(\underline{\underline{\epsilon}}= \underline{\underline{\epsilon}}^{\dagger})$ , we obtain a symmetrical tensor ( diagonalizable).

I do not assume that the suggested Lagrangian is the right mold in which to pour the physical optics, but it has the merit of containing more information.

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    $\begingroup$ please don't add one answer after the other; the way to use this site is to improve your existing answer; the OP knows that their Lagrangian is mistaken for lossy media (complex $\epsilon$), a useful answer would show how to modify it such that it applies when $\epsilon\neq \epsilon^\dagger$; your suggestion to replace $\epsilon-1$ by $\epsilon-\epsilon^\dagger$ does not seem helpful in that respect. $\endgroup$ Sep 22, 2021 at 11:44
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It seems to me that your Lagrangian does not reflect the reality of physical phenomena and especially the absorption phenomenon of an EM wave, in the case of the equality of your result, there is no absorption (see, Landau, Lifchitz, tome VIII and also tome V), and all the results derive from the relations of Kramers – Kronig :https://en.wikipedia.org/wiki/Kramers%E2%80%93Kronig_relations

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 20, 2021 at 20:24
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In the Lagrangian he got, there is a term that reflects the interaction of the electromagnetic field with matter which has the form of equation (2) in this article ( to within a multiplicative factor : derivation of the energy per unit of volume)) but with a different factor which contains the tensor and its Hermitian conjugate. we do not know how he obtained this Lagrangian. For more information, there is this link: https://www.worldscientific.com/doi/pdf/10.1142/9789811203145_0001 where the relations (1,94), (1,95) express the negligent (absence) of the absorption.

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  • $\begingroup$ Sorry, I forgot the first link :journals.aps.org/pra/pdf/10.1103/PhysRevA.92.053847 $\endgroup$
    – The Tiler
    Sep 21, 2021 at 15:52
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    $\begingroup$ Instead of adding this new answer, please delete it and add its content to the other one. Furthermore, it is not clear how this provides an answer. $\endgroup$
    – Alex M.
    Sep 21, 2021 at 16:31
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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 21, 2021 at 16:31

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