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According to this source, the divergence of the Poynting vector is related to the total energy density of an electromagnetic wave, which is (locally) expressed as

$$-\nabla\cdot S=EJ+(E\frac{\partial D}{\partial t}+H\frac{\partial B}{\partial t})$$

I see, that the first term on the R.H.S. corresponds to Joule heating. But it's only present if there are free charge carriers. The 2nd term is called "radiative loss" in the source I cited above. I'm not sure how to interpret it. I believe it's just "locally" a loss. The wave moves along and carries momentum and energy with it, which implys that this "loss" does not correspond to a loss globally.

So how is it then, that radiation heats media with no free charge carriers? Or does the radiation induce free carriers by photo-ionization first?

EDIT: I'm also wondering if attenuation of intensity, as an EM-wave travels through a medium, is always necessarily caused by absorption of light. Is the 2nd term able to attenuate the EM-wave in absence of the first one on the RHS?

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    $\begingroup$ Do not forget that there exist rotational and vibrational levels in the molecules and solids , the energy differences there are in the infrared which is how infrared radiation is absorbed, Visible light does not transit through opaque solids, it leaves the energy on the surface , converted to infrared through inelastic scatterings and excitations.. $\endgroup$
    – anna v
    Oct 6, 2018 at 18:06
  • $\begingroup$ If this question is about microwave cooking, it is because the polar molecules get polarized in alternating directions, and relax to random orientation. $\endgroup$
    – user137289
    Oct 6, 2018 at 18:43

4 Answers 4

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So how is it then, that radiation heats media with no free charge carriers? Or does the radiation induce free carriers by photo-ionization first?

This is an excellent question, very well reasoned. As other answers have described there are various microscopic processes whereby non-conductors can absorb EM energy. But there should also be a macroscopic equation that reflects the situation where a material is filled with such microscopic processes. Indeed, there is.

As you have ascertained, the expression your source gave is not the most general expression for macroscopic EM. A more general expression is given in this MIT electromagnetism textbook: https://web.mit.edu/6.013_book/www/book.html

In section 11.2 they derive the following more general form: $$ -\nabla \cdot (E\times H)=\frac{1}{2}\epsilon_0 \frac{\partial}{\partial t} E^2 + \frac{1}{2}\mu_0 \frac{\partial}{\partial t} H^2 + E \cdot J_u + H \cdot \mu_0 \frac{\partial}{\partial t} M + E \cdot \frac{\partial}{\partial t} P$$

In particular, note the last term $E\cdot \frac{\partial}{\partial t}P$. As an E field acts on a non-conductor there are no free charges to produce free current or charge, but there are bound charges that can be polarized and create bound charge density. Work can be done on non conductive matter in this way. Thereby resolving your dilemma.

Notice also the other new term $H \cdot \mu_0 \frac{\partial}{\partial t} M$. This a major macroscopic means of energy transfer for an induction stovetop.

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  • $\begingroup$ Great answer, although this formula is not more general, but actually the same, isnt it? (with D=E+P and B=H+M) $\endgroup$
    – lalala
    May 6, 2021 at 7:20
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The non-conductive medium does not conduct at zero frequency, but it does conduct at non zero frequency. The joule heating is due to the first term in your equation.

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  • $\begingroup$ Ok.so the 2nd term does not contribute to heating,right? $\endgroup$
    – OD IUM
    Oct 6, 2018 at 18:20
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First, we need to clarify three ways how light can interact with the atoms of the material:

  1. elastic scattering, reflection, is on the surface (mostly with visible light)

  2. inelastic scattering, is how thermal heat is transferred to the vibrational motion of molecules (heats up material), this is with mostly non-visible light

  3. absorption, re-emission, this can mostly give color (with visible light) to the surface of the material

Now you are asking how non-conducting material can heat up because of EM waves. This is because non-visible light elastically scatters on the molecules, usually infrared's energy matches the energy level differences in the vibrational motion levels of the material even with non-conductive material. When infrared light interacts with the material, it is absorbed by the molecules, and transfers energy to the vibrational motion of these molecules in the material (heats up).

You do not need free charge carriers this way.

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  • $\begingroup$ Ok.but how could this be expressed mathematically? $\endgroup$
    – OD IUM
    Oct 6, 2018 at 18:41
  • $\begingroup$ Do you have a reference for thus absorption reemission concept? How does it conserve coherency? How does it take the collective nature of optical excitations in solids into. account? Please clarify. Also please clarify how molecular vibrations and rotations survive in dielectric media. $\endgroup$
    – my2cts
    Oct 6, 2018 at 21:15
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Mechanical motion caused by electron orbitals or cloud being highly distorted and pulling atoms in such a way that physical friction causes dielectric heating. If you put Pyrex onto a hot plate it is mechanical conduction, of course the low tan delta (dielectric loss) of Pyrex will prevent dielectric heating, but water, a polar substance will heat rapidly, because the weird shaped polar molecules get tangled with one another as electric fields act on the positive coulomb charges that the hydrogen molecules have exposed. so mechanical friction is at play internally in the liquid. this internal heating drops with distance exponentially (skin depth) due to energy loss (attenuation). Conventional conduction then takes over. Your microwave oven cooks chicken better than turkeys, because the skin depth at standard microwave frequencies is too shallow for turkeys the deeper parts are heated by conduction (a slow process) rather than dielectric loss, a long standing time helps. A lower frequency microwave is preferred for turkeys.

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