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I was wondering how to calculate the power dissipation density (electromagnetic losses) when two waves of different frequencies are used simultaneously to heat a dielectric object. Of course, this material would present different permittivities depending on the frequency and also on the temperature. But just focusing on the frequency here:

In a sinusoidal steady state, for a point in the dielectric, the time average of the power loss can be calculated as

$$P_d = \mathbf E \cdot \frac{\partial \mathbf D}{\partial t}$$

Which, in view of the time average theorem, can be expressed as

$$P_d = \frac{1}{2}Re(jw \mathbf D)\cdot \mathbf E$$

Knowing that the displacement flux density vector is related to the electric field by a complex permittivity or dielectric constant:

$$\mathbf D = \epsilon \mathbf E=(\epsilon'-j\epsilon'')\mathbf E $$

Then:

$$P_d = \frac{\omega}{2} |\mathbf E|^2\epsilon'' $$

If using the superposition principle, the combined electric field would be the sum of the electric field caused by the two waves, how can one calculate the combined flux density vector? I guess this would be the way to then use the expression for the power loss (rather than just adding up the two individual power losses), but I am not certain as I lack knowledge in the matter at hand.

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Indeed, you take the electric field to be the sum of the two fields at different frequencies, calculate instantaneous power and carry out frequency averaging.

Averaging might be somewhat tricky here: in one-frequency case the averaging is performed over a period of oscillations, but this is difficult to do in case of two frequencies, unless they are commensurate, $n_1\omega_1=n_2\omega_2$, so that combined oscillations have a well-defined period. Assuming that the two frequencies are commensurate is a good practical way to solve this problem. Alternative is averaging over a long period of time: $$ \frac{1}{T}\int_0^Tdt\left|E_1e^{i\omega_1 t} + E_2e^{i\omega_2 t}\right|^2 = \frac{1}{T}\int_0^Tdt\left[|E_1|^2 + |E_2|^2 + 2\Re\left(E_1E_2^*e^{i(\omega_1 -\omega_2) t}\right)\right]=\\ |E_1|^2 + |E_2|^2 + 2\Re\left[E_1E_2^*\frac{1}{T}\int_0^Tdte^{i(\omega_1 -\omega_2) t}\right]= |E_1|^2 + |E_2|^2 + 2\Re\left[E_1E_2^*\frac{1}{T}\frac{e^{i(\omega_1 -\omega_2) T}-1}{i(\omega_1 -\omega_2)}\right] $$ The last term vanishes in the limit $T\rightarrow +\infty$, leaving us with sum $|E_1|^2 + |E_2|^2$.

Remark: For the question in the OP this calculation has to be generalized taking into account the time-dependence of $D(t)$, which in time domain is a convolution of $\epsilon$ and $E$.

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  • $\begingroup$ Hi Vadim, and thanks for your quick reply! I think I do understand what you mean, though the interesting part that I don't grasp is how the electric displacement vectors add up. While, by the superposition principle the electic field would be the sum of both electric fields, would the combined electric displacement vectors be the sum of both individual vectors? Taking into account that the imaginary part of the permittivities will be different for each wave as this depends on frequency. In this case I could use bot combined E and D vectors in the expression for the power density. $\endgroup$
    – cosmogato
    Commented Apr 29, 2021 at 16:02
  • $\begingroup$ I think the simplest thing to do would be to consider $E(t)=E_1e^{i\omega_1 t}+E_2e^{i\omega_2 t} + c.c$, $D(t)=D_1e^{i\omega_1 t}+D_2e^{i\omega_2 t} + c.c$, use the formula for power in the OP and perform the derivation similar to the one that I have in my answer. Then in the end one can set $D_{1,2}=\epsilon_{1,2}E_{1,2}$. $\endgroup$
    – Roger V.
    Commented Apr 29, 2021 at 17:01

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