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Suppose the electromagnetic field at a point is a superposition to two plane waves $(\vec E_1, \vec H_1)$ and $(\vec E_2, \vec H_2)$. If the two plane-waves have different frequencies, the (time-averaged)Poynting vector is given by $$\frac{1}{2} \space \left\{\vec E_1 \times \vec H_1^* \space + \space \vec E_2 \times \vec H_2^*\right\}$$ If the two waves have the same frequency and same direction of propagation, the answer would change to $$\frac{1}{2} \space \left(\vec E_1 + \vec E_2\right) \times \left(\vec H_1 + \vec H_2\right)^*$$ If the two waves have the same frequency but different directions of propagation, then what is the right formula for (average) Poynting vector?

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If plane wave linearly polarized in direction $\hat e$ has propagation vector $\mathbf k = k \hat k$, where $\hat k$ denotes a unit vector, it can be written as $$\mathbf E = \hat e E^0 e^{\mathfrak j (\omega t - \mathbf k \cdot \mathbf r)} \tag{1}$$ $$\mathbf H = \hat h H^0 e^{\mathfrak j (\omega t - \mathbf k \cdot \mathbf r)}\tag{2}$$ where the unit vector $\hat h$ is defined as $\hat h = \hat k \times \hat e$ and $H^0 =\sqrt{\frac{\epsilon_0}{\mu_0}}E^0$.

Now let two plane waves move in the directions $\mathbf k_{1,2} = \hat k_{1,2}$ but have the same frequency and wave number: $$\mathbf E_{1,2} = \hat e_{1,2} E^0_{1,2} e^{\mathfrak j (\omega t - \mathbf k_{1,2}\cdot \mathbf r)} \tag{3}$$ $$\mathbf H_{1,2} = \hat h_{1,2} H^0_{1,2} e^{\mathfrak j (\omega t - \mathbf k_{1,2} \cdot \mathbf r)}\tag{4} $$

The Poynting vector of the sum of the two waves is: $$\mathbf S = (\mathbf E_1+\mathbf E_2)\times (\mathbf H^*_1+\mathbf H^*_2)\\ =(\hat e_1 E^0_1 e^{\mathfrak j (\omega t - \mathbf k_1 \cdot \mathbf r)} +\hat e_2 E^0_2 e^{\mathfrak j (\omega t - \mathbf k_2 \cdot \mathbf r)} )\times (\hat h_1 H^{0*}_1 e^{\mathfrak j (-\omega t + \mathbf k_1 \cdot \mathbf r)} +\hat h_2 H^{0*}_2 e^{\mathfrak j (-\omega t + \mathbf k_2 \cdot \mathbf r)} ) \tag{5}$$ Expand (5): $$\mathbf S =\hat e_1\times \hat h_1 E^0_1H^{0*}_1 +\hat e_2\times \hat h_2 E^0_2H^{0*}_2\\ +\hat e_2 E^0_2 e^{\mathfrak j (\omega t - \mathbf k_2 \cdot \mathbf r)} \times \hat h_1 H^{0*}_1 e^{\mathfrak j (-\omega t + \mathbf k_1 \cdot \mathbf r)} + \hat e_1 E^0_1 e^{\mathfrak j (\omega t - \mathbf k_1 \cdot \mathbf r)} \times \hat h_2 H^{0*}_2 e^{\mathfrak j (-\omega t + \mathbf k_2 \cdot \mathbf r)} \tag{6}$$ or $$\mathbf S =\hat e_1\times \hat h_1 E^0_1H^{0*}_1 +\hat e_2\times \hat h_2 E^0_2H^{0*}_2\\ +\hat e_2 \times \hat h_1 E^0_2H^{0*}_1 e^{-\mathfrak j (\mathbf {k_2 -k_1})\cdot \mathbf r)} + \hat e_1 \times \hat h_2 E^0_1H^{0*}_2 e^{\mathfrak j (\mathbf {k_2-k_1}) \cdot \mathbf r} \tag{7}$$

Now $\hat e_{1,2}\times \hat h_{1,2}=\hat k_{1,2}$ and therefore $$ \mathbf S = \sqrt{\frac{\mu_0}{\epsilon_0}}\left(\hat k_1 |E^0_1|^2 +\hat k_2 |E^0_2|^2\right) +\mathbf R \tag{8}$$ where the vector $\mathbf R $ is defined by $$\mathbf R = \sqrt{\frac{\mu_0}{\epsilon_0}}\left(\hat e_2 \times \hat h_1 E^0_2E^{0*}_1 e^{-\mathfrak j (\mathbf {k_2 -k_1})\cdot \mathbf r)} + \hat e_1 \times \hat h_2 E^0_1E^{0*}_2 e^{\mathfrak j (\mathbf {k_2-k_1}) \cdot \mathbf r}\right)\tag{9}.$$

In eq (8) the first two terms are the Poynting vectors of the individual plane waves pointing in their respective propagating directions, resp. The remaining term represented by the vector $\mathbf R$ is the interference term between the two waves. Notice that the frequency term falls out so (8) is indeed a true time average but because the waves point in different directions it has both an active, real, and a reactive, imaginary, term, the latter being in time quadrature relative to the former.

The interference term $\mathbf R$ depends both on the relative directions, $\hat e_1 \times \hat h_2$ and $\hat e_2 \times \hat h_1 $, and relative time phases, $E^0_1E^{0*}_2$ and $E^0_2E^{0*}_1$, of the two waves and it is plane periodic in a pair of opposite directions $\pm (\mathbf {k_2-k_1})$.

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  • $\begingroup$ Thank you very much. My follow-up question is, under what circumstances can be neglect $\vec R$? For instance, if we average $\vec R$ over all space, it vanishes, as long as $\vec k_1 \neq \vec k_2$. Perhaps it is negligible whenever $\vec R$ is integrated over a region (whose dimensions are) much larger than $\frac{1}{||\vec k_1 - \vec k_2||}$. Are there some upper-bounds on such an average? Hopefully, the bound can incorporate the impact of $\left(\hat e_1 \times \hat h_2\right)$ as well.... $\endgroup$ Sep 17, 2023 at 15:55
  • $\begingroup$ I do not think it is ever negligible unless they are nearly orthogonal to each other. Since $|\hat e_{1,2}|=|\hat h_{1,2}|=1$ the various terms have size $|E^0_{1}|^2$, $|E^0_{2}|^2$, $|E^0_{1}E^0_{2}|$, multiplied by the sine angle between $\hat e_1$ and $\hat h_2$ so if say $|E^0_{1}|^2 \ll |E^0_{2}|^2$ you still have the mixed term to deal with. Since these are plane waves the interference term is also homogeneous aside form the spatial fluctuation with periodcity $\frac{1}{|\mathbf k_1-\mathbf k_2|}$ $\endgroup$
    – hyportnex
    Sep 17, 2023 at 18:12
  • $\begingroup$ Do you agree that the spatial average of $\vec R$ over all space (all of $\mathbb{R}^3$) is zero (whenever $\vec k_1 \neq \vec k_2$)? Or am I missing something there as well? $\endgroup$ Sep 19, 2023 at 17:57
  • $\begingroup$ yes, that average is zero, it is a "sinusoid" after all, and this is how energy is conserved globally. $\endgroup$
    – hyportnex
    Sep 19, 2023 at 18:03
  • $\begingroup$ So, if we average $\vec R$ over a large sphere, doesn't the average approach zero as the radius $A$ of the sphere goes to infinity? Perhaps at a rate proportional to $\frac{1}{||\vec k_1 - \vec k_2|| A}$? $\endgroup$ Sep 19, 2023 at 18:36

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