Poynting vector and plane waves

Question

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?

Answer 1

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 $\pm (\mathbf {k_2-k_1})$.