Poynting vector of two orthogonal oscillating dipoles Why is the Poynting vector of two orthogonal oscillating dipoles, located to the origin of the axes, the sum of the Poynting vector of the two dipoles when separated? 
 A: It's not the sum of the individual Poynting vectors, unless we make three assumptions:

*

*We're in the radiation zone;

*The dipoles are oscillating 90° out of phase, and

*We average the Poynting vector over time.

But quite often that's the case we're concerned with.

Let's denote our dipoles by $\vec{p}_1(t)$ and $\vec{p}_2(t)$, and the fields they each create by $\vec{E}_1$, $\vec{B}_1$, $\vec{E}_2$, and $\vec{B}_2$ respectively.  The electric and magnetic fields obey superposition, so the Poynting vector of the total fields should be
$$
\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} = \frac{1}{\mu_0} (\vec{E}_1 + \vec{E}_2) \times (\vec{B}_1 + \vec{B}_2) \\
= \frac{1}{\mu_0} \left( \vec{E}_1 \times \vec{B}_1 + \vec{E}_2 \times \vec{B}_2 + \vec{E}_1 \times \vec{B}_2 + \vec{E}_2 \times \vec{B}_1 \right) \\
= \vec{S}_1 + \vec{S}_2 + \frac{1}{\mu_0} \left(\vec{E}_1 \times \vec{B}_2 + \vec{E}_2 \times \vec{B}_1 \right) 
$$
The first two terms in the second line give the Poynting vectors for $\vec{p}_1$ alone and $\vec{p}_2$ alone.  The question is why the cross terms (the second two terms in the second line) should vanish.  Indeed, for a general set of electromagnetic fields, these terms will remain present, since the Poynting vector is not linear in the fields and so doesn't obey a naïve superposition principle.
However, in the radiation zone, the electric and magnetic fields are given by
$$
\vec{E}_1 = \frac{\mu_0}{4 \pi r} \hat{r} \times (\hat{r} \times \ddot{\vec{p}}_1) \qquad \vec{B}_1 = - \frac{\mu_0}{4 \pi r} \hat{r} \times \ddot{\vec{p}}_1,
$$
where the time derivatives are evaluated at the retarded time, and similarly for $\vec{E}_2$ and $\vec{B}_2$.  If we work through the cross products, we find that both of these terms work out to be equal:
$$
\vec{E}_1 \times \vec{B}_2 = \vec{E}_2 \times \vec{B}_1 = \frac{\mu_0^2}{16 \pi^2 r^2 c}\left[\ddot{\vec{p}}_1 \cdot \ddot{\vec{p}}_2 - (\hat{r} \cdot \ddot{\vec{p}}_1) (\hat{r} \cdot \ddot{\vec{p}}_2)\right]\hat{r} 
$$
Now, it's not too hard to see that the first term vanishes, since by assumption $\vec{p}_1(t)$ and $\vec{p}_2(t)$ are orthogonal.  But the second term does not in general vanish;  for example, if $\vec{p}_1 \propto \hat{x}$, $\vec{p}_2 \propto \hat{y}$, and $\hat{r} = (\hat{x} + \hat{y})/\sqrt{2}$, then it's not hard to see that this term doesn't go away.  I think this term does vanish when integrated over a sphere, so that the net power radiated remains the same; but the difference in the Poynting vector leads to this radiated power being distributed differently over the sphere.  This can be thought of in terms of interference between the waves from the two dipoles:  depending on where you are in space, the waves constructively or destructively interfere.
However, if the dipoles are oscillating sinusoidally and they are 90° out of phase with each other, then it's not too hard to see that one of $\ddot{\vec{p}}_1$ and $\ddot{\vec{p}}_2$ will be a sine function while the other will be a cosine.  The overall time dependence of the cross terms in $\vec{S}$ will be of the form $\sin(\omega t) \cos (\omega t)$, and when we average this over one period of oscillation, it will average out to zero.  So in this case, while it is not true that
$$
\vec{S} = \vec{S}_1 + \vec{S}_2,
$$
it is true that
$$
\langle\vec{S} \rangle = \langle\vec{S}_1 \rangle+ \langle\vec{S}_2\rangle.
$$
More generally, if the phase difference between the two oscillations is $\delta$, then the cross term averaged over a period will be proportional to $\cos \delta$;  only when the two are 90° out of phase will this quantity vanish.
Finally, if we're not in the radiation zone, the electric and magnetic fields will also contain pieces proportional to $\vec{p}_1$, $\vec{p}_2$, $\dot{\vec{p}}_1$, and $\dot{\vec{p}}_2$, all evaluated at the retarded time.  These pieces will have different phases than $\ddot{\vec{p}}_1$ and $\ddot{\vec{p}}_2$, and whether or not they vanish (even when time-averaged) will depend on the precise phase and amplitude differences between all the pieces that make up the field.  It's possible that there could be some fortuitous cancellations, but in general I would think that the Poynting vector does not obey superposition if you're closer in.
