Magnetic monopoles and the Poynting vector Assume that magnetic monopoles exist. Let us have one charge(for example, an electron) of electric charge $q_e$ at the origin and one magnetic monopole particle of magnetic charge $q_m$ separated by a vector $\vec{a}$.
Then the electric field generated by the charge will be of the form :
$\vec{E}(\vec{r})=\frac{q_e}{4\pi\epsilon_0}\frac{\vec{r}}{||\vec{r}||^3}$
and the magnetic field generated by the magnetic monopole :
$\vec{B}(\vec{r})=\frac{q_m}{4\pi\mu_0}\frac{\vec{r}-\vec{a}}{||\vec{r}-\vec{a}||^3}$.
So by considering the standard definition of the Poynting vector $\vec{S} = \vec{E}\times\vec{B}$, we have $\vec{S}(\vec{r}) = -\frac{q_eq_m}{(4\pi)^2\epsilon_0\mu_0}\frac{\vec{r}\times\vec{a}}{||\vec{r}||^3||\vec{r}-\vec{a}||^3}$, which is non-zero for all vectors $\vec{r}$ non parallel to $\vec{a}$.
According to my understanding, this means that EM energy is radiated away, which cannot happen due to energy conservation.
My question is: which of the following statement is correct and why :
(1) The poynting vector formula is modified when considering that magnetic monopoles exist.
(2) The magnetic field of a magnetic monopole is not radial as I assumed.
(3) Energy would indeed be radiated away, making the 2 particles disappear in EM radiation, similar to matter-antimatter interaction.
(4) Something else.
 A: No, you've gotten your fields right.
What you are forgetting is that we can only conclude emission of energy from the system if the Poynting vector has a nonzero integral over a closed surface containing the system. 
As long as our closed surface contains the sources, Poynting's theorem guarantees our result will be independent of the surface we choose.
So let's choose a sphere $\mathbb{S}(R)$ centered halfway between the sources, and make its radius $R$ increase without limit. Our result will be independent of radius, so it has to be equal to the limiting value $\lim\limits_{R\to\infty} \int_{\mathbb{S}(R)} \vec{S} \cdot \hat{n} \,d A$. As our sphere becomes very big, the electric and magnetic fields are parallel, with zero cross product. It should be clear that our limit is nought. Of course, you should work the limit out more rigorously using first order approximations to the fields at finite radiusses and overbounds on the error. But the answer is indeed that no power is being radiated.
