Radiations in static fields by poynting Vector Why is the Poynting vector $\vec S=\frac{1}{\mu_0}(\vec E\times \vec B)$ not able to produce radiation when fields are static?
 A: Firstly, the Poynting vector does not produce radiation, accelerated charges do. The poynting vector is a measure of the flow of energy. By definition, a static field is time independant, Meaning no net gain or loss of energy at any point in space(as E and B are unchanged, and thus is the energy density)
I assume you really want to know the condition on the poynting vector that makes it such that the field has no net gain or loss of energy anywhere.
This condition, in regions where there are no charges is that
$\nabla \cdot \vec{S} = 0$
Or  more intuitively
$\iint \vec{S} \cdot \vec{da} = 0$
About some closed surface.
This means the flux about some closed surface is 0 meaning a net gain/ loss of zero energy in any region in question.
This comes directly from Poyntings theorem.
Edit: and as Rob said, radiation in time dependant. So  no need to even look at the poynting vector. If E and B are unchanging. There is no radiation.
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A: To provide an answer based on Maxwell's equations you can start with defining the Poynting vector in time domain as $\vec{\mathcal{P}}=\vec{\mathcal{E}}\times \vec{\mathcal{H}}$ where $\vec{\mathcal{E}}$ and $\vec{\mathcal{H}}$ are respectively the electric and magnetic field. Using the divergence theorem we have
$$\int_{V}(\nabla\cdot \vec{\mathcal{P}}) dV'=\int_{\partial V}(\vec{\mathcal{P}}\cdot\hat n) dS'$$
We also have the identity $\nabla\cdot(\vec A\times \vec B)=\vec B\cdot(\nabla\times \vec A)-\vec A\cdot(\nabla\times \vec B)$ for any two vector $\vec A$ and $\vec B$. From Maxwell's equations we know that in the static case and in a region where there is no source $\nabla\times \vec{\mathcal{E}}=\nabla\times \vec{\mathcal{H}}=0$ and consequently, $\nabla\cdot \vec{\mathcal{P}}=\nabla\cdot(\vec{\mathcal{E}}\times \vec{\mathcal{H}})=0$. Thus,
$$\int_{\partial V}(\vec{\mathcal{P}}\cdot\hat n) dS'=0$$
This result has interesting implications as well for the directions of the magnetic and electric field on the surface. Using the triple product identity for $\vec a\cdot(\vec b\times \vec c)$ we have
$$0=\int_{\partial V}(\vec{\mathcal{P}}\cdot\hat n) dS'=\int_{\partial V}[(\vec{\mathcal{E}}\times \vec{\mathcal{H}})\cdot\hat n] dS'=\int_{\partial V}[\vec{\mathcal{E}}\cdot (\vec{\mathcal{H}}\times\hat n)] dS'=\int_{\partial V}[\vec{\mathcal{H}}\cdot (\vec{\mathcal{E}}\times\hat n)] dS'$$
Note that e.g., $\vec{\mathcal{E}}\times\hat n$ denotes the component of the electric field tangential to the surface $\partial V$.
A: I will just supplement @jensen paull 's answer here.
Consider the Maxwell equation in vacuum (for simplicity):
$$\nabla \times \textbf{E} = - \frac{\partial \textbf{B}}{\partial t} $$
$$\nabla \times \textbf{B} = \mu_0\epsilon_0\frac{\partial \textbf{E}}{\partial t} $$
Finding the curl of the above two equations,
$$\nabla \times (\nabla \times \textbf{E}) = \nabla \times -\frac{\partial \textbf{B}}{\partial t}=-\frac{\partial }{\partial t}( \nabla \times \textbf{B}) = -\mu_0\epsilon_0\frac{\partial^2 \textbf{E}}{\partial t^2}$$
$$\nabla \times (\nabla \times \textbf{B}) = \nabla \times \mu_0\epsilon_0\frac{\partial \textbf{E}}{\partial t}=\mu_0\epsilon_0\frac{\partial }{\partial t}( \nabla \times \textbf{E}) = -\mu_0\epsilon_0\frac{\partial^2 \textbf{B}}{\partial t^2}$$
Furthermore, gradient of divergence for $\textbf{B}$ and $\textbf{E}$ are zero,
$$\nabla \times (\nabla \times \textbf{B}) = \nabla (\nabla \cdot \textbf{B}) - \nabla^2\textbf{B}  = - \nabla^2\textbf{B}$$
$$\nabla \times (\nabla \times \textbf{E}) = \nabla (\nabla \cdot \textbf{E}) - \nabla^2\textbf{E}  = - \nabla^2\textbf{E}$$
Thus,
$$\nabla^2\textbf{E} = \mu_0\epsilon_0\frac{\partial^2 \textbf{E}}{\partial t^2}$$
$$\nabla^2\textbf{B} = \mu_0\epsilon_0\frac{\partial^2 \textbf{B}}{\partial t^2}$$
The solution for this set of differential equation is the standard three-dimensional wave equation. Thus, if $\frac{\partial^2 \textbf{E}}{\partial t^2}$ or $\frac{\partial^2 \textbf{B}}{\partial t^2}$ is zero the Maxwell equation does not satisfy the wave equation thus no radiation is generated. Only an accelerating charge can produce non zero $\frac{\partial^2 \textbf{E}}{\partial t^2}$ and $\frac{\partial^2 \textbf{B}}{\partial t^2}$.
