Is the Poynting vector continuous? Helmholtz equation for the magnetic field is
$$∇^2 \mathbf{H} + k^2 \mathbf{H} = \mathbf{0}$$
Assuming time-harmonic waves Ampere's law can be rearranged
$$\mathbf{E} = \frac{∇\times\mathbf{H}}{σ+jωε}$$
Thus the time averaged power (derived from the Poynting vector) is
$$P_{\text{avg}} = \frac{1}{2}\,\operatorname{Re}\, (\mathbf{E}\times\mathbf{H}^*)$$
$$P_{\text{avg}}  = \frac{1}{2}\,\operatorname{Re}\,(\frac{∇\times\mathbf{H}}{σ+jωε}\times\mathbf{H}^*)$$
For the sake of simplicity let $σ=0$
$$P_{\text{avg}} = \frac{1}{2ωε}\,\operatorname{Im}\,(\mathbf{H}\times(∇\times\mathbf{H})^*)$$
If am em wave travels from free space into a region that has a different permittivity, this formula seems to say that magnitude of power will change. Where did the power go? Is the Poynting vector continuous?
 A: The Poynting vector is the directional energy flux of the EM field.
It has a normal and tangential component.
S = E x H
Now at the edge of the media, it is the normal component that is continuous.
Why is the tangential component not continuous?


The Poynting vector appears in Poynting's theorem (see that article for the derivation), an energy-conservation law:


$${\frac {\partial u}{\partial t}}=-\mathbf {\nabla } \cdot \mathbf {S} -\mathbf {J_{\mathrm {f} }} \cdot \mathbf {E}$$
The first term in the right-hand side represents the electromagnetic energy flow into a small volume, while the second term subtracts the work done by the field on free electrical currents, which thereby exits from electromagnetic energy as dissipation, heat, etc.
https://en.wikipedia.org/wiki/Poynting_vector


The rate of energy transfer (per unit volume) from a region of space equals the rate of work done on a charge distribution plus the energy flux leaving that region.


https://en.wikipedia.org/wiki/Poynting%27s_theorem
At the edge of the two media, some of the energy is reflected or inelastically scattered, or absorbed.
Energy must be conserved, so some of the photons will be reflected (returning to the original medium), some will be inelastically scattered (giving some of their energy to the atoms in the new medium), and some will be absorbed (giving all their energy to the new media). 
A: No. It is the same reason why the electric field component perpendicular to an interface appears to decrease by a factor $1/\epsilon$ when going from free space to the material with different permittivity. Remember, it is the perpendicular electric displacement field which is continuous, not the perpendicular electric field. You may think of this as an "electric field screening" effect. 
A: Let me first remark that the Poynting vector is defined in terms of field $H$ rather than in terms of the "true" magnetic field $B$. Same is true for the energy density in a medium:
$$
u=\frac{1}{2}\left(\mathbf{E}\cdot\mathbf{D}+\mathbf{B}\cdot\mathbf{H}\right).
$$
Obviously, if the medium is non-conducting, there are will be no energy dissipating and the energy of the electromagnetic field should be conserved. There are two palces where this energy could go:

*

*the polarization and magnetization of the medium - indeed, $\epsilon$ and $\mu$ are the coefficients which actually relate the polarization and magnetization arising in the media in response to the electromagnetic field - the fictitious fields $D$ and $H$ being introduced merely for coenvenience.

*the reflected wave - if a wave is incident on a medium with different permittivity and permeability, part of it will be transmitted into the medium and part of it will be reflected. That is, the Poynting vector outside of the medium will be not the same, as if the medium was not there.

The Poynting vector will be continuous - as I mentioned in the beginning, the Poynting vector and the energy density are expressly defined as to take into account the dielectric and magnetic properties of a medium. Solving the scattering prolem of a wave incident on a dielectric half-space and demonstrating the continuity of the pointing vector could be a good excercise. As an extreme example one could consider a wave incident on a metal, where all the wave is reflected and the energy is conserved. In a case of a wave incident on a metallica half-space perpendicularly to its surface, it becomes a standing wave, and the net Poynting vector is zero!
A: By the Maxwel Faraday equation, when we are dealing with a plane wave normal to the interface, it is $ \vec{E}$ and not $\vec{D}$ which is continuous: $\frac{{\vec{D}}_2}{\epsilon_2}=\frac{{\vec{D}}_1}{\epsilon_1}$ and, by Maxwell Ampère's equation, we deduce that $\vec{\nabla}\bigwedge\vec{H}$ is also discontinuous: $ \frac{{(\vec{\nabla}\bigwedge\vec{H})}_2}{\epsilon_2}=\frac{{(\vec{\nabla}\bigwedge\vec{H})}_1}{\epsilon_1}$.
Since$ \vec{H}$ is continuous, the averaged Poynting vector is therefore continuous.
