Poynting Vector Always Perpendicular to Planes of same phase? I'm calculating this in a linear isotropic medium. For plane waves, the poynting vector is perpendicular to planes with the same phase: Let
$$
\vec{E}(\vec{x}, t) = \vec{E}_0 e^{i(\vec{k}\vec{x} - \omega t)}
$$
Then $\vec{S}$, the poyntingvector, will be parallel to $\vec{k}$. What about the case of a more general, monochromatic wave:
$$
\vec{E}(\vec{x}, t) = \vec{E}_0(\vec{x}) e^{i(k \chi(\vec{x}) - \omega t)}
$$
Will $\vec{S}$ still be paralell to $\nabla \chi$ in this case?
 A: According to https://en.wikipedia.org/wiki/Wave_vector,
"The direction in which the wave vector points must be distinguished from the "direction of wave propagation". The "direction of wave propagation" is the direction of a wave's energy flow, and the direction that a small wave packet will move, i.e. the direction of the group velocity. For light waves, this is also the direction of the Poynting vector. On the other hand, the wave vector points in the direction of phase velocity. In other words, the wave vector points in the normal direction to the surfaces of constant phase, also called wave fronts.
In a lossless isotropic medium such as air, any gas, any liquid, or some solids (such as glass), the direction of the wavevector is exactly the same as the direction of wave propagation. If the medium is lossy, the wave vector in general points in directions other than that of wave propagation. The condition for wave vector to point in the same direction in which the wave propagates is that the wave has to be homogeneous, which isn't necessarily satisfied when the medium is lossy. In a homogeneous wave, the surfaces of constant phase are also surfaces of constant amplitude. In case of inhomogeneous waves, these two species of surfaces differ in orientation. Wave vector is always perpendicular to surfaces of constant phase."
This sounds reasonable, but you may wish to look for a more reliable source.
