# Can a wave's Poynting vector be in the opposite direction compared to its direction of propagation?

Can a wave's Poynting vector be in the opposite direction compared to its direction of propagation, and if so, what physical implications does it have?

As I understand, the poynting vector s can be represented as: s = E x H, both E and H being vectors, and since E and H are always perpendicular, they will form 90º and then the cross product will always be maximum and positive, but since I have been asked what the implications of the poynting vector having the opposite direction are, I am not sure.

• How would you define a positive or negative vector? A vector can point into any direction in space (opposed to numbers on the number line which can be left or right of the zero). Jan 14, 2019 at 9:16
• Depends on your coordinate system. If $\vec E=\hat i$ & $\vec H=\hat k$ then $\vec S=-\hat j$ Jan 14, 2019 at 9:25
• I define positive as in the same direction as the wave´s propagation direction Jan 14, 2019 at 9:41
• @harshit54: No, even if you fix a coordinate system, there is no way to define positive and negative vectors, only positive and negative components.
– user4552
Jan 14, 2019 at 14:21
• @BenCrowell Yes you are right. A negative vector does not make sense. The negative of a vector makes sense. Sorry for my wrong comment. Jan 14, 2019 at 14:39

In general the Poynting vector is the in the direction in which the wave propagates. Usually, for university level work etc, for the most part we already know the direction of the EM wave. Thus the Poynting vector comes out positive, and in the direction we have predefined. However, if, for instance, we rotate our coordinates, the Poynting vector will point in the direction of propagation in the new coordinate system.

Now the direction of propagation can for instance be in the direction of the negative z axis, which would yield a 'negative' vector.

Having said that, if by 'negative' one means that the amplitude is negative, i.e. the energy propagation is negative, this would imply the 'source' of the wave is actually gaining energy as opposed to radiating it. This is possible for instance when you have an antenna that receives radiation. What is meant by that is that the external wave is oscillating the charge in the antenna, thus deposits energy into the antenna.

In that case though you need to take the viewpoint of the charge at the antenna to get the so-called negative Poynting vector.

• Yes, by negative I mean that the poynting vector is in the opposite direction of the wave. So I understand from your reply that the poynting vector will ALWAYS be in the same direction of the wave´s propagation, and that it can´t be opposed to it. Jan 14, 2019 at 9:39
• @JesusChueca Actually the Poynting vector defines the direction of propagation. Think about it: Is there another way to define direction of propagation, than by the direction of energy flow, which the Poynting vector specifies? Jan 14, 2019 at 9:52
• I understand what you mean... and I do agree with you that the direction of propagation is the direction of energy flow, but in an electromagnetic wave where E and H are both always positive or negative at the same time, would it be possible to have a Poynting vector in the opposite sense of the propagation? Jan 14, 2019 at 9:56
• @Andreas H. the direction of propagation (wave vector) is not always the same direction as that of Poynting's vector. In some crystals and plastics (calcite, polystyrene), there can be EM wave propagating in one direction but Poynting vector pointing in a different direction. Jan 14, 2019 at 13:24

In negative index materials (where $$\varepsilon$$ and $$\mu$$ are both negative) the Poynting vector $$S$$ is antiparallel to $$k$$. Such negative index materials have been realized with metamaterials where the effective electric permittivity and effective magnetic permeability are both negative.

• What’s does it mean physically? I find it hard to interpret. That the direction of energy flow is not in the direction of the wave? What does that even mean? Jun 27, 2019 at 14:17

We need clear definitions to answer the question. The Poynting vector is $${\bf S} = {\bf E} \times {\bf H}$$ and the wave vector is in the direction of the gradient of the phase, such that the wave motion is $$\exp(i({\bf k} \cdot {\bf x} - \omega t))$$. The physical meaning is that energy flux travels in the direction of $$\bf S$$ and wavefronts move in the direction of $$\bf k$$. 4th Maxwell's eqn in dielectric medium (in absence of conduction current) is $$c \nabla \times {\bf H} = \dot{\bf D}$$ and plugging in a plane wave solution gives $$c i {\bf k} \times {\bf H} = -i \omega {\bf D}$$ so we have $${\bf H} \times {\bf k} = (\omega/c) {\bf D}.$$ I obtained this in order to be clear about the relative directions of $$\bf H$$, $$\bf k$$ and $$\bf D$$. We thus find that the Poynting vector $${\bf E} \times {\bf H}$$ can be in the opposite direction to $$\bf k$$ if $$\bf E$$ is in the opposite direction to $$\bf D$$. This can happen in certain unusual materials, as Max Lein says. In order to understand what is happening, you need to keep in mind that $$\bf D$$ includes polarization combined with $$\bf E$$, and also all the fields $$\bf E,D,H,B$$ are spatial averages. In such a wave as the energy moves in one direction the wavefronts are moving back in the other direction, which seems odd, but wavefronts are only statements about the evolution of the phase; they do not themselves constitute the movement of anything at all in the direction of $$\bf k$$.

In terms of the Fourier transform $$\sum_{\vec{k}}A_{\vec{k}}\text{e}^{i{\vec{k}\cdot\vec{x}}}$$, a wave from some sorce is a superposition of plane waves with different wave vectors $$\vec{k}$$, including the opposite directions too but with different amplitudes and phases, especially if $$\vec{x}$$ is "close" to the source. At the same time the Poynting's vector $$\propto \vec{E}(\vec{x})\times\vec{B}(\vec{x})$$ at a given point of space $$\vec{x}$$ is unique.

Each plane (running) wave has the propagation direction along its wave vector $$\vec{k}$$, so does the wave energy flow. But generally the EMF is not a plane wave, it "spreads" or "focuses" thanks to the presence of different spatial harmonics in its Fourier decomposition.