Poynting vector direction

Question

Just a quick question if I may.

The Poynting vector, or the energy flux density, is given by:

$mathbf{S} = frac{1}{mu_{0}}(mathbf{E} times mathbf{B})$

So it's the cross product between the $mathbf{E}$-field and $mathbf{B}$-field. So depending on the direction of the fields, the Poynting vector will point in some direction. So lets say the $mathbf{E}$-field has the direction $mathbf{e}_{y}$ and the $mathbf{B}$-field has the direction $mathbf{e}_{z}$, then the resulting direction for $mathbf{S}$ will be $mathbf{e}_{z}$.

So my question is, is that the direction of which the energy is flowing, or is there some fancy thing I need to know, like it's the opposite or something like that ?

Thanks in advance :)

Ondřej Černotík · Accepted Answer

The Poynting vector was defined as directional energy flux density. Therefore, it naturally shows the way energy flows and you do not have to switch the direction or anything. So, if you have an $mathbf{E}$-field in the direction $mathbf{e}_y$ and $mathbf{B}$-field in the direction $mathbf{e}_z$, Poynting vector is in the direction $mathbf{e}_x$ and that is the direction in which energy flows.

Correct answer by Ondřej Černotík on August 1, 2020

haneka.w · Answer

The Poynting vector always points in the same direction of propagation as the fields. In this case would be $textbf{e}_x$. Note that $textbf{e}_y$ is the polarization of the $textbf{E}$-field, and $textbf{e}_z$ is the polarization of the $textbf{B}$-field. In other words,
$$text{direction of Poynting vector} = text{direction of propagation of the electromagnetic wave} $$

Poynting vector direction

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