# Poynting vector direction

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 ?

You are going to keep on collecting downvotes unless you change the result. It is in the $e_x$ direction, not $e_z$ – Eduardo Guerras Valera May 27 '13 at 23:17
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.