Consider that we have the electric field, $$\mathbf{E}=E_0\cos(kz-\omega t)\hat{\mathbf{x}}\tag{1}$$ and the magnetic field, $$\mathbf{B}=\frac{B_0}{c} \cos(kz-\omega t)\hat{\mathbf{y}}\tag{2}$$ These are the plane wave solution in vacuum, for an electromagnetic wave moving in the $z$ direction. The Maxwell stress tensor for this electromagnetic field has only one component given by, $$T_{zz}=-u=-\epsilon_0E_0^2\cos^2(kz-\omega t)\tag{3}$$ Where $u$ is energy density of the field. What this means is that the momentum transported by the field should be in the $z$ direction only. That should be evident. However I have some confusion regarding this.
Say a static electron free from any other forces is placed in the path of the given electromagnetic wave. Should this electron move along the $z$ direction then because the stress tensor says that a force acts perpendicular to the $z$ surface?
However from the Lorentz force relation, $$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})\tag{4}$$ We know that the force should be in the direction of the electric field. So which direction will this electron move towards?
I know I have some sort of misconception. In addition to this I would also like to know what sort of force are we talking about here when we are dealing with the stress tensor? Is it the same force as the Lorentz force? If it's not then what exactly is this momentum that the electromagnetic wave is carrying?
