Equation of a reflected electromagnetic wave

Question

According to my textbook, for an electromagnetic wave travelling to the left: $$ E = E_0 \cdot \cos(kx - \omega t) $$ and $$ B = -B_0 \cdot \cos(kx - \omega t) $$ where E is in the y-direction and B is the z direction (but I'm assuming this is just an arbitrary coordinate system chosen, and as long as they are perpendicular it is okay). If the electromagnetic wave is sent across a perpendicular conductor and reflected back then the reflected wave is flipped. To flip it we would need the Poynting vector to point in the opposite direction and $kx-\omega t$ to become $kx+\omega t$. To change the Poynting vector, in the book, they flip the electric field vector so it becomes $$ E = - E_0 \cdot \cos(kx + \omega t) $$

and the magnetic field becomes $$ -B = B_0 \cdot \cos(kx + \omega t) $$, where the magnetic field doesn't flip. Would it be equivalent if we flipped the magnetic field and not the electric field? The cross product (Poynting vector) would still face in the new direction of motion and as far as I can tell, the shape of the em-wave would still be the same so I don't see why not...

Answer 1

No, changing direction of the magnetic field would not be valid for the reflected wave.

You are forgetting the boundary conditions that must be satisfied for the fields at a conductor surface. In particular, the sum of tangental E fields of the incident and reflected waves must be zero at the surface.

Your suggested E and B field directions are valid for propagation in the opposite direction, but would have to be generated by another source, not by reflection.