## In the vicinity of a perfect reflector, there will be incoming and outgoing waves of equal and opposite momentum. Does this imply the Poynting vector is locally zero, and if so, how do we account for the presence of radiation pressure on the reflector, which is proportional to the Poynting vector?

The Poynting vector doesn't tell you anything about the pressure. The Poynting vector tells you about the flow of electromagnetic

Now that we've cleared that up, one key thing to remember about electromagnetic plane waves is that they satisfy the right-hand rule, that is, [math]\mathbf{E} \times \mathbf{B}[/math] points in the direction of propagation. That implies that if you have another wave going in the opposite direction, then wherever you have constructive interference in E, you have destructive interference in B, and

The Poynting vector does indeed vanish, and you might be wondering how the reflector can then pick up kinetic energy as electromagnetic radiation falls upon it. The answer is rather interesting. When the reflector is stationary, the time derivative of its kinetic energy is zero, since kinetic energy is

*energy,*but it doesn't tell you about the flow of*momentum*. The object describing the flow of momentum is called the*Maxwell stress tensor*(MST). Like the Poynting vector, the Maxwell stress tensor is quadratic in the fields.Now that we've cleared that up, one key thing to remember about electromagnetic plane waves is that they satisfy the right-hand rule, that is, [math]\mathbf{E} \times \mathbf{B}[/math] points in the direction of propagation. That implies that if you have another wave going in the opposite direction, then wherever you have constructive interference in E, you have destructive interference in B, and

*vice versa*. Unlike the Poynting vector, the MST doesn't vanish when only one of the fields vanishes. So there is no problem; even though the electric field is perfectly cancelled right outside the surface, the magnetic field isn't.The Poynting vector does indeed vanish, and you might be wondering how the reflector can then pick up kinetic energy as electromagnetic radiation falls upon it. The answer is rather interesting. When the reflector is stationary, the time derivative of its kinetic energy is zero, since kinetic energy is

*quadratic*in velocity. So there is no problem with conservation of energy. When the reflector is moving away from the incident wave, the reflected wave is red shifted relative to the incident wave, so the energy flux of the incident wave is only partially cancelled by the energy flux of the reflected wave, and the net flow of energy is in the direction of the incident wave, toward the reflector, which is just what you'd expect since the reflector will accelerate. When the reflector is moving toward the incident wave, the reflected wave is blue shifted relative to the incident wave, so the net flow of energy is in the direction of the reflected wave, away from the reflector, which is again exactly what you'd expect since the reflector will decelerate. This is another example of how classical electrodynamics and relativity depend upon each other for consistency.