# Poynting vector and intensity in scalar diffraction theory

1. Jun 27, 2014

### ashita

The power density of an electromagnetic wave is proportional to the absolute square of the electric field |E|^2 (assuming a plane wave). Here, E is a vector so the absolute square involves all three of Ex, Ey, and Ez.

In homogeneous, linear media, it's easy to show that each component of E follows its own Helmholtz equation. This decouples the three components and allows one to define a unified scalar wave (usually U) that can represent any of the field components. This is the foundation of scalar diffraction theory.

In scalar diffraction theory, when people are interested in finding the intensity distribution at an image, they simply find |U|^2. A separate U is not found for Ex, Ey, and Ez. How is this an accurate representation of |E|^2, which includes all three field components? I have a Fourier optics book that claims these two quantities are directly proportional to each other, but I don't know how to show this.

Doesn't polarization play an important role in interference? When we apply Huygen's principle, why don't we worry about the polarization of the spherical waves at a point?

2. Jul 7, 2014

### rigetFrog

I remember reading some where that the lack of polarization role was very lucky for humanity, otherwise it would've taken many more decades to figure out interference.

The subtlety here is that the interference patterns doesn't arise from two particles. It's the wave function of a single particle that interferes with itself. I.e. the wave is interfering with something that has the same polarization.

3. Jul 9, 2014

### Andy Resnick

Usually when scalar theory is presented, the electric field is taken to be a plane wave (at least locally), and underlying that simplification the field is linearly polarized, and E ≠ 0 for only 1 component. Polarization is then 'recovered' by having 2 linearly independent field components.

The generalization of scalar theory, vector diffraction theory, is less commonly encountered but well-developed nonetheless. Vector theory becomes important for high numerical aperture systems, like microscope objectives.

I'm not sure I understand what the Poynting vector has to do with your question, although to be honest, I always associate the Poynting vector with the momentum of the field, not the energy density or intensity.