# Relation between Hamiltonian of light ray and that of mechanics

1. Jun 23, 2014

### genxium

I'm learning ray optics and feeling so confused by the definition of "Hamiltonian of light".

What I learned was that the "Hamiltonian of light" defined by $H = n-|\vec{p}| = 0$ indicates the momentum conservation, where $n$ is refractive index and $\vec{p}$ here is the canonical momentum. The canonical momentum is defined by $\vec{p}=\frac{dL}{d\vec{r}'}=\frac{dL}{d(\frac{d\vec{r}}{ds})}$ where $\vec{r}$ is the position vector, $s$ is the path length and $L = n*|\vec{r}'|$ is the Lagrangian.

My questions are

1. $H$ of light is conserved, but is momentum of light conversed? If so how is it indicated in the equations?

2. $H$ of classical mechanics is $K+V$=kinective energy+potential energy, this is a clear physical meaning, but what does $H$ of light mean?

(Sorry for the long definition statement, I wanna make sure that people hold the same definition of things otherwise they can point out where I went wrong)

2. Jun 25, 2014

### UltrafastPED

3. Jun 25, 2014

### Andy Resnick

http://en.wikipedia.org/wiki/Hamiltonian_optics

Buchdahl's book is good, too.

1) Yes (both Snell's law and specular reflection conserve momentum)
2) It means rays are perpendicular to wavefronts.

4. Jun 27, 2014

### genxium

@Andy, thank you so much for the reply! It's informative but may I ask for more of question 1? What do you mean by "Snell's Law conserves momentum"?

I'm actually not clear about the definition of light momentum here, is it "momentum of ray" or "momentum of photon"? What is the instance that "owns" momentum? When I learned the canonical momentum mentioned in my question it was derived from Fermat's Principle thus I take it as "momentum of ray".

5. Jun 27, 2014

### Andy Resnick

This thread is predicated on using the geometrical (ray) model of light propagation; momentum of the ray is represented by the wavevector k, with |k|= 2*pi/λ.