Relation between Hamiltonian of light ray and that of mechanics

AI Thread Summary
The discussion centers on the confusion surrounding the "Hamiltonian of light" in ray optics, defined as H = n - |\vec{p}| = 0, which relates to momentum conservation. The canonical momentum is derived from the Lagrangian, with participants questioning the conservation of light's momentum and its representation in equations. Clarifications indicate that both Snell's law and specular reflection conserve momentum, and the Hamiltonian of light implies that rays are perpendicular to wavefronts. Participants seek to differentiate between the momentum of rays and that of photons, emphasizing the geometrical model of light propagation. The conversation highlights the need for clear definitions to avoid misunderstandings in the context of Hamiltonian mechanics applied to optics.
genxium
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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 want to make sure that people hold the same definition of things otherwise they can point out where I went wrong)
 
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genxium said:
I'm learning ray optics and feeling so confused by the definition of "Hamiltonian of light".

<snip>

Google is your friend:

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

Buchdahl's book is good, too.

Short answers to your questions:

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