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Relation between Hamiltonian of light ray and that of mechanics 
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#1
Jun2314, 09:28 PM

P: 77

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 [itex]H = n\vec{p} = 0[/itex] indicates the momentum conservation, where [itex]n[/itex] is refractive index and [itex]\vec{p}[/itex] here is the canonical momentum. The canonical momentum is defined by [itex]\vec{p}=\frac{dL}{d\vec{r}'}=\frac{dL}{d(\frac{d\vec{r}}{ds})}[/itex] where [itex]\vec{r}[/itex] is the position vector, [itex]s[/itex] is the path length and [itex]L = n*\vec{r}' [/itex] is the Lagrangian. My questions are 1. [itex]H[/itex] of light is conserved, but is momentum of light conversed? If so how is it indicated in the equations? 2. [itex]H[/itex] of classical mechanics is [itex]K+V[/itex]=kinective energy+potential energy, this is a clear physical meaning, but what does [itex]H[/itex] 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
Jun2514, 09:23 AM

Sci Advisor
Thanks
PF Gold
P: 1,908

I don't have an immediate answer for you, but here is a lecture which goes over the same material. Look particularly at p. 34 and following:
https://www.fields.utoronto.ca/progr...esJuly2012.pdf 


#3
Jun2514, 06:31 PM

Sci Advisor
P: 5,510

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. 


#4
Jun2714, 02:12 AM

P: 77

Relation between Hamiltonian of light ray and that of mechanics
@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
Jun2714, 07:30 AM

Sci Advisor
P: 5,510

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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