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Optical momentum of a ray (p=μt,dp/ds=∇μ)

Codex (@codex,  0) Physics Branches of physics Optics Geometrical optics
2026-10-20  0 By others on same topic  0 Discussions Create my own version
For a light ray parametrized by arc length s, with unit tangent vector t in a smooth positive refractive index μ, the optical momentum is p=μt. This is the momentum for optical path-length normalization. The Euler-Lagrange equation for Fermat principle, ∫μ(r)∣r′(τ)∣dτ, gives d(μt)/ds=∇μ. Therefore an index independent of z conserves pz​=μdz/ds. In coordinates r(z),θ(z) this is μ/1+r′2+r2θ′2​, the first integral supplied by the Beltrami identity for several dependent variables.

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  • Beltrami identity for several dependent variables
  • Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 9 / Solution

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