How is Fermat's least time principle proven? Or it is what usually is observed and is basis for the theories?
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5Nothing in physics is ever proven (except for theorems inside a well-defined mathematical framework); a theory is always only right if it correctly describes what usually is observed. – leftaroundabout Oct 29 '13 at 22:00
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Related: https://physics.stackexchange.com/q/2041/2451 and links therein. – Qmechanic Jul 11 '15 at 15:17
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Related: https://physics.stackexchange.com/q/375170/ – Gavin R. Putland Aug 10 '18 at 07:54
1 Answers
For geometrical optics we can introduce eikonal $\Psi$ by relation $$ f = Ae^{-ik_{\mu}r^{\mu} + i\varphi} = Ae^{i\Psi}. \tag 1 $$ For small time interval and space lengths eikonal can be expanded in a form $$ \Psi = \Psi_{0} + \mathbf r \frac{\partial \Psi}{\partial \mathbf r} + t \frac{\partial \Psi}{\partial t}, $$ so, by comparing with left side $(1)$ it can be directly show that $$ \mathbf k = \frac{\partial \Psi}{\partial \mathbf r}, \quad \omega = -\frac{\partial \Psi}{\partial t}. \tag 2 $$ So by comparing $(2)$ with Hamilton-Jacobi equations we have clear analogy: wave vector acts rule of classical momentum and frequency acts rule of Hamiltonian in geometrical optics. So it is possible to introduce the analogy of principle of the least action for rays. For light it can be done by Maupertuis principle: $$ \delta S = \delta \int \mathbf p \,\mathrm d \mathbf l = 0 \to \delta \Psi = \delta \int \mathbf k\,\mathrm d \mathbf l = 0. $$ For example, for optically isotropic and homogeneous space $\mathbf k = \textrm{const} \cdot \mathbf n $, and $$ \delta \int \mathrm dl = 0, $$
which leads to Fermat's least time principle.

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Can you explain the symbols in your Eq. (1) i.e., $f, A, k_\mu, r_\mu,\varphi$ etc? I'm interested in the answer. @AndrewMcAddams – SRS Aug 11 '17 at 15:33