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I noticed from the units of $S$ that despite the notational similarity Fermat's principle $$ \delta S= \delta\int_{\mathbf{A}}^{\mathbf{B}} n \, ds =0 $$ is not a principle of least action but a principle of least length. Confusingly one often writes this principle even using a so called "optical Lagrangian" $L= n \frac{ds}{dx_3}$ as $$ \delta S= \delta\int_{\mathbf{A}}^{\mathbf{B}} L \, dx_3 =0 $$ However this Lagrangian doesn't have units of energy as the usual Lagrangian $L= T-V$ has. Instead the optical Lagrangian has no units. So I wondered what is missing to make Fermat's principle into a principle of least action and figured from the units that one has to multiply the principle with some "optical momentum" $p_O$ so $S$ becomes an action: $$ \delta S= p_O\ \delta \int_{\mathbf{A}}^{\mathbf{B}} n \, ds =0 $$ But what should this $p_O$ be? Since we are talking about light we could set $p_O=\hbar k = \frac{h}{\lambda}$. But Fermat's principle is the basis of geometrical optics, which can be derived as the limit of zero wavelength $\lambda \rightarrow 0$ from Maxwells wave equations of light. And a zero wavelength means light of infinite momentum according to $p_O= \frac{h}{\lambda}$. So it doesn't seem to make sense to multiply Fermat's principle with a momentum, because that would yield a infinite value for the action.

So is the reason why we cannot express Fermat's principle as principle of least action the fact that geometrical optics implies an infinite momentum for light?

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  1. Firstly, units & dimensions are not an issue, since a variational principle can always be multiplied with an appropriate constant.

  2. Secondly, the classical mechanical analogy of Fermat's principle of stationary time in geometrical optics does (at least superficially) not look like a stationary action principle, since initial and final times, $t_i$ and $t_f$, are kept fixed in the latter.

  3. Rather the classical mechanical analogy is Jacobi's formulation of Maupertuis' principle for the abbreviated action. Here the index of refraction $n({\bf r})$ should be identified with the momentum $p({\bf r})=\sqrt{2m(E-V({\bf r}))}$.

  4. Nevertheless, looking a bit deeper, Fermat's principle can in fact be formulated as a stationary action principle, see e.g. my Phys.SE answer here.

References:

  1. H. Goldstein, Classical Mechanics, 2nd ed.; Sections 8.6 & 10.8.
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