Lagrangian's formalism and Fermat's principle

Based on Lagrangian's formalism one can write Eq.(2.9) as

$$\tau = \displaystyle{ \int\limits_{A}^{B}} {\cal L} \; {ds} ,$$ (2.10)

where ${\cal L}$ represents the system's Lagrangian:

$${\cal L} = \frac{1}{c} .$$ (2.11)

Within the context of the formalism ${\cal L}$ is supposed to be a function of coordinates and generalized velocities[5]:

$${\cal L}(x,y,z,\dot{x},\dot{y},\dot{z}) = \frac{1}{c}\sqrt{ \dot{x}^2 + \dot{y}^2 + \dot{z}^2 } ,$$ (2.12)

where

$$\dot{x} = \frac{dx}{ds} , \hskip5mm \dot{y} = \frac{dy}{ds} , \hskip5mm \dot{z} = \frac{dz}{ds} .$$

In this way, the perturbation of the travel time corresponds to

$$\delta \tau = \displaystyle{ \int\limits_{A}^{B}} \left\{ \... ...L}}}{\partial{\dot{z}}}\delta \dot{z} \right] \right\} ds =$$

$$= \underbrace{\left. \frac{\partial{{\cal L}}}{\partial{\dot... ... L}}}{\partial{\dot{z}}}\delta z \right\vert _{A}^{B}}_{=0} +$$

$$+ \displaystyle{ \int\limits_{A}^{B}} \left\{ \left[ \fra... ...partial{\dot{z}}} \right) \right] \delta z \right\} ds .$$

According to Fermat's principle

$$\delta\tau = 0 ,$$

which implies that

$$\begin{array}{ccccc} \displaystyle{ \frac{d}{ds}\left( \frac... ...frac{\partial{{\cal L}}}{\partial{z}}} & = & 0 . \end{array}$$ (2.13)

On the other side, using Eq.(2.12) one can obtain that

$$\frac{\partial{{\cal L}}}{\partial{\dot{x}}} = \frac{\dot{x}... ...}} {\cal L} = {\cal L}\dot{x} = \frac{1}{c}\frac{dx}{ds} ,$$

$$\frac{\partial{{\cal L}}}{\partial{\dot{y}}} = \frac{\dot{y}... ...}} {\cal L} = {\cal L}\dot{y} = \frac{1}{c}\frac{dy}{ds} ,$$

$$\frac{\partial{{\cal L}}}{\partial{\dot{z}}} = \frac{\dot{z}... ...}} {\cal L} = {\cal L}\dot{z} = \frac{1}{c}\frac{dz}{ds} .$$

Therefore, one can conclude that the solution of the Eikonal equation requires the solution of the following system of equations:

$$\frac{d}{ds}\left( \frac{1}{c}\frac{dx}{ds} \right) = \frac{... ...partial{}}{\partial{y}}\left( \frac{1}{c} \right) \hskip5mm ,$$

$$\frac{d}{ds}\left( \frac{1}{c}\frac{dz}{ds} \right) = \frac{\partial{}}{\partial{z}}\left( \frac{1}{c} \right) .$$ (2.14)