deletions | additions       diff --git a/The_absorption_of_the_laser__.tex b/The_absorption_of_the_laser__.tex  index c9e5c04..0e28bde 100644  --- a/The_absorption_of_the_laser__.tex  +++ b/The_absorption_of_the_laser__.tex 
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The absorption of the laser beam by the medium gives then rise to an absorbed optical energy $\gamma I$. Assuming that all the optical energy is converted to heat, a local increase of temperature appears. Temperature distribution $T$ in absence of convection and of phase transition can be computed using the heat equation:  \begin{equation} %\begin{equation}  %  k \nabla ^2 T = \rho C \frac{\partial T}{\partial t} - \gamma I %  \label{eq:eqChaleur} \end{equation}  where %\end{equation}  %where  $\rho$ is the density, $C$ the heat capacity and $k$ the thermal conductivity. Calculating the exact solution of this equation is behind the scope of this article, but during laser emission, we can roughly approximate  the first termcan be approximated  by $k T / \gamma^2$ and the second by $\rho C T / \tau$. Taking $k \approx$ $k$ =  0.6 W.m$^{-1}$.K$^{-1}$, $\rho$ = 1000 kg.m$^{-3}$, $C$ = 4180 J.kg$^{-1}$.m$^{-3}$ (water thermal conductivity, density and heat capacity respectively), $\gamma \approx$ 40 $\mu$m and $\tau$ = 10 ns, the first term is negligible compared to the second one, so that equation \ref{eq:eqChaleur} can be simplified as:%The thermal diffusion path, equal to $\sqrt{4 k \tau}$, with $\tau$ = 10 ns the laser emission duration and $k$ = 1.43 10$^{-7}$ m$^2$.s$^{-1}$ for water \cite{Blumm_2003}, is approximately equal here to 80 nm. As $\gamma^{-1} \gg \sqrt{4\kappa t}$, propagation of heat is negligible during laser emission, so that equation \ref{eq:eqChaleur} can be simplified as: \begin{equation} %\begin{equation}  %  \frac{\partial T}{\partial t} = \frac{\gamma I}{\rho C} = \frac{\gamma}{\rho C S} \frac{dE}{dt} %  \label{eq:eqChaleurApprox} \end{equation} %\end{equation}  Substituting low-energy experimental parameters ($\gamma^{-1} \approx$ 40 $\mu$m$^{-1}$, $S$ = 20 mm$^{2}$, $E$ ($E$  = 10 mJ, $\rho$ = 1000 kg.m$^{-3}$, $C$ $S$  = 4180 J.kg$^{-1}$.K$^{-1}$) 20 mm$^{2}$)  leads to a maximum increase of temperature of 3 K. This increase of temperature gives rise to a local dilatation of the medium. The induced displacements can then generate to shear waves: this constitutes the \textit{thermoelastic regime}. To describe physically this regime, we supposed the medium as homogeneous and isotropic. And as the depth of absorption (about 40 $\mu$m) is hundred times smaller than the beam diameter (5 mm), we adopted a 1D model. The stress $\sigma_{zz}$ is the sum between the axial strain component and the thermal expansion component \cite{scruby1990laser}:  \begin{equation}