deletions | additions       diff --git a/The_absorption_of_the_laser__.tex b/The_absorption_of_the_laser__.tex  index cb32243..c9e5c04 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}  k \nabla ^2 T =  \rho C \frac{\partial T}{\partial t} = \kappa \nabla ^2 T + -  \gamma I \label{eq:eqChaleur}  \end{equation}  where $\rho$ is the density, $C$ the heat capacity and $\kappa$ $k$  the thermal diffusivity. The conductivity. Calculating the exact solution of this equation is behind the scope of this article, but during laser emission, the first term can be approximated by $k T / \gamma^2$ and the second by $\rho C T / \tau$. Taking $k \approx$ 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\kappa $\sqrt{4 k  \tau}$, with $\tau$ = 10 ns the laser emission duration and $\kappa$ $k$  = 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}  \frac{\partial T}{\partial t} = \frac{\gamma I}{\rho C} = \frac{\gamma}{\rho C S} \frac{dE}{dt}  \label{eq:eqChaleurApprox}