deletions | additions       diff --git a/The_absorption_of_the_laser__.tex b/The_absorption_of_the_laser__.tex  index 8791e04..a999351 100644  --- a/The_absorption_of_the_laser__.tex  +++ b/The_absorption_of_the_laser__.tex 
...
k \nabla ^2 T = \rho C \frac{\partial T}{\partial t} - \gamma I  \label{eq:eqChaleur}  \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 term by $k T / \gamma^2$ and the second by $\rho C T / \tau$. Taking $k$ = 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), $\gamma^{-1} \approx$ 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 simplifiedas:%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}  \frac{\partial T}{\partial t} = \frac{\gamma I}{\rho C} = \frac{\gamma}{\rho C S} \frac{dE}{dt}  \label{eq:eqChaleurApprox}