deletions | additions       diff --git a/Introduction.tex b/Introduction.tex  index ef3f7e8..de07a68 100644  --- a/Introduction.tex  +++ b/Introduction.tex 
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u_z = \frac{(3\lambda + 2\mu)}{(\lambda + 2\mu)} \frac{\alpha E \zeta}{\rho C S \zeta}  % \label{eq:deplUnidim}  \end{equation}  As in a biological soft tissues, $\mu \gg \lambda$, $u_z \approx 3 the displacement $u_z$ can be approximated as $3  \alpha \frac{E}{\rho C S}$. Taking as an order of magnitude $\alpha$ = 70.10$^{-6}$ K$^{-1}$ (water linear thermal dilatation coefficient), $E$ = 200 mJ, $\rho$ = 1000 kg.m$^{-3}$ (water density), $C$ = 4180 kg.m$^{-3}$ (water calorific capacity) and $S$ = 20 mm$^2$, we obtain a displacement $u_z$= 0.5 $\mu$m. This value is slightly smaller than the typical ultrasound resolution of displacement, typically of a few micrometers. In a tridimensional model, displacements along X and Y axis would also occurs, as the local expansion acts as dipolar forces parallel to the surface. In the ablative regime, the local increase of temperature is so high that the surface of the medium is vaporized. This phenomenon creates a stress $\sigma$ in the medium, given by \cite{scruby1990laser}:  \begin{equation} 
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