deletions | additions       diff --git a/Introduction.tex b/Introduction.tex  index 49b9eda..5de5d89 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:deplThermo}  \end{equation}  As in a biological soft tissues, $\mu \gg \ll  \lambda$, the displacement $u_z$ can be approximated as $\frac{3 \alpha E}{\rho C S}$. 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 3D 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 $\sigma_{zz}$ in the medium, as the sum of the P-wave modulus and a term given by the second law of motion \cite{scruby1990laser}:  \begin{equation} 
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