deletions | additions       diff --git a/Simu disp maps.tex b/Simu disp maps.tex  index 03e5093..a7bb22f 100644  --- a/Simu disp maps.tex  +++ b/Simu disp maps.tex 
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Similarly to the thermoelastic regime, this leads to a displacement $u_z$:  \begin{equation}  u_z = \frac{\zeta}{\rho (\lambda + 2 \mu)}\frac{I^2}{(L+C(T_V-T_0))^2}\approx \frac{\zeta}{\rho \lambda}\frac{E^2/S^2 \tau^2}{(L+C(T_V-T_0))^2} \mu)}\frac{I^2}{(L+C(T_V-T_0))^2}  \label{eq:deplAbla}  \end{equation}  As in a biological soft tissues, $\mu \ll \lambda$, the displacement $u_z$ can be approximated as $\frac{\zeta}{\rho \lambda}\frac{E^2/S^2 \tau^2}{(L+C(T_V-T_0))^2}$. \lambda}\frac{I}{(L+C(T_V-T_0))^2}$.  Estimating $\zeta$ equal to 50 $\mu$m (average depth of absorption), $\lambda$ = 2 GPa (first Lamé's coefficient of water), $L$ = 2.2 MJ.kg$^{-1}$ (vaporization latent heat of water), $C$ = 4180 J.kg$^{-1}$.K$^{-1}$ (water heat capacity), $T_V-T_0$ = 373-293 = 80 K (water vaporization temperature minus laboratory temperature), $\rho$ = 1000 kg.m$^{-3}$ (water density), $E$ = 200 mJ, $S$ = 20 mm$^2$ and $\tau$ = 10 ns, we obtain a displacement $u_z$ approximately equal to 3.9 $\mu$m. This is in the order of magnitude of the experimental measured displacement (2 $\mu$m). A numerical calculation was then performed. The vaporization was modeled as a point force directed along Z direction with a depth of 50 $\mu$m and increasing linearly from -2.5 to 0 mm and decreasing symmetrically from 0 to 2.5 mm, to simulate an approximate Gaussian shape. Propagation as a shear wave was calculated using Green operators $G_{zz}$ \cite{aki1980quantitative}:  \begin{equation}