deletions | additions       diff --git a/Introduction.tex b/Introduction.tex  index eb74eed..b8a6104 100644  --- a/Introduction.tex  +++ b/Introduction.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} \lambda}\frac{E^2/S^2 \tau^2}{(L+C(T_V-T_0))^2}  \label{eq:deplAbla}  \end{equation}  Estimating $\zeta$ equal to 100 $\mu$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$^2$, $\tau$ = 20 ns, we obtain a displacement $u_z$ approximately equal to 2.0 $\mu$m. This value is four times higher than the one obtained with thermoelastic expansion. 
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