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The absorption of the laser beam by the medium gives then rise to an absorbed optical energy $\gamma I$. Assuming that all the optical energy is converted to heat, a local increase of temperature appears. Temperature distribution $T$ in absence of convection and of phase transition can be computed using the heat equation:  \begin{equation}  \frac{\partial T}{\partial t} = \kappa \nabla ^2 T + \frac{\gamma I}{\rho C}  \label{eq:eqChaleur}  \end{equation}  where $\rho$ is the density, $C$ the heat capacity and $\kappa$ the thermal diffusivity. The thermal diffusion path, equal to $\sqrt{4\kappa \tau}$, with $\tau$ = 10 ns the laser emission duration and $\kappa$ = 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}  \label{eq:eqChaleurApprox}  \end{equation}  Substituting the experimental parameters ($\gamma^{-1} \approx$ 50 $\mu$m$^{-1}$, $S$ = 20 mm$^{2}$, $E$ = 0.2 J, $\rho \approx$ 1000 kg.m$^{-3}$, $C \approx$ 4180 J.kg$^{-1}$.K$^{-1}$) lead to a maximum increase of temperature of 60 K.  This local increase of temperature lead to a local dilatation of the medium occurs. We suppose that the medium is homogeneous and isotropic, and as the depth of absorption (about 40 $\mu$m) is small compared to the beam diameter (5 mm), we have adopted a 1D model. The stress $\sigma_{zz}$ is the sum between the axial strain component and the thermal expansion component \cite{scruby1990laser}:  \begin{equation}  \sigma_{zz} = (\lambda + 2 \mu) \frac{\partial u_z}{\partial z} - 3(\lambda + \frac{2}{3}\mu) \frac{\alpha E}{\rho C S \zeta}  \label{eq:stressThermo}  \end{equation}  where $\lambda + 2 \mu$ is the P-wave modulus, $\lambda + \frac{2}{3}\mu$ the bulk modulus with $\lambda$ and $\mu$ respectively the first and second Lamé's coefficient, $\alpha$ the thermal dilatation coefficient and $\zeta$ the depth over which there is a temperature rise. In the absence of external constraints normal to the surface, the stress across the surface must be zero, i.e. $\sigma_{zz} (z=0) = 0$, so that equation \ref{eq:stressThermo} can be integrated, giving a displacement $u_z$ from the surface:  \begin{equation}  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, $\mu \ll \lambda$, the displacement can be approximated as:  \begin{equation}  u_z = \frac{3 \alpha E}{\rho C S}  \label{eq:deplThermoApprox}  \end{equation}  Substituting $\alpha$ = 70.10$^{-6}$ K$^{-1}$ (water linear thermal dilatation coefficient), $E$ = 0.2 J, $\rho$ = 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 experimental displacement (about 3 $\mu$m). This local displacement can lead to shear waves because of the limited size of the source. 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, but calculus is beyond the scope of this article.  With stronger or more focused laser pulse, the local increase of temperature could also vaporize a part of the surface of the medium. However in our case, the temperature did not increase enough (about 60 K) to vaporize the medium.