deletions | additions       diff --git a/Physical model.tex b/Physical model.tex  index 7a2cc1c..2728c2b 100644  --- a/Physical model.tex  +++ b/Physical model.tex 
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\label{eq:expontentialDecay}  \end{equation}  where $R$ is the reflection coefficient of the material (supposed negligible on a black mat material as the one used here) and $\gamma$ the absorption coefficient of the medium. The absorption coefficient $\gamma$ can be estimated in two ways: by measuring the fraction of light going through different thickness of the medium or by calculating the skin depth $\delta = \frac{2}{\gamma}$ (the factor 2 is due to the fact that $\delta$ is related to magnitude of the electrical field while $\gamma$ is related to the magnitude of the optical energy, which is the square of the electrical field magnitude). Skin depth $\delta$ is equal to $(\pi \sigma \mu_r \mu_0 \nu)^{-\frac{1}{2}}$, where $\sigma$ is the electrical conductivity of the medium, $\mu_r \mu_0$ its permeability and $\nu$ the frequency of the radiation. Substituting $\sigma \approx$ 0.1 S.m$^{-1}$, $\mu_r \mu_0$ = 4 $\pi \times 10^{-7} H.m^{-1}$ and $\nu$ = 3 10$^8$ / 532 10$^{-9}$ = 5.6 10$^{14}$ Hz, the skin depth for our medium is about 70 $\mu$m: it means that about 63\% of the radiation energy is absorbed in the first 70 micrometers of the sample. We have validated experimentally this value by measuring the fraction of light which go through different thicknesses of the medium (respectively 0, 30, 50 and 100 $\mu$m) with a laser beam power measurement device (QE50LP-S-MB-D0 energy detector, Gentec, Qu\'ebec, QC, Canada). We found respective transmitted powers of 100\%, 88\%, 71\% and 57\%, as illustrated in the Figure \ref{figAbsorpExp}. An exponential fit indicated that $\gamma^{-1} \approx$ 50 $\mu m$ in our sample.  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.