Pol Grasland-Mongrain edited The_absorption_of_the_laser__.tex  over 8 years ago

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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 2  $\mu$m). This local displacement can lead to shear waves because of the limited size of the source. To verify this physical model, a numerical calculation was performed. The thermal dilatation was modeled as two opposite forces directed along Y direction with a depth of 50 $\mu$m and decreasing linearly from 2.5 to 0 mm (respectively -2.5 to 0 mm), to simulate an approximate Gaussian shape \cite{Davies_1993}. Propagation as a shear wave along Z axis was calculated using Green operators $G_{yz}$ as calculated by Aki Richards \cite{aki1980quantitative}:  \begin{equation}