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Pol Grasland-Mongrain edited The_absorption_of_the_laser__.tex
over 8 years ago
Commit id: 1342f8889c1cc1f1aa40b1fa2201c693318510bc
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k \nabla ^2 T = \rho C \frac{\partial T}{\partial t} - \gamma I
\label{eq:eqChaleur}
\end{equation}
where $\rho$ is the density, $C$ the heat capacity and $k$ the thermal conductivity. Calculating the exact solution of this equation is behind the scope of this article, but during laser emission, we can roughly approximate the first term by $k T / \gamma^2$ and the second by $\rho C T / \tau$. Taking $k$ = 0.6 W.m$^{-1}$.K$^{-1}$, $\rho$ = 1000 kg.m$^{-3}$, $C$ = 4180 J.kg$^{-1}$.m$^{-3}$ (water thermal conductivity, density and heat capacity respectively),
$\gamma $\gamma^{-1} \approx$ 40 $\mu$m and $\tau$ = 10 ns, the first term is negligible compared to the second one, so that equation \ref{eq:eqChaleur} can be simplified as:%The thermal diffusion path, equal to $\sqrt{4 k \tau}$, with $\tau$ = 10 ns the laser emission duration and $k$ = 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} = \frac{\gamma}{\rho C S} \frac{dE}{dt}
\label{eq:eqChaleurApprox}