Equação da Transferência Radiativa

Lei de Beer

Lei de Beer

Lei de Beer - coeficientes

\[ \beta_\lambda(s) = \beta_{a\lambda} + \beta_{e\lambda}, \]

\[ \beta_\lambda(s) = k_\lambda(s)\rho(s) = \sigma_\lambda(s) N(s),\]

Lei de Beer - Integral

\[ dL_\lambda(\Omega,s) = -L_\lambda(\Omega,s)\beta_\lambda(s)ds \rightarrow \frac{dL_\lambda(\Omega,s)}{L_\lambda(\Omega,s)} = -\beta_\lambda(s)ds,\] \[ \int_{L(s_1)}^{L(s_2)}\frac{dL_\lambda(\Omega,s)}{L_\lambda(\Omega,s)} = -\int_{s_1}^{s_2}\beta_\lambda(s)ds,\] \[ \ln L_\lambda(\Omega,s_2) - \ln L_\lambda(\Omega,s_1) = -\int_{s_1}^{s_2}\beta_\lambda(s)ds,\] \[ \ln\left(\frac{L_\lambda(\Omega,s_2)}{L_\lambda(\Omega,s_1)}\right) = -\int_{s_1}^{s_2}\beta_\lambda(s)ds \rightarrow L_\lambda(\Omega,s_2) = L_\lambda(\Omega,s_1) e^{-\int_{s_1}^{s_2}\beta_\lambda(s)ds}.\]

Definindo \(\tau = \int_{s_1}^{s_2}\beta_\lambda(s)ds\), temos a Lei de Beer: \(L=L_0e^{-\tau}\).

Exercício: