deletions | additions       diff --git a/subsection_Mode_Visibility_Here_we__.tex b/subsection_Mode_Visibility_Here_we__.tex  index 2c847c5..fea0860 100644  --- a/subsection_Mode_Visibility_Here_we__.tex  +++ b/subsection_Mode_Visibility_Here_we__.tex 
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The energy of a mode within the envelope is proportional to its surface amplitude squared, hence, the visibility of a mode scales as $V_{\alpha} \propto E_{\alpha,{\rm ac}}$. Then the ratio of the visibility of the suppressed mode to that of the normal mode is  \begin{equation}  \frac{V_{\rm sup}^2}{V_\alpha^2} sup}^2}{V_{\rm norm}^2}  = \frac{E_{\rm ac}}{E_{\alpha,{\rm ac}}} \, . \end{equation}  Then equation \ref{eqn:ebalance} leads to  \begin{equation}  \frac{V_{\rm sup}^2}{V_\alpha^2} sup}^2}{V_{\rm norm}^2}  = \frac{\gamma_{\rm ac}}{\gamma_{\rm ac} + T^2/(2 t_{\rm cross})} \, . \end{equation}  Using the fact that the large frequency separation is $\Delta \nu \simeq (2 t_{\rm cross})^{-1}$ \citep{Chaplin_2013} and defining $\tau_{\rm ac} $\tau_0  = \gamma_{\rm ac}^{-1}$, we have our final result: \begin{equation}  \frac{V_{\rm sup}^2}{V_\alpha^2} sup}^2}{V_{\rm norm}^2}  = \bigg[1 + \Delta \nu \tau_{\rm ac} \tau_0  T^2 \bigg]^{-1} \, . \end{equation}