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Jim Fuller edited Mode Visibility.tex
almost 9 years ago
Commit id: 5a11c5abc03ad4c33931a87f6fd656f96e243d75
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The degree of wave transmission between the core and envelope is determined by the tunneling integral through the intervening evanescent zone. The transmission coefficient is
\begin{equation}
\label{eqn:integral2}
T
\simeq \sim \bigg( \frac{r_1}{r_2} \bigg)^{\sqrt{l(l+1)}} \, ,
\end{equation}
where $r_1$ and $r_2$ are the lower and upper boundaries of the evanescent zone, respectively. For waves of the same frequency, larger values of $\ell$ have larger values of $r_2$, thus equation \ref{eqn:integral2} demonstrates that high $\ell$ waves have much smaller transmission coefficients through the evanescent zone. The fraction of wave energy transmitted through the evanescent zone is $T^2$.
To estimate the reduced mode visibility due to energy loss in the core, we assume that all mode energy which leaks into the g mode cavity is completely lost. The mode then loses a fraction $T^2$ of its energy in a time $2 t_{\rm cross}$, where $t_{\rm cross}$ is the wave crossing time of the acoustic cavity.
{\bf Due to the larger energy loss rate, the mode has less energy $E_{\rm ac}$ within the acoustic cavity and produces a smaller luminosity fluctuation $V$ at the stellar surface, whose amplitude scales as $V^2 \propto E_{\m ac}$.} We show (supplementary online text) that the ratio of visibility between a suppressed mode and its non-suppressed counterpart is
\begin{equation}
\label{eqn:vis}
\frac{V_{\rm sup}^2}{V_{\rm norm}^2} = \bigg[1 + \Delta \nu \, \tau \, T^2 \bigg]^{-1} \, ,
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