Tobias C Hinse edited untitled.tex  over 8 years ago

Commit id: 88de92dc60bf509ddab66ffaf59e4b616dc20cd3

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Based on the above result I cannot see that the residuals relative to a linear ephemeris allow the inclusion of a secular term accounting for a quadratic ephemeris. The $\chi^2$ increases with an extra parameter which is not what is expected. I will continue now and fit a 1- and 2-companion model.  \section{Linear + 1-companion LTT model using MPFIT}  We have considered a linear + 1-LTT model (excluding secular changes as described in a quadratic ephemeris). We have again used MPFIT for this task. The model is taken from Irwin (19??). We considered $10^7$ initial guesses. The initial guess for the reference epoch and binary period were taken from the best-fit obtained from a linear ephemeris model. Inital guesses for the semi-amplitude of the light-time orbit were taken from an estimate of the amplitude as shown in Fig. 2. Initial guesses for the eccentricity covered the interval [0,0.9995]. Initial guess for the argument of pericenter covered the interval [0,360] degrees. Initial guess for the orbital period was also estimated from Fig. 2. Initial guess for the time of pericenter passage were obtained from T0 and the orbital period of the light-time orbit. Initial guesses were drawn at random. The methodology follows the same techniques as described in Hinse et al. (2012). Best-fit parameters were obtained from the best-fit solution covariance matrix as returned by MPFIT. Parameters errors should be considered as formal. The best-fit had a $\chi^2=185.2$ with (42-7) degrees of freedom resulting in a reduced $\chi^2_{\nu}=5.3$. The corresponding RMS scatter of data points around the best-fit is 15.7 seconds. The  best-fit parameters are shown listed  in Table \ref{BestFitParamsLinPlus1LTT}. \ref{BestFitParamsLinPlus1LTT} and shown in Fig.~\ref{BestFitModel_LinPlus1LTT}. Recalling the average timing error (of 42 timing measurements) to be 6 seconds, that means that the RMS residuals are on a $2.6\sigma$ level.  \begin{table}   \begin{tabular}{ c c }  $T_0$ (BJD) & $2,450,021.77924 \pm 3 \times 10^{-5}$ \\