Accuracy of orbital period measurement

Let’s say you have measured the HJD_0 at times t_1 and t_2, separated by N orbital cycles. Then the orbital period, P, becomes:

P=\dfrac{t_{2} - t_{1}}{N}…(1)

From error propagation theory we know that if:



\sigma^2_P=\left(\dfrac{\partial P}{\partial t_{1}} \sigma_{t_{1}}\right)^2+\left(\dfrac{\partial P}{\partial t_{2}} \sigma_{t_{2}}\right)^2

in which we are ignoring the partial derivative of P with respect to N because we consider \sigma_N to be zero. The equation becomes:

\sigma^2_P=\left(\dfrac{-\sigma_{t_1}}{N}\right)^2+ \left(\dfrac{\sigma_{t_2}}{N}\right)^2=\dfrac{\sigma_{t_1}^2+ \sigma_{t_2}^2}{N^2}

and thus:

\sigma_P=\dfrac{1}{N}\sqrt{\sigma_{t_1}^2+ \sigma_{t_2}^2}…(2)

Then using (1) in (2):

\sigma_P=\dfrac{P}{t_2 - t_1}\sqrt{\sigma_{t_1}^2+ \sigma_{t_2}^2}…(3)

In real life  t_1 and t_2 are typically reported with uncertainties of the order 1 \times 10^{-4} d. Therefore, for a system with period \sim6 h and t_1, t_2 separated by one year we expect an uncertainty of \sim 1 \times 10^{-7} d for the orbital period.

Roche lobe plots

For all the following I’m using gnuplot.

Isopotential curves


f(x,y)=2/(((x**2+y**2)**2)*(1+q)) +2*q/((((x-a)**2+y**2)**2)*(1+q)) +(x – q/(1+q))**2 +y**2
set view 0,0,1.3
set xlabel “x”
set ylabel “y”
set contours
set isosamples 100,100
set size ratio -1
unset surface
set cntrparam levels disc -2.2,-2.5,-3.55,-29.7,-100,-1000
unset ztics
splot [x=-6:6] [y=-6:6] -f(x,y)