robphy said:
In particular, (if my calcuation is correct) it seems that for large orbital speeds, the eclipse does not appear in line with the axis of rotation... but on the receding side of the axis. Does that agree with your calculation?
Yes, I think so. This is because, roughly, the light from the back star has to traverse the diameter of orbit, which takes time, while the light from the front star doesn't.
Here is my analysis - I hope there aren't too many mistakes.
Consider a binary star system that consists of two equal-mass stars that orbit their center of mass on a circle that has radius [itex]R[/itex]. Assume that the centre of the orbits is on the x-axis a distance [itex]D[/itex] from the origin, and that an observer is at the origin of an inertial coordinate system. In the observer's coordinate system, the positions of the two stars are
[tex]
\begin{align*}<br />
x_{1}\left( t\right) & =D+R\cos\omega t\\<br />
y_{1}\left( t\right) & =R\sin\omega t\\<br />
x_{2}\left( t\right) & =D-R\cos\omega t\\<br />
y_{2}\left( t\right) & =-R\sin\omega t.<br />
\end{align*}[/tex]
Now Consider an imaginary screen that is a small distance [itex]d[/itex] (where is [itex]d[/itex] on the order of 1 metre or less) in front of the observer, i.e., that lies in the plane [itex]x=d[/itex]. Light that travels from the stars to the observer passes through this imaginary screen. Using similar triangles, and assuming the speed of light is infinite, the height [itex]h[/itex] of the image on the screen of a star is given by
[tex]
h\left( t\right) =\frac{d}{x\left( t\right) }y\left( t\right).[/tex]
Of course, the speed of light is not infinite, so the actual times of reception [itex]t_{r}[/itex] of the images on the screen have to be determined. If [itex]t_{e}[/itex] is the time of emission, then the above equation becomes
[tex]
h\left( t_{r}\right) =\frac{d}{x\left( t_{e}\right) }y\left( t_{e}\right).[/tex]
The time of travel for light is just the difference between the time of reception [itex]t_{r}[/itex] and the time of emission [itex]t_{e}[/itex], and consequently
[tex]
\begin{align*}<br />
t_{r} & =t_{e}+\sqrt{x\left( t_{e}\right) ^{2}+y\left( t_{e}\right) ^{2}<br />
}/c\\<br />
& =t_{e}+\frac{x\left( t_{e}\right) }{c}\sqrt{1+\left( \frac{y\left(<br />
t_{e}\right) }{x\left( t_{e}\right) }\right) ^{2}}.<br />
\end{align*}[/tex]
Since [itex]D\gg R[/itex], to a first approximation [itex]x\thickapprox D[/itex] and [itex]y/x\thickapprox0[/itex], so [itex]t=t_{r}-D/c[/itex]. This results in, e.g.,
[tex]
h_{1}\left( t_{r}\right) =\frac{d}{D}R\sin\left[ \omega\left(<br />
t_{r}-D/c\right) \right][/tex]]
for star 1. When no approximations are made, [itex]h[/itex] versus [itex]t_{r}[/itex] can be plotted parametrically using [itex]t_{e}[/itex] as the parameter, e.g., plot
[tex]
\begin{align*}<br />
\left( t_{r},h_{1}\right) & =\left( t_{e}+\frac{x_{1}\left(<br />
t_{e}\right) }{c}\sqrt{1+\left( \frac{y_{1}\left( t_{e}\right) }<br />
{x_{1}\left( t_{e}\right) }\right) ^{2}},\frac{d}{x_{1}\left(<br />
t_{e}\right) }y_{1}\left( t_{e}\right) \right) \\<br />
& =\left( t_{e}+\frac{D+R\cos\omega t_{e}}{c}\sqrt{1+\left( \frac<br />
{R\sin\omega t_{e}}{D+R\cos\omega t_{e}}\right) ^{2}},\frac{d}{D+R\cos\omega<br />
t_{e}}R\sin\omega t_{e}\right)<br />
\end{align*}[/tex]
for star 1.
