# Eclipsing binaries (references to data and models?)

1. May 27, 2006

### robphy

For some reason, I developed a sudden interest in eclipsing binary stars today and observational data from them.

Specifically, I'm trying to understand what would be "seen" from
a pair of identical sources
rotating with the same constant speed on opposite sides
in a common circular orbit
when viewed edge-on by an observer in the plane of the orbit.

I'm interested in both the low-orbital-speed case and the high-orbital-speed [i.e. relativistic] case.

Can anyone point me to references to models and observational data?

Thanks.

2. May 27, 2006

### tony873004

I wouldn't imagine relativity would play a significant role since stars with fast orbits have velocities that are in the 100's of km/s, maybe low 1000's for some really special cases. But this still well below the speed of light.

If your view is edge-on, you'll have 1 total eclipse, and 1 annular eclipse (unless the stars are exactly the same size), as the larger star can completely cover the smaller star, but the smaller star can not completely cover the larger star.

You'll have a light curve caused by a disk (or ellipsoid if they're spinning fast) passing in front of another disk. The area of 1 circle is covering the area of another circle. Limb darkening will play a role as well. A star is not uniformily bright over the entire disk.

Just my guesses... Sorry for no references.

3. May 28, 2006

### robphy

My underlying question is this:
do the "observed star positions in the sky" vary sinuosoidally (assuming a common circular orbit for the two identical stars) with respect to the axis of rotation?

4. May 28, 2006

### George Jones

Staff Emeritus
Yes, viewed edge-on, the (distant, slow-moving) stars look like they move with simple harmonic motion along a line.

[Edit: I might have a go at calculating what is "seen" as orbital speed increases, i.e., when departures from observed SHO are noticeable.]

Note that, even in this slow-moving case, relativity does play a role in the relationship between the "seen" locations of the stars.

Suppose that relativity is not true, and that the velocity of observed light depends on the relative velocity between source and observer. If star A is moving towards us and, consequently, star B is moving away from us, than the speed of light from A is larger than the speed of light from B. If B is moving towards us and A away, than the speeds are the other war round.

The foumula for time of light travel, t = d/v, then shows that the motion that we "see" will be quite complicated.

Last edited: May 28, 2006
5. May 28, 2006

### robphy

Please do have a go at the calculation.
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, this is precisely the issue that I am looking at (de Sitter's argument).

6. Jun 3, 2006

### George Jones

Staff Emeritus
I've attached a couple of graphs that resulted from some calculations that I made. The top graph shows the image amplitude vs observer time (in years) for a slow-moving binary system. The bottom graph zooms in on a portion of the graph where the images coincide.

The coincidence event in this case occurs off-axis at an amplitude that is roughly 1/30000 of the full amplitude of the observed motion.

Later, I will post the details of the calculation so that it can be checked. If this doesn't happen today, it might not happen for a few days.

File size:
19.3 KB
Views:
64
7. Jun 9, 2006

### George Jones

Staff Emeritus
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 $R$. Assume that the centre of the orbits is on the x-axis a distance $D$ 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

\begin{align*} x_{1}\left( t\right) & =D+R\cos\omega t\\ y_{1}\left( t\right) & =R\sin\omega t\\ x_{2}\left( t\right) & =D-R\cos\omega t\\ y_{2}\left( t\right) & =-R\sin\omega t. \end{align*}

Now Consider an imaginary screen that is a small distance $d$ (where is $d$ on the order of 1 metre or less) in front of the observer, i.e., that lies in the plane $x=d$. 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 $h$ of the image on the screen of a star is given by

$$h\left( t\right) =\frac{d}{x\left( t\right) }y\left( t\right).$$

Of course, the speed of light is not infinite, so the actual times of reception $t_{r}$ of the images on the screen have to be determined. If $t_{e}$ is the time of emission, then the above equation becomes

$$h\left( t_{r}\right) =\frac{d}{x\left( t_{e}\right) }y\left( t_{e}\right).$$

The time of travel for light is just the difference between the time of reception $t_{r}$ and the time of emission $t_{e}$, and consequently

\begin{align*} t_{r} & =t_{e}+\sqrt{x\left( t_{e}\right) ^{2}+y\left( t_{e}\right) ^{2} }/c\\ & =t_{e}+\frac{x\left( t_{e}\right) }{c}\sqrt{1+\left( \frac{y\left( t_{e}\right) }{x\left( t_{e}\right) }\right) ^{2}}. \end{align*}

Since $D\gg R$, to a first approximation $x\thickapprox D$ and $y/x\thickapprox0$, so $t=t_{r}-D/c$. This results in, e.g.,

$$h_{1}\left( t_{r}\right) =\frac{d}{D}R\sin\left[ \omega\left( t_{r}-D/c\right) \right]$$]

for star 1. When no approximations are made, $h$ versus $t_{r}$ can be plotted parametrically using $t_{e}$ as the parameter, e.g., plot

\begin{align*} \left( t_{r},h_{1}\right) & =\left( t_{e}+\frac{x_{1}\left( t_{e}\right) }{c}\sqrt{1+\left( \frac{y_{1}\left( t_{e}\right) } {x_{1}\left( t_{e}\right) }\right) ^{2}},\frac{d}{x_{1}\left( t_{e}\right) }y_{1}\left( t_{e}\right) \right) \\ & =\left( t_{e}+\frac{D+R\cos\omega t_{e}}{c}\sqrt{1+\left( \frac {R\sin\omega t_{e}}{D+R\cos\omega t_{e}}\right) ^{2}},\frac{d}{D+R\cos\omega t_{e}}R\sin\omega t_{e}\right) \end{align*}

for star 1.

Last edited: Jun 9, 2006
8. Jun 9, 2006

### robphy

I'm off to a conference... so I might not be able to respond so easily.
I'll check your calculation against the calculations in my simulation.
Here's a crude clip