Parameters of binary simulations

Stellar rotation speed is important in this simulation. Research in the rotation period and gyrochronology has been down in M35 (S., Meibom, R., D., Mathieu and K., G., Stassun, 2009). The picture in the paper shows the period distribution of K and M type of stars.

The sun's rotation period is 24 days and the orbital period at sun's surface is 0.12 days. Therefore the picture above shows that:

\[\frac{\Omega}{\omega_{orb}}=\frac{P_{orb}}{P}\in(0.001,1)\]

and Skumanich's law is the relation between angular speed and the age of the star:

\[\Omega\sim t{-0.5}\]

The older the star, the slower it rotate. K and M type of stars may be low-mass orange dwarfs or K type giants, but the giants can not have such short rotation period, it must be main sequence low-mass orange stars. That's the tool the paper use to determine the age of the star but it also provide us some data.

Basically, the period of AGB or giants cannot be measured at this time because the Doppler effect would be too small at its photon sphere and they indeed rotation very slow. Thus its rotation speed is negligible compared to the break up speed. Here we assume there is a third star that is so close to the primary and it spin up the envelop of the primary's.

So we need some other mechanism to give the envelop angular momentum. In N., Soker and A., Harpaz, 2000, they give a formula of the ratio of the final angular frequency and break up angular frequency for horizontal branch stars:

\[\frac{\Omega}{\omega_{Kep}}=0.1(\frac{M_p}{0.01M_{env}})(\frac{a_0}{R}){0.5}(\frac{\alpha}{0.1}){-1}\]

where , is the coefficient that determined by the envelop's density profile.

is the mass of the planet (Jupiter like or BD), is the mass of the envelop's. is the initial separation of the planet and the star. is the radius of the star.

An example can be:

, , and

then the envelop will reach half of the break up speed.

I just realize that

\[\frac{\Omega}{\omega_{orb}}\]

should incorporate the effect of for gas. (Or should I? Because observation can not determine the )

# separationequatorial open anglemass ratio temperaturelinks to movie
1 4AU 2pi/9 0.5 100 K 0.67 0.005 1.mov​
2 4AU pi/3 0.5 100 K 0.67
3 4AU pi/3 0.5 100 K 0.05 0.8 contour​4AUtopview​​4AUsideview
4 4AU pi/3 0.5 200 K 0.2
5 3AU pi/3 0.5 100 K 0.05 1.08 3AUsideview​3AUtopview​
6 3AU pi/3 0.8 100 K 0.2 1.08
7 4AU pi/3 0.5 100 K 0.05 1.08 ​4AUsideview
8 6AU pi/3 0.5 100 K 0.05 1.08 ​6AUtopview6AUsideview​
9 4AU pi/2 0.5 100 K 0.67 0.0005 ​contour9_sideview.gif​
10 4AU pi/3 0.5 400 K 0.2 0.8 ​https://www.youtube.com/watch?v=tefDvaO4o9k
11 4AU 7pi/9 0.5 100 K 0 1.14
12 3 AU pi 0.5 100 K 0 1.08

(1, 2, 9) can investigate the effect of open angle.

(2, 3, 7) can investigate the effect of angular momentum of the equatorial outflow.

(3, 4, 10) can investigate the effect of temperature.

(3, 5, 8) can investigate the effect of separation.

(5, 6) can investigate the effect of mass ratio.

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