Return to Meeting Page
Useful Links
Project Pages (Stuff currently on my plate - ordered by priority)
Recent Blog Entries
Journal Club for March 22
Thoughts on binary sims
First we have position of primary
\vec{X_p}=R_p [\cos{\Omega t}, \sin{\Omega} t, 0]
and velocity of primary as
\vec{v_p}=R_p \Omega [-\sin{\Omega t}, \cos{\Omega t}, 0]
The velocity v_0 at the surface of the primary x_0 can be written as
\vec{v}_0=\vec{v_p}+\hat{x}_0 v_w
where
\hat{x}_0 = \frac{\vec{x}_0-\vec{X_p}}{S} and |\vec{x}_0-\vec{X}_p| = S
And then the velocity is constant along characteristics given by \vec{x}-\vec{x}_0 = \vec{v}_0 (t-t_0)
Given \vec{x} and t we wish to find \vec{x}_0 and t_0 (and \vec{v}_0=v) so that \vec{x}-\vec{x}_0 = \vec{v}_0 (t-t_0)
substituting the expression for \vec{v}_0 we have
\vec{x}-\vec{x}_0(t_0) = (\vec{v_p}(t_0)+\hat{x}_0(t_0) v_w) (t-t_0)
or
\vec{x}-\vec{x}_0(t_0) = (\vec{v_p}(t_0)+\frac{\vec{x}_0(t_0)-\vec{X_p}(t_0)}{|\vec{x}_0(t_0)-\vec{X_p}(t_0)|} v_w) (t-t_0)
which gives two non-linear equations which along with the constraint that |\vec{x}_0-\vec{X}_p| = S can be solved for \vec{x_0} and t_0.
Approximations
We can simplify things if we assume that v_w >> R_p\Omega
\vec{x}-\vec{x}_0(t_0) = \hat{x}_0 v_w (t-t_0)
and that |x| >> |x_0| since we can then solve for t-t_0=|x|/v_w.
Then we can calculate \vec{v_p}(t_0)= R_p\Omega[-\sin{(\Omega t+\phi)}, \cos{(\Omega t +\phi)}, 0] where \phi=-\Omega |x|/v_w
and \vec{X_p(t_0)}=R_p [\cos{(\Omega t + \phi)}, \sin{(\Omega t + \phi)}, 0]
The wind arriving at x(t) left the surface of the primary when it was at X_p(t_0) travelling at a speed somewhere between v_w-R\Omega and v_w+R\Omega. We can make the assumption that the wind speed goes like v_w+R\Omega\sin(\gamma) where \gamma=\alpha-\Omega t_0 where \alpha is the angle towards x and \gamma is the angle between \vec{x}(t) and X_p(t_0)
We then have that \vec{j}=\vec{x}\times\vec{v} = |x|(v_w+R\Omega\sin{\gamma})\sin{\gamma}
and [[latex($\gamma =
The wind that is arriving at x(t) left from the surface of the primary when it was at a position X_p(t_0) travelling at a velocity of v_p(t_0) + v_w \hat{n}
where \hat{n}=\frac{x(t)-X'(t)}{|x(t)-X'(t)|} where X'(t)=X_p(t_0)+v_p(t_0)(t-t_0).
If we consider the specific angular momentum \vec{j}=\vec{x} \times \vec{v} we can identify three components:
\vec{j}=\vec{x} \times \vec{v}_p(t_0)+v_w \vec{x} \times \hat{n}
now \vec{n} = |x|[\cos{\alpha}, \sin{\alpha}, 0]-R_p[\cos{\Omega t_0}, \sin{\Omega t_0}, 0] - R_p \Omega (t-t_0) [-\sin{\Omega t_0}, \cos{\Omega t_0}, 0]
Furthermore, \vec{x}-\vec{v}_p(t_0)(t-t_0)=\vec{x}_0(t_0)+\hat{x}_0v_w(t-t_0)
And since \vec{x_0}=\vec{X}_p+S\hat{x}_0
we have
\vec{x}-\vec{v}_p(t_0)(t-t_0)-\vec{X}_p(t_0)=S\hat{x}_0+\hat{x}_0v_w(t-t_0)
and \hat{x}_0 = \frac{\vec{x}-\vec{v}_p(t_0)(t-t_0)-\vec{X}_p(t_0)}{S+v_w(t-t_0)}
so that \vec{v}=\vec{v}_0=\vec{v_p}+\hat{x}_0 v_w=R_p\Omega[-\sin{(\Omega t+\phi)}, \cos{(\Omega t +\phi)}, 0] +\frac{\vec{x}-\vec{v}_p(t_0)(t-t_0)-\vec{X}_p(t_0)}{S+v_w(t-t_0)}v_w
\vec{v}=R_p\Omega[-\sin{(\Omega t+\phi)}, \cos{(\Omega t +\phi)}, 0] \left ( \frac{1}{1+\frac{|x|}{S}} \right) + \hat{x} v_w \left(\frac{1}{1+\frac{S}{|x|}} \right ) - \frac{\vec{X}_p}{|x|} v_w \left ( \frac{1}{1+\frac{S}{|x|}}\right )
or
\vec{v} \approx \hat{x}v_w-\frac{R_p}{|x|} v_w [\cos{(\Omega t+\phi)}, \sin{(\Omega t +\phi)}, 0] + \frac{S}{|x|} R_p\Omega [-\sin{(\Omega t+\phi)}, \cos{(\Omega t +\phi)}, 0]
and if we express \vec{x}=|x|[[\cos(\alpha), \sin(\alpha), 0] then j=\vec{x} \times \vec{v}=-R_p v_w \sin(\beta)+R_p S\Omega \cos(\beta) where \beta=\alpha - \Omega t - \phi and in particular - if we consider the orbit of the secondary we have \alpha=\Omega t + \pi so that j=-R_p v_w \sin(\pi-\phi)
