4 Kernels for the eclipsing case

In this section we derive the kernels for the case when one star is partially eclipsed by another (a future paper (Coleman et al. in preparation) will discuss the case of a gravitational lens). The geometry of this situation is as shown in Fig. 1.

  

Figure 1: The eclipsed star has radius 1, and the occultor radius $\rho$. $(r,\chi)$ are polar coordinates in the projected plane of the source, and $(s(t),\phi(t))$ are the coordinates of the centre of the occultor; both $\chi$ and $\phi$ are taken from the radius which touches the point of closest approach. When the centres of the two stars are a distance s apart, the occultor cuts off an angle $\psi(r;s,\rho)$ of an annulus of radius r, centred on the eclipsed star

We can take the occultor to be opaque, so that its "transfer function'' A₀(r₀), where r₀ is the (projected) radius from the centre of the occultor , is  

$$A_0(r_0) = \cases{0&for $r_0<\rho$\cr 1&for $r_0\gt\rho$}. \relax$$ (14)

This makes $A(r,\chi;s)$, the kernel in terms of r, a function of $\psi(r;s)$. In terms of i(r), the intensity of the star as a function of (projected) radius, the total flux from the eclipsed star is  

$$F_\mathrm{I}(s) = \int_{\mbox{\scriptsize area}} i(r) A(r,\... ... r\,\relax {\rm d}\relax r\,\relax {\rm d}\relax \chi). \relax$$ (15)

Thus define $\tilde A(r;s)=r\int_0^{2\pi} A(r,\chi;s)\,\relax {\rm d}\relax \chi$, so that  

$$F_\mathrm{I}(s) = \int_0^1 i(r) \tilde A(r;s)\,\relax {\rm d}\relax r, \relax$$ (16)

putting Eq. (15) into the form of Eq. (4).

It is easy to see that $\tilde A(r;s)=r\int_0^{2\pi-\psi(s)} 1\,\relax {\rm d}\relax \chi$. Writing $\gamma\equiv\cos[\psi(r;s)/2]$ we find that  

$$\gamma = \cases{ 1& , $\phantom{-}1<\bar\gamma$\cr \bar\g... ...\gamma\le1$\cr -1& , $\phantom{-1\le}\bar\gamma<-1$ } \relax$$ (17)

We can now write  

$$\tilde A(r;s) = \cases{2r(\pi-\psi/2) & {} \cr = 2r\arccos... ...ho$\cr 0 & , $r<-(s-\rho)$\cr 2\pi r & , otherwise } \relax$$ (18)

defined for $r\ge 0$.

The calculation is a little more intricate for the Stokes parameters. The light from each point on the star's disk must be linearly polarized in the tangential direction. Using the angle $\chi$ defined in Fig. 1, the Stokes parameters must therefore be $F_\mathrm{U}(r,\chi)=-P(r)\sin 2\chi$ and $F_\mathrm{Q}(r,\chi)=-P(r)\cos 2\chi$, for some function P(r) which we wish to recover (note that we use the unnormalised Stokes parameters, since the normalised ones have contributions to the noise from the intensity as well as the polarization measurements). Defining  

$$\tilde A_\mathrm{Q}(r;s,\phi)= -r\int_0^{2\pi}\cos2\chi A(r,\chi;s,\phi)\,\relax {\rm d}\relax \chi, \relax$$ (19)

we therefore find that the total polarized flux in the Q direction, measured when the centres are a distance s apart, is  

$$F_\mathrm{Q}(s(t),\phi(t)) = \int_0^1 P(r) \tilde A_\mathrm{Q}(r;s,\phi)\,\relax {\rm d}\relax r, \relax$$ (20)

and similarly for $A_\mathrm{U}(r,\chi;s,\phi)$ and $F_\mathrm{U}(s,\phi)$. This is now in the form of Eq. (4). Setting $l=s\cos\phi$, we can thus see that

$$\begin{eqnarray} \relax \tilde A_\mathrm{Q}(r;s,\phi) &=& r\cos2\phi(t)\sin\p... ...\frac{l^2}{s^2}}\right)^{1\over2} \gamma\sqrt{1-\gamma^2} \relax \end{eqnarray}$$ (21) (22)

These kernels are broad and smooth, hence the ill-conditioning of the inverse problem.