# Scattering by multiple linear screens¶

## Multiple screens¶

Consider screens with linear features between the telescope and the pulsar. Consider a coordinate system in which z is along the line of sight (with direction $$\hat{z}$$), and the lines on a given screen $$s$$ describing the features are given by,

$d_{s}\hat{z} + \vec{r} + s \hat{u}$

where $$\vec{r}$$ is a cylindrical radius from the line of sight to to the line (ie., $$\hat{r}\cdot\hat{z}=0$$) and $$\hat{u}=\hat{z}\times\hat{r}$$ a unit vector perpendicular to it in the plane of the screen.

Imagine now going from the observer to some point along a line on a first screen. It is easiest to work in terms of angles relative to the observer, so we use $$\rho=r/d$$ and $$\varsigma=s/d$$ to write this trajectory as,

$d(\hat{z} + \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1}).$

When it hits the line, light can be bent only perpedicular to the line, by an angle which we will label $$\alpha$$ (with positive $$\alpha$$ implying bending closer to the line of sight; equal to $$\hat\alpha$$ in Simard & Pen). Hence, beyond the screen it will travel along

$d(\hat{z} + \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1}) - (d-d_{1})\alpha_{1}\hat{r}_{1}.$

If it then hits a line on the second screen, it will again be bent, and then follow,

$d(\hat{z} + \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1}) - (d-d_{1})\alpha_{1}\hat{r}_{1} - (d-d_{2})\hat{r}_{2}.$

In order to specify the full trajectory, we need to make sure that it actually intersects the line on the second screen, and ends at the pulsar, i.e.,

$\begin{split}\begin{eqnarray} d_{2}(\hat{z} + \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1}) - (d_{2}-d_{1})\alpha_{1}\hat{r}_{1} &=& d_{2}(\hat{z} + \rho_{2}\hat{r}_{2} + \varsigma_{2}\hat{u}_{2}),\\ d_{p}(\hat{z} + \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1}) - (d_{p}-d_{1})\alpha_{1}\hat{r}_{1} - (d_{p}-d_{2})\alpha_{2}\hat{r}_{2} &=& d_{p}\hat{z}. \end{eqnarray}\end{split}$

This can be simplified to,

$\begin{split}\begin{eqnarray} \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1} - (1-d_{1}/d_{2})\alpha_{1}\hat{r}_{1} &=& \rho_{2}\hat{r}_{2} + \varsigma_{2}\hat{u}_{2},\\ \rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1} - (1-d_{1}/d_{p})\alpha_{1}\hat{r}_{1} - (1-d_{2}/d_{p})\hat{r}_{2}) &=& 0. \end{eqnarray}\end{split}$

Obviously, this can be extended to multiple screens, and for $$n$$ screens one generally winds up with $$n$$ equations for two-dimensional vectors with $$2n$$ unknowns, the $$n$$ bending angles $$\alpha_{i}$$ and the $$n$$ angular offsets $$\varsigma_{i}$$ along the lines.

### Direct solutions for one or two screens¶

If there is only a single screen, the situation is simple: we just need to make sure the trajectory after the first screen ends at the pulsar. Following the above simplification, one needs,

$\rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1} - (1-d_{1}/d_{p})\alpha_{1}\hat{r}_{1} = 0.$

Thus, taking inner products with $$\hat{r}_{1}$$ and $$\hat{u}_{1}$$, one finds $$\alpha_{1}=\rho_{1}d_{p}/(d_{p}-d_{1})$$ and $$\varsigma_{1}=0$$, i.e., the ray will always be bent along the point closes to the line of sight.

For two screens, the above still can be solved fairly easily by substitution, though it becomes less obvious this is still useful. We start by considering inner products with $$\hat{r}_{1}$$ and $$\hat{u}_{1}$$, giving the following four equations (writing $$s_{12}\equiv1-d_{1}/d_{2}$$, etc., for brevity),

$\begin{split}\begin{eqnarray} \rho_{1} - s_{12}\alpha_{1} &=& \rho_{2}\hat{r}_{2}\cdot\hat{r}_{1} + \varsigma_{2}\hat{u}_{2}\cdot\hat{r}_{1} = \rho_{2}\cos \delta - \varsigma_{2}\sin \delta,\\ \varsigma_{1} &=& \rho_{2}\hat{r}_{2}\cdot\hat{u}_{1} + \varsigma_{2}\hat{u}_{2}\cdot\hat{u}_{1} = \rho_{2} \sin \delta + \varsigma_{2} \cos \delta,\\ \rho_{1} - s_{1p}\alpha_{1} &=& s_{2p}\alpha_{2}\hat{r}_{2}\cdot\hat{r}_{1} = s_{2p}\alpha_{2}\cos \delta\\ \varsigma_{1} &=& s_{2p}\alpha_{2}\hat{r}_{2}\cdot\hat{u}_{1} = s_{2p}\alpha_{2}\sin \delta, \end{eqnarray}\end{split}$

where in the second equalities we used $$\hat{r}_{2}\cdot\hat{r}_{1}=\hat{u}_{2}\cdot\hat{u}_{1}=\cos \delta$$ and $$\hat{r}_{2}\cdot\hat{u}_{1}=-\hat{u}_{2}\cdot\hat{r}_{1}=\sin \delta$$.

Eliminating $$s_{2p}\alpha_{2}$$ between the third and fourth,

$\rho_{1} - s_{1p}\alpha_{1} = \varsigma_{1}/\tan \delta.$

Solving for $$\alpha_{1}$$ and inserting this in the first,

$\rho_{1} - (s_{12}/s_{1p})(\rho_{1}-\varsigma_{1}/\tan \delta) = \rho_{2}\cos \delta - \varsigma_{2}\sin \delta.$

Collecting $$\rho_{1}$$ terms and inserting the second,

$\rho_{1}(1 - s_{12}/s_{1p}) + (s_{12}/s_{1p})(\rho_{2} \sin \delta + \varsigma_{2} \cos \delta)/\tan \delta = \rho_{2}\cos \delta - \varsigma_{2}\sin \delta.$

Bringing $$\varsigma_{2}$$ and $$\rho_{2}$$ terms together, this simplifies to

$\varsigma_{2}(\sin \delta + (s_{12}/s_{1p}) \cos^{2}\delta/\sin \delta) = (\rho_{2}\cos \delta - \rho_{1})(1-s_{12}/s_{1p}),$

and thus

$\varsigma_{2}(1 - \cos^{2}\delta(1-s_{12}/s_{1p})) = (\rho_{2}\cos \delta -\rho_{1})(1-s_{12}/s_{1p})\sin \delta.$

The other unknowns then follow from substitution.

### General solution¶

For $$n$$ screens, it has to hold for all screens $$i>1$$, that,

$\rho_{1}\hat{r}_{1} + \varsigma_{1}\hat{u}_{1} - \sum_{j=1}^{i-1} \alpha_{j}(1-d_{j}/d_{i})\hat{r}_{j} = \rho_{i}\hat{r}_{i} + \varsigma_{i}\hat{u}_{i}.$

The same has to hold for the pulsar, and one can incorporate this by giving it an index $$p$$ for which $$p-1=n$$ in the summation. Above, we effectively assumed $$\rho_{p}=\varsigma_{p}=0$$, but in the general case one will have a spatial offset $$\vec{r}_{p}$$ corresponding to an angular offset $$\rho_{p}=r_{p}/d_{p}$$ in direction $$\hat{r}_{p}$$ (and $$\varsigma_{p}=0$$).

One can also have the telescope be offset (e.g., from the solar system barycentre), by some $$\vec{r}_{t}$$ with norm $$r_{t}$$ and direction $$\hat{r}_{t}$$. Essentially the same equation will hold, with summation starting at $$j=t=0$$, $$d_{t}=0$$, $$\rho_{ti}=r_{t}/d_{i}$$, and $$\alpha_{t}$$ and $$\varsigma_{t}$$ angles towards the line of sight and perpendicular to $$\hat{r}_{t}$$, respectively. Bringing unknowns to one side and the knowns to the other, and again writing $$s_{ji}=1-d_{j}/d_{i}$$ (where all $$s_{ti}=1$$), one finds,

$\varsigma_{t}\hat{u}_{t} - \varsigma_{i}\hat{u}_{i} - \sum_{j=0}^{i-1} \alpha_{j}s_{ji}\hat{r}_{j} = \rho_{i}\hat{r}_{i} - \rho_{t}\hat{r}_{t}$

Instead of taking inner products with $$\hat{r}_{1}$$ and $$\hat{u}_{1}$$, it seems easier to allow one to chose particular $$x$$ and $$y$$ directions and then define angles $$\phi_{i}$$ such that $$\hat{r}_{i}=\cos \phi_{i} \hat{x} + \sin \phi_{i}\hat{y}$$ and thus $$\hat{u}_{i}=-\sin \phi_{i} \hat{x} + \cos \phi_{i} \hat{y}$$. With that, the equations in matrix form are,

$\begin{split}\left(\begin{matrix} -\sin \phi_{t} & -\cos \phi_{t} & \sin \phi_{1} & 0 & \ldots & 0 & 0\\ \cos \phi_{t} & -\sin \phi_{t} & -\cos \phi_{1} & 0 & \ldots & 0 & 0\\ -\sin \phi_{t} & -\cos \phi_{t} & 0 & -s_{12}\cos \phi_{1} & \ldots & 0 & 0\\ \cos \phi_{t} & -\sin \phi_{t} & 0 & -s_{12}\sin \phi_{1} & \ldots & 0 & 0\\ \vdots & \vdots &\vdots & \vdots & \ddots & \vdots &\vdots \\ -\sin \phi_{t} & -\cos \phi_{t} & 0 & -s_{1n}\cos \phi_{1} & \ldots & \sin \phi_{n} & 0\\ \cos \phi_{t} & -\sin \phi_{t} & 0 & -s_{1n}\sin \phi_{1} & \ldots & -\cos \phi_{n} & 0\\ -\sin \phi_{t} & -\cos \phi_{t} & 0 & -s_{1p}\cos \phi_{1} & \ldots & 0 & -s_{np}\cos \phi_{n}\\ \cos \phi_{t} & -\sin \phi_{t} & 0 & -s_{1p}\sin \phi_{1} & \ldots & 0 & -s_{np}\sin \phi_{n}\\ \end{matrix}\right) \left(\begin{matrix} \varsigma_{t}\\ \alpha_{t}\\ \varsigma_{1}\\ \alpha_{1}\\ \vdots\\ \varsigma_{n-1}\\ \alpha_{n-1}\\ \varsigma_{n}\\ \alpha_{n} \end{matrix}\right) = \left(\begin{matrix} \rho_{1}\cos \phi_{1} - \rho_{t}\cos \phi_{t}\\ \rho_{1}\sin \phi_{1} - \rho_{t}\sin \phi_{t}\\ \rho_{2}\cos \phi_{2} - \rho_{t}\cos \phi_{t}\\ \rho_{2}\sin \phi_{2} - \rho_{t}\sin \phi_{t}\\ \vdots\\ \rho_{n}\cos \phi_{n} - \rho_{t}\cos \phi_{t}\\ \rho_{n}\sin \phi_{n} - \rho_{t}\sin \phi_{t}\\ \rho_{p}\cos \phi_{p} - \rho_{t}\cos \phi_{t}\\ \rho_{p}\sin \phi_{p} - \rho_{t}\sin \phi_{t} \end{matrix}\right).\end{split}$

These can be solved by by inverting the matrix $$A$$. Recognizing that on the right-hand side, the terms are just $$x$$ and $$y$$ projections of $$\vec{\theta}\equiv(\vec{r}_{i}-\vec{r}_{t})/d_{i}$$, one finds,

$\begin{split}\left(\begin{matrix} \varsigma_{t}\\ \alpha_{t}\\ \varsigma_{1}\\ \alpha_{1}\\ \vdots\\ \varsigma_{n-1}\\ \alpha_{n-1}\\ \varsigma_{n}\\ \alpha_{n} \end{matrix}\right) = A^{-1} \left(\begin{matrix} \theta_{1,x}\\ \theta_{1,y}\\ \theta_{2,x}\\ \theta_{2,y}\\ \vdots\\ \theta_{n,x}\\ \theta_{n,y}\\ \theta_{p,x}\\ \theta_{p,y}\\ \end{matrix}\right).\end{split}$

### Velocities¶

In principle, the telescope, screens and pulsar will all have velocities. One sees that the entries in the matrix involve only angles of the telescope and screens and ratios of distances, which will change with time much slower than everything else. Hence, one can solve for the time derivatives of the parameters by applying the matrix inverse to the time derivatives of the entries of the right-hand side vector, which are simply the $$x$$ and $$y$$ components of the proper motions motions of the screens and the pulsar relative to the telescope, i.e.,

$\begin{split}\left(\begin{matrix} \dot{\varsigma}_{t}\\ \dot{\alpha}_{t}\\ \dot{\varsigma}_{1}\\ \dot{\alpha}_{1}\\ \vdots\\ \dot{\varsigma}_{n-1}\\ \dot{\alpha}_{n-1}\\ \dot{\varsigma}_{n}\\ \dot{\alpha}_{n} \end{matrix}\right) = A^{-1} \left(\begin{matrix} \mu_{1,x}\\ \mu_{1,y}\\ \mu_{2,x}\\ \mu_{2,y}\\ \vdots\\ \mu_{n,x}\\ \mu_{n,y}\\ \mu_{p,x}\\ \mu_{p,y}\\ \end{matrix}\right).\end{split}$