Quasar Alignments - Derivation of Expectation

What follows is based on section 5.2 of the Ph.D. thesis “The Alignment and Clustering of Quasars”, by Glyn George (1983).

N points are distributed randomly over a field of area A.   Around any given point, one would expect to find, on average, in the distance range (x, x + dx), a number of other points, given by
The pivotal quasar forms one end of a candidate alignment.   Each of these other quasars, in turn, will form the other end of a candidate alignment of length x.   If an intermediate quasar falls within a distance pmax of the line joining the first two quasars, then it is in the acceptance zone shown here and the candidate alignment will be accepted.

Having taken one point as the pivot and another as the far end of the alignment, there are (N – 2) quasars remaining in the field, each of which has an equal chance of falling in the acceptance zone.   Hence, in each acceptance zone, we expect to find, on average, a number of points given by

We expect the number of alignments of “tolerance” p < p_max and of span   d ∈ (x, x + dx)   around a given point to be

Integrating over all values of the alignment span d up to d_max, we obtain the number of alignments expected around each point to be

Now we take each quasar in turn as the pivotal quasar and sum the expectations.   However, each distinct alignment will be counted twice:   once with the quasar at one end as the pivot, then again with the quasar at the other end as the pivot.   We must therefore divide this sum by 2 in order to give the total number of alignments of distinct triples of points expected, E[n_P], in a field with periodic boundaries:

Boundary Correction

Let us now determine the boundary correction for the L×L square, for which we assume d_max < L / 2 .

The square divides into three obvious regions, (centre, edges and corners), whose extent varies with the alignment span x. Elementary search annuli whose centres fall in region (1) will lie entirely within the square and will be unaffected by edge effects.   That is, C1 = 1   and   A1 = (L – 2x)2, where   Ai = the area of region (i)   and Ci = the edge-effects correction factor for region (i).

In region (2), one of the four edges will cause some of the search area to be lost, as shown here.   s   is the distance from the centre of the search annulus to the nearest edge of the square and x is the inner radius of the annulus.   Let A(s) be the proportion of the search area remaining inside the field.   Then, using the notation in this figure, s = x cos θ Recalling that the search area is an elementary annulus, we have The probability of finding a randomly placed point in the range (s, s + ds), given that it is somewhere in (0, x), is

For a given span x, the mean proportion of area remaining in a search area placed at random in region (2) is

Therefore, in region (2),

In region (3), part of the elementary search area will be lost over two of the four edges.   s   is the distance from the centre of the search annulus to the nearer vertical edge of the square, t   is the distance from the centre of the search annulus to the nearer horizontal edge of the square and x is the inner radius of the annulus.   Let A(s, t) be the proportion of the search area remaining inside the field.   Then, using the notation in this figure,           s = x cos θ and           t = x cos φ , which leads to

The probability of finding a randomly placed point in the range s < X < s + ds ,   t < Y < t + dt , given that the point is somewhere in the range 0 < X, Y < x , is

For a given span x, the mean proportion of area remaining in a search area placed at random in region (3) is

Therefore, in region (3),

Bringing the correction factors for the three regions together, we arrive at the overall correction factor   C(x)   for edge effects on an alignment of span x:

At the start of the derivation of E[n_P], we replace
         
by
         
With this amendment, we proceed as before, to obtain

For the standard 5°×5° field in the thesis, with   L = 5°,   d_max = 1°   and   p_max = 30" , we have

and

E[n_F]   =   0.8156 × E[n_P]   =   2.2775×10^–5 N (N – 1) (N – 2)

In the thesis, these expressions were verified by Monte-Carlo simulations and by comparison with the few published works available at that time.