2 Calculating ${\vec z} {_2'}$

In the following I outline the method used to calculate the redshift of an object as seen by another, distant observer. I use the notation of L00 where I wrote the Robertson-Walker line element as

$${\rm d}s^2 = -c^2 {\rm d}t^2 + a^2(t)\left[{\rm d}\chi^2 + \... ...chi)({\rm d}\theta^2 + \sin^2\theta \: {\rm d}\phi^2)\right].$$ (1)

The radial coordinate $\chi$ is dimensionless and $\Sigma(\chi)$ is defined as $\sin \chi$, $\chi$ or $\sinh \chi$ if k = +1, 0 or -1. Length dimensions are included in a(t) by setting a₀ as the curvature radius for $k = \pm1$ or a₀ = c H₀^-1 for k = 0.

Consider an object 1 observed by us today at z₁ and an object 2 at z₂ separated by an angle $\alpha$ on the sky (cf. Fig. 1 of L00). Object 2 emits a photon towards object 1 which is received by object 1 at the same time as object 1 emits the photon we receive from it today. I denote the redshift of object 2 as observed by object 1 as z₂' and the comoving coordinate distance between them as $\chi_2'$. We have (L00)

 

$$\Sigma^2\left(\frac{\chi_2'}{2}\right) = \Sigma^2\left(\frac{\chi... ...ha}{2} + \Sigma^2\left(\frac{\chi_2 - \chi_1}{2}\right) \cos^2\frac{\alpha}{2},$$ (2)

where $\chi_{1,2} = \chi(z_{1,2})$. In L00 I set $\Omega_\Lambda= 0$ and derived analytic expressions for the right and left-hand sides of Eq. (2) in terms of z₁, z₂ and z₂'respectively. In the case $\Omega_\Lambda\neq 0$ we relate the right-hand side of Eq. (2) to z₁ and z₂ by using

$$\chi(z) = \sqrt{\vert\Omega_k\vert} \int_0^z \left[ E(x) \right]^{-1/2} {\rm d}x,$$ (3)

where $\Omega_k= 1 - \Omega_{\rm M}- \Omega_\Lambda$ and

$$E(z) = \Omega_\Lambda+ (1+z)^2 \Omega_k+ (1+z)^3 \Omega_{\rm M}.$$ (4)

When $\Omega_k= 0$ then $\chi(z)$ is given by just the integral (i.e. we set a₀ = c H₀^-1). We relate the left-hand side of Eq. (2) to z₂' by noting that an observer at z₁ would write the above as

$$\chi_{z_1}(z) = \sqrt{\vert\Omega_k(z_1) \vert} \int_0^{z} \left[ E_{z_1}(x) \right]^{-1/2} {\rm d}x,$$ (5)

where

$$E_{z_1}(z) = \Omega_\Lambda(z_1) + (1+z)^2 \Omega_k(z_1) + (1+z)^3 \Omega_{\rm M}(z_1)$$ (6)

and

$$\Omega_\Lambda(z_1) = \frac{\Omega_\Lambda}{E(z_1)} \nonumber$$

$$\Omega_k(z_1) = \frac{(1+z_1)^2 \Omega_k}{E(z_1)}$$ (7)

$$\Omega_{\rm M}(z_1) = \frac{(1+z_1)^3 \Omega_{\rm M}}{E(z_1)}\cdot$$

Hence we have $\chi_2' = \chi_{z_1}(z_2')$. It is straightforward to show that

$$\chi_2' = \chi_{z_1}(z_2') = \chi(\tilde z_2) - \chi(z_1),$$ (8)

where $\tilde z_2 = (1+z_1)(1+z_2') - 1$ is today's redshift of the photon emitted by object 2 and observed by object 1 to have a redshift of z₂'. Finally, to calculate z₂' for a given z₁, z₂ and $\alpha$ we substitute Eq. (8) into Eq. (2) and numerically solve for $\tilde z_2$ using the Newton-Raphson method (Press et al. 2002), where the derivative of the function to be solved is simply given by

$$\frac{{\rm d}\chi}{{\rm d}z}(z) = \sqrt{\frac{\vert \Omega_k\vert}{E(z)}}\cdot$$ (9)