<p>It turns out that using the <span class="math-container">$\Lambda = 0$</span> formula for our universe does not work, as eshaya pointed out. Going to the section where <span class="math-container">$\Lambda \neq 0$</span>, the paper presents this integral: <span class="math-container">$$\chi\left(z\right)=\int_{1}^{1+z}\left(\Omega_{m}x^{3}+\left(1-\Omega_{m}-\Omega_{\Lambda}\right)x^{2}+\Omega_{\Lambda}\right)^{-\frac{1}{2}}dx$$</span></p>
<p>Putting this into equation <span class="math-container">$(15)$</span>, we cannot use the closed-form solution presented in the second line as that only works if <span class="math-container">$\Lambda = 0$</span>. <span class="math-container">$P^2=\left(\frac{1}{1+z_{1}}q_{1}\right)^{2}\cdot2\left(\Sigma^{2}\left(\frac{\chi\left(z_{2}\right)+\chi\left(z_{1}\right)}{2}\right)^{2}\sin^{2}\left(\frac{a}{2}\right)+\Sigma^{2}\left(\frac{\chi\left(z_{2}\right)-\chi\left(z_{1}\right)}{2}\right)\cos^{2}\left(\frac{a}{2}\right)\right)^{\frac{1}{2}}$</span></p>
<p>Then <span class="math-container">$z_2^{\prime}=\frac{2P^2}{\left(q_{1}-2P^2\right)^{2}}\left(1+q_{1}-2P^2+\sqrt{\frac{q_{1}^{2}}{P^2}+1-2q_{1}}\right)$</span>.</p>
<p>A key issue I noticed with this formula is the provided equation for <span class="math-container">$q_1 = q_0 \frac{1+z_1}{1+2q_0z_1}$</span>, which has a vertical asymptote at <span class="math-container">$ z_1 = -\frac{1}{2q_0}$</span>. As a result, I substituted equation <span class="math-container">$(27)$</span> from <a href="https://arxiv.org/pdf/2204.12097" rel="nofollow noreferrer">https://arxiv.org/pdf/2204.12097</a>, where <span class="math-container">$q_1\left(z\right)=-1+\frac{3\Omega_{m}\left(1+z_1\right)^{3}}{2\left(\Omega_{m}\left(1+z_1\right)^{3}+\left(1-\Omega_{m}\right)\right)}$</span> (assuming that <span class="math-container">$\nu = \tilde{\xi_1} = 0$</span>, which is a reasonable approximation).</p>
<p>This resolves the imaginary number problem and seems to be a proper solution to the question. Please let me know if I have made any errors in my calculations.</p>

CC BY-SA 4.0
2024-11-04T18:05:51.377Z
2024-11-04T18:05:51.377Z
WarpPrime
<p>I was looking at <a href="https://astronomy.stackexchange.com/questions/11348/how-to-find-the-distance-between-two-quasars">this question</a> and read through the linked article <a href="http://arxiv.org/pdf/astro-ph/0007341v1.pdf" rel="nofollow noreferrer">http://arxiv.org/pdf/astro-ph/0007341v1.pdf</a>. When I was attempting to compute the distance between two galaxies using the following formulae in the paper:</p>
<p><a href="https://i.sstatic.net/H5r0oSOy.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/H5r0oSOy.png" alt="enter image description here" /></a> <a href="https://i.sstatic.net/Cbq0mUxr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Cbq0mUxr.png" alt="enter image description here" /></a> <a href="https://i.sstatic.net/MBFkVlxp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MBFkVlxp.png" alt="enter image description here" /></a> <a href="https://i.sstatic.net/k90rXQb8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/k90rXQb8.png" alt="enter image description here" /></a></p>
<p>When plugging in <span class="math-container">$q_0 = \frac{1}{2}\Omega_M - \Omega_\Lambda = -0.55$</span> and arbitary values of <span class="math-container">$z_1, z_2, \alpha$</span>, I keep running into the issue where <span class="math-container">$z'_2$</span> is a complex number! Have I performed an error in my calculations, or am I misinterpreting the value of <span class="math-container">$q_0$</span>?</p>

CC BY-SA 4.0
2024-11-02T10:15:38.960Z
Did you read section 2.1 on $\Lambda \neq 0$?
eshaya
<distances><redshift><mathematics>
Cosmological redshift to distance calculation: Why are imaginary numbers involved?
WarpPrime
# Cosmological redshift to distance calculation: Why are imaginary numbers involved?

(Asked by: username_0 on 2024-11-02)

I was looking at [this question](https://astronomy.stackexchange.com/questions/11348/how-to-find-the-distance-between-two-quasars) and read through the linked article <http://arxiv.org/pdf/astro-ph/0007341v1.pdf>. When I was attempting to compute the distance between two galaxies using the following formulae in the paper:

[![enter image description here](https://i.sstatic.net/H5r0oSOy.png)](https://i.sstatic.net/H5r0oSOy.png) [![enter image description here](https://i.sstatic.net/Cbq0mUxr.png)](https://i.sstatic.net/Cbq0mUxr.png) [![enter image description here](https://i.sstatic.net/MBFkVlxp.png)](https://i.sstatic.net/MBFkVlxp.png) [![enter image description here](https://i.sstatic.net/k90rXQb8.png)](https://i.sstatic.net/k90rXQb8.png)

When plugging in $q\_0 = \frac{1}{2}\Omega\_M - \Omega\_\Lambda = -0.55$ and arbitary values of $z\_1, z\_2, \alpha$, I keep running into the issue where $z'\_2$ is a complex number! Have I performed an error in my calculations, or am I misinterpreting the value of $q\_0$?

**Tags:** distances, redshift, mathematics
> **Comments:**
> **username_1**: Did you read section 2.1 on $\Lambda \neq 0$?  

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## 1 answers

### Answer by username_0 on 2024-11-04. Score: 1

It turns out that using the $\Lambda = 0$ formula for our universe does not work, as eshaya pointed out. Going to the section where $\Lambda \neq 0$, the paper presents this integral: $$\chi\left(z\right)=\int\_{1}^{1+z}\left(\Omega\_{m}x^{3}+\left(1-\Omega\_{m}-\Omega\_{\Lambda}\right)x^{2}+\Omega\_{\Lambda}\right)^{-\frac{1}{2}}dx$$

Putting this into equation $(15)$, we cannot use the closed-form solution presented in the second line as that only works if $\Lambda = 0$. $P^2=\left(\frac{1}{1+z\_{1}}q\_{1}\right)^{2}\cdot2\left(\Sigma^{2}\left(\frac{\chi\left(z\_{2}\right)+\chi\left(z\_{1}\right)}{2}\right)^{2}\sin^{2}\left(\frac{a}{2}\right)+\Sigma^{2}\left(\frac{\chi\left(z\_{2}\right)-\chi\left(z\_{1}\right)}{2}\right)\cos^{2}\left(\frac{a}{2}\right)\right)^{\frac{1}{2}}$

Then $z\_2^{\prime}=\frac{2P^2}{\left(q\_{1}-2P^2\right)^{2}}\left(1+q\_{1}-2P^2+\sqrt{\frac{q\_{1}^{2}}{P^2}+1-2q\_{1}}\right)$.

A key issue I noticed with this formula is the provided equation for $q\_1 = q\_0 \frac{1+z\_1}{1+2q\_0z\_1}$, which has a vertical asymptote at $ z\_1 = -\frac{1}{2q\_0}$. As a result, I substituted equation $(27)$ from <https://arxiv.org/pdf/2204.12097>, where $q\_1\left(z\right)=-1+\frac{3\Omega\_{m}\left(1+z\_1\right)^{3}}{2\left(\Omega\_{m}\left(1+z\_1\right)^{3}+\left(1-\Omega\_{m}\right)\right)}$ (assuming that $\nu = \tilde{\xi\_1} = 0$, which is a reasonable approximation).

This resolves the imaginary number problem and seems to be a proper solution to the question. Please let me know if I have made any errors in my calculations.


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