<p>This formula is exact if the expansion is linear (<span class="math-container">$a(t) = H_0 t$</span>) and all peculiar velocities are zero. In that case, the comoving distance to the object is <span class="math-container">$$\int_{t_\text{then}}^{t_\text{now}} \frac{c\,\mathrm dt}{a(t)} = \frac{c}{H_0} \ln \frac{t_\text{now}}{t_\text{then}} = \frac{c}{H_0} \ln (1{+}z)$$</span> and the present recessional velocity is <span class="math-container">$H_0$</span> times that.</p>
<p>In the real world, the expansion has been not too far from linear, and recessional velocities are not too large, so it's reasonably accurate.</p>

CC BY-SA 4.0
2020-08-08T18:52:22.020Z
2020-08-08T18:52:22.020Z
benrg
<p>From IOAA 2013 (Greece) Theory question no. 15, they stated that the approximate formula to find velocity from cosmological redshift is <span class="math-container">$$v = c*log_e (1+z)$$</span> and that it is often used by cosmologists. I did a quick google search but found nothing similar to this formula.</p>
<p>So, where did this equation come from and is it really often used by cosmologists?</p>

<cosmology><redshift>
Approximate formula to find velocity from cosmological redshift
Arnab Chowdhury
# Approximate formula to find velocity from cosmological redshift

(Asked by: username_0 on 2020-08-08)

From IOAA 2013 (Greece) Theory question no. 15, they stated that the approximate formula to find velocity from cosmological redshift is $$v = c\*log\_e (1+z)$$ and that it is often used by cosmologists. I did a quick google search but found nothing similar to this formula.

So, where did this equation come from and is it really often used by cosmologists?

**Tags:** cosmology, redshift


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## 1 answers

### Answer by username_1 on 2020-08-08. Score: 1 | Accepted answer

This formula is exact if the expansion is linear ($a(t) = H\_0 t$) and all peculiar velocities are zero. In that case, the comoving distance to the object is $$\int\_{t\_\text{then}}^{t\_\text{now}} \frac{c\,\mathrm dt}{a(t)} = \frac{c}{H\_0} \ln \frac{t\_\text{now}}{t\_\text{then}} = \frac{c}{H\_0} \ln (1{+}z)$$ and the present recessional velocity is $H\_0$ times that.

In the real world, the expansion has been not too far from linear, and recessional velocities are not too large, so it's reasonably accurate.


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