<p>I think you are missing a minus sign on the left side.<br />
Substituting <span class="math-container">$v_p = R(t)\frac{d \chi}{dt}$</span> in the metric equation:
<span class="math-container">$$
ds^2 = (c^2 - v_p^2)dt^2
$$</span>
Take the square root of both sides and integrate from point 1 to point 2:
<span class="math-container">$$
\Delta s_{12} = \int^2_1 \sqrt{c^2 - v_p^2} dt
$$</span></p>

CC BY-SA 4.0
2023-01-11T22:29:24.013Z
2023-01-11T23:22:41.757Z
2023-01-11T23:22:41.757Z
eshaya
<p>Using the following FRW metric,</p>
<p><span class="math-container">$$ds^2=-c^2dt^2+R^2(t)(d\chi^2+f(\chi)d\Omega)$$</span></p>
<p>and considering <span class="math-container">$d\Omega=0$</span> from now on, it is clear that for a light-like trajectory, the equation must read <span class="math-container">$ds^2=0$</span> and so:</p>
<p><span class="math-container">$$d\chi=\frac{cdt}{R(t)}$$</span></p>
<p>However, if an observer wanted to travel from one galaxy to another, for instance, at a peculiar, velocity <span class="math-container">$v_p$</span>, it would be a time-like interval. However, I do not know how I could plug <span class="math-container">$v_p$</span> into the metric above (for <span class="math-container">$d\Omega=0$</span>). What would be the relation between <span class="math-container">$d\chi$</span> and <span class="math-container">$dt$</span> in this case?</p>
<p>Many thanks for your response.</p>

<cosmology><distances><friedmann-equations>
How can a time-like trajectory be calculated using the Friedmann-Robertson-Walker metric?
user9867
# How can a time-like trajectory be calculated using the Friedmann-Robertson-Walker metric?

(Asked by: username_0 on 2023-01-11)

Using the following FRW metric,

$$ds^2=-c^2dt^2+R^2(t)(d\chi^2+f(\chi)d\Omega)$$

and considering $d\Omega=0$ from now on, it is clear that for a light-like trajectory, the equation must read $ds^2=0$ and so:

$$d\chi=\frac{cdt}{R(t)}$$

However, if an observer wanted to travel from one galaxy to another, for instance, at a peculiar, velocity $v\_p$, it would be a time-like interval. However, I do not know how I could plug $v\_p$ into the metric above (for $d\Omega=0$). What would be the relation between $d\chi$ and $dt$ in this case?

Many thanks for your response.

**Tags:** cosmology, distances, friedmann-equations


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## 1 answers

### Answer by username_1 on 2023-01-11. Score: 1 | Accepted answer

I think you are missing a minus sign on the left side.  
Substituting $v\_p = R(t)\frac{d \chi}{dt}$ in the metric equation:
$$
ds^2 = (c^2 - v\_p^2)dt^2
$$
Take the square root of both sides and integrate from point 1 to point 2:
$$
\Delta s\_{12} = \int^2\_1 \sqrt{c^2 - v\_p^2} dt
$$


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