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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.

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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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