How to put $c$ back into relativistic equations? Many books set the speed of light $c=1$ for convenience. For example, Weinberg in his textbook "Gravitation and Cosmology" (though $G$ is still left as a constant):
$$\begin{align}
\mathrm{d}\tau^2 &= \mathrm{d}t^2 - R^2(t)~[f(r,\theta,\phi)] \tag{11.9.16} \\
t &= \frac{\psi + \sin\psi}{2\sqrt k} \tag{11.9.25}
\end{align}$$
If I now want to run real numbers in SI units, do I


*

*Assume $t'[\mathrm{sec}] \rightarrow t = t'/c = t'/299792458$ where the new time unit is $3.34\ \mathrm{ns}$

*Multiply each $t$ and $\tau$ with $c$:
$$\begin{align}
c^2 \mathrm{d}\tau^2 &= c^2 \mathrm{d}t^2 - R^2(t)~[f(r,\theta,\phi)] \tag{11.9.16} \\
ct &= \frac{\psi + \sin\psi}{2\sqrt k} \tag{11.9.25}
\end{align}$$

*Something else?
 A: Putting the factors of $c$ and $G$ back can be a tedious business. Many of us do it by (educated) guesswork, followed by checking that the guesses give sensible results.
The rigorous way to do this is using dimensional analysis, that is checking that when adding quantities they must all have the same dimensions, and that if you have some equation:
$$ t = \text{something} $$
The dimension of the $\text{something}$ must be time.
If you take the equation $11.9.16$ for the metric Weinberg writes this as:
$$ d\tau^2 = dt^2 - R^2(t)\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right) $$
where the scale factor $R(t)$ is dimensionless. The right hand side has time + space so to make this dimensionally consistent we have to either multiply the time term by $c^2$ or divide the distance term by $c^2$. Typically we would do the former to get:
$$ c^2 d\tau^2 = c^2 dt^2 - R^2(t)\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right) $$
A similar argument applies to his equation $11.9.25$:
$$ t = \frac{\psi + \sin\psi}{2\sqrt{k}} $$
In this equation $\psi$ is dimensionless and $k$ has dimensions of $L^{-2}$ so the right hand side has dimensions of $L$. To make this dimensionally consistent either multiply by $1/c$ on the right hand side or $c$ on the left hand side.
A: Setting c = 1 essentially causes quantities with dimension of time to be expressed in units of length. If you know the length dimension of a calculated quantity, you can re-express it in terms of time by plugging in the corresponding amount of factors of $c$, which just act as conversion factors.
This amounts to doing (1) at the very end of the calculation. 
Alternatively, you can add in the $c$'s like in option (2) before starting your calculations. This might be more convenient if you are calculating complicated objects in which the dimension of the end result is difficult to see.
