# The spacetime metric and my past-lightcone

The spacetime separation of any two events in a generic expanding universe is given by the RW metric:

$$ds^2 = -c^2dt^2 + a(t)^2\left[ dr^2 + S_\kappa(r)^2d\Omega^2\right].$$ Here, $S_\kappa(r)$ defines the curvature of space, and $d\Omega^2$ includes the change of variables from Cartesian to spherical coordinates as

$$d\Omega^2 \equiv d\theta^2 + \sin^2\theta d\phi^2.$$

Now, the observable universe, with respect to an arbitrary observer, is defined as the set of all events which lie on her past-lightcone, i.e. those events which are spacetime-separated from the obsever by a null geodesic, satisfying

$$0 = -c^2dt^2 + a(t)^2\left[ dr^2 + S_\kappa(r)^2d\Omega^2\right].$$

And this is the part that I'm trying to make sense of... We can decide to place the observer at the origin. In this case, all geodesics which intersect the observers position are constant in the angle ($\theta$, $\phi$), and therefore give

$$d\Omega^2 = 0.$$

That would imply that the set of all events lying on the observers lightcone are simply given by

$$0 = -c^2dt^2 + a(t)^2dr^2$$

which evaluates to

$$\int_{t_\text{event}}^{t_\text{obs}}c\dfrac{dt}{a(t)} = ||\textbf{r}||\>.$$

where $||\textbf{r}||$ is the magnitude of the event's spatial position vector.

Am I missing something obvious by being fixated on the fact that this condition nowhere contains the curvature of the universe? I seem to have arrived at the result that the contents of the past-lightcone of an observer are independent of the curvature of spacetime. In other words, the universe would look identical no matter the true form of $S_\kappa(r)^2$.Obviously this cannot possibly be correct. My math is right, though.

• The curvature of the universe influences the form of $a(t)$. – Javier Apr 23 '18 at 21:34
• @Javier Hmm... Then why write it this way? The form changes to $a(t)$ don't come from the $a(t)^2S_\kappa(r)^2d\Omega^2$ term? Even besides that, we can consider a static universe in which $a(t) = 1$, and my question will still remain exactly as it is. – Anonymous Apr 23 '18 at 21:44
• The spatial curvature appears in the Einstein equations, which in turn determine $a(t)$. You cannot just choose any curvature and scale factor. – Javier Apr 23 '18 at 21:50
• @Javier I see. So we cannot talk about this in terms of a non-flat Minkowski space? – Anonymous Apr 23 '18 at 21:51
• What do you mean by non-flat Minkowski space? Minkowski is flat by definition. – Javier Apr 23 '18 at 21:59