Viable values for the $K$ parameter in the FLRW metric

Question

The FLWR metric is sometimes given as $$c^2 d\tau^2 = c^2 dt^2 - \frac{a(t)^2}{(1-KX^2)} dX^2. $$

I am not interested in the tangential motion so I set $d \Omega = 0$ although it is of interest in certain 'circles'. As I understand it $K$ is the curvature geometry of the universe where K<0 is open, K=0 is flat and K>0 is closed and finite. If I rearrange the above equation I can obtain $$a(t)\frac{dX}{dt} = c\sqrt{(1-KX^2)} \sqrt{1 - \frac{d\tau^2}{dt^2}}.$$

and for a photon $d\tau/dt = 0$ so:

$$a(t)\frac{dX}{dt} = c\sqrt{(1-KX^2)}.$$

The problem is, if I choose K=1 I get the local peculiar velocity of a photon to be imaginary. Does this rule out a universe having a closed (K=1) geometry as a viable model?

Answer 1

First, define a new time coordinate $\eta$ through integrating $$ d\eta = \frac{c dt}{a(t)}.$$ This time coordinate is often called conformal time. With this coordinate, the condition for a radial, null trajectory is $$ \frac{dX}{d\eta} = \sqrt{1-K X^2}.$$ So far I haven't done anything new, just simplified my life by defining a new time coordinate. But this is a differential equation: we can solve it the physicists way of rearranging the differentials as $$ \int \frac{dX}{\sqrt{1-K X^2}} = \int d\eta = \eta$$ where have chosen the boundary conditions $(X_0,\eta_0)=(0,0)$ to fix the integration constants. But the LHS is solvable! In particular, it says that $$\eta = \begin{cases} \sin^{-1}(X) & K =1 \\ X & K = 0 \\ \sinh^{-1}(X) & K = 1 \end{cases}$$

Equivalently, inverting these expressions,

$$X = \begin{cases} \sin(\eta) & K = 1 \\ \eta & K = 0 \\ \sinh(\eta) & K = 1 \end{cases}$$

So we can see the interpretation of your observation. When $K=1$, the X coordinate has a maximum value, and is more like an angle than a true radial coordinate. That's why the velocity was imaginary for large enough $X$: the manifold just doesn't exist there! In fact, changing coordinates $X=\sin(\varphi)$, the spatial part of the metric you wrote is exactly the metric of a three sphere when you restore the $d\Omega$. That's not a coincidence, and deeply related to the fact that $K=1$ is the case of de Sitter spacetimes. Check out the Wikipedia page on de Sitter space for more information.

Answer 2

$$c^2 d\tau^2 = c^2 dt^2 - \frac{a(t)^2}{(1-KX^2)} dX^2. $$

For this metric expression, if you set $K=1$, then the spatial surfaces are hyperspheres of radius $a$. Every position on these hyperspheres has $|X|\leq 1$, so you will never obtain an imaginary quantity. Objects at $X=1$ lie at a distance of $\pi a/2$ (which you can see by integrating the metric or by realizing that this is 1/4 of the way around the hypersphere). Nearby objects with $|X|\ll 1$ lie at distance $a|X|$.