Is the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?

Question

I've mostly been considering the closed FLRW universe in this. It seems that any point on the FLRW universe at a given cosmological time would have the same invariant interval between it and the singularity. Is that right?

From another perspective (on the three sphere), One can claim that, via spherical symmetry, the interval should be the same between the origin (singularity) and any point on the three sphere.

PS. apologies for the lack of math, I'm tired.

EDIT: Here's a general argument regarding this: The invariant interval is defined as:

$$\mid ds\mid=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

Because it is a scalar it may be written as an exact differential form:

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}$$

Where summation over $\mu$ is implied. Note that $\frac{\partial s}{\partial x_{\mu}}$ can be identified with the metric:

$$\mid ds\mid=\sqrt{\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}dx^{\mu}dx^{\nu}}$$

The second equation can also be written as:

$$ds=\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Integrating now over some arbitrary interval:

$$S=\intop_{a}^{b}\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Via the fundamental theorem of calculus, it is clear that the interval is independent of the path taken between the points $\{a,b\}$. I don't see how this is wrong (although once again I'm admittedly tired).

As Rennie says below this is true for all comoving observers in the FLRW universe regardless of position. Then shouldn't any observer (regardless of frame) agree on the interval S? Of course different frames will measure different times for the universe since it's inception, but I'm speaking of the interval.

EDIT For a chosen set of coordinates, the interval between two points appears to be independent of the path taken. The twin paradox is a perfect example (for flat space).

Consider one observer at “rest”, whilst the other one speeds off. Considering just the inertial (stationary in free space) coordinate system, the (stationary) observer witnesses a (proper) time pass $\Delta\tau$.

Let us denote the interval ($S$) witnessed by the observer to have units of distance. $$S^{2}=c^{2}\Delta\tau$$

For simplicity let us assume the other observer (twin if you will) to travel some arbitrary (obviously not traveling faster than light) distance $\Delta x$ before the two observers position again coincides. Then clearly in the chosen coordinate system:

$$S^{2}=c^{2}\Delta\tau^{2}=c^{2}\varDelta t^{2}-\triangle x^{2}$$

Where $\Delta t$ is the time measured by the traveling observer. Obviously, for a given coordinate choice, the invariant interval is independent of the path taken between two points since the motion of the traveling observer was entirely arbitrary.

Generalizing to an FLRW (Friedmann-Lemaitre-Robertson-Walker) universe, we know that all comoving frames will measure the same (proper) time (interval) regardless of position.

It follows that for any particular choice of coordinate frame, an observer will witness all other (observable) bodies in the universe as having traversed the same interval.

I got a lot of flack for this question, but it seems pretty elementary, apparently I had to specify that I'm only using one set of coordinates (though isn't that typical?) Is this right guys? maybe I'm missing something.

Answer 1

This is true for comoving observers, but only for comoving observers.

If we write the FLRW metric in the most general form:

$$ c^2d\tau^2 = c^2dt^2 - a^2(t)\left(\frac{dr^2}{1 - kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right) \tag{1}$$

then for a comoving observer $dr = d\theta = d\phi = 0$ and equation (1) becomes:

$$ d\tau = dt $$

And obviously the proper time is equal to the coordinate time. All comoving observers agree on the time and therefore on the time since the Big Bang.