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.

  • $\begingroup$ I can't add much to Rennie answer below. A comoving observer is one that is inertial and that is frame dragged with the expansion of space. I wrote a comment to your question physics.stackexchange.com/questions/295303/… $\endgroup$ Dec 12, 2016 at 11:13
  • $\begingroup$ This question does not make sense, there is no such thing as an "invariant interval between points", there are only lengths of paths, and your function $s$ doesn't exist. $\endgroup$
    – ACuriousMind
    Dec 13, 2016 at 23:57
  • $\begingroup$ @ACuriousMind Please see edit $\endgroup$
    – R. Rankin
    Dec 15, 2016 at 9:16
  • $\begingroup$ "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."...then how are the proper times that have passed for the twins different when they meet again (which is after all the whole point of the twin paradox* - the proper time of paths is an invariant, which resolves the paradox because it is unambiguous whose path is "longer"!)? You clearly need to go back and treat both the math and the physical concepts at play here much more carefully. $\endgroup$
    – ACuriousMind
    Dec 15, 2016 at 13:49
  • $\begingroup$ @ACuriousMind The last equation above clearly shows both objects having experienced different times. One could instead integrate along the path of the "moving" party above and call this proper time, that is however a different coordinate system. In that system then, the observed interval would still be the same for both parties (though not the same as the other coordinate system). $\endgroup$
    – R. Rankin
    Dec 16, 2016 at 0:41

1 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.

  • $\begingroup$ A non comoving observer would have to agree on the interval they see the comoving observer traverse though by definition. I'd been thinking a bit about this as it's related to another question of mine: physics.stackexchange.com/questions/295303/… $\endgroup$
    – R. Rankin
    Dec 12, 2016 at 10:36
  • $\begingroup$ apologies, I meant to say invariant interval (not time interval), I've edited the above question. $\endgroup$
    – R. Rankin
    Dec 12, 2016 at 11:14
  • 1
    $\begingroup$ @R.Rankin Does John's answer not also imply that? If the invariant intervals are all equal to the co-ordinate time intervals for every timeslice, then surely their integrals co-incide too? Or am I misunderstanding you? $\endgroup$ Dec 12, 2016 at 12:59
  • $\begingroup$ @WetSavannaAnimalakaRodVance I was thinking that all observers regardless of frame would have to agree on the value of the scale parameter a. Is that wrong? Which defines a given cosmological time. $\endgroup$
    – R. Rankin
    Dec 13, 2016 at 0:06
  • 1
    $\begingroup$ Hey John, can you please say what exactly is true here? The math in the question is ill-defined and there is no such thing as the "invariant interval" between points in an arbitary spacetime. The proper time is the length of the world line of an observer, but this is a property of the world line, not of the points. $\endgroup$
    – ACuriousMind
    Dec 13, 2016 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.