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The Friedmann-Lemaitre-Robertson-Walker metric (FLRW metric) is described as:

$$ds^2 = dt^2 - a^2(t) (\frac{d \bar{r}^2}{1-K\bar{r}^2} + \bar{r}^2 d\Omega^2)$$

What does $a(t)$ represent?

I know that FRLW metric comes from embedding a 3D sphere in a 4D flat space. a is described as the radius of the sphere. Does it mean that a represents somehow the radius of our Universe (if FRLW metric is used to describe it)? Would it also mean that FLRW metric implies that our Universe would be embedded in higher dimensional world? Or is this just mathematical?

I also found that in my textbook that: (for a lightpulse emitted at $t_1$ and detected at $t_0$) (btw, why do they need to coincide in some way to lead to the next expression?)

$$\lambda_0 = \lambda_1 \frac{a(t_0)}{a(t_1)}$$

so $a(t_0)>a(t_1)$ means redshift and $a(t_0)<a(t_1)$ means blueshift. Which would mean that it describes the expansion of the Universe?

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$a(t)$ is the scale factor, and roughly corresponds to the size of the universe. Conventionally $a_{right\_now} \equiv 1$, and $a_{big\_bang} \equiv 0$. The variation of $a$ over time is a measure of how fast the universe is expanding. If $da/dt > 0$, then the scale factor is increasing with time, which implies the universe is expanding. Conversely, if the universe's expansion ever stops and reverses, $da/dt$ will be less than zero. If you're studying cosmology you will soon encounter expressions for how fast $a(t)$ varies during the eras of radiation & matter domination.

The scale factor is not the same as redshift, although the two are related by $a(t) = 1/(1+z)$.

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    $\begingroup$ Measurements indicate that the universe is probably infinite in spatial extent, so it doesn’t really make sense to say that $a$ corresponds to the size of the universe. $\endgroup$
    – J. Murray
    Jan 19, 2021 at 4:44
  • $\begingroup$ Does it mean that any redshift is related to the scale factor? I thought that redshift was due to the Doppler effect, which in turm is related to the velocity of the galaxies. But isn't there a contribution to redshift that comes from acceleration of expansion? Can you still related it to the scale factor with the same formula? Is it contained in that formula? $\endgroup$
    – Maj
    Jan 19, 2021 at 10:08
  • $\begingroup$ @J.Murray right, but it's a way to guide one's intuition (I certainly think about the scale factor that way). $\endgroup$
    – Allure
    Jan 19, 2021 at 10:57
  • $\begingroup$ @Maj only the cosmological redshift (the redshift arising from the expansion of the universe) is related to scale factor by that relation. At cosmological distances, the Doppler effect arising from so-called peculiar velocities is negligible. Some people still interpret cosmological redshift as a Doppler redshift, though. It depends on what text you're using. $\endgroup$
    – Allure
    Jan 19, 2021 at 11:01
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    $\begingroup$ @safesphere I am curious how you could possibly justify the claim that the universe must be spatially finite. It’s certainly possible - maybe it is topologically closed, or maybe it has some kind of edge - but the idea that the alternative should be rejected out of hand is a viewpoint I’ve never encountered before. $\endgroup$
    – J. Murray
    Feb 7, 2021 at 15:19
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Think of any surface with a rotational symmetry like that of a pot shaped on a potter's wheel. You can cover it with coordinates analogous to latitude and longitude. (In fact Earth, idealized as an oblate spheroid, is a surface with this symmetry.) The "latitude" is the distance from the bottom-center of the pot, measured along the pot, on a line located in a plane with the axis of symmetry. The "longitude" is the angle of that plane, with an arbitrary zero point.

Writing $t$ for the latitude and $x$ for the longitude, distances measured on this pot are $\sqrt{dt^2 + a(t)^2 dx^2}$, where $a(t)$ is the radius of the circle at a distance $t$ from the bottom-center.

You can generalize this to $n$-dimensional pots by replacing the circle at distance $t$ with a $(n{-}1)$-sphere. You need additional coordinates to cover the sphere. You can do this by a similar distance-from-reference point trick, but using a fixed function like $\sin(x/R)$ in place of the general $a(t)$.

You can generalize to "open pots" where the surface at distance $t$ is not a sphere but a Euclidean or hyperbolic space (with $x$ or $\sinh(x/R)$, respectively, in place of $\sin(x/R)$).

Finally, you can generalize to Lorentzian signature by flipping the sign of $dt^2$. That's the FLRW metric.

Any space(time) with a symmetry similar to the pot can be covered by coordinates like this. In standard cosmology spacetime has this shape at large scales, and it's convenient to use these coordinates.

Although I started with a pot in a flat background, that was just for visualization purposes. The pot surface is the only thing that actually exists, at least in cosmology.

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  • $\begingroup$ So I am guessing there is no way of gaining knowledge of the flat background solely from the pot surface. Is this right? The pot surface is all that exists and we cannot infer anything about the background? $\endgroup$
    – Maj
    Jan 19, 2021 at 10:11

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