# Why does spatial curvature become observable only in late universe?

1. Sep 10, 2014

### Lapidus

So the universe starts with an amount of matter, radiation, a fixed spatial curvature constant and a cosmic constant. Due to the expansion of the universe matter and radiation dillute and the spatial curtvature decreases whereas the cosmic constant remains fixed.

Radiation dillutes faster than matter, and curvature decreases slower than both radiation and matter dillute. But nonetheless the spatial curvature decreases, it gets smaller the older the universe gets. (Right? Like a ballon that gets blown and which surface becomes less and less curved.)

Why then does spatial curvature become (better) observable in the latter stage of the universe?

THANK YOU

2. Sep 11, 2014

### Chalnoth

The reduction in curvature over time has to compete with the fact that as our universe ages, we can observe distances further away. Being able to observe further away provides us with a longer lever-arm with which to measure the curvature. For example, our current best estimates of the spatial curvature of our universe stem from the combination of cosmic microwave background data (which comes from the furthest edge of our observable universe) and baryon acoustic oscillation data (which comes, primarily, from galaxies within a few billion light years).

As the Baryon Acoustic Oscillations which result in the details of the statistics of the typical separation between galaxies in the nearby universe are the exact same process which results in the hot and cold spots on the CMB, it is relatively easy to compare the distance scales on the CMB to distance scales in the (relatively) nearby universe, providing a highly accurate measure of spatial curvature. As our universe expands, that lever arm gets larger.

Furthermore, as we look further away, we are looking back in time. The CMB is now and always will be an image of our universe when it was a specific age (about 300,000 years old, if I recall). So the curvature at the time the CMB was emitted will always be the same. But as it gets further away, we'll have a longer lever arm with which to measure.

That said, I'm not 100% certain that this argument is correct. It makes sense to me, but I can't be sure it's right just now.

3. Sep 11, 2014

### Jorrie

Regarding the first part of the statement, I have always thought that any non-zero spatial curvature that existed at early times would increase as time goes on, i.e. evolve away from flatness. I'm not sure if the presence of the cosmological constant might change that, but it was definitely the case in the era when lambda was thought to be zero.

4. Sep 11, 2014

### Chalnoth

Ahh, right. Can't believe I made that mistake.

The effect of curvature does dilute with expansion, at a rate of $1/a^2$. But that rate is slower than the dilution of matter or radiation. So yes, at the time the CMB was emitted the curvature was much smaller in its effect than it is at present (the recent accelerated expansion hasn't gone on long enough to dilute the curvature sufficiently to make this statement false).

5. Sep 12, 2014

### Lapidus

Thanks for the answers, Chalnoth!

But what are the physics behind that? Normally you would think that the longer something blows up the flatter it will get. But obviously the Friedman equations say it is otherwise: the larger the universe grows, the more it gets spatially curved. That's counter-intuitive to me. Is there some physical explanation for it?

THANKS again!

Addendum: As I understand in inflationary models the (extremely) rapid expansion just does flat-out the universe. Again, why is that not the case for the Big Bang cosmology?

Last edited: Sep 12, 2014
6. Sep 12, 2014

### Chalnoth

The effect of expansion should be rather intuitive. I think it's easiest to think of it in terms of the radius of curvature. It turns out that the radius of curvature is simply proportional to the scale factor: double the average distances between objects, and you also double the radius of curvature.

So yes, in a sense, the universe does get "flatter" whenever it expands, because the radius of curvature increases with that expansion.

But the curvature also has an impact on the expansion, set by the first of the FRW equations:

$$H^2 = {8 \pi G \over 3} \rho - {kc^2 \over a^2}$$

Here $\rho$ is the energy density of the universe (note that I'm omitting the cosmological constant, because you can wrap that into $\rho$). To make this easier to understand, you can rewrite the above equation in terms of the current expansion rate $H_0$:

$$H^2 = H_0^2\left({\Omega_r \over a^4} + {\Omega_m \over a^3} + \Omega_\Lambda + {\Omega_k \over a^2}\right)$$

As $H_0 = H(a = 1)$, the various density fractions all sum to one ($\Omega_r, \Omega_m, \Omega_\Lambda, \Omega_k$). If you don't have any cosmological constant ($\Omega_\Lambda$), then the curvature fraction ($\Omega_k$) dilutes the slowest. So even though the curvature gets smaller, it gets smaller more slowly than the energy density, and so becomes more noticeable.

But if you have a cosmological constant, the reverse happens. The impact of the curvature dilutes, but the constant doesn't. During inflation, this is what occurred (and is also occurring at present), as the energy density during inflation was nearly constant.

7. Sep 13, 2014

### Lapidus

Thanks again for answering!

I see what the Friedmann equations are saying, but I still do not see the physical mechanism behind all this. I understand when radiation dilutes faster than matter, then matter becomes more noticeable. But curvature is not "stuff" like radiation, matter or vacuum energy. It is the bending of space, that in turn is determined by the stuff that is distributed in space (and time).

How and why does the curvature of space become more noticeable when stuff in this space becomes more spread out? What is the physical and causal connection here?

Stuff (i.e. matter, energy) curves space(time). Why does space curve more when it expands even though the amount of stuff remains the same and just thins-out?

8. Sep 13, 2014

### Jorrie

I do not think anyone here said space curves more as it expands. Just that it should become more observable in later times. Except that the observed curvature is so close to zero that we cannot really say that we observe it. We can just set limits to it and those limits still include zero curvature. Add to that the fact that dark energy should be driving spatial curvature even closer to zero.

9. Sep 13, 2014

### Lapidus

OK! So a rough analogy would be that if the universe is a sphere then the Big-Bang expansion with matter and radiation makes us see more and more of the sphere. Whereas in an inflationary expansion with only a vacuum constant the sphere would only get blown up.

10. Sep 13, 2014

### Jorrie

Roughly yes, except that we will "see less and less" with accelerated "inflationary" expansion.

11. Sep 13, 2014

### Chalnoth

Perhaps one way of understanding this is thinking about a somewhat different universe: a closed, matter-dominated universe. Such a universe will expand up to a point, slow down, then collapse back inward. In some sense, this is a little bit like throwing a ball up in the air (and forward somewhat) and letting it fall down. As the ball is closest to its apex, the curvature of its path is most apparent. Its path gets straighter as it falls down further.

Similar with the universe that collapses on itself: the curvature is most apparent at the furthest amount it expands.

Does that make any sense?

12. Sep 13, 2014

### Lapidus

That helped very much! I think things getting clearer now.

Only for clarification: the curvature of the universe is set by the beginning of the universe. So are the amounts of the other components.

But are they independent? Is the value and sign of the curvature parameter independent of the matter and radiation content of the universe?

Thanks once again!

13. Sep 13, 2014

### Jorrie

No, the curvature parameter depends on the sum of the density parameters: $\Omega_k = 1- (\Omega_m + \Omega_\Lambda + \Omega_r)$ (Chalnoth did mention it before). I think it is reasonable to assume that the density parameters were set shortly after inflation ended and that they may well sum to unity; hence the curvature parameter was either zero or very, very close to zero.

14. Sep 13, 2014

### Chalnoth

I think the best way of thinking about the curvature is in terms of the relationship between density and expansion:

$$H^2 = {8 \pi G \over 3} \rho - {kc^2 \over a^2}$$

Or, alternatively we could write:

$$k = {a^2\over c^2}\left({8\pi G\over 3} \rho - H^2\right)$$

You can ignore the $8\pi G/3$ part. That's just a conversion factor. So we can just imagine changing our units to where $8\pi G/3 = 1$, $c = 1$, and also just say that we're talking about the time when $a = 1$ (usually considered to be right now). In that situation, we can just write:

$$k = \rho - H^2$$

This, I think, clarifies what's going on. The spatial curvature is related to the balance between the rate of expansion and how much stuff there is. If the two are equal (with the right conversion factors), then there is no spatial curvature. If you have too much matter around for the specific expansion rate, then the universe has positive spatial curvature. If it has too little, it has negative spatial curvature.

This is easiest to understand in the context of a matter-only universe. In the matter-only universe, the expansion is slow enough that the self-gravity of the matter slows, stops, and then causes the universe to collapse back on itself. This is like the ball that you toss up into the air and then comes back down.

If expansion is fast enough, then the expansion just continues forever. This is an open universe (negative spatial curvature). It is equivalent to the ball that you throw upward so fast that it escapes the Earth's gravity and escapes out into space. The escape velocity from the Earth's surface is about 11km/s (about 24,000 miles per hour). So if, say, you throw it up at 20km/s, then it will slowly approach about 17km/s as it gets far away (this is the kinetic energy it has remaining after leaving the Earth's pull, and I'm taking the easy way out and assuming the atmosphere doesn't slow it down). The gravitational pull of the Earth will slow it down forever, but the speed is so much faster that it doesn't matter too much: it just keeps going at high speeds.

If the expansion rate is just right, then the matter-only universe will have just barely enough enough momentum to continue expanding forever. That expansion rate will approach zero as time goes to infinity, but it will always continue to expand.

Incidentally, this leads to another reason why the effect of curvature becomes more pronounced: imagine if, right now, the expansion rate and matter density were almost exactly balanced (and we lived in a matter-only universe). But the density was just a teensy, tiny bit higher. Observers in our universe might not notice for billions of years that we live in a positive curvature universe, but eventually the expansion will stop and start going back. That will be a really big effect, and is completely different from the expansion continuing on forever. But you might not even be able to measure that effect today, because it might be way too small.

Of course, if the universe has dark energy, this analogy is broken: the final fate of the universe is eternal expansion no matter what the spatial curvature is (as long as it isn't so big that it causes recollapse really early....but now that dark energy is dominant, the curvature can't be that big). So you can't really use the ball analogy. The curvature is still about the balance between the rate of expansion and how much stuff there is, but it just no longer has the recollapse/eternal expansion behavior that makes it work sorta kinda like a ball you're throwing upward at different speeds.

15. Sep 14, 2014

### Lapidus

Thanks Jorrie, and a very big thank you to Chalnoth for this long and very clear last post. Much appreciated!