Size of Universe after inflation

Question

I read in some website that during the period of inflation, the expansion of the universe underwent incredibly fast, and its size increased by a factor of $10 ^{50}$, see this link

In this field, I think, there is nothing for sure, but if there was really inflation, what does it mean that the size increased by a factor of $10 ^{50}$, but from what extent? From 1 mm? Or from the Planck length? Or something else?

Has it been established that (whatever was the original measure) the factor is increased by $10 ^{50}$, and not for example of 40 or $10 ^{50^{60}}$?

Answer 1

In the phrase used in the article you link:

inflated the size of the cosmos by a factor of $10^{50}$

the word size is misleading and should be replaced by scale factor. Whether the universe has a size or not isn't clear. The universe may well be infinite, in which case its size isn't defined. However the scale factor is precisely defined, and it's the scale factor that changed by $10^n$ during inflation (where the value of $n$ depends on the model you use).

You probably learned Pythagoras' theorem at school, and this tells you that if you move a distance $dx$ in the $x$ direction and a distance $dy$ in the $y$ direction then the total distance you've moved, $ds$, is given by:

$$ ds^2 = dx^2 + dy^2 $$

General relativity is basically a theory for calculating the distance $ds$ as in the equation above, but the expression used is rather more complicated than Pythagoras' theorem because (a) it includes movements in time and (b) spacetime can be curved. If you make a few simplifying (but physically reasonable) assumptions about the distribution of matter universe general relativity tells us that the analogous expression for calculating $ds$ is:

$$ ds^2 = -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2) $$

This equation is called the FLRW metric if you want to investigate it further. As promised, this expression includes time (with a minus sign!) but for our purposes the important bit is $a(t)$, which is called the scale factor.

If you ignore $dt$ for the moment, the expression looks much like Pythagoras' theorem, but the total distance $ds$ is multiplied by $a(t)$ so if $a(t)$ increases with time then the distance $ds$ increases with time by the same amount. What happened during inflation is that $a(t)$ increased by $10^{50}$, or $10^{60}$ or whatever number your favoured theory of inflation predicts.

So when the article says the size increased by $10^{50}$ what it really means is that if you choose any two points then during inflation the distance between those points increased by $10^{50}$.

Answer 2

The best way to think about inflation (according to my University's cosmology expert) is to think as if you were measuring the universe with an ever-shrinking ruler.

So if you were measuring 1 meter, then your two reference points would now be separated by $10^{50}$ meters, because your meter is shrinking. If you were measuring things 1 mm apart, now they would be $10^{47}$ meters apart. To directly answer your question, there is no definite size of the Universe, because it is infinite.

The definition of a factor implies that there doesn't need to be an original measure.

If there was an original measure (and the universe was not infinite), I bet you scientists wouldn't use a factor but would give the actual size!