I have a question about the size of "singularity"
This question has already been asked here, Question about Schwarzschild radius
But what I want to know is this density thing that I'd like a confirmation.

1. But what if it were squeezed to 3.1 KM, the "sun" radius is still 3.1KM, right. It's density will be 18400 trillion of water. Can we, theortically, "measure" the sun with a ruler if it's radius is 3.1KM?
Once it goes below 2.951 KM, it's radius will be 0 KM, singularity. Is that right?
2. Still according to Schwarzschild calculator, for an object (star? nebulae?) 1 G solar mass it's Scharwzshild radius is 2.95E9 KM. It's density is 1.84 percentthat of water.
Did I make a mistake in my calculation?
If that objects is squeezed to 2.94E9, will it disappear and become a black hole?
3. Still using that calculator above. According to Wikipedia
The universe mass is 1E53Kg
So it's Schwarzshild radius is 1.48E26 metres. Dividing it by seconds/minutes/hours/days/years
I find this: 15.6 billion light years. And our universe is 13.6 billion years old. So??
Are we living in a black hole?
If the question number 1 and 2 are true, about "squeezing object below it's Schwarzshild radius", why aren't we squeezed into a singularity?
Did I make a mistake in my calculation, again?
4. Is it possible for a very massive object below it's schwarzshild radius, but doesn't become a black hole? Doesn't become a singularity?

The "singularity" does not have a "size" in a sensible way.
The word is a mathematical way of saying that something is "interesting" or odd about the maths.

1. depends on the ruler ... you usually want to do some sort of triangulation with maybe a relativistic correction.
... once the Sun is squashed below the Schwarzschild radius, there is nothing in the theory to stop it collapsing all the way to nothing. This condition is referred to as "singular". Note: all theoretical point masses have a singularity at their locations.

2. You have certainly made a mistake.
Consider - your density figure for the Sun as a black hole is much less than the density of water at STP. About 2-hundredths the density (1.8%).
Yet, the Sun is currently about 144% or one and a half times the density of water.
So your calculations are saying that you can compress something and end up with a lower density ... does this make sense?

3. Continuing to use a bad method will continue to get you nonsensical results. You have certainly made a mistake in your calculation again.
Note: When you read about size, mass, and age of the Universe you have to be careful - for instance, the "edge" of the visible universe is much farther than it's age multiplied by the speed of light.

Bottom line: Q1 and Q2 are not both true.

4. No. That is pretty much what "Schwarzschild radius" means.
We do not expect that the singularity is a physical object that exists in Nature though.
The gravitational singularity is just the locus of points where the known laws of physics don't work and nobody knows how to fix it.
For a Schwarzschild black hole, this is a single point at the center of mass.

Dear Simon Bridge, I'd like some verification here.

And if we do the math, sun radius 696342^{3} x π x [itex]\frac{4}{3}[/itex] divided by it's mass is exactly 144% that of water as you pointed out. But the sun at 3KM is 184 trillion of water.
What I want know is for an object 1 billion solar mass and it's Schwarzshild radius. Its density is just 1.84% that of water?? Is that right?

Thanks Simon Bridge.
Sorry, I hastily typed. Density is mass/volume. But I calculated an object about 1 billion solar mass, then I find the Schwarzshild radius. But, if I divide it's mass with its volume, I found that it's 1% density of water.
Okay.., I'l do it again.
Mass: 1 billion solar mass, 1.989 x 10^{39}
Radius: 2.951 x 10^{12}m = 2.951 x 10^{13}dm
Volume: 2.951 x 10^{13} x 2.951 x 10^{13} x 2.951 x 10^{13} x [itex]\frac{4}{3}[/itex] x [itex]\pi[/itex] = 1.0766 x 10^{41}l
----------------------------
So,
Mass: 1.989 x 10^{39}Kg
Volume: 1.0766 x 10^{41}l
Density: 1.85% of water?

If that so, so this black hole has density less than neutron star?

If you look at ##V=(4\pi/3)R^3## and ##R=2GM/c^2## before you start plugging in numbers (this is a generally good habit - the numbers go in last when you're trying to understand the relationships between things) you'll see that as M increases so does R, but V increases as the cube of R. Double M and V will increase by a factor of eight, meaning that the density will decrease by a factor of four. If you feel like doing a bit of algebra, you can even work out the formula for the "density". It will look like ##\rho=K/M^2## where ##K## is a constant whose value depends on your units.
So it should be clear that you can make the "density" as small as you like just by choosing a sufficiently large M. Yes, if the black hole is large enough it's "density" will be less than that of water, or air, or whatever.

Whether the result of this calculation is actually the density of anything is a different question. The mass inside the black hole is not spread uniformly throughout its "volume" so you won't find that "density" anywhere... and worse, although the surface area of a sphere around the black hole is ##4\pi{R}^2##, thanks to spacetime curvature the diameter across that sphere is not ##2R## and the volume of the space contained within it is not ##(4\pi/3)/R^3##.

Yes, thanks Nugatory. Just by looking at the formula at a glance
R =2GM/c^{2}
R =M x 2G/c^{2}
R =M x some constant ,
the comparison between mass and volume is exponential to three degree. It's just that I just learned that the mass of a black hole divided its volume from Schwarzshild Radius can be less water density. I'm afraid that I made a mistake in my calculation, that's why I need to verify this in PF forum.
And triple the mass of the black hole, the density will decrease by a factor of nine...?

For a fixed volume, increasing the mass also increases the density. But I see what you are asking now.
For a black hole, increasing the mass also increases it's Schwarzschild radius.
So black holes do not have to be dense.

$$V=\frac{4\pi}{3}R^3,\; R=\frac{2GM}{c^2}\; \implies V=\frac{4\pi}{3}\frac{8G^3}{c^2}M^3\\
\implies \rho = M/V = \frac{32}{3\pi G^3 c^2}\frac{1}{M^2}$$ ... so, triple the mass of the black hole and the density (when you average it across the volume implied by the Schwarzschild radius) reduces by a factor of nine - you are correct.

You can do the calculation a bit differently - work out the size that a blob of water density needs to be so that the escape velocity at it's surface is the speed of light.
Just keep piling mass on and you will eventually get there even just using classical physics.

Caveat: take in Nugatory's note about how the mass inside the Schwarzschild radius is not uniform ... the math puts all the mass at a "singularity", which gives it infinite density. So what ppl are talking about would be the critical mean density just before collapse or something like that. In this sense it is actually a bit misleading to think about black hole formation in terms of density. We are not thinking of a supermassive black hole as the sort of object that would float on water.

Not only that, as Nugatory noted, the spatial volume inside the horizon is not ##( 4 \pi / 3 ) R^3##. More precisely, it isn't unless you choose a particular coordinate chart (Painleve coordinates). In other coordinates it's different; in fact, it's possible to choose coordinates in which it's infinite. So the physical meaning of the "density" being calculated for a black hole is tenuous at best.

Shouldn't it
$$V=\frac{4\pi}{3}\frac{8G^3}{c^\textbf{6}}M^3$$
I replace c^{2} with c^{6}
Btw, you have confirmed my suspicion.
"How can a black hole be more sparse than neutron star, or water for that matter (Is it 'matter', right. Not antimatter', but that should belong to another thread. Can we detect if this black hole comes from matter or antimatter) "?
I think I made a mistake in my calculation.
Thanks a lot everybody.

Yep - I leave these little errors so I can tell if people are paying attention ... either that or I just mess up ;)

Note: there is, as yet, no way to tell an antimatter black hole from a matter black hole.
This question gets asked a lot ... see:
https://sciencequestionswithchris.wordpress.com/2014/05/16/how-can-you-tell-a-black-hole-made-out-of-antimatter-from-a-black-hole-made-out-of-matter/ [Broken]

If this threatens to be a derail we can ask a mod to move it.