Gravitation, light and black holes I have few basic misunderstandings about black holes.
I know that heavy mass causes a curved space time around it, what i don't understand is if this mass is a star and it becomes a black hole, why its space time becomes more curved?
is the density of the mass is a part of the function?
In addition, I know that as you get closer to the singularity of the black hole there is a stronger force which pulls you to the center, is that force is gravity? and if it does, how is that possible that the gravitation is the weakest force in nature? 
Furthermore, black holes called this name because even light can't escape them after the event horizon, I know that as things happens closer to the center of a black hole the outside viewer will see them slower, is that the reason that the light can't escape the black hole or that's the effect of the "gravitation"? 
*equations would really help, thanks for advanced :) 
 A: Event horizon for Static non rotating charge less black hole (Schwarzschild black hole) is defined as spherical surface of radius $$r_s=\frac{2GM}{c^2}\sim 2.95\frac{M}{M_{\bigodot}} Km$$ Where $G$ is gravitational constant and $M$ is mass of the black hole and $M_{\bigodot}$ is the solar mass.So as the mass contained inside increases the event horizon increases in size. Nothing can escape from inside the event horizon.
Sun would  also have a black hole with  the $r_s \sim 3$ Km if all of its masses were concentrated inside a radius $r<r_s$.This is important to understand that while finding the schwarzschild solution of Einstein equation we assumed that we are at a source free region and the curvature of space time is only due to mass that is contained inside the volume of radius $r$ at which we are solving the equation. So a black hole of $r_s\sim 3Km$ would be there only if the one solar mass in contained inside a volume of radius less than 3 Km.
A hand waving argument can be given in the following way .The space time metric for the above mentioned case can be written as 
$$ds^2=(1-\frac{r_s}{r})c^2dt^2-\frac{1}{1-\frac{r_s}{r}}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2)$$
So inside the horizon where $r<r_s$ the coefficients of $dt^2$ and $dr^2$ changes sign implying that space has becomes unidirectional (towards black hole) and everything moves towards black hole center.
$ds^2$ is the interval between two events in space-time and $(t,r,\theta,\phi)$ is co-ordinate of an event in space-time. $ds^2=g_{ij}dx^idx^j$ where $g_{ij}$ is the space-time metric.Of course there is force which we know as gravity but the the actual intention of General relativity is not to see gravity as a force but rather a geometry of space-time caused by the mass.Presence of mass causes surrounding space-time to be curved. Far away from the mass the space-time is approximately flat and object can move in any direction but as we approach say black hole the direction an object can move become oriented to the center of black hole.
For the sake of simplicity let us take the interval in special relativity which is 
 $$ds^2=-c^2dt^2+(dx^2+dy^2+dz^2)$$ We distinguish the time coordinate from the spatial coordinate by the fact the the coefficient of $dt^2$ is $-1$ whereas spatial parts have sign $+1,+1,+1$.Otherwise how will you distinguish them for the sake of  understanding? While writing the metric for the general relativity in the above case we took "$t$" to be time and "r" to be radial distance and time has unique direction(towards future) but once we cross $r=r_s$ where $r<r_s$ coefficient of $dt^2$ becomes positive and that of $dr^2$ becomes negative which implies $t$ and $r$ has interchanged their role.Inside event horizon "$r$" is unidirectional (towards future that is the center of the black hole).
One can explicitly calculate that once an object is inside event horizon it will hit the singularity (center of black hole) in finite time.
A: The region outside a star of mass $M$ has a certain type of curvature. That type of curvature gets stronger the deeper in it goes.
If you had a star of mass $M$ and then had a shell of mass $m$ around it, then the curvature would be of type $M+m$ outside the shell. And it would be of type $M$ inside the shell but outside the star of mass $M.$
So a big ball of gas basically is a series of shells. So the type of curvature gets smaller and smaller the deeper you go.
So long ago when our sun was a big ball of gas, spacetime outside the whole ball was of the same type as is outside the sun now, of type $M_\odot.$ But the type of curvature deeper inside was of smaller and smaller type. So the curvature where the earth is now, wasn't as large back then.
Then, as the gas collapsed, the type $M_\odot$ that was outside got to go deeper, because the type only changes when it meets energy, momentum, pressure, or stress. So eventually, that stronger type got to exist where we are, and that's how strong curvature comes about.
Now as the sun got smaller and smaller, that type $M_\odot$ curvature got to go deeper, but still only goes as far as the outer layer of the sun. And it would have to get much smaller than the sun before a black hole formed and it doesn't go that deep.
To get that deep, the same mass of the sun, $M_\odot$ would have to be concentrated very densely. So the denser you are, the farther in your outer layer can be, and the more space that larger type curvature can extend itself. And since that curvature type is stronger the deeper it goes, you get stronger curvature when you are denser. But the curvature is only stronger when you are closer.
A black hole with the same mass as the sun, would feel just as strong way out here. It's just that when you get close to the sun you aren't aren't as close to the black hole. Which means you can keep moving inwards towards the balck hole without passing into deeper layers of a weaker type of curvature.
That's what makes things strongly curved near  black holes. Of course you are right that a black hole looks black solely because it is slow motion and red. And we never actually see an event horizon form if we stay outside. So its entirely theoretical unless you want to go inside.
What we see is each layer of the star we see it from back before any event horizon formed. Moving slower and looking redder. And then even slower and even redder. Never forming.
It like if you knew an employee was about to say "I quit" over the phone over a second. And you decided to record it and play the first 1/2 second of the recording over a decade and play the next 1/4 second of the recording over the next decade and then next 1/8 second of a recording over the next decade. And so on. You could claim you haven't finished hearing them say they quit. What you hear would be slower and deeper sounding and fainter. And of course your ear would stop being able to hear such deep sounds after a while.
This time dilation ends up having other consequences. That curvature of type $M$ has time tick at a different rate than the curvature of type $M+m$ outside the shell. So they disagree at the speed the matter falls. And so they disagree about how far it fell too.
So the people on the outside think it fell farther than the people on the inside do. So the shell ends up falling downwards more than it gets closer to the star. The surface of the star itself ends up farther away from the distance people when the shell contracts.
Creating space is natural when stars form. As natural as creating the curvature and the time dilation. 
