Why does a star collapse under its own gravity when the gravity at its centre is zero? The gravity at the centre of a star is zero as in the case of any uniform solid sphere with some mass. When a massive star dies, why does it give rise to a black hole at it's centre? 
I know how to derive the field equations for gravity inside a star assuming the star as a uniform solid sphere of mass M and radius R. I need to know how to find the expression for the total pressure due to gravity at the centre.
 A: The condition for creation of a black hole is:
$$ \text{gravitational potential} \le -\frac{ c^2 }{ 2 } $$
I won't go into the details of how to calculate the potential.  But for the center of a star, suffice it to say that it's slightly more complicated than $-GM/r$.
You can see that this makes no reference to the gravitational field itself.  It comes from the integral of the gravitational field.  What's more, it's subjective.  If I'm at a different gravitational potential than you (practically, I am, somewhat), then you and I will disagree about where the event horizons are, and even which objects may be black holes.  And yet, this is what the physics tells us.
Light cannot escape from below the event horizon, so we're tempted to think of it as a matter of the acceleration there.  But this isn't quite the case.  The conflict is resolved in the subtleties of the mathematics of general relativity.  I find it more accurate to think of an accumulated current of spacetime, but formally, this is a "geodesic".  A geodesic is one of the lines you can travel if you undergo no acceleration.  At the event horizon, there are no geodesics that more away from the singularity.  So even light "stands still".  The light cones are tilted.  This tilting isn't the same as acceleration.  It's something different entirely.  This is truly strange, and it's what happens between different gravitational potentials.
A: Since every particle attracts all other particles, there is a net force directed towards the center of the star (or any object), for any particle not at the center.  Therefore, the particles will move towards the center (collapse), unless some opposing force prevents it.  In the case of a star, the kinetic energy of the particles creates the opposing force, until the energy "runs out" and the collapse fallows.   
A: It's because the value of the gravitational field at the center of a star is not the relevant quantity to describe gravitational collapse.  The following argument is Newtonian.
Let's assume for simplicity that the star is a sphere with uniform density $\rho$. Consider a small portion of the mass $ m$ of the star that's not at its center but rather at a distance $r$ from its center.  This portion feels a gravitational interaction towards the other mass in the star.  It turns out, however, that all of the mass at distances greater than $r$ from the center will contribute no net force  this portion.  So we focus on the mass at distances less than $r$ away from the center.  Using Newton's Law of Gravitation, one can show that the net result of this mass is to exert a force on $ m$ equal in magnitude to
\begin{align}
  F = \frac{G( m)(\tfrac{4}{3}\pi r^3 \rho)}{r^2} = \frac{4}{3}G m\pi\rho r
\end{align}
and pointing toward the center of the star.  It follows that unless there is another force on $m$ equal in magnitude to $F$ but pointing radially outward, the mass will be pulled towards the center of the star.  This is basically what happens when stars exhaust their fuel; there no longer is sufficient outward pressure to counteract gravity, and the star collapses.
A: Well, you're right that a particle sitting at the centre of a star (or generally the centre of any spherical distribution of matter) feels no net gravitational force. So, in the absence of other forces, it will simply continue to sit at the centre. But every other particle in the spherical distribution will feel a gravitational force pulling it toward the centre. There is a distinction here; there is no net force at the centre, but there is a lot of force toward the centre.
Now forming a black hole is much more complicated, because gravity is not the only force. Typically there is some form of pressure force that opposes collapse. The standard picture of a star is when the outward pressure balances the inward gravity, and is called hydrostatic equilibrium. If the star loses pressure support (often happens as it runs out of fuel for whatever nuclear reaction is ongoing), it will start to collapse due to gravity. Then either some other source of pressure will stabilize the star at a new equilibrium (could be a new nuclear reaction starting, typical in post-main sequence evolution of stars, or quantum mechanical effects like "electron degeneracy pressure" supporting a white dwarf, or "neutron degeneracy pressure" for neutron stars). Rotation can also help stabilize the star. If no mechanism provides sufficient pressure to oppose gravity, then you get a black hole.
A: what causes the star to collapse is pressure. what causes the pressure is gravity, but even though the strength of the gravitational field in the center of the star is zero, the pressure at the center of a star sure-as-hell ain't zero.
