Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
Note that not all massive stars transform into black holes, just the ones larger than about 25 solar masses. The ones between around 8 and 25 solar masses turn into neutron star. –  Kyle Kanos Feb 3 at 19:19
1  
A stretched balloon doesn't have any notable forces at it's middle either, but when I pop it, it also collapses to (near) it's center –  Mooing Duck Feb 4 at 1:31
    
It's the pressure at the center that causes collapse. That pressure is brought about by gravity, but it is the pressure at the center that causes further squishing. –  Olin Lathrop Feb 10 at 14:56

5 Answers 5

up vote 17 down vote accepted

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.

share|improve this answer
2  
As we're considering the limit of r to zero, I find the equation (which is a common model) to be unconvincing. As we get to the center, the force you mention tapers off to zero. However, the pressure continues to rise, and this is what stars can't maintain. As the fuel is depleted, the pressure will have no choice but to fall. I believe there is a pressure term in the momentum "currents" of the general relativity field equation. So that's certainly relevant, but still much more difficult than looking at it from the perspective of a far-away observer. –  Alan Rominger Feb 3 at 19:35
    
@AlanSE: but that depends entriely on the radial dependence of the density. Assume that it falls off like $\frac{1}{r^{n}}$ for some $ n > 0$. Then, the force diverges. And the argument for why real stars collapse is based in general relativity, and the stability of stars under perturbations in full general relativity. newtonian mechanics can't help you there. –  Jerry Schirmer Feb 3 at 19:40
4  
@AlanSE I felt that the OP was just confused by the general idea that collapse can happen to gravitational bodies given that there is a vanishing field at the center, so I decided to give the simplistic Newtonian argument above. I certainly have not addressed details of black hole formation. If that's what the OP is actually looking for, then I certainly haven't answered the question. –  joshphysics Feb 3 at 19:42
2  
@joshphysics: I think even the newtonian argument is already good enough to get the feeling behind the colapse, although GR does change the picture a lot. For completness one could add the notion of Jeans instability en.wikipedia.org/wiki/Jeans_instability which I think shows that collapse goes on inevitably in certain conditions. For spherically symmetric cases one can map the GR argument in the newtonian one fairly easily, and thus should suffice for the Schwarzschild Black Hole –  cesaruliana Feb 4 at 3:05
    
@cesaruliana Cool interesting! Had never learned about the Jeans instability. Thanks for the link. –  joshphysics Feb 4 at 3:34

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
2  
This is laughably false. Pressure pushes outwards. –  user54609 Feb 8 at 20:31
    
The weight of the overlying material creates the pressure, but if you want to credit the pressure for a role in the dynamics of the star that role is supporting the outer layers. –  dmckee Feb 8 at 23:30
    
um, @user54609, pressure in a fluid pushes in all directions. inwards, outwards, sideways, whatever. and dmckee is correct that it's the weight of the overlying material that creates the pressure. and if the pressure gets unbelievably intense, interesting things might happen to atoms in the material. –  robert bristow-johnson Feb 9 at 0:37
1  
But the fact is that the higher the pressure, the less likely the star collapses. The reason why the balloon collapses when you break it is because when you break it, the pressure goes away. The lack of pressure causes gravity (in the star) or the springiness of the balloon to take over and make the thing collapse. –  user54609 Feb 9 at 18:46
    
This answer is highly relevant in the case of black holes. In General Relativity, the pressure of the gas contributes to the curvature of space-time and means that in GR a larger pressure gradient is required to support a given star. Ultimately this is why black holes exist, because there is no way to can keep ramping up the pressure gradient inside the star without increasing the pressure at its centre and therefore increasing the required pressure gradient and so on... So, whilst most of the comments here are true for "normal" stars they are not in the field of neutron stars/BHs. –  Rob Jeffries Mar 30 at 9:33

protected by Qmechanic Feb 8 at 20:31

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.