(I could be phrasing my question incorrectly, so bear with me as I try to explain it.) Recently I've been very interested in the geometry and physics of black holes, so I started messing around with various gravity-related equations and geometric equations pertaining to n dimensions, only to find myself wanting more. I wondered if there was an equation for the amount of pressure (energy) on the center point of the gravitational force of a collapsing star, eventually creating a black hole. Does this equation exist? If not, who wants to figure it out?

  • $\begingroup$ A star doesn't quietly collapse into a black hole. A stellar mass black hole is the remnant left after a supernova explosion, if the original star's mass is large enough (and if the explosion doesn't completely disupt the star), as described in the Supernova Wikipedia article. So there's a lot more energy involved than the pressure that creates the black hole. $\endgroup$ – PM 2Ring Feb 7 at 6:26
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    $\begingroup$ Pressure $\neq$ energy. Pressure is (related to) kinetic energy density. The pressure immediately prior to collapse would be similar to the central pressure of a fully relativistic polytrope consisting of iron and of mass close to the General Relativistic limit for stability. But these things are not calculated by an equation, they are worked out in numerical computations of stellar structure. $\endgroup$ – Rob Jeffries Feb 7 at 7:12
  • $\begingroup$ The static stellar structure equations are often dealt with at length in astrophysics textbooks. In particular, often the central pressure is what you start with and then calculate the rest of the parameters outwards (in textbooks analytically for some special cases, in reality numerically using a computer) until you find the surface and can determine all the properties of the star. This way one can develop the relationship between things like mass, radius and pressure for various stellar models. But a star imploding into a black hole is not static: the calculations become much different. $\endgroup$ – Anders Sandberg Feb 7 at 9:14

There is no such equation I'm afraid.

Even for a static star the calculation is hard. The pressure is calculated from the equation of state but while we have a pretty good handle on the equation of state for ordinary stars the equation of state for neutron stars is not fully understood. The problem is it's impossible to reproduce such extreme conditions in the laboratory so we have no way of determining the equation of state experimentally. In practice we use a best guess and the calculation is done numerically on a (very) large computer.

For a collapsing star the calculation is even more complicated. The problem is that for a collapsing star we also need to consider the momentum of the infalling gas because the rate of change of momentum of this gas creates an additional force and hence contributes to the pressure. As with the static star we have to resort to numerical models and a great deal of computer time.

Googling will find you lots of articles on modelling stars, both static and collapsing, but be warned that this has evolved into a very complex discipline that will be impenetrable for the beginner.


At the final fraction of a nanosecond before a collection of particles of, say, mass = M, collapses to form a black-hole, one may perhaps assume that the initial mass, M, has been transformed into E = mc^2. If the math (below) is correct, then the force of gravity at the Schwarzschild boundary would be constant for all black-holes and all black-holes would have the same energy-density...2E/r:

Since r = 2GM/c^2, and c^2 = 2GM/r(Schwarzschild) = constant, (and M/r = constant), one might assume that the escape velocity for all Black-holes should be constant; and the gravitational force at the Schwarzschild boundary should also be constant:

r(s) = 2GM/c^2 *radius of BH; or

c^2 = 2GM/r = constant

Thus if c = constant, then M/r = constant, and multiplying be C^2…

c^4 = 2GMc^2/r = 2GE/r = constant; and dividing by G….

c^4/G = 2E/r = constant; note that c^4/G = the Planck force (F) = 1.2 Newtons X 10^44: then

F = 2E/r = constant. = The gravitational force of a black-hole.

Thus it would appear that all black-holes have a maximum force (very close to the EH) equal to the Planck force. This would suggest that all objects located at (near) the surface of a BH would "weigh" the same, regardless of the radius (size) of the BH.

Additionally...Multiplying by 2pi gives:

F = 4pi E/Schwarzschild circumference (C) = constant. Thus it would also appear that the energy density per unit (Schwarzschild) circumference is the same for all black-holes, regardless of their size. This would suggest that all black-holes have the same "temperature"...the Planck temperature.

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    $\begingroup$ As you've been told previously, your theories regarding black holes are non-mainstream. $\endgroup$ – PM 2Ring Mar 14 at 5:21
  • $\begingroup$ @PM Dear Sir: The equations you have deleted do not comprise a theory; I have attempted to support the assertions made in the comment by a mathematical argument. If you have really read the material (above) you must have some idea of inconsistencies, if there are any. I would greatly appreciate understanding any inconsistencies you may find in the math and any argument you may have regarding the assertions contained in the comment. Thanks for your time and your attention to this matter...RobertO $\endgroup$ – RobertO Mar 15 at 19:17
  • $\begingroup$ What do you mean? I haven't deleted any of your equations. (And I didn't downvote this answer). However, in comments on another post of yours I have mentioned an incorrect assumption of yours. You said that relativity stops particles that are moving close to c from increasing their momentum. That's untrue: momentum (& kinetic energy) can grow without limit, even though speed of particles with nonzero mass must remain below c. $\endgroup$ – PM 2Ring Mar 15 at 19:29
  • $\begingroup$ PM: Dear Sir, mille pardons for having falsely accused you of an action that you were not responsible for. With respect to the limits of KE....my assumption is NOT that "relativity stops particle...close to 'c' ". My argument is about the possible transformation of particles as they approach hyper-relativistic values may undergo a (as yet undetermined) QM process that produces E= mc^2 radiation. All of this is, of course 'up for grabs'. $\endgroup$ – RobertO Mar 16 at 20:19

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