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Mathematical definition of a weak gravity is simple $g=\frac{GM}{r^2}$ but what is mathematical definition of a strong gravity? (blackhole-like or close to a blackhole-like object)

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it's far more complicated than just that, what you're looking for is the framework of general relativity! –  Schlomo Steinbergerstein Apr 20 '13 at 10:18
    
"mathematical definition of a stronger gravity?" the simple answer is, a larger M. –  Ataraxia Apr 20 '13 at 13:56
    
@ZettaSuro: or the strong coupling limit (in the ADM formalism) when $G\to\infty$ --- i.e., gravity becomes "really, really strong". –  Alex Nelson Apr 20 '13 at 16:46

1 Answer 1

Gravity is described by Einstein's field equation. The Newton equation that you mention isn't the mathematical definition of anything, it's just an approximate solution for a spherically symmetric body when gravity isn't too strong. The corresponding exact solution is the Schwarzschild metric, and this is what we use to calculate what happens near a black hole.

Even the Schwarzschild metric isn't a mathematical definition, it's just the solution to the field equations that we get given various conditions. For example, if the black hole is rotating the Schwarzschild metric no longer describes the gravity around the black hole and we have to use a different solution. For a rotating black hole the solution is the Kerr metric. For more complicated system, e.g. two black holes orbiting each other, there are no analytical solutions and we have to calculate the gravitational field numerically using computers.

In the sense in which (I think) you mean the mathematical definition, that definition would be the field equation. All the solutions I mentioned above, including Newton's equation, can be calculated from the field equation.

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