# Singularity in a black hole [duplicate]

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This might be a naive question, but how can an object such as a black hole singularity have infinite density but finite mass? (For example, we can approximate the mass of a black hole based on Kepler's Laws and use info from surrounding movements of stars to determine the central mass, but the black hole, excluding the event horizon, has infinite density.)

## marked as duplicate by Qmechanic♦Sep 28 '13 at 17:26

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• Possible duplicate: physics.stackexchange.com/q/25802/2451 More questions on density of black holes: physics.stackexchange.com/… – Qmechanic Aug 8 '13 at 2:55
• To speak about (mass) density, you need a mass $M$ and, admitting a spherical symmetry, a radius $R$. The only quantity you may use is the radius of the horizon, or Schwartzschild radius $R_s$. For a Schwartzschild black hole, you have $R_s = \frac{2GM}{c^2}$. So you can calculate a "density". You will find, that, more the black hole is big, more its "density" is low – Trimok Aug 8 '13 at 7:39
• Please see my answer in physics.stackexchange.com/q/72927 . You can apply it to the "infinite density" as well as to an infinitessimal volume. – anna v Aug 8 '13 at 8:25

## 1 Answer

The simple way to think about it is the following limit: $$\rho = \lim_{V\to0}\frac{M}{V}$$

As $V\to0$, $\rho\to\infty$ while $M$ remains constant.