Posted on 05/07/2016 8:24:24 AM PDT by MtnClimber
The shape of a Schwarzschild black hole is that of a sphere, and so its area is $A=4\pi r_ h^2$. We might be tempted to use our ordinary intuition about geometry, and deduce that the volume of a Schwarzschild black hole must be $V=4/3 \pi r_ h^3$. This is, however, not necessarily the case. It turns out that the volume of a black hole is not a well-defined notion in general relativity. The reason is that general relativity is a geometric theory of a four-dimensional spacetime, that is, three dimensions of space and one dimension of time. In order to specify a spatial volume, one has to specify a specific moment in time......
Even though a Schwarzschild black hole looks the same forever to an outside observer (ignoring something called Hawking radiation that would shrink the black hole), its volume actually gets larger with time.
When the time $\nu $ is sufficiently large, the volume is given by:
\[ V \approx 3\sqrt{3}\pi M^2 \nu . \] From this expression one sees that the volume of a black hole continues to grow larger and larger, although its surface area never changes! To give an idea of how large the interior of a black hole could become, this formula estimates that the volume for Sagittarius A*, the supermassive black hole at the centre of our Milky Way Galaxy, can fit a million solar systems, despite its Schwarzschild radius being only about 10 times the Earth-Moon distance. (Sagittarius A* is actually a rotating black hole, so its geometry is not really well-described by the Schwarzschild solution, but this subtlety does not change the result by much.) Growing with dimensions
(Excerpt) Read more at plus.maths.org ...
A very well written article. I had to cut and cut to get the exerpt to fit in 300 words.
We can verify all of that next time we actually go see one in person ...
That sounds vaguely familiar...
must not post picture of wookie’s derriere
No, no. The volume is well known to be $V=2/3 \rhu r_ lsmft_egbd%%sf h<> 4c
The key equation is the following; When the time $\nu $ is sufficiently large, the volume is given by:
\[ V \approx 3\sqrt{3}\pi M^2 \nu . \]
Untangling this mashup of ill-conversion leaves us with the following;
When the time ν is sufficiently large, the volume is given by;
V≅ 3√3 π M^2 ν
Remember the 'distributive property'
Unicode, love it or ...
Black Hole = racist and sexist.
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