# 黑洞一定密度很大么？

\begin{eqnarray}
r\le \frac{2MG}{c^2}=\frac{2G}{c^2} \frac{4}{3}\pi r^3 \rho
\end{eqnarray}

\begin{eqnarray}
r\ge\sqrt{\frac{3c^2}{8\pi \rho G}}
\end{eqnarray}

Professor Cahill他突然觉得这个很奇怪。我说这个是对的，按照下面来解释。

r\le \frac{2G M}{c^2}

\begin{eqnarray}
\rho&=& \frac{M}{\frac{4}{3}\pi r^3} \\
&\ge & \frac{M}{\frac{4}{3}\pi (2MG/c^2)^3} \\
&=&{\text{Const}} \frac{1}{M^2}
\end{eqnarray}

update:2011-10-27

Some guy mentioned that the calculation for volume is done improperly. I think here is the trick.

Think about the solar first. When we are going to calculate the density, we should know it’s volume. Then we should calculate the following integration:

\int h_1 h_2 h_3 \mathrm dx_1\mathrm dx_2\mathrm dx_3

where $$h_1$$,$$h_2$$,$$h_3$$ are the scale factors of the space part (make a projection or just take the space part out if the metric is orthonormal) space-time manifold inside the sun. However, to calculate the metric inside the sun, we should know the exact equation of state of the inner sun. This work can be done under these normal stars or planets, not black holes. This means we can never do such a calculation to get the volume of black holes because we know nothing about the inner space-time of them. Besides, even we are going to use the extended Schwarzschild metric, which is not exact here, the space and time exchanges inside the horizon and this leads to a weird phenomena.

Then we would like to make a little convention. How about use the space-time structure at the horizon? I think it’s fine. Since it is just about the same size of the a classical (not exactly) volume, we finally adopt the so called classical calculation, shown in the preceding part.

Well, if we are going to calculate the volume assuming a uniform black hole, the work is fairly easy. I am not going to show the calculation here because it is nearly nonsense.