黑洞一定密度很大么？

相关文章：球状尘埃云的密度

今天Professor Cahill听说我是做宇宙学的，于是让我帮他检查一下他书里面的一个小节。

这个小节中有一部分是这样的：

设想一团弥散的无旋的尘埃，密度为\(\rho\)，半径为\(r\)，质量为\(M\)，那么如果半径满足下述关系，那么这团尘埃会具备黑洞的性质：
\begin{eqnarray}
r\le \frac{2MG}{c^2}=\frac{2G}{c^2} \frac{4}{3}\pi r^3 \rho
\end{eqnarray}

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

这也就是说，不管给定多小的密度，那么只要半径足够大，那么这团尘埃也会具备黑洞的性质，因为这团尘埃完全在它的的Schwarzschild半径外面，等价于是一个Schwarzschild黑洞。

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。

其实真的不奇怪，看一下下面的这个就明白了。

现在考虑一团弥散的无旋的尘埃，质量为\(M\)，半径为\(r\)，我们来看看对它的密度有什么要求。
同样的，首先，
\begin{equation}
r\le \frac{2G M}{c^2}
\end{equation}

那么密度
\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}

这个式子告诉我们，如果天体的质量很大，那么这个天体是个Schwarzschild黑洞的密度的下限会很小。

这个更上面第一个推导是等价的，你们如果说，我们有一团密度很小的尘埃云，那么这团尘埃云是黑洞的前提是半径要很大很大，又因为质量要正比半径三次方呢，所以密度小半径大，质量很大。

我以前从来没想过这个问题……

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:
\begin{equation}
\int h_1 h_2 h_3 \mathrm dx_1\mathrm dx_2\mathrm dx_3
\end{equation}

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.