所谓特征值与平均值，就是发散级数在取不同截断时得到的值与级数平均值的关系。

$\ell_p=\sqrt{\frac{\hbar G}{c^3}$

Let’s calculate the simplest black hole which is the Schwarzschild black hole. The metric of the black hole reads

ds^2=-\bigg(1-\frac{2M}{r}\bigg)dt^2+\bigg(1-\frac{2M}{r}\bigg)^{-1}dr^2+r^2d\theta+r^2\sin^2\theta d\Phi^2

where M is the mass of the black hole. We can see that this metric describes a statistical black hole. We put it into Klein-Gordon equation

\frac{1}{\sqrt{-g}}\frac{\partial}{\partial x^\mu}\bigg(\sqrt{-g}g^{\mu\nu}\frac{\partial \phi}{\partial x^\nu}\bigg)-\frac{\mu_0 ^2 c^2}{\hbar^2}\phi=0

where $$g=-r^4\sin^2\theta$$ is the determinant of the metric. then we can get
\label{eq:CG-LONG}
-\bigg(1-\frac{2M}{r}\bigg)^{-1}\frac{\partial^2\phi}{\partial t^2}+\frac{1}{r^2}\frac{\partial}{\partial r}\bigg[(r^2-2Mr)\frac{\partial \phi}{\partial r}\bigg]+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\bigg(\sin\theta\frac{\partial \phi}{\partial \theta}\bigg)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2\phi}{\partial\Phi^2}=\frac{\mu_0^2}{\hbar^2}\phi

We set the wave function as

\phi=e^{-i\omega t}Y_{lm}(\theta,\Phi)\psi(r)

put it into \eqref{eq:CG-LONG},then we get
\label{eq:jingxiangfangcheng}
\frac{d}{dr}\bigg[(r^2-2Mr)\frac{d\psi}{dr}\bigg]+\bigg[\frac{r^3\omega^2}{r-2M}-\frac{\mu_0^2}{\hbar^2}r^2-l(l+1)\bigg]\psi=0

Using Wentzel-Kramers-Brillouin (WKB) approximation,let

\psi(r)=e^{\frac{i}{h}s(r)}\\

put it into \eqref{eq:jingxiangfangcheng}, then we have

k^2=\bigg(1-\frac{2M}{r}\bigg)^{-1}\bigg[\omega^2\bigg(1-\frac{2M}{r}\bigg)^{-1}-\frac{\mu_0^2}{\hbar^2}-\frac{l(l+1)}{r^2}\bigg]

which is the wave vector. Here we use the semi-classical quantization condition
\label{eq:semi-classical-vactor}
n\pi=\int_{r_{H}+h}^Lk(r,l,\omega)dr,n \in \mathbb{N_+}

2条评论

1. 文中一个行间公式$2\pi$没有显示出来

1. 你看的真仔细，谢谢！！