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ergospherical

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[This thread can be considered the A-level footnote to https://www.physicsforums.com/threads/is-there-an-inside-to-a-black-hole.1007588/]

For a

For ##r > r_g## there is ##-g_{tt} = 1 - \frac{r_g}{r} < 1##. The proper time ##d\tau = \sqrt{-g_{tt}} dt < dt## between two nearby events at the same position is less than the coordinate time; at finite distances from the source, time "slows down" compared to infinity.

Now consider an in-falling particle, that is, one with ##\dfrac{dr}{dt} < 0##. The proper time to fall from ##r_a \ (> r_g)## to ##r_b \ (> r_g)## is\begin{align*}

\tau(a\rightarrow b) = \int \sqrt{-g_{\mu \nu} dx^{\mu} dx^{\nu}} = \int_{r_a}^{r_b} dr \sqrt{-g_{tt} \left(\dfrac{dt}{dr}\right)^2 - g_{rr}}

\end{align*}The quantity ##\dfrac{E}{m} \equiv - u \cdot k = -g_{\mu \nu} u^{\mu} k^{\nu}

= -g_{tt} \dfrac{dt}{d\tau}##is

g_{\mu \nu} u^{\mu} u^{\nu} = g_{tt} \left( \dfrac{dt}{d\tau} \right)^2 + g_{rr} \left( \dfrac{dr}{d\tau} \right)^2 = -1\end{align*}denoting ##f = 1 - \dfrac{r_g}{r}## such that ##g_{tt} = -f## and ##g_{rr} = f^{-1}##, then this reduces to\begin{align*}

\dfrac{dt}{dr} = f^{-1} \left( 1 - \dfrac{m^2 f}{E^2}\right)^{-\frac{1}{2}}

\end{align*}The original integral is therefore

\begin{align*}

\tau(a\rightarrow b) = \int_{r_a}^{r_b} dr f^{-\frac{1}{2}} \sqrt{\left(1- \dfrac{m^2 f}{E^2} \right)^{-1} - 1}\end{align*}Putting the energy ##E## equal to its initial value at point ##a##, as ##E \equiv E_0 = m\sqrt{f_0}##, then \begin{align*}

\left(1 - \dfrac{m^2 f}{E^2} \right)^{-1} - 1 = \left( 1 - \dfrac{f}{f_0} \right)^{-1} - 1 = \dfrac{f}{f_0 - f}

\end{align*}which finally gives

\begin{align*}

\tau(a\rightarrow b) = \int_{r_a}^{r_b} \dfrac{dr}{\sqrt{f_0 - f}} = \int_{r_a}^{r_b} dr \left( \dfrac{r_g}{r} - \dfrac{r_g}{r_a} \right)^{-\frac{1}{2}}

\end{align*}which

For a "distant" observer (i.e. ##r_{\mathcal{O}} \rightarrow \infty##), processes occurring at the event horizon appear "frozen" since ##\sqrt{-g_{tt}} \rightarrow 0## [i.e. finite proper time intervals ##\delta \tau## correspond to "infinite" coordinate time intervals ##\delta t \rightarrow \infty##].

stevil said:Since time slows down as an object gets nearer to the EH, would an outside observer, living and observing for an almost infinite amount of time ever see an object/particle cross the event horizon threshold?

For a

**static**[admits a hypersurface orthogonal timelike Killing field ##k##], spherically symmetric spacetime, a time coordinate ##t## can be chosen as the parameter distance along the integral curves of ##k## from some initial hypersurface. Meanwhile the radial coordinate ##r## is the**area-radius****function**##r(p) = \sqrt{\dfrac{A(p)}{4\pi}}##, with ##A(p)## being the area of the ##\mathbf{S}^2## orbit through ##p##.For ##r > r_g## there is ##-g_{tt} = 1 - \frac{r_g}{r} < 1##. The proper time ##d\tau = \sqrt{-g_{tt}} dt < dt## between two nearby events at the same position is less than the coordinate time; at finite distances from the source, time "slows down" compared to infinity.

Now consider an in-falling particle, that is, one with ##\dfrac{dr}{dt} < 0##. The proper time to fall from ##r_a \ (> r_g)## to ##r_b \ (> r_g)## is\begin{align*}

\tau(a\rightarrow b) = \int \sqrt{-g_{\mu \nu} dx^{\mu} dx^{\nu}} = \int_{r_a}^{r_b} dr \sqrt{-g_{tt} \left(\dfrac{dt}{dr}\right)^2 - g_{rr}}

\end{align*}The quantity ##\dfrac{E}{m} \equiv - u \cdot k = -g_{\mu \nu} u^{\mu} k^{\nu}

= -g_{tt} \dfrac{dt}{d\tau}##is

**conserved**along the radial geodesics, since ##\dfrac{1}{m} \dfrac{dE}{d\tau} = u^{\nu} \nabla_{\nu}(u_{\mu} k^{\mu}) = (u^{\nu} \nabla_{\nu} u_{\mu})k^{\mu} + u^{\mu} u^{\nu} \nabla_{[\nu} k_{\mu]} = 0## by the geodesic equation and Killing's equation respectively. Therefore ##\dfrac{dt}{d\tau} = -\dfrac{E}{mg_{tt}}##. Furthermore, since ##u \cdot u = -1##,\begin{align*}g_{\mu \nu} u^{\mu} u^{\nu} = g_{tt} \left( \dfrac{dt}{d\tau} \right)^2 + g_{rr} \left( \dfrac{dr}{d\tau} \right)^2 = -1\end{align*}denoting ##f = 1 - \dfrac{r_g}{r}## such that ##g_{tt} = -f## and ##g_{rr} = f^{-1}##, then this reduces to\begin{align*}

\dfrac{dt}{dr} = f^{-1} \left( 1 - \dfrac{m^2 f}{E^2}\right)^{-\frac{1}{2}}

\end{align*}The original integral is therefore

\begin{align*}

\tau(a\rightarrow b) = \int_{r_a}^{r_b} dr f^{-\frac{1}{2}} \sqrt{\left(1- \dfrac{m^2 f}{E^2} \right)^{-1} - 1}\end{align*}Putting the energy ##E## equal to its initial value at point ##a##, as ##E \equiv E_0 = m\sqrt{f_0}##, then \begin{align*}

\left(1 - \dfrac{m^2 f}{E^2} \right)^{-1} - 1 = \left( 1 - \dfrac{f}{f_0} \right)^{-1} - 1 = \dfrac{f}{f_0 - f}

\end{align*}which finally gives

\begin{align*}

\tau(a\rightarrow b) = \int_{r_a}^{r_b} \dfrac{dr}{\sqrt{f_0 - f}} = \int_{r_a}^{r_b} dr \left( \dfrac{r_g}{r} - \dfrac{r_g}{r_a} \right)^{-\frac{1}{2}}

\end{align*}which

**converges**to some finite value as ##r_b## approaches ##r_g##. In other words, the particle reaches the event horizon in finite proper time.For a "distant" observer (i.e. ##r_{\mathcal{O}} \rightarrow \infty##), processes occurring at the event horizon appear "frozen" since ##\sqrt{-g_{tt}} \rightarrow 0## [i.e. finite proper time intervals ##\delta \tau## correspond to "infinite" coordinate time intervals ##\delta t \rightarrow \infty##].

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