Math Treatment of Approaching Event Horizon

In summary, for a static, spherically symmetric spacetime, a distant observer would never see an object or particle cross the event horizon of a black hole, as it would appear "frozen" at the horizon. However, in the particle's frame of reference, it does cross the event horizon in finite proper time due to extreme time dilation caused by the strong gravity near the event horizon. This is only possible for objects with a negative radial velocity, as objects with positive radial velocity would never reach the event horizon.
  • #1
ergospherical
1,055
1,346
[This thread can be considered the A-level footnote to https://www.physicsforums.com/threads/is-there-an-inside-to-a-black-hole.1007588/]

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##].
 
Last edited by a moderator:
  • Like
  • Informative
Likes dextercioby, Delta2, Astronuc and 3 others
Physics news on Phys.org
  • #2
Therefore, an outside observer would never see the particle cross the event horizon, as it appears to be "frozen" at the horizon. However, in the particle's frame of reference, it does indeed cross the event horizon in finite proper time. This is due to the extreme time dilation caused by the strong gravity near the event horizon.
 

FAQ: Math Treatment of Approaching Event Horizon

1. What is an event horizon?

An event horizon is a boundary in space-time beyond which no light or information can escape due to the strong gravitational pull of a black hole.

2. How is math used to study approaching event horizons?

Math is used to describe the behavior of matter and light as it approaches the event horizon, as well as the curvature of space-time caused by the black hole's gravity.

3. What is the significance of the Schwarzschild radius in the math treatment of approaching event horizons?

The Schwarzschild radius is the distance from the center of a black hole at which the event horizon is located. It is a crucial parameter in the mathematical equations used to describe the behavior of matter and light near the event horizon.

4. Can math be used to predict what happens to matter and light once they cross the event horizon?

No, math can only describe the behavior of matter and light as they approach the event horizon. Once they cross the event horizon, the laws of physics as we know them break down and we cannot make predictions using traditional mathematical models.

5. How does the math treatment of approaching event horizons contribute to our understanding of black holes?

The math treatment of approaching event horizons allows us to make predictions and calculations about the behavior of matter and light near black holes, which helps us to better understand the properties and effects of these mysterious objects in our universe.

Back
Top