# Would the event horizon of a black hole shrink as you approach?

While light cannot escape an event horizon, external light should still be observable from within. Would "entering" an event horizon cause it to apparently shrink away from you as you neared the singularity?

The event horizon will appear to do just the opposite. There is a radial distance at which a photon will in fact orbit the black hole. We can find this by working with the Schwarzschild metric $$ds^2~=~\left(1~-~\frac{2m}{r}\right)dt^2~-~\left(1~-~\frac{2m}{r}\right)^{-1}dr^2~-~r^2d\Omega^2.$$ where $m~=~GM/c^2$ and $d\Omega^2~=~sin^2\theta d\phi^2~+~d\theta^2$. We now consider an orbit that is circular, and so $dr~=~0$, and we put the orbit on a plane with $\theta~=~\pi/2$ so that $$ds^2~=~\left(1~-~\frac{2m}{r}\right)dt^2~-~r^2d\phi^2.$$ Before considering the orbit of a photon we can look at the circular orbit of a massive particle with the angular velocity $\omega~=~d\phi/dt$ so the metric is $$ds^2~=~\left(1~-~\frac{2m}{r}~-~r^2\omega^2\right)dt^2.$$ This can be thought of as a Lagrangian that computes the orbit, and the inclusion of the $dr$ can generalize this for non-spherical orbits. We also have that there is a generalized Lorentz gamma factor $\Gamma~=~dt/ds$, which for $m~=~0$ reduces to the gamma factor in special relativity $\gamma~=~1/\sqrt{1~-~v^2/c^2}$.
The vanishing of $1/\Gamma$ means that we have the angular velocity $$\omega^2~=~\left(\frac{d\phi}{dt}\right)^2~=~1~-~\frac{2m}{r}.$$ Now compute the radius for the circular orbit of a photon. For simplicity let $A~=~1~-~2m/r$ and $A'~=~dA/dr$. We now look for the radial geodesic equation with $$\Gamma^r_{tt}~=~AA'/2,~\Gamma^r_{\phi\phi}~=~-Ar,$$ again for $\theta~=~\pi/2$ and put this in the geodesic equation to get $$\frac{m}{r^3}~=~\left(\frac{d\phi}{dt}\right)^2~=~1~-~\frac{2m}{r}$$ and find the radius is $r~=~3m$