\(E = m{c}^{2}\) was interpreted as kinetic energy along the time dimension.
Unless photons have the same time speed as us, photons do not exist in the same way as us and its mass is to be interpreted differently. Photons will have a time speed of \(c\) only when \({v}_{s}\) is zero in free space. This would otherwise be called rest mass.
From the post "A Right Turn, A Wrong Turn" we have the equation for photon deceleration under gravity,
\({g}_{B} =\frac{1}{2}\cfrac{c}{r}\cfrac{1}{\sqrt{1-{ e }^{ -\cfrac { x }{ r } }}}{ e }^{ -\cfrac { x }{ r } }\)
and the velocity field,
and the velocity field,
\({v}_{r}(x)=c\sqrt{1-e^{-\cfrac{x}{r}} }- {v}_{or}\)
When \({v}_{or}\) is \(c\), that is light approaching directly towards the center of the black hole, on the surface of the black hole \({v}_{s}\) is zero. At that point \({v}_{t}\) = \(c\) and photon manifest its true mass. Thereafter still under deceleration, photon bounces out of the back hole. BUT light cannot escape from a black hole!
Light inside a black hole cannot escape the black hole and light outside the black hole cannot enter. Light insight the black hole at \({v}_{s}\) = \(c\) and \({v}_{t}\) = 0 cannot escape the black hole. Light from outside the black hole, on reaching the black hole \({v}_{s}\) = 0, \({v}_{t}\) = \(c\) can not go further but is under high deceleration and returns with increasing \({v}_{s}\).
So if you shine a light at a black hole it will reflect. A black hole is shiny under illumination.
