Describe a nuclear bomb near a black hole relative to a bomb in open space, due to $E=mc^2$ The energy of a nuclear bomb is described by $E=mc^2$.  Near a black hole we have tremendous time dilation, so the rate of time and the speed of light is slower relative to open space. To an outside observer, what would a bomb near a black hole look like compared to a bomb in open space?  Would the bomb have the same blast radius and same destructive power, just take longer to cover that radius? Or would it have a smaller blast radius and less destructive power when we look at $E=mc^2$ as a whole?
 A: Blast radius and destructive power are characteristics of a shockwave, which requires a dense medium like air. The characteristic white-hot fireball associated with nuclear explosions is also a property of air, ionized into plasma by the x-ray hot light from the primary detonation. In empty space a nuclear bomb is a flashbulb with an approximately blackbody spectrum whose peak is in the lower x-ray band.
So, it would look like a bright point of light from whatever angle you look at it. The strong curvature of space (gravitational lensing) will make the apparent location of the light source shift as you move around it, but it will remain a bright point of light.
A time-dilated x-ray flashbulb is a lower frequency (redshifted) flashbulb that lasts longer and is dimmer (fewer photons per second).
The number of photons emitted by the bomb does not change.
The energy per photon is $h\nu$, hence the total energy measured by a distant observer is less than the energy measured by a nearby observer (measuring $\nu'$) who occludes the same $m$ photons by $\Delta E = h\sum{_{n=1}^m} (\nu_n - \nu'_n)$
If you adopt a set of conventions that allows you to have a well-defined local gravitational potential energy (that is, you aren't doing General Relativity any more), you will find that the missing kinetic energy from the reduced photon wavelength has been spent doing work on the gravitational field, exactly as the kinetic energy of a ball tossed skyward goes missing as the ball gets higher. This is probably a less true way of looking at it, but it's useful and the math is easy.
