0
$\begingroup$

I watched a video recently about using the ergosphere of a spinning black hole as a source of virtually infinite energy by surrounding it with mirrors and shining electromagnetic waves in there. In the video, it visualized the increase in energy as the light spinning faster around the black hole in the mirror chamber and an increase in brightness.

Now I don't really know much about this, but from what I understand the light can theoretically go faster only within the ergosphere but is otherwise going around the black hole at c, and it can't get any brighter as the actual amount of light is decreasing as some of it falls into the black hole. I just want to know whether I'm right in thinking the actual increase in energy is found in the frequency of the waves or energy of the photons. If that's the case, is there a limit to the frequency of the wave? What, if anything, would be more energetic than ultra high energy gamma rays? Also, if you were to shoot infrared light into the chamber, and if you were able to see inside, would it turn into visible light and shift from red to purple before disappearing into ultraviolet rays? Cause that's sorta what I'm visualizing in my head.

$\endgroup$

1 Answer 1

0
$\begingroup$

..., and it can't get any brighter as the actual amount of light is decreasing as some of it falls into the black hole.

This is not true. The number of photons does not have to be (and is not) preserved in nature.

The process of super-radiance (as the effect you are referring to is known). Can be understood purely at the level of classical waves. In that setting the exact statement is that certain monochromatic (i.e. with a single frequency) waves can scatter off a black hole and return to infinity with the same frequency but larger amplitude (and thus energy.)

In the context of quantum mechanics, this means that this process must increase the number of photons. It is a variant of stimulated emission.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.