Re: How much energy would the decay of 10 microgram mini black hole produce?

Question: How much energy would the decay of 10 microgram mini black hole 
produce?
From: Yevgeniy
Grade: undergrad
City: Staten Island , State/Prov.: NY Country: USA
Area: Physics Message ID Number: 1124903742.Ph

How much energy would the decay of 10 microgram mini black hole produce in 
joules?

Yevgeniy,

Ouch…you are definitely way out there on the leading edge!  The first 
question is can such a thing as a mini black hole exist?  

“A micro black hole, also called a quantum mechanical black hole and 
inevitably a mini black hole is simply a black hole around which quantum 
mechanical effects play an important role.”

The usual way to make a black hole is to allow nature to accumulate 
enormous amounts of mass until the gravity is so great that even light can 
not escape.  But total mass is not the complete story.  What if there was 
less mass but it was concentrated into a smaller volume?  Wouldn’t this 
create the same effect?  The answer is yes…theoretically.  

So if we can compress matter to such an extent that gravitational effects, 
on a very small region of the universe, would be great enough to keep 
light from escaping, then we would have a small black hole (micro, mini, 
or whatever)

Hawking and others tell us that such a phenomenon would be very unstable 
and decay giving off energy in the form of radiation.  Any mini black 
holes created during super nova class explosions in the universe would 
quickly dissipate.  In theory, however, we might create mini black holes 
in ultra high energy accelerators.  So perhaps we can create them, and 
therefore your question is valid, and they would decay giving off energy…
so you want to know how much.    

Carl Wolf of  looked into the problem.  Here are some pertinent extracts 
from his paper:

The Decay of a Black Hole as a Quantum Process
Carl WOLF
Department of Physics, North Adams State College,
North Adams, MA (01247) USA

By considering the decay of a black hole as a result of the interaction of 
a quantum system with a dissipative vacuum, or in terms of string theory, 
the interaction of localized string modes with non-local string modes, we 
demonstrate how the single parameter representing dissipation can be 
calculated in terms of known fundamental constants. In view of the 
uncertainty embodied in all the present theories of black hole decay it is 
suggested that the approach of this note might provide fresh insight to 
the problem of black hole decay.

And in conclusion Carl stated:

“It once more lends evidence to the fact that space, time, and matter are 
inseparable and questions such as CP and T violation might only be 
understandable when we understand how matter is truly intertwined with 
geometry through perhaps string theory or perhaps a theory of gravitation 
beyond the structure of general relativity.”



So it looks as if you have to use a “retarded” Schoedinger equation!  
That’s how I felt when I had to work with it back in the sixties.  But if 
you look at his energy equation you see a familiar expression:

E = mc^2.  

So maybe Einstein has answered the question for us.   Convert 10 
micrograms to kilograms:
 
10 ug = 1E-10 kg.  

Plug this into Einstein’s general theory of relativity:

E = (1E-10 kg)*(3E8 m/s)^2 = 9E6 Joules

So neglecting complications from time/space distortions it appears as if 
10 ug of matter can be converted into about 9 million Joules of energy 
whether it is done in a nuclear explosion or mini black hole decay.

Seems simple enough but I suspect that there are complications brought on 
by quantum mechanical effects in a system existing in the Planck scale of 
dimensions.  

Keep asking the tough questions,

Regards,

E Stammel
stammeew@delhi.edu