REM%IMSSS@SU-AI.ARPA (Robert Elton Maas) (03/22/86)
REM>... dropping matter into a black hole is both efficient and REM>relatively safe/simple once you have a black hole handy. D> Date: 20 Mar 86 10:12:30 GMT D> From: brahms!desj@ucbvax.berkeley.edu (David desJardins) D> Subject: Re: antimatter -> black hole D> I give up. I don't have a clue what you are talking about. How D> is dropping things into black holes supposed to create energy (at D> "close to 100% efficiency")?? To put it simply, one doesn't create energy, one converts gravitational potential energy into knetic or other form of energy. It's the same as letting rain water fall through a turbine. "How does letting water fall create energy?" It doesn't, it transforms it from potential to knetic. One of the paradoxes of classical theory is that two point masses have an infinite amount of gravitational potential energy, because each's gravitational well is infinitely deep (force is R**(-2), potential is integral of that which is (R**(-1))/(-1) so as R approaches zero the potential approaches minus infinity). But the event horizon around point masses limits this to a finite value; you can draw useful energy from a falling object only as long as you can hold onto it so the potential energy is transferred from the falling object to your apparatus for collecting the energy. If you just let the object fall into a black hole, all the knetic energy is lost in the hole. But if you tie it to a rope and wrap the other end of the rope around a shaft that drives a generator or somesuch, you can convert the fall of potential into useful energy up to when the rope breaks or the falling object reaches the event horizon, whichever happens first. So it's obvious you can get energy by lowering something into a black hole, providing you don't just let it freefall, you have it's falling do useful work on some mechanical contraption. The hard part is computing how much energy you get for a given mass being lowered almost all the way to the event horizon before the rope breaks. I haven't done the calculations myself, but I read in some journal that the answer is you get exactly E=M*C**2 out of it, i.e. 100% mass-to-energy conversion as per Einstein's equation. (Would Gene Salamin or Hans Moravec or some other expert on physics & relativity & quantum mechanics who has actually done the calculation please confirm or correct the answer I quoted? (I wish Hawking were on this list, he's the real expert on black holes!!))
weemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (03/22/86)
In article <8603220245.AA06331@s1-b.arpa> REM%IMSSS@SU-AI.ARPA (Robert Elton Maas) writes: >So it's obvious you can get energy by lowering something into a black >hole, providing you don't just let it freefall, you have it's falling >do useful work on some mechanical contraption. The hard part is >computing how much energy you get for a given mass being lowered >almost all the way to the event horizon before the rope breaks. I >haven't done the calculations myself, but I read in some journal that >the answer is you get exactly E=M*C**2 out of it, i.e. 100% >mass-to-energy conversion as per Einstein's equation. (Would Gene >Salamin or Hans Moravec or some other expert on physics & relativity & >quantum mechanics who has actually done the calculation please confirm >or correct the answer I quoted? (I wish Hawking were on this list, >he's the real expert on black holes!!)) The calculation is rather simple. Assuming a 100% conversion rate of the infinitesimal gain in potential energy to photons emitted back, and then accounting for the gravitational red shift on the way back to the starting height, and then integrating from initial height to the Schwarzchild radius, one gets that M*c**2 is the energy received at the top. Of course, there is an infinite delay at the end, but never mind that. ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720