Re: Stealthing your M-Brain

From: Robert J. Bradbury (bradbury@aeiveos.com)
Date: Fri May 17 2002 - 12:56:22 MDT


On Fri, 17 May 2002, Hal Finney wrote:

> Ignoring black holes for the moment (which are super-cold),

Not if you are using them as power sources -- dumping matter
into them is going to generate X-rays or Gamma-rays(?).

> It seems to me that in equilibrium, the same amount of power
> has to be radiated from the shell.

Right. But depending on how much of the "effective" power
you want to suck off at each shell level (determined by the
coolant chosen and radiator material for that level) each
successive shell has to radiate the entire power output (or
at least that fraction it has "captured) at a cooler temperature
(effectively shifting the entire blackbody radiation curve to cooler
and cooler ranges as it moves away from the star.

Of course you *could* choose to radiate at a higher temperature
(i.e. concentrate IR radiation to 10,000 suns and use that to
harvest electricty), but I'm pretty sure you take a huge efficiency
hit (less effective power available at each level) by taking that
approach. I think there is going to be an "optimal" power conversion
strategy for the temperature of the radiation coming from the next
innermost level and you had best use that if you want to harvest
the most power.

> So in practice the same amount of power comes from outside the shell
> as from the star, and the only difference is that because the shell is
> larger, the temperature is less.

Yep.

> The outgoing shell temperature depends solely on the temperature
> of the star, and the ratio of the size of the shell to the size of the star.

Actually "power" of the star, since that can vary a lot with star type,
element composition, etc.

> I don't see where "cooling" comes into play here, except that you could
> redistribute the outgoing radiation to be stronger in some directions
> and weaker in others.

True. That's why I consider the JBrain and Asteroid Brain approaches
somewhat inefficient. In densely populated systems it will be difficult
to prevent them from radiating onto each other (making their radiators
less efficient) or the star (potentially swelling its atmosphere as
was discussed).

> But it seems like for a given star, the radiated
> temperature will depend solely on the size of the shell, and you can't
> cool it.

Can't cool the star without star-lifting (or perhaps stealthing it with
a layer of black holes).

> Now, adding black holes changes the equation, because they are effectively
> black bodies at the Hawking temperature, which will probably be very low.
> So if you can get your black hole to absorb large quantities of energy,
> then you can reduce your radiation.

This raises an interesting question -- black holes are always pictured
like swirling tornados with the matter going down the funnel. What
happens to radiation coming at the black hole from the side or bottom?

Or is this a false image -- black holes are completely omnidirectional
when they are not being fed matter (only radiation?). Indeed black
holes surrounding a star would always be feeding on the solar wind
produced by the star so its doubtful they would be completely free
of X-ray emissions.

> Beaming power back at the star, from this perspective, amounts to
> reducing the net amount of power radiated by the star. It is equivalent
> to the star being partially shut down and radiating less. But you could
> accomplish the same thing by building your shell around a cooler star,
> so it's not clear that this accomplishes anything.

Huh? Beaming power (radiation?) back to the star presumably increases
the temperature of its atmosphere.

> One other idea, based on what I wrote above, is to store much of your
> excess energy. I said it couldn't go on forever, but theoretically you
> could convert the energy to matter/antimatter and it might be practical
> to store it almost forever. So this is one other kind of cooling that
> could work.

I'm pretty sure you could never do this with 100% efficiency. You
have to dump the energy you "can't" harvest someplace. So for example
the UV photons solar cells can't collect either have to be reflected
back (to the sun) or absorbed by the material coating the cell, in
which case they are presumably reradiated as IR photons.

Robert



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