From: Robin Hanson (rhanson@gmu.edu)
Date: Fri Sep 17 1999 - 18:50:33 MDT
Robert J. Bradbury wrote:
> > >a) It is possible that at the time we are at they are already harvesting
> > > 90% of the photons.
> > This doesn't answer the question about the 10% we see.
>
> Because they are at the point of "diminishing" returns.
> You and I don't eat all of the food on our plate if we
> are sufficiently "full".
>
> ... > why let all those photons go?
>
> Because you don't have the metal to harvest them! ....
> In our solar system, we do not have enough material (even after all the
> planets, comets, asteroids, etc.) to construct radiators that can
> radiate near the background temperature of the universe. We (in our
> solar system) would probably end up 10-40K above the background temp.
> That is, thermodynamically & computationally, not the most efficient
> place to be. The only solutions are to harvest material remotely
> and ship it back and that is very expensive or breed it locally
> (from energy) and that takes a long time. We still see stars because
> there isn't enough metal in the Galaxy to "optimally" hide them all yet.
Consider the function P(m) which describes the most P(ower)you can extract
from a star given a certain amount of available M(etal).
Since you can choose not to use metal, P must be an increasing function.
If you have two stars, and are trying to extract the most power from
them, and you have a certain amount M of metal available, your problem is
max P(m1) + P(m2) such that m1 + m2 = M
m1,m2 and m1 >=0, m2 >=0
You are suggesting that an optimum is m1 = 0 and m2 = M, putting all
the metal at one star so as to get every last photon and reradiate at near
3K, while completely leaving the other star alone. This requires that
P''(m) > 0 on average, with *increasing*, not diminishing, returns.
Instead I expect any plausible model of metal-limited Dyson sphere to
show decreasing returns: The first few tons gives lots of power while
the last few tons gives a lot lot less. This implies that to get the
most power one shoul spread the metal evenly across the two stars:
m1 = m2 = M/2 .
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