Why Would Aliens Hide? (was: Dyson shells are possible)

Robin Hanson (rhanson@gmu.edu)
Fri, 17 Sep 1999 20:50:33 -0400

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 .