# Thermodynamically possible to hide a Dyson sphere?

You build a Dyson sphere around a star to capture all its energy. The outer surface of the Dyson sphere still radiates heat at much higher temperature than the cold space background, so you're easy to detect.

But you'd like to stay hidden. So you cool the outer surface of the Dyson sphere to near cold space background. Of course you still need to radiate your excess heat somewhere, so you plan to radiate it off in directed beams, away from the directions of the nearby solar systems, to stay hidden from your neighbors at least.

Questions:

1. Is such directed radiation of excess heat allowed by known laws of physics?
2. Would the energy of the star be sufficient for running the cooling system?
-
At least for #2: Probably not, as the point of the Dyson sphere is that the builders energy demands have exceeded the output of the start itself. –  Kyle Kanos Jun 30 '13 at 23:32
You could always somehow convert the excess energy to mass. :) –  Jerry Schirmer Jul 1 '13 at 5:15

Let's begin with the limitations of the second law of thermodynamics - how cool could the Dyson sphere itself be? Realistically, the only way it itself can dump heat is likely to be radiatively. So, let its radius be $R_D$, and that of the star $R_S$. We suppose we design for a radiation dumping temperature of $T_D$ hoping to make this low enough that the neighbours won't see it. Then, the power $P$ (rate of winning of useful work from the Dyson sphere) is limited by the reversible case $\left(1-\frac{T_D}{T_S}\right) Q_S$ and $Q_D = \frac{T_D}{T_S} Q_S$, where $T_S$ is the star's surface temperature, $Q_S$ the star's output and $Q_D$ the rate of waste heat dumping at the Dyson sphere. Now the Stefan-Boltzmann law means that we also have $Q_D = \frac{R_D^2 T_D^4}{R_S^2 T_S^4} Q_S$. Whence, $R_D^2 T_D^3 = R_S^2 T_S^2$. Curiously, the Dyson sphere's minimum temperature varies exactly inversely to the orbital period in Kepler's law (pure co-incidence, of course)! If the sphere is of the order of Pluto's orbital radius and the star like our Sun, we get $\frac{T_D}{T_S} = \left(\frac{7\times10^5\mathrm{km}}{6\times10^9\mathrm{km}}\right)^{\frac{2}{3}} = 0.01$, so we can theoretically get it down to an order of magnitude above the background radiation temperature with a sphere of comparable size to the solar system. This may or may not be a problem, depending on how your neighbours are watching! You can see that the sphere has to be awfully big. Now, I have heard some talk of "blacker than black" metamaterials ($i.e.$ those which radiate more power than a blackbody http://arxiv.org/pdf/1109.5444.pdf - I've not seen a peer reviewed version of this and do not pretend to undestand it yet), so we might do better than the above calculation with a Dyson sphere made out of such a metamaterial, if they are indeed possible (or use Bill Oddie's Ecky Thump Black Pudding material (noone blacker) :) )