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Q&A How efficient can a Dyson sphere be?

I think a reasonable assumption is that you don't want to accumulate entropy from the Dyson sphere. That is, the entropy you get from the star must not be lower than the entropy you send to outer s...

posted 3y ago by celtschk‭  ·  edited 3y ago by celtschk‭

Answer
#3: Post edited by user avatar celtschk‭ · 2020-11-12T17:19:15Z (over 3 years ago)
#2: Post edited by user avatar celtschk‭ · 2020-11-12T17:17:14Z (over 3 years ago)
Thinking again, the efficiency will likely be lower
#1: Initial revision by user avatar celtschk‭ · 2020-11-12T16:48:26Z (over 3 years ago)
I think a reasonable assumption is that you don't want to accumulate entropy from the Dyson sphere. That is, the entropy you get from the star must not be lower than the entropy you send to outer space (indeed, you'll have to send out more entropy because on one hand you have other sources of entropy like the cosmological microwave background, energy released by radioactive substances, inefficiencies of the machinery used inside the sphere, and last not least the entropy produced by the humans living inside the sphere).

I'm going to assume that the star's entropy is predominantly emitted in the form of (thermal) radiation. I'm also going to assume that thermal radiation is the only significant way for the Dyson sphere to get rid of entropy.

According to the Stefan-Boltzmann law, the total power emitted by a hot body is given by
$$P = \sigma A T^4$$
where $\sigma = 5.67\cdot 10^{-8}\ \rm W m^{-2} K^{-4}$ is the Stefan Boltzmann constant. Since Power is Energy over time, and the stars radiation is thermal radiation (so it's all thermal energy), the corresponding entropy rate is
$$\dot S = \frac{P}{T} = \sigma A T^3$$
Since we are talking about spherical objects (the star and the Dyson sphere), we have in both cases
$$A = \frac{4\pi}{3} r^3$$
where $r$ is the corresponding object (star or Dyson sphere).

Now according to the condition above, we have
$$\dot S_{\rm DS} \ge \dot S_* $$
Inserting the formulas above and cancelling all the constants then gives us
$$ r_{\rm DS}^2 T_{\rm DS}^3 \ge r_\*^2 T_\*^3 $$
which implies for the ideal efficiency
$$\eta_{\rm Carnot} = 1-\frac{T_{\rm DS}}{T_\*} \le 1-\left(\frac{r_\*}{r_{\rm DS}}\right)^{\frac{2}{3}}$$
Assuming the sun as star, $1\ \rm AU$ is [about 215 times the radius of the sun](https://www.translatorscafe.com/unit-converter/en-US/length/23-92/astronomical%20unit-Sun%E2%80%99s%20radius/) and therefore the corresponding efficiency is $97\\%$.

I think that the outside material of the Dyson sphere does not matter much here, except that it should be a good heat conductor and its outside should be as black as possible (but we are [already quite good in that regard](https://www.dezeen.com/2019/09/24/blackest-black-mit-material-news-vantablack/) and by the time we can make Dyson spheres, we'll surely have improved on that).

Now of course there will also be inefficiencies in whatever machinery is used inside the Dyson sphere, and *that* will lower the overall efficiency (and the amount of entropy to get rid of). Unfortunately I have no idea how efficient/inefficient the technology of a Type II civilisation will be.