# Brown Dwarves: Dyson Spheres in disguise?

Would a Dyson sphere make a red dwarf appear to be a Brown Dwarf? Would it disguise a star enough to misidentify it what size it is? I'm just wondering if it could be possible that some Dyson spheres are out there drifting around, camouflaged like a rather innocuous and innocent star?

• I'd expect the emissions to change a lot, be interesting to see what the answers say though. Jan 12, 2015 at 15:05
• red dwarf could still have flares Jan 12, 2015 at 15:58
• Disguise it from who? You have an anthropomorphic notion that brown dwarves are useless and not worth investigating. They might be tremendously useful to a space-faring society, all that warm hydrogen, helium and lithium just lying around. Better off building matrioshka spheres as in @SerbanTanasa's answer and lower your radiated energy so nobody sees you from a distance at all. Jan 12, 2015 at 17:57
• Disguise it from how far away and from what technology? The unaided eye (or an IR camera) at 100km can easily distinguish the difference. At 100 light-years, it's a whole different story... Jan 12, 2015 at 18:10
• What would it look like if it was not a total sphere? Dyson's original idea was based on a continuous evolution of adding more and more satellites to a star. A swarm of independently orbiting industry, solar collectors and living space. Jan 12, 2015 at 21:40

No, you could not. Temperature probably isn't an issue, but a Dyson sphere shouldn't show the proper spectral lines.

## The-best case scenario

This site gives the formula for the temperature of a Dyson Sphere as $$T=\left( \frac{E}{4 \pi r^2 \eta\sigma} \right)^{\frac{1}{4}}$$ where $E$ is the star's energy output, $r$ is the Dyson Sphere's radius, $\eta$ is the emissivity and $\sigma$ is the Stefan-Boltzmann constant. The Stefan-Boltzmann law says that the energy output (luminosity, $L$) of a star is $$L=4 \pi \sigma R^2 T_*^4$$ The energy output is $L$, so substituting this into the first expression gives $$T=\left( \frac{4 \pi \sigma R^2 T_*^4}{4 \pi r^2 \eta\sigma} \right)^{\frac{1}{4}}$$ Which is $$T=\left( \frac{R^2 T_{\odot}^4}{r^2\eta} \right)^{\frac{1}{4}}$$ $$T=T_*\left( \frac{R^2}{r^2\eta} \right)^{\frac{1}{4}}$$ Wikipedia gives the emissivity of concrete - my Dyson-Sphere-building-material of choice - as $0.91$. Let's say the Dyson Sphere has a radius of $1.5$ times that of the star. That gives me $$T=T_*(1.5^2 \times 0.91)^{-\frac{1}{4}} \approx 0.836 T_{\odot}$$ Wikipedia and Wikipedia say that a temperature of a red dwarf could be as low as 2300 K, and a brown dwarf could have a temperature of about 1900 K - $0.826$ times the temperature of a red dwarf and so in the acceptable range for our Dyson Sphere.

A more commonly-used radius is $r\approx1\text{ AU}$ - the distance from Earth to the Sun. when substituted in, this gives $r=215R$ and $T=0.07T_*$, a much lower value. Interestingly, this fits with previous results. Slysh (1985) looked at things from the perspective of thermodynamic efficiency. The efficiency, $\eta_T$, is given by $$\eta_T=1-\frac{T}{T_*}$$ It should be expected, at best, that $\eta_T\approx0.95$, so we get $T=0.05T_*$ - pretty close to what our result above was.

As Serban Tanasa rightfully pointed out, there are some problems with concrete. Steel or iron would be a better choice. Their emissivities are given here: $$\begin{array}{|c|c|}\hline \text{Material} & \text{Emissivity}\\\hline \text{Concrete} & 0.81\\\hline \text{Cement} & 0.54\\\hline \text{Galvanized steel} & 0.88\\\hline \text{Iron} & 0.87\text{-}0.95\\\hline \end{array}$$ Note the lower value for concrete than the one I used above. The difference in values turns out to have little effect. At any rate, if we use iron, and choose the lower limit for $\eta$, we get $T=0.071T_*$ - essentially the same as above.

Let's do some recalculations, using both the derivation from scratch and Slysh's results. We'll use a number of stars: $$\begin{array}{|c|c|c|c|c|}\hline \text{Star} & \text{Spectral type} & T_*\text{ (K)} & T\text{ (K)}\text{ (via emissivity)} & T\text{ (K)}\text{ (Slysh)}\\\hline \text{Zeta Puppis} & \text{O4} & 40000 & 2840 & 2000\\\hline \text{Eta Aurigae} & \text{B3} & 17200 & 1220 & 860\\\hline \text{Fomalhaut} & \text{A3} & 8590 & 610 & 430\\\hline \text{Tau Boötis} & \text{F6} & 6360 & 450 & 320\\\hline \text{Sun} & \text{G2} & 5770 & 410 & 290\\\hline \text{Alpha Centauri B} & \text{K1} & 5260 & 370 & 260\\\hline \text{Gliese 581} & \text{M3} & 3480 & 250 & 170\\\hline \end{array}$$ Here, I assume $r=1\text{ AU}$ and $\eta=0.87$.

These temperatures are reasonable values. If we accept a lower temperature limit of $300\text{-}400\text{ K}$ for a brown dwarf, Slysh's rule lets us choose stars roughly as hot as the Sun, or hotter. The emissivity calculations let us choose, in general, any star hotter than a red dwarf.

From a temperature perspective alone, there shouldn't be serious issues.

## The spectral line problem

There have been questions as to whether or not the emission spectrum of a Dyson Sphere would match that of a brown dwarf. It is certainly the case that the peak wavelengths would match that of a brown dwarf, with the most light radiated in the infrared. In other words, if you looked at a Dyson Sphere and a brown dwarf with an infrared telescope, you would see two similar sources.

If you measured the emission lines, though, you would definitely see different materials in the two objects - there's no way around that. And yes, the radius of the Dyson Sphere would be much larger than that of a red dwarf - so certainly larger than that of a brown dwarf, as pointed out by JDlugosz.

Here are some lines you'd expect to see in a brown dwarf:

• Lithium
• Titanium monoxide
• Ammonia
• Methane
• Heavier molecules a la titanium monoxide

Not all of these are necessarily going to be present in a brown dwarf's spectrum, but the absence of all of them in the spectrum of a Dyson sphere is going to raise some red flags. That's you main problem.

Thanks to all those who commented and pointed out inaccuracies and errors; the answer is the better for that.

No. A Dyson sphere would emit something closely matching black-body radiation. A star, while also emitting something close to black-body radiation, has tell-tale spikes in its spectrum. Below is the sun's spectrum compared to what its ideal black-body spectrum would look like: Brown and red dwarfs have their own "fingerprint" signatures, which differ from that of both the sun and an ideal black body. This fingerprint is the first thing astronomers look at, so I would not expect them to be fooled for long.

• So how hard would it be to emit enough energy to in the right wave lengths to fake out distance observers? Jan 12, 2015 at 21:00
• @bowlturner, it's conceivable to do so. He who can build a Dyson Sphere could also build an ACME "Spectral Shaper Pro". But then again ckersch's issues of wrong mass apply. Jan 12, 2015 at 21:54
• cool, hadn't even thought of emission absorbtion bands! Jan 12, 2015 at 22:56
• @bowlturner easy to fake; just don't. Have a cloud of dust around your engineered stuff, and distant observers will see only a dust cloud in thermal equilibrium with the energynfrom the sun. That it got used for energy and changed to different frequency doesn't matter as long as it's still absorbed by the dust. For extra stealth, make sure the composition of the cloud is natural, not your tailings and trash. Jan 13, 2015 at 3:23

I don't think so, because a dyson sphere would not have the same emission spectrum of a star.

Consider two cases: we can have a translucent dyson sphere that lets out some of the light of the star, or we can have a dyson sphere that is opaque and emits light as blackbody radiation due to being heated by the star.

The light initially emitted by the star will have an emission spectrum that is dependent on its temperature. Elements which are either fully ionized in a star or which are too cold to absorb energy will not absorb light. Based on this, we can identify the temperature of a star, not based on its luminosity, but based on its emission spectrum.

Now consider the light emitted by the dyson sphere. If we emit as a black body, we will not have the same emission spectrum as our star unless the dyson sphere has the same elemental makeup as the star in question. Since stars are made up mostly of gasses, this would be difficult to achieve. If we emit through transparency, we will still have the same emission spectrum as a red dwarf, but appear less luminous. Most materials also have a transparency that varies based on spectrum, so we'll see the spectrum of the red dwarf reduced at different frequencies dependent on the material of the sphere.

A search for Dyson spheres has actually been carried out under a similar theoretical framework. Under the assumption that most terrestrial objects radiate mostly in the infrared spectrum, astronomers searched for stars that had a spectrum shifted more towards that part of the spectrum than the star would otherwise be expected to emit under. The search did not successfully find anything that looked like even a partial Dyson sphere.

Even if we manage to get our Dyson sphere to emit the same spectrum of a brown dwarf, though, it will still appear too big and too massive to be a brown dwarf. Brown dwarves are smaller than red dwarves, and a Dyson sphere would need to be significantly bigger than a red dwarf.

• How do you know how big or far away it is if there isn't near another star to compare it too? all of our classifications are based on assumptions of other measurements. Jan 12, 2015 at 17:44
• Depends how far it is. If it's very far away, redshift can be used to assess size. For closer stars, parallax. If nobody looks to closely at the star, it may well go unnoticed, but under close observation I would expect that the Dyson sphere would look noticeably different. Jan 12, 2015 at 17:51
• IIRC, there's a lot of gap between parallax limits (few hundred LY, I think), and the beginning of effective redshift (has to be at least outside our own galaxy, so tens of thousands of LY). Jan 12, 2015 at 20:13
• It would emit in infrared, probably moreso than a regular star in the case of a partial Dyson sphere. This is because a partial Dyson sphere would absorb energy and emit blackbody radiation, which would be mostly infrared if the Dyson sphere was around the same temperature of earth. For a hotter Dyson sphere, the blackbody radiation would be dominated by higher energy wavelengths. Either way, the emission spectrum of the Dyson sphere would lack the absorption spectra of the stellar atmosphere. Jan 12, 2015 at 22:32
• @RBarryYoung paralex works up to 20k lightyears; still within our galaxy. Red shift of local group galaxies won't show Hubble's relationship (Andramada is heading right for us and is blue shifted), so takes effect at much larger scales than millions of light years. An important mechanism you didn't mention is Cepheid variables. Trying to find the spelling for that I came upon a great diagram: en.m.wikipedia.org/wiki/Cosmic_distance_ladder Jan 12, 2015 at 23:17

Not an expert on blackbody radiation, so you might get a more competent answer on (Astro)Physics, but conceptually if you have a matrioshka system, where each layer captures and uses the radiation of the more inward layer, you can bring the blackbody radiation level down to an arbitrarily low level >= CMB.

The only "disguise" issue would be that brown dwarfs are generally capped around $80 M_J$ (Jupiter Masses), while red dwarfs are generally at $[0.1-0.5] M_S =[100-500]M_J$. There might be a small overlap at the lowest of the low red dwarfs. But generally, your apparent stellar radii might mismatch, and the orbits of any remaining planets would be anomalous, upon closer inspection.

• But don't we compute the masses by the radiation put out? Jan 12, 2015 at 16:09
• Jan 12, 2015 at 16:13
• Reading that answer basically told me that a star by it's lonesome is mostly guessed at by it's radiation. So if it looks like a brown dwarf, we classify it as such...? Jan 12, 2015 at 16:23
• So it could be a great disguise unless someone is actually looking for you! Jan 12, 2015 at 16:32
• @HDE226868 right, but that is assuming nothing like a Dyson Sphere is altering it's luminosity. We'd likely mistake it's distance from us right? I suspect a serious study and tracking of one of these would show up irregularities, but first someone would have to care, and there are a lot of brown dwarfs out there. Jan 12, 2015 at 16:40

First - Brown Dwarfs are small. A Dyson Sphere for our Solar System to duplicate Earth like environments would have to be about 2 AU in diameter to give Earth-like solar radiation incoming. A Dyson Sphere would be BIG.

The mass of the Dyson Sphere would be large - the sun itself and the mass of the sphere - so in a double star would look very different than a light brown dwarf.

The Dyson Sphere, in the end, would have to re-radiate all the incoming solar energy out the shell as heat. So the total luminosity would be the equivalent of the star inside, not the small amount reported. If you trapped heat in, you bake your sphere until it glows on its own.

EDIT:

I just thought of another test. A Dyson sphere, due to its large size, will rotate very slowly if at all. A planet or sun rotates about its axis in hours or days. This speed difference between the limbs of the object give splits to the spectral lines that can be seen. So a visible Dyson sphere would look anomalous due to its slow rotation. I'd think that to get a sphere to revolve about its axis in months or less would require unreasonably strong materials.

• A red dwarf is much smaller than our sun. Jan 16, 2015 at 3:04
• True enough...but a Dyson Sphere still has to have the diameter of the habitable zone orbit of said star...which is much larger than the diameter of a normal star. Jan 16, 2015 at 16:51
• @bowlturner, such Dyson Disguise is going to be enormously large brown dwarf just because radius of "successful" low-mass star is bigger than "failed" one's. Aug 22, 2015 at 12:55

Look on YouTube for the weekly SETI seminars. Not too long ago they discussed exactly that, and what instruments are needed to be able to tell if that were the case!

I seem to recall that it's not brown dwarfs that it looks like (a brown dwarf is only just larger than Jupiter) but some types of dusty systems or star formation. Models of spectra show that a particular spectral band that is not distinguished in current readings would show the difference between Dyson spheres and imposters.

Most of the answers above assume a shell of material located at a radius around a star which would support Earth-like habitats. However, colonization by robotic life would produce a significantly different architecture. And, the other respondents assume that a dyson sphere (or Dyson swarm, rather) would be orbiting a luminous star - if, instead, a swarm was harvesting material from a Jupiter, it may be dim enough to go unnoticed. If I was a space-faring artificial life form, I would prefer to orbit and mine a Jupiter - lower gravity would make siphoning gas much simpler, and there is no risk of a nova (with a star, loss of gasses would lead to cessation of fusion at its core, followed by collapse, then explosion as fusion re-ignites...). We do not have the sophistication to detect cool, wayward Jupiters, so this possibility may fit your goal of a 'hidden Dyson'. For example, the recently discovered super-magnetic Jupiter: https://phys.org/news/2018-08-vla-extrasolar-planetary-mass-magnetic-powerhouse.html