Brown Dwarves: Dyson Spheres in disguise?

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 $$L=4\pi \sigma R^2{T}_{\ast }^4$$ The energy output is $L$, so substituting this into the first expression gives $$T={\left(\frac{4\pi \sigma R^2{T}_{\ast }^4}{4\pi r^2\eta \sigma }\right)}^{\frac{1}{4}}$$ Which is $$T={\left(\frac{R^2{T}_{\odot }^4}{r^2\eta }\right)}^{\frac{1}{4}}$$ $$T=T_{\ast }{\left(\frac{R^2}{r^2\eta }\right)}^{\frac{1}{4}}$$ 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 $$T=T_{\ast }({1.5}^2\times 0.91{)}^{-\frac{1}{4}}\approx 0.836T_{\odot }$$ 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 realistic radius

A more commonly-used radius is $r\approx 1\text{ AU}$ - the distance from Earth to the Sun. when substituted in, this gives $r=215R$ and $T=0.07T_{\ast }$, 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 $${\eta }_T=1-\frac{T}{T_{\ast }}$$ It should be expected, at best, that ${\eta }_T\approx 0.95$, so we get $T=0.05T_{\ast }$ - 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: $$\begin{array}{|cc|}\hline \text{Material}& \text{Emissivity}\\ \text{Concrete}& 0.81\\ \text{Cement}& 0.54\\ \text{Galvanized steel}& 0.88\\ \text{Iron}& 0.87\text{-}0.95\\ \hline\end{array}$$ 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 $T=0.071T_{\ast }$ - 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: $$\begin{array}{|ccccc|}\hline \text{Star}& \text{Spectral type}& T_{\ast }\text{ (K)}& T\text{ (K)}\text{ (via emissivity)}& T\text{ (K)}\text{ (Slysh)}\\ \text{Zeta Puppis}& \text{O4}& 40000& 2840& 2000\\ \text{Eta Aurigae}& \text{B3}& 17200& 1220& 860\\ \text{Fomalhaut}& \text{A3}& 8590& 610& 430\\ \text{Tau Boötis}& \text{F6}& 6360& 450& 320\\ \text{Sun}& \text{G2}& 5770& 410& 290\\ \text{Alpha Centauri B}& \text{K1}& 5260& 370& 260\\ \text{Gliese 581}& \text{M3}& 3480& 250& 170\\ \hline\end{array}$$ 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^[1]
• Titanium monoxide^[2]
• Ammonia^[2]
• Methane^[2]
• 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.

posted about 10 years ago

HDE 226868‭

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