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 $$L=4 \pi \sigma R^2 T_*^4$$ 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 realistic radius

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^[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 over 9 years ago

HDE 226868‭

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