Okay I did the same calculation but with Martian orbit as dyson sphere size, and 50% efficiency I got a wavelength of 3.4um so nicely in the infrared range of JWST.

I think the sphere would need to be like 99.95% efficient to be undetectable by JWST at Martian orbit radius.

Edit to add disclaimer: this is shitpost level math here guys, I’m just spitballing.

It would definitely emit thermal radiation. if it was 99% efficient and the size of pluto’s orbit, around a star like the sun, and the energy was used to create matter, I think it would radiate the remaining energy as 0.009 W/m2 with a peak emission wavelength of 150micrometers. The James Webb telescope has infrared capabilities that max out at 28.5 micrometers so def not detectable.

But probably a dyson sphere would be smaller than pluto’s orbit, which would greatly increase the apparent power, and shorten the wavelength. idk it’s all imaginary.

I won’t subject you to my hand writing but I did (power of sun × 0.01)/(surface area of sphere with Pluto’s orbital radius) to get radiation intensity (0.009 W/m2). Then rearranged Stefan-Boltzmann law to solve for temperature (19.8K). Then used Wien’s Displacement Law to calculate the peak wavelength (1.5×10^-4 m).

Maybe I’ll run the numbers again with a martian orbit radius, and 50% efficiency.

Like I’m sure it’s not but I don’t know if it’s a worse explanation than any of the other ideas being considered. But I don’t know enough to even know how wrong I am.

Is dark matter just Dyson sphered stars?

What unhealthy eating habit are you indulging in at 21:45?

It’s actually super interesting to look at the seasonality of solar. It’s significant in summer and negligible in winter.