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See [1] and [2] where the idea of hiding a Dyson sphere is covered in depth. It essentially means breaking the laws of thermodynamics, specifically entropy. Hit

by cletus 8y ago

See [1] and [2] where the idea of hiding a Dyson sphere is covered in depth. It essentially means breaking the laws of thermodynamics, specifically entropy. Hit and object with light from a star and it heats up. It radiates away that heat in a predictable way. [1] https://www.youtube.com/watch?v=QfuK8la0y6s https://www.youtube.com/watch?v=QfuK8la0y6s [2] https://www.youtube.com/watch?v=HlmKejRSVd8 https://www.youtube.com/watch?v=HlmKejRSVd8

Sargos 8y ago
> It radiates away that heat in a predictable way. Unless that heat energy itself is also caught and used.
- dodobirdlord 8y ago
  That is exactly what is meant by "It essentially means breaking the laws of thermodynamics".
ChrisFoster 8y ago
Waste heat must be radiated away, though thermodynamics would be quite content if that radiation is anisotropic. For example, consider a perfectly insulating black box with small hole. Observers on the opposite side from the hole will not observe any radiation from the interior of the box. I don't think there's any physical limit on collimating the outgoing radiation to hide from everyone outside a narrow beam. It's "just an engineering problem".
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- hexane360 8y ago
  Actually, I believe the law of Etendue[1] covers that. Basically, any focusing of the area of light must be balanced by a widening of the angle of that light. 1: https://en.wikipedia.org/wiki/Etendue https://en.wikipedia.org/wiki/Etendue
  - ChrisFoster 8y ago
    In terms of hiding an emitter, the relevant quantity is the beam solid angle in the far field. This is wavelength dependent which implies temperature dependence if we're talking thermal radiation. The lower the temperature, the longer the wavelength and the wider the beam (given fixed emitter geometry). So there's a tradeoff between hiding your Dyson sphere vs extracting the most energy from the source.

by cletus 8y ago

Sargos 8y ago

> It radiates away that heat in a predictable way. Unless that heat energy itself is also caught and used.

dodobirdlord 8y ago
That is exactly what is meant by "It essentially means breaking the laws of thermodynamics".

ChrisFoster 8y ago

Waste heat must be radiated away, though thermodynamics would be quite content if that radiation is anisotropic. For example, consider a perfectly insulating black box with small hole. Observers on the opposite side from the hole will not observe any radiation from the interior of the box. I don't think there's any physical limit on collimating the outgoing radiation to hide from everyone outside a narrow beam. It's "just an engineering problem".

deleted 8y ago
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deleted 8y ago
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hexane360 8y ago
Actually, I believe the law of Etendue[1] covers that. Basically, any focusing of the area of light must be balanced by a widening of the angle of that light. 1: https://en.wikipedia.org/wiki/Etendue https://en.wikipedia.org/wiki/Etendue
- ChrisFoster 8y ago
  In terms of hiding an emitter, the relevant quantity is the beam solid angle in the far field. This is wavelength dependent which implies temperature dependence if we're talking thermal radiation. The lower the temperature, the longer the wavelength and the wider the beam (given fixed emitter geometry). So there's a tradeoff between hiding your Dyson sphere vs extracting the most energy from the source.