Even a Dyson sphere, which is technically unlikely anyway, would be possible to spot. You would look for something very bright in the infrared spectrum with almost no light in the visible spectrum. It would also be larger than a normal star of the same energy, but that would be hard to tell given all the other issues.
A partial swarm is easier because it will have variability towards more infrared and then back to a more normal spectrum.
And, of course, all this is speculation until we find a candidate and determine it doesn’t have a natural source for that behavior.
Why would there necessarily be strong infrared emissions? Since a Dyson Sphere is meant to harvest all energy produced by a star, any leakage would be unnecessary inefficiency, wouldn’t it?
Thermodynamics says that energy can’t be destroyed (mass-energy, but generally that won’t matter). So after the work of running your stellar civilization is done, you will radiate out waste heat. There is no real way around this without breaking thermodynamics or having a handy black hole to dump all your waste heat into. Therefore, the energy of the star will still be released, but it will be released as infrared.
If you’re using the Dyson sphere purely as a power plant and e.g. charge batteries, the thermal radiation will be distributed over the whole area covered by the civilization.
A solar panel, or any other power generator we use, doesn’t radiate away all the generated energy either. It’s radiated from the point of use.
So you heat habitats, which radiate heat. And run computers, which radiate heat. And move objects around, which radiates heat (among other things). And if you merely absorb energy from your star…it radiates as heat. This is the whole idea of entropy. Unless your lasers are particularly efficient and you use them to beam the energy elsewhere, your Dyson swarm is going to radiate heat equivalent to the energy your star puts out.
You’re ignoring my example - what if you charge up batteries at the Dyson sphere, and use the energy anywhere else? There’s no physical reason the energy must be used around the Dyson sphere.
So all you need is a perfect charging system. We don’t have those, and physics doesn’t allow for them. This would be no different than the laser example I gave, and this only makes sense after you have a second Dyson swarm.
Why perfect? As long as the efficiency is high enough, you wouldn’t see the sphere itself as very bright, it would be quite dim. Do we know any hard, physical limitations for this, like we do for speed?
A partial answer to your question is that there’s a minimum amount of heat necessarily radiated when doing computation, given by the Landauer principle.
Furthermore, I also do not think that we will detect dyson spheres, because if a civilisation wishes to hide, they won’t radiate heat uncontrollably by extracting all possible energy, but rather send that energy elsewhere, for example by dumping it into a black hole. But I could be wrong and such a civilisation might care more about energy than remaining undiscovered.
Even a Dyson sphere, which is technically unlikely anyway, would be possible to spot. You would look for something very bright in the infrared spectrum with almost no light in the visible spectrum. It would also be larger than a normal star of the same energy, but that would be hard to tell given all the other issues.
A partial swarm is easier because it will have variability towards more infrared and then back to a more normal spectrum.
And, of course, all this is speculation until we find a candidate and determine it doesn’t have a natural source for that behavior.
Why would there necessarily be strong infrared emissions? Since a Dyson Sphere is meant to harvest all energy produced by a star, any leakage would be unnecessary inefficiency, wouldn’t it?
Thermodynamics says that energy can’t be destroyed (mass-energy, but generally that won’t matter). So after the work of running your stellar civilization is done, you will radiate out waste heat. There is no real way around this without breaking thermodynamics or having a handy black hole to dump all your waste heat into. Therefore, the energy of the star will still be released, but it will be released as infrared.
If you’re using the Dyson sphere purely as a power plant and e.g. charge batteries, the thermal radiation will be distributed over the whole area covered by the civilization.
A solar panel, or any other power generator we use, doesn’t radiate away all the generated energy either. It’s radiated from the point of use.
So you heat habitats, which radiate heat. And run computers, which radiate heat. And move objects around, which radiates heat (among other things). And if you merely absorb energy from your star…it radiates as heat. This is the whole idea of entropy. Unless your lasers are particularly efficient and you use them to beam the energy elsewhere, your Dyson swarm is going to radiate heat equivalent to the energy your star puts out.
You’re ignoring my example - what if you charge up batteries at the Dyson sphere, and use the energy anywhere else? There’s no physical reason the energy must be used around the Dyson sphere.
So all you need is a perfect charging system. We don’t have those, and physics doesn’t allow for them. This would be no different than the laser example I gave, and this only makes sense after you have a second Dyson swarm.
Why perfect? As long as the efficiency is high enough, you wouldn’t see the sphere itself as very bright, it would be quite dim. Do we know any hard, physical limitations for this, like we do for speed?
A partial answer to your question is that there’s a minimum amount of heat necessarily radiated when doing computation, given by the Landauer principle.
Furthermore, I also do not think that we will detect dyson spheres, because if a civilisation wishes to hide, they won’t radiate heat uncontrollably by extracting all possible energy, but rather send that energy elsewhere, for example by dumping it into a black hole. But I could be wrong and such a civilisation might care more about energy than remaining undiscovered.