# Light, Work, and the 2nd Law

I am trying to figure out what restrictions there are on converting light energy into work.

I understand that solar energy can be converted to electrical energy with roughly 90% Carnot efficiency. This can be derived from the temperatures of the Sun and the panels on Earth. What about low-temperature thermal radiation? What determines when light energy can be converted to work?

• I’m voting to close this question because it shoud be posted on the world building site. Commented Sep 22, 2023 at 8:42
• "I have been designing a physically compatible magic system for a story." (Emphasis added.) We deal with mainstream physics on this site. See: physics.stackexchange.com/help/on-topic
– hft
Commented Sep 22, 2023 at 15:30
• Next time I will not include that sentence. The question was one of purely physical nature.
– Buff
Commented Sep 22, 2023 at 16:38
• That is not the only problem. Your question is also pretty broad and seems to include multiple different question as one post. Please don't be discouraged, but please do read up on how to ask a good question ("good" is based on the conventions of this community, so you must read the FAQ to know what "good" means): physics.stackexchange.com/help/how-to-ask
– hft
Commented Sep 22, 2023 at 16:47

When thinking of making light to produce work, don't think of the source, think of the light itself. So you have three components:

1. Light (carrying energy $$U$$ per work cycle)
2. Engine (that will have done work $$A$$ by the end of the work cycle)
3. Heat bath (at temperature $$T_b$$)

Can $$A=U$$? At the end of the day everything boils down to the Second Law of Thermodynamics. You can't decrease the entropy of the Universe, so that is what sets your upper limit on efficiency. The thing to notice here is that light itself has entropy $$S_{light}$$. This is an old subject dating to Landau himself: https://doi.org/10.1016/B978-0-08-010586-4.50067-5. he was pondering there on questions similar to yours and came up to realize that light itself is also a thermodynamic body.

So the ultimate limit on efficiency would be $$U=A+T_b S_{light}$$. You can check Landau's paper above on how to calculate $$S_{light}$$ for a general situation. In the special case of perfectly coherent light (coming from a perfect laser) $$S_{light}=0$$; then $$A=u$$

• @hyportnex Yes. You're right. I will fix it now.
– John
Commented Sep 22, 2023 at 13:27
• @hyportnex Indeed
– John
Commented Sep 22, 2023 at 14:27
• but not anymore!!! Commented Sep 22, 2023 at 14:28