3 ms·
The article notes that the LED "acts as a very low temperature object". I interpret this to mean "the LED radiates as though its temperature is lower", rather t
by timmaxw 8y ago
The article notes that the LED "acts as a very low temperature object". I interpret this to mean "the LED radiates as though its temperature is lower", rather than "the LED radiates as though its emissivity is lower". In other words, AFAICT the effect can still exist even if the LED has 100% emissivity in both absorption and emission, which would make it a black-body. So the fact that the real LED's emissivity is not 100%, is not relevant to the effect. But we're splitting hairs here.
- the8472 8y agoThe paper is talking about effective temperature for specific photon energies. Additionally black body emission refers to thermal equilibrium conditions which the LED clearly is not. And then there's the near-field coupling they're using: > Recent experimental advances in near-field radiation have shown that heat-transfer rates on the nanoscale can exceed the blackbody limit by several orders of magnitude14–17, owing to contributions from evanescent and surface modes18,19. As a result, energy conversion rates can be greatly enhanced on this scale20. So this whole system relies on not being a black body on several levels.
- baybal2 8y agoEven that will only work if they don't produce more heat doing that, than they pump away. Same issue as with anti-Stokes scattering that was once held promise of "self-cooling LEDs" or using gas dynamic lasers for spaceship cooling. There is no trick against thermodynamics after all.