36 ms·

I think where the light is being sent is open space so what ever temperature the local background radiation is at; a common local minimum being the cosmic backg

by davorak 14y ago

I think where the light is being sent is open space so what ever temperature the local background radiation is at; a common local minimum being the cosmic background radiation.

ars 14y ago
> a common local minimum being the cosmic background radiation No, that's only if you are in outer space. On earth it's the temperature of the air, or the walls. It doesn't matter if the light eventually gets to outer space - what matters is that there is something hotter near you (namely the air). i.e. it's not where the photons end up that matters, it's the total of all the photons that are being sent in your (the lamp's) direction that counts.
- davorak 14y ago
  Sorry if what I said was unclear. > it's not where the photons end up that matters, it's the total of all the photons that are being sent in your (the lamp's) direction that counts. This is what I was referencing when I siad "the local background radiation." The common minimum was supposed convey high frequency when choosing a random spot in the known universe. > On earth it's the temperature of the air, or the walls. The temperature of the air only comes into play if it is emiting radiation at the frequency of light produced by the led and at greater intensity. I do not see why this would be required.
  - ars 14y ago
    > The temperature of the air only comes into play if it is emiting radiation at the frequency of light produced by the led and at greater intensity. Blackbody radiation. Air (everything) always emits light. > I do not see why this would be required. It's a fundamental law of thermodynamics that requires a heat source and a cold source in order to convert heat into some other form of energy. The air around you is the cold source, and the LED lamp is the hot source. If the LED is colder than the air it will not work - the LED must be hotter (in this particular case they put it in an oven).
    - jules 14y ago
      If you put a piece of metal in a very hot oven it will emit light without any electric power input whatsoever. That's one way to see that this is indeed not magic. The accomplishment here appears to be to get the same result at lower temperatures.
    - Cushman 14y ago
      I think I understand what you're saying now. In vacuum, an object will radiate until it reaches absolute zero; in air, however, you're constantly being heated to the ambient temperature, meaning you can't radiate your way down until the air has lost its heat too. The thing I think you're not taking into account is that this device has the effect of slightly increasing your black-body radiation. So you're at 1300K, radiating at 1300K, but we apply a little electricity and you start radiating a little more-- maybe you now look like you're at 1301K. Your heat hasn't actually increased, just apparent radiation, and as a consequence you cool down a little faster than your black-body profile would suggest. So say you're at room temperature, radiating at room temperature, being warmed by the air, and not losing any heat. We apply current again, and you start radiating at a little over ambient temperature. The air is still warming you, but you're cooling a little faster than before, with the extra heat being deposited wherever the radiation is absorbed. To the air, you now feel cool. Why should the absolute temperature of the ambient air have an effect on the process?

by davorak 14y ago

I think where the light is being sent is open space so what ever temperature the local background radiation is at; a common local minimum being the cosmic background radiation.

ars 14y ago

> a common local minimum being the cosmic background radiation No, that's only if you are in outer space. On earth it's the temperature of the air, or the walls. It doesn't matter if the light eventually gets to outer space - what matters is that there is something hotter near you (namely the air). i.e. it's not where the photons end up that matters, it's the total of all the photons that are being sent in your (the lamp's) direction that counts.

davorak 14y ago
Sorry if what I said was unclear. > it's not where the photons end up that matters, it's the total of all the photons that are being sent in your (the lamp's) direction that counts. This is what I was referencing when I siad "the local background radiation." The common minimum was supposed convey high frequency when choosing a random spot in the known universe. > On earth it's the temperature of the air, or the walls. The temperature of the air only comes into play if it is emiting radiation at the frequency of light produced by the led and at greater intensity. I do not see why this would be required.
- ars 14y ago
  > The temperature of the air only comes into play if it is emiting radiation at the frequency of light produced by the led and at greater intensity. Blackbody radiation. Air (everything) always emits light. > I do not see why this would be required. It's a fundamental law of thermodynamics that requires a heat source and a cold source in order to convert heat into some other form of energy. The air around you is the cold source, and the LED lamp is the hot source. If the LED is colder than the air it will not work - the LED must be hotter (in this particular case they put it in an oven).
  - jules 14y ago
    If you put a piece of metal in a very hot oven it will emit light without any electric power input whatsoever. That's one way to see that this is indeed not magic. The accomplishment here appears to be to get the same result at lower temperatures.
  - Cushman 14y ago
    I think I understand what you're saying now. In vacuum, an object will radiate until it reaches absolute zero; in air, however, you're constantly being heated to the ambient temperature, meaning you can't radiate your way down until the air has lost its heat too. The thing I think you're not taking into account is that this device has the effect of slightly increasing your black-body radiation. So you're at 1300K, radiating at 1300K, but we apply a little electricity and you start radiating a little more-- maybe you now look like you're at 1301K. Your heat hasn't actually increased, just apparent radiation, and as a consequence you cool down a little faster than your black-body profile would suggest. So say you're at room temperature, radiating at room temperature, being warmed by the air, and not losing any heat. We apply current again, and you start radiating at a little over ambient temperature. The air is still warming you, but you're cooling a little faster than before, with the extra heat being deposited wherever the radiation is absorbed. To the air, you now feel cool. Why should the absolute temperature of the ambient air have an effect on the process?