6 ms·

An idea I was told about relatively recently that kind of blew my mind: - Space is cold (~3K) - The atmosphere is really quite transparent in the wavelegth 8-

An idea I was told about relatively recently that kind of blew my mind: - Space is cold (~3K) - The atmosphere is really quite transparent in the wavelegth 8-13 μm - If one can construct an optical filter (e.g. grating) which is highly reflective except for a bandpass at 8-13μm, it is possible to reflect most solar energy but still "see" the cold of deep space - It's possible to build a box with sufficiently low thermal conduction, and this 'magic mirror' on top, to effectively refrigerate its contents by radiation to space. This uses no power, and works in direct sunlight [^1]. So weird, probably not very practical (what about cloudy days?), but very cool! [^1] in fact, if you put it in the shade then it would warm up because it could no longer radiate to space.

MengerSponge 6y ago
https://www.nature.com/articles/ncomms13729 https://www.nature.com/articles/ncomms13729
Aperocky 6y ago
Space is without temperature. Temperature is an attribute of things, space is a lack of things. What you're referring to is the temperature of cosmic background radiation, which is akin to how you emit infrared radiation at your body temperature. But you certainly don't classify the space where this radiation move through as having your body temperature. The two things are not the same imo.
- heisenzombie 6y ago
  "Temperature" is a slippery word that has many different definitions depending on what area of physics you're working in. In this case it is perfectly fine to refer to "the temperature of space" since you're explicitly using it as a cold sink. As evidence, see the paper (published in Nature no less!) that MengerSponge posted. The authors say "On the other hand, outer space, at a temperature of 3 K, provides a much colder heat sink."
  - Aperocky 6y ago
    Here's something I don't understand though, if we put a blackbody in deep space, would it eventually be cooled/heated to the temperature of the cosmic background radiation? The space around us is filled with photon coming from Sun at ~5500K, but it doesn't heat anything to anywhere near that. So given the current density of cosmic background radiation at 411 photon cm -3, would it be sufficient to place a blackbody to the equilibrium temperature? I am not a physicist by training, and this is a genuine question.
    - pdonis 6y ago
      > if we put a blackbody in deep space, would it eventually be cooled/heated to the temperature of the cosmic background radiation? No. As you note, in space in our vicinity there is a hot object, the Sun, in the sky, so the "temperature" of space in our vicinity is not really the 3K of the CMB, because an object placed in space in our vicinity will not come to thermal equilibrium at that temperature, but at some higher temperature. Exactly what higher temperature, if we assume the object itself is a black body, will depend on the object's surface area and its distance from the Sun. We could, of course, theoretically place an object outside our solar system, or even outside our galaxy, or even outside our local group of galaxies, or even outside the galaxy cluster of which our local group is a part. Doing those things would move the object farther and farther, on average, from high temperature radiation sources like the Sun. But there would still be radiation from those sources in the sky, meaning that the object would still reach thermal equilibrium at some temperature higher than the 3K CMB temperature. In short, an object would come to thermal equilibrium in space at the 3K CMB temperature only if the CMB were the only radiation source in the object's sky. But in actual fact it never will be.
      - heisenzombie 6y ago
        Nice. Seems related to: https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature https://en.wikipedia.org/wiki/Planetary_equilibrium_temperat... The temperature of a perfect black body in earth’s orbit would be ~255K (-18C) (This is not the case for the technology under discussion because it’s very much not a black body) I will point out that even at modest (astronomically speaking) distances, the solid angle of a star gets preeeeetty small. For example, a black body at just 1 light year from the Sun (just a quarter of the way to Alpha Centauri) would come to an equilibrium temperature of ~1K (assuming an otherwise empty universe). So that’s already a smaller effect than the CMB. I think in intergalactic space, the thermal equilibrium of a black body would be absolutely dominated by the CMB.
hoseja 6y ago
This feels like tricking Maxwell's daemon.
- taneq 6y ago
  It really, really does. Then again, it's not a 'closed system' - if the original Maxwell's daemon was trying to just cool down all the atoms instead of sorting them, then all it'd need to do is just go into a dark corner and radiate the heat away. All this system is doing is providing a readily accessible "dark corner".
Tade0 6y ago
There's even a cost-effective solution available: https://mse.umd.edu/news/story/cooling-wood-an-ecofriendly-building-material https://mse.umd.edu/news/story/cooling-wood-an-ecofriendly-b...
mkl 6y ago
> [^1] in fact, if you put it in the shade then it would warm up because it could no longer radiate to space. If I understand correctly, this isn't quite right. If it's in the shade, then whatever is shading it is radiating more heat onto it (because the shading object is hotter than space) and possibly reflecting its own radiated heat back. If it's under a clear sky then it's receiving less heat, so what it loses by radiating has more effect. Does that sound right?
- hajimemash 6y ago
  I agree with your description, OP possibly concised his words a little too much as to imply the wrong thing.
  - heisenzombie 6y ago
    I agree too :)

MengerSponge 6y ago

https://www.nature.com/articles/ncomms13729 https://www.nature.com/articles/ncomms13729

Aperocky 6y ago

Space is without temperature. Temperature is an attribute of things, space is a lack of things. What you're referring to is the temperature of cosmic background radiation, which is akin to how you emit infrared radiation at your body temperature. But you certainly don't classify the space where this radiation move through as having your body temperature. The two things are not the same imo.

heisenzombie 6y ago
"Temperature" is a slippery word that has many different definitions depending on what area of physics you're working in. In this case it is perfectly fine to refer to "the temperature of space" since you're explicitly using it as a cold sink. As evidence, see the paper (published in Nature no less!) that MengerSponge posted. The authors say "On the other hand, outer space, at a temperature of 3 K, provides a much colder heat sink."
- Aperocky 6y ago
  Here's something I don't understand though, if we put a blackbody in deep space, would it eventually be cooled/heated to the temperature of the cosmic background radiation? The space around us is filled with photon coming from Sun at ~5500K, but it doesn't heat anything to anywhere near that. So given the current density of cosmic background radiation at 411 photon cm -3, would it be sufficient to place a blackbody to the equilibrium temperature? I am not a physicist by training, and this is a genuine question.
  - pdonis 6y ago
    > if we put a blackbody in deep space, would it eventually be cooled/heated to the temperature of the cosmic background radiation? No. As you note, in space in our vicinity there is a hot object, the Sun, in the sky, so the "temperature" of space in our vicinity is not really the 3K of the CMB, because an object placed in space in our vicinity will not come to thermal equilibrium at that temperature, but at some higher temperature. Exactly what higher temperature, if we assume the object itself is a black body, will depend on the object's surface area and its distance from the Sun. We could, of course, theoretically place an object outside our solar system, or even outside our galaxy, or even outside our local group of galaxies, or even outside the galaxy cluster of which our local group is a part. Doing those things would move the object farther and farther, on average, from high temperature radiation sources like the Sun. But there would still be radiation from those sources in the sky, meaning that the object would still reach thermal equilibrium at some temperature higher than the 3K CMB temperature. In short, an object would come to thermal equilibrium in space at the 3K CMB temperature only if the CMB were the only radiation source in the object's sky. But in actual fact it never will be.
    - heisenzombie 6y ago
      Nice. Seems related to: https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature https://en.wikipedia.org/wiki/Planetary_equilibrium_temperat... The temperature of a perfect black body in earth’s orbit would be ~255K (-18C) (This is not the case for the technology under discussion because it’s very much not a black body) I will point out that even at modest (astronomically speaking) distances, the solid angle of a star gets preeeeetty small. For example, a black body at just 1 light year from the Sun (just a quarter of the way to Alpha Centauri) would come to an equilibrium temperature of ~1K (assuming an otherwise empty universe). So that’s already a smaller effect than the CMB. I think in intergalactic space, the thermal equilibrium of a black body would be absolutely dominated by the CMB.

hoseja 6y ago

This feels like tricking Maxwell's daemon.

taneq 6y ago
It really, really does. Then again, it's not a 'closed system' - if the original Maxwell's daemon was trying to just cool down all the atoms instead of sorting them, then all it'd need to do is just go into a dark corner and radiate the heat away. All this system is doing is providing a readily accessible "dark corner".

Tade0 6y ago

There's even a cost-effective solution available: https://mse.umd.edu/news/story/cooling-wood-an-ecofriendly-building-material https://mse.umd.edu/news/story/cooling-wood-an-ecofriendly-b...

mkl 6y ago

> [^1] in fact, if you put it in the shade then it would warm up because it could no longer radiate to space. If I understand correctly, this isn't quite right. If it's in the shade, then whatever is shading it is radiating more heat onto it (because the shading object is hotter than space) and possibly reflecting its own radiated heat back. If it's under a clear sky then it's receiving less heat, so what it loses by radiating has more effect. Does that sound right?

hajimemash 6y ago
I agree with your description, OP possibly concised his words a little too much as to imply the wrong thing.
- heisenzombie 6y ago
  I agree too :)