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I've been thinking about ways to cloak ships in space, since they all are warmer than their background they can be seen in the IR spectrum (given a large enough telescope).

So I was thinking, could you direct the waste heat in one direction, using a laser? For example, drive a heat pump where the hot side is a ruby crystal, which gets hot enough to glow, but not hot enough to melt (say 1700 C). Would this be equivalent to "flashing" the crystal how a normal laser works? Thus allowing you to direct the waste heat while keeping the hull of your ship at the same temperature as space.

Basically, do the crystal still behave in the same way at higher temperatures, and can you use the black body radiation from the crystal itself to drive the laser?

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I read about a solar-powered laser several decades ago, which isn't too far from what you are suggesting. There is another issue, though, that is much more fundamental: Usually a laser - almost by definition - emits very low-entropy light. In most cases that means it's monochromatic; in some cases it's broadband but highly organized in the frequency domain. In almost all cases it is highly spatially coherent (can be focused to a small point), which means low entropy in the spatial domain.

The very close relationship between heat and entropy implies that when you're removing heat you are removing entropy. If your laser beam is very low entropy you are removing energy but not removing entropy, and are therefore not removing heat. More heat & entropy will be generated in the process of powering the laser than the laser beam can remove. Simple black body radiation, perhaps masked in most directions so it's emitted primarily in one direction (away from whoever you're hiding from), would be better at removing heat.

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  • $\begingroup$ Re, "...Simple black body radiation, perhaps masked in most directions so it's emitted primarily in one direction..." FWIW: You can accomplish that by hiding behind a mirror. $\endgroup$ Apr 11, 2019 at 13:44
  • $\begingroup$ That is of course right. $\endgroup$
    – S. McGrew
    Apr 11, 2019 at 14:23
  • $\begingroup$ I've considered the "mirror masked black body" approach, but that only works if you know where your enemy is, the point of the laser was to get an extremely narrow risk of detection. But I guess you could get a very small risk of detection by letting the two holes you send your thermal radiation through be far apart :) $\endgroup$ Apr 11, 2019 at 15:10
  • $\begingroup$ Some paints have a really high emissivity and it actually results in cooling like 5degC over ambient $\endgroup$
    – ChemEng
    Jul 15, 2021 at 12:10
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To have a laser, you need to achieve an occupation-inversion of your energy-levels. This is done by pumping, i.e. selective excitations of electrons from the ground state $E_g$ to the pump state $E_p$, where $E_g < E_p$. However, thermal excitation may never achive occupation inversion, since then the electrons are distributed wrt. to the Bolzmann-distribution, where the occupation of a state $\rho$ with energy $E$ obeys $\rho(E)\propto \exp(-\tfrac{E}{k_BT})$, which is decreasing with $E$.

The prior only holds in a static case, where all the thermodynamic observables are constant. There are lasers which function out of thermodynamical equilibrium. For more information on this please refer to https://en.wikipedia.org/wiki/Gas_dynamic_laser. However, this is equivalent to a gas-leak in your ship. Also, the gases themselves are heated and may make it even more visible, since it will leave behind a glowing trail.

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  • $\begingroup$ Thanks for the more in depth thermochemical explanation :) The Gas Dynamic Laser looks really cool, could you just capture the exhaust in a container (possibly reuse it) or would recompressing the gas insert more heat into the system than the laser can get rid of? :S Like is it theoretically not possible or just practically very difficult? $\endgroup$ Apr 11, 2019 at 15:13
  • $\begingroup$ Ops, I think I just caught myself trying to break the second law of thermodynamic, in order for the laser to get rid of heat, it would need an efficiency above 100%, right? $\endgroup$ Apr 11, 2019 at 15:22
  • $\begingroup$ I can't tell that for sure, but you certainly would need to break some law :-( $\endgroup$
    – denklo
    Apr 12, 2019 at 5:58

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