1
$\begingroup$

In an incandescent light bulb pretty much almost all of the input power is radiated as photons. While perhaps 10% of the power is useful for human vision and the rest are longer wavelengths, it's all light.

Using a good projector light bulb with a few mm source size and say a 10 cm mirror, or an array of them, I'm guessing you could be 95% efficient in converting electrical input power to photon propulsion.

Conversely lasers are usually less efficient than this aren't they?

Question: Would an incandescent light bulb actually be more efficient than a laser for photon propulsion?

Improve this question
$\endgroup$
6
  • $\begingroup$ While it's likely that a focused bulb can be more efficient than laser light, 95% efficiency is highly questionable. That would imply that you could build a video projector without active cooling because almost all heat would be radiated out through the lens. $\endgroup$
    – asdfex
    Apr 26, 2020 at 11:58
  • $\begingroup$ @asdfex you've missed the point of the first sentence. Projectors for humans can only make use of visible light so they can only use a small fraction of the power of an incandescent bulb; most power is radiated in infrared. But for this purpose those photons are just as good as the visible ones. Of course there's the problem that light bulbs burn out after a while. $\endgroup$
    – uhoh
    Apr 26, 2020 at 15:04
  • $\begingroup$ No, I didn't. For propulsion you need a narrow beam width and optics that can focus the whole wavelength range into such a narrow beam. For 95% efficiency you need near perfect mirrors and lenses. To my knowledge no such materials exist. Something like 70-80% seems quite feasible though. $\endgroup$
    – asdfex
    Apr 26, 2020 at 15:27
  • $\begingroup$ @asdfex have you checked the spectral reflectivity of gold between say 600 nm and say 10 um? This is just a random link: tydexoptics.com/optical_coatings/metal_mirrors It really can be gold mirrors, and cosine losses for a half angle of say 6 degrees are well below 1%. I wouldn't have guessed 95% if I hadn't thought about the physics first at least a little bit. The hard part is the filament lifetime. For tungsten you'd like to use a halogen gas inside a fused silica tube but that glass has some IR absorption... $\endgroup$
    – uhoh
    Apr 26, 2020 at 16:09
  • $\begingroup$ ..., and it will re-radiate from a size much larger than the size of the filament, so be somewhat out of focus, so have some larger cosine loss. For that reason you'd like to run a little cooler so the filamnents don't evaporate so fast so you can just leave them exposed to the vacuum of space, sans halogène $\endgroup$
    – uhoh
    Apr 26, 2020 at 16:12

1 Answer 1

Reset to default
5
$\begingroup$

It would have more TWR, the same efficiency, but the matter is almost irrelevant anyway.

(If you're trying to build a weapon, not a thruster, the laser is better because it is collimated in a narrow beam and the lamp isn't.)

Lasers are inefficient... and their inefficiency means they output heat. Which can only be disposed of in space by radiation. So if your radiator is at the focus of a much larger parabolic mirror sail, or even is just one-sided, you get photon thrust from the waste heat, same as the light bulb. However, the light bulb would be a lot lighter and simpler and cheaper.

However, the whole idea here is just... unbalanced. Photon thrusters really only make sense when you have antimatter fuel, a black hole reactor, or other highly speculative near-total-conversion reactor, in which case you are going to be directly converting matter into electromagnetic radiation, not generating electricity (Ineficient! Heavy!) and then turning that electricity into light.

(Converting any kind of heat source or other energy source, into a well-ordered "work" source such as torque on a shaft or electric current, inevitably has thermodynamic limits to its efficiency, unless it's already a well-ordered energy source, which few are. Often efficiency will be limited to less than 50 percent. If heat is what you want, you should just start with heat.)

If the thruster is being powered by beamed power from outside, you would do better with a simple solar sail. If you are powering it with any near-term (or even far-term) power source such as a fission or fusion reactor, you should at most be ejecting spent reactor fuel, and in most more likely designs, you will be ejecting a larger quantity of inert reaction mass accelerated with power from the reactor.

Improve this answer
$\endgroup$
2
  • $\begingroup$ +1 might make more explicit the thermodynamic implications of generating the electricity to power all those bulbs/whatever. $\endgroup$
    – antlersoft
    Apr 23, 2020 at 13:43
  • 1
    $\begingroup$ Yep, the TL;DR is "minimize the number ot steps from stored energy to output thrust" and don't forget momentum. $\endgroup$
    – Carl Witthoft
    Apr 23, 2020 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.