Can you use a magnifying glass and moonlight to light a fire? (2016)

Again you are "explaining" when you should be asking.

> None of the fibers would "look at the moon". Fiber isnt the same as lenses.

I think you meant to ask "how is a fiber different from a lens?" instead of asserting "fiber isn't the same as lenses," which turns out to be completely wrong.

To a physicist, a lens is a shaped piece of refractive medium-- that is exactly what a fiber is. There is no magic inside the fiber; it does not add any photons. You are looking through a shaped piece glass, through which you will see a (possibly very distorted) image of what's on the other side.

In this case, it's the moon that is on the other side. The most you could do is surround your match/ant/whatever with fibers/lenses which are showing the moon on the other side of them. And from this fact it is unavoidable that you cannot make your surrounded subject hotter than the surface of the moon. If you are trying to say anything to the contrary, I'm sorry, but you are Flat Wrong.

Any other configuration of fiber would give you less moonlight than completely surrounding the subject, so it's not going to improve the odds of starting a fire.

> Note too that i said match. Getting tiny bit of a matchhead hot enough to decompose isnt the same as burning an ant.

I think you meant to ask "does the material or shape change anything?" instead of asserting "it isn't the same," which again, in the context of this question, turns out to be completely wrong.

The maxiumum temperature that can be imparted by a lens/mirror system doesn't depend on what it's focused on at all. In the long run, the ant/match/whatever will reach that maximum temperature and not get hotter. From a thermodynamics perspective, it is like putting an object in an oven of a particular temperature-- It doesn't matter if you put in a brick, or a cake, or match, or an ant; eventually they will all be 250 degrees if the oven is set to 250 degrees. In this case, the "oven" is the moon, and it's set to about 120C. A match combusts at about 600C. There will be no combusting of the match by putting it in the moon-oven. Sorry. End of story.

And if I can give some advice: If what you're hearing doesn't make sense to you, it is always safer to ask a question than to blindly assert your gut-guess of how you think it is. Someone who thinks they're an expert but spouts nonsense looks like a fool. Someone who asks a question, though, looks curious, which is smart. These two people have the same degree of knowledge, but one comes off looking much worse than the other. That's why asking questions is better, especially when you don't know the knowledge of the people who are listening.

I don’t buy the thermodynamic argument. Here’s a version I would believe: if you have a gadget that, exposed only to the sun and to empty space, heats some target hotter than the sun, then that gadget must not work if you take away the empty space part. This is because your gadget could be used to drive a heat engine, which is impossible without a temperature difference, and the sun is more or less a blackbody emitter. Lenses and mirrors aren’t magically taking advantage of the cold parts of the sky, so there you go.

But the moon is not blackbody, and I think the whole argument falls apart. Here’s a thought experiment: go stand on the moon, and assume the moon is made of rock that diffusely reflects, say, half of the indicent 500nm light. Stand somewhere that’s in shadow, so you can’t see the sun. Wrap a piece of paper and some air in perfectly insulating, perfectly reflecting material, except that the material lets 100% of 499-501nm light through, but only on the moon side. The target will be in a bath of 499-501nm light at 1/2 the intensity (energy density per unit volume) of the sun, which is far more than half the temperature of the sun. It’ll catch fire after a while.

Now do the same experiment on the Earth, at night, with lenses to bathe it in moonlight from all sides. Fire! So I claim that lenses+mirrors+filters can start a fire with moonlight.

Another interesting question: can you use a luminescent solar concentrator or other fluorescent material to pull this off without taking such egregious advantage of the spectrum of moonlight? These types of materials can violate conservation of étendue.

I think you'd need a different argument to show that's impossible, since the surfaces of stars are very hot, and their light is unmediated.

Consider this:

Say a light source has a T temperature resulting in X photons emitted. I redirect all the photons to a single point. I see arguments mentioning that that single point cannot be hotter than the source because there's no more photons to make it hotter.

I now add a second light source of the same T temperature, that emits the same amount of photons and also focus all of them on that same point. I now have more photons, but temperature source of all photons is the same. How does adding more photons not make my point hotter?

As I thought more of why one might get confused with this question is because one might miss that there are two effects here, not one.

To have a combustion we need two things - light of high enough temperature and light of high enough energy concentration. The two are not the same. All visible light has high enough temperature. However the concentration is the problem

Why do we need high concentration for combustion? If we don't supply enough energy to compensate for heat loss, the area will never get hot enough.

By the way - interesting corollary of this is that even with sun lights - if we have a mechanism that takes away heat fast enough (say by blowing cold air at the area) we would not be able to reach combustion with large mirror in direct sunlight.