15 ms·

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

• dr_orpheus 8y ago
  This is great, I always love the xkcd "What if" explanations. I even have the book sitting on my desk next to me.
• detaro 8y ago
  EDIT: I'm not sure about this anymore.
  • jstanley 8y ago
    I think TFA refutes your argument, but I'm not smart enough to know for sure. Can you please expand on this?
    • randyrand 8y ago
      What if the sun was a bunch of parallel laser beams? Then you could focus them all into a single point.
      • jstanley 8y ago
        Right, but that point would be exactly as hot as the surface that the light was coming from, and no hotter. (Is the argument as I understand it).
        • jplee 8y ago
          If this is true, how do laser cutters work? The focused beam can melt through steel, but surely the temperature of the gain medium is much lower than the melting point of steel? But the thermodynamics argument seems like it makes sense too...
          • landryraccoon 8y ago
            Lasers have negative temperature. The light emitted from the sun is nothing like laser light. https://en.wikipedia.org/wiki/Negative_temperature#Lasers https://en.wikipedia.org/wiki/Negative_temperature#Lasers
          • whatshisface 8y ago
            Like the other commenter said, lasers are not black-body emitters. They "pump" light out and are very far from equilibrium.
        • DuskStar 8y ago
          Wouldn't that kind of make the National Ignition Facility [0] impossible? 0: https://en.wikipedia.org/wiki/National_Ignition_Facility https://en.wikipedia.org/wiki/National_Ignition_Facility
          • jstanley 8y ago
            From what I've read in the other comments, the argument doesn't apply to lasers.
      • whatshisface 8y ago
        Lasers are kept far from equilibrium by design, unlike blackbody radiators which are in equilibrium with the electromagnetic field they are radiating in to.
    • detaro 8y ago
      On second look, the thing I thought of doesn't work, so I retract the statement...
• dang 8y ago
  Discussed at the time: https://news.ycombinator.com/item?id=11211454 https://news.ycombinator.com/item?id=11211454
• deleted 8y ago
  [deleted]
• ohiovr 8y ago
  I had the presumption that you could make a mega hot spot with a large enough lens back in the 90s. Fortunately usenet set me straight. It was a long and facinating journey to understand all the whys about it.
• robertelder 8y ago
  I'm a bit skeptical of the conclusion and the reasoning process used to arrive there. For example, the article states "In other words, all a lens system can do is make every line of sight end on the surface of a light source, which is equivalent to making the light source surround the target." If you forget about optics for a second, imagine that the outer surface of the sun were wrapped around a point (think of the image shown in the article). If you consider conservation of energy for the energy flux from the surface of the sun being entirely directed to a single body of matter that absorbs this heat (assume it's a penny), the steady-state blackbody emission of the penny would have to equal the energy flux from the entire surface of the sun. I think this situation would end up making the 'temperature' of the penny much higher than the surface of the sun for the same reason that the center of the sun is hotter than the surface: There is energy expended, and it comes from the fusion of light elements inside the sun, so there is no violation of entropy as stated in the article: "you'd be making heat flow from a colder place to a hotter place without expending energy."
  • aqme28 8y ago
    You cannot use a magnifying glass to make something hotter than the source of the light. The thermodynamics principle is unyielding. This is a tough tricky thermodynamics question because it really seems like you can use a really big magnifying glass to make say an object that is hotter than the sun. I remember working through it in a physics class.
  • JimboOmega 8y ago
    If you directed all the input to the penny, then you've created a close system that can't lose energy, and doesn't the whole system - including the sun - keep rising in temperature anyway? Ultimately the outside of this inverted sun system needs to vent the whole of the energy, the energy can't just fall inwards. What you're describing is a reversal though; it requires the penny to be much hotter than the sun, only as a result of the sun's energy flowing into it.
  • landryraccoon 8y ago
    That doesn't work because when the penny is hotter than the sun it radiates an equal amount of energy back to the sun. The sun is a blackbody, it radiates light because it's hot. Once the penny is the same temperature it will radiate back at the sun until the two are in equilibrium.
• parliament32 8y ago
  > You can't use lenses and mirrors to make something hotter than the surface of the light source itself. This is an interesting argument. Can I not reflect some sunlight off a mirror, then do the magnifying-glass-to-start-a-fire trick in daytime? Doesn't the mirror stay cool? Isn't the moon just a (poor) mirror for the sun's light?
  • nealabq 8y ago
    If the light is being radiated from a black body surface, then you cannot make something hotter than that surface. But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. The moon also radiates like a black body, but virtually all that light has longer wavelengths than the reflected visible light. You could start a fire from the light of a single star if you had a big enough lens. You could also start a fire with the light from a hand mirror at the distance of the moon if it reflected the sun's light at you. But you'd need a very big lens.
    • baddox 8y ago
      > But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. That's my impression as well. A blackbody has albedo 0. The moon has an albedo of around 0.12. While I suspect you can't start a fire from moonlight in practice, I don't think the arguments in this article are correct.
      • stcredzero 8y ago
        While I suspect you can't start a fire from moonlight in practice, I don't think the arguments in this article are correct. I guess if you're reading that as some kind of absolutely logical argument. I read stuff like that as an abstraction which just kinda works in the messy real world. Pretty much like how the typical explanation for how a wing works turns out to be an over-simplification. It only partly works that way. Actual wings are complicated, but in the aggregate, they just get enough air molecules to go downward to net out the forces to keep the plane from going downward. It turns out that there are a lot of mechanisms contributing to this all at once. (Which is something else he discusses in that series.)
• DenisM 8y ago
  Thermodynamics argument seems iffy to me. I can cover the entire surface of the earth with solar panels to harvest the moon light and use the combined electricity to melt iron. If this is ok with thermodynamics then so is using lenses. The rest of the argument seems to be that light cannot be optically condensed to a single point as there will always be some dispersion due to diffraction, and the size and shape of the dispersion is dictated by that of the source. That is you can't make the target denser then source, hence the temperature must be less. EDIT: Several commenters submitted that lenses are reversible while solar panels are not, and this makes all the difference. My retort is that I can make non-reversible lenses by covering them in a thin layer of dust. Since the lense system is now non-reversible can I use these sub-par lenses to create higher temperature than I could with clear lenses?
  • marcinzm 8y ago
    >If this is ok with thermodynamics then so is using lenses. It's okay because solar panels are not perfectly efficient while lenses are, in theory, perfectly efficient. That efficiency loss is in essence the "cost" of moving heat from a colder to a hotter place. I'm guessing, but don't quote me on it, that solar panel efficiency is related to the temperature of the sun and the local environment just like any other heat engine.
    • dr_orpheus 8y ago
      You are correct about the solar panels. Solar panels are more efficient when they are colder, i.e. the temperature delta between the sun and the panel is larger.
    • sampo 8y ago
      > solar panel efficiency is related to the temperature of the sun and the local environment just like any other heat engine You are correct, https://en.wikipedia.org/wiki/Solar_cell_efficiency#Thermodynamic_efficiency_limit_and_infinite-stack_limit https://en.wikipedia.org/wiki/Solar_cell_efficiency#Thermody...
  • schoen 8y ago
    I was wondering about this too. After all, human beings have produced temperatures that were hotter than the sun's surface and almost all of the energy that we used to do that ultimately came from the sun (except for very small components that came from nuclear reactions on Earth and geothermal energy, which also comes from nuclear reactions inside Earth). I think that this issue is addressed by footnote 2 > And, more specifically, everything [lenses and mirrors] do is fully reversible—which means you can add them in without increasing the entropy of the system. Presumably we can't say the same of the electrical devices that we use to collect, store, and transmit sunlight, or to create high temperatures from this stored energy. For example, if you use a "lunar panel" to store energy in a battery and then heat something on an electric stove, many of the components in this process will not be reversible, differently from mirrors and lenses. (Putting a hot object on top of the stove won't cause the lunar panel to emit light back in the direction of the moon!) So I think footnote 2 is actually very important, because it's not that we can never use any energy source to create something hotter than that source, it's that we can never do so using only reversible processes, including purely passive optics.
• dooglius 8y ago
  One thing that makes me uncertain about this is the fact that the sun is generating the light from a fusion reaction, thus expending energy. In other words, the system entropy does not decrease because the fusion reaction makes up for any lost entropy by the cold-to-hot temperature flow. This is the same reason why a system consisting of a battery and a fridge would work.
• wefarrell 8y ago
  I've often wondered how big of a lens you would need to grow a plant using a light source from outside of the solar system, if it's even possible.
• FiatLuxDave 8y ago
  I love xkcd, but this is completely wrong. It is well known that the spectral temperature of the moon is about 4000K. See for example : http://www.lumec.com/newsletter/architect_06-10/the_sun_the_moon.html http://www.lumec.com/newsletter/architect_06-10/the_sun_the_... or https://physics.stackexchange.com/questions/244922/why-does-moonlight-have-a-lower-color-temperature https://physics.stackexchange.com/questions/244922/why-does-... . That is the maximum temperature that you can achieve with light from the moon, no matter how concentrated. 4000 K is plenty hot enough to start a fire.
  • saagarjha 8y ago
    Randall has a response to this argument in the article.
    • thatcherc 8y ago
      Does he though? Here's all I see - > "But wait," you might say. "The Moon's light isn't like the Sun's! The Sun is a blackbody—its light output is related to its high temperature. The Moon shines with reflected sunlight, which has a "temperature" of thousands of degrees—that argument doesn't work!" > It turns out it does work, for reasons we'll get to later. But the rest of the article is about etendue, and I don't see how the issue of reflected light is addressed (though possibly the answer is implied with an etendue argument I missed). I'm very curious now - it seems to me that if the Moon was a perfect mirror, you should be able to start a fire with it. Maybe the Moon's low albedo is the reason? Fun fact about the Moon: its albedo (roughly the fraction of incident light it scatters) is about the same as asphalt - not very reflective at all! [0] It just looks bright to us because the Sun is so tremendously bright. [0] - https://www.lcas-astronomy.org/articles/display.php?filename=albedo_effects&category=observing https://www.lcas-astronomy.org/articles/display.php?filename... , also learned this in an astronomy class
  • deleted 8y ago
    [deleted]
  • dr_orpheus 8y ago
    Color temperature is a representation of the shape of the spectrum as compared to a blackbody of the same temperature. It is not necessarily representative of the surface emitting the light. Because the moon absorbs some of the wavelengths of light more than others it shifts the spectrum to represent a blackbody object of a lower temperature. So depending on what wavelengths are absorbed the color temperature can actually be shifted up or down. For another example, the color temperature of standard incandescent bulbs is ~2700K, and this is similar to blackbody radiation. However, LED lights can have the same color temperature, but the surface of the LED is not at 2700K because the light emitted is being created through a different process than blackbody radiaton. That said, I am still wrapping my head around the xkcd explanation because blackbody radiation of the moon at 100K would not be enough but there is also reflected sunlight at a .16 albedo.
• adammunich 8y ago
  I'm not so sure about this. Imagine you had a large cloud of planar mirrors, each, specifically can be aimed at any given point --even points that overlap. While I agree that you cannot focus a whole image to a smaller area than the diffraction limit allows for a continuous lens surface, if you omit diffraction, mirrors could certainly do it.
  • wrycoder 8y ago
    You are describing a parabolic concentrator.
• zeristor 8y ago
  So what about non-linear optics and optical frequency doubling? https://en.wikipedia.org/wiki/Second-harmonic_generation https://en.wikipedia.org/wiki/Second-harmonic_generation It was just an option on my MSc in LASERs but I thought it was cool (with the potential to be very warm). Although my frequency has been halved and I've been working in software for decades
  • petermcneeley 8y ago
    I thought of that as well. Once you have a different frequency you can also do more bizarre fun things. http://optics.org/news/7/1/11 http://optics.org/news/7/1/11
• newnewpdro 8y ago
  I'm pretty sure if I were to go pick up a number of magnifying glasses and focus them on the same point using moonlight, the temperature at that point would increase with every additional magnifying glass. Am I to accept that the additional magnifying glasses would cease increasing the temperature once the temperature matched that of the moon's surface?
  • umanwizard 8y ago
    I don't know enough about physics to judge whether Randall's argument is correct, but in general it is possible for a sequence to be strictly increasing yet bounded. For example, 1/2, 3/4, 7/8, 15/16, ...
  • jarfil 8y ago
    You can't focus light from a single source with just magnifying glasses at different positions onto the same point. You can try it with the sun, get two magnifying glasses and try to focus them on the same point at once, it's not possible. With the addition of some well placed mirrors though, that's possible.
• IgorPartola 8y ago
  I don’t buy the argument that you can’t concentrate two beams on the same spot. Sure you might not be able to do that with one lens, but concentrating on a small area is as good an approximation. But if you require that it really be a point, why can’t I do that with N lenses and mirrors that are fully reversible but all align to aim at the same point from different angles?
  • hinkley 8y ago
    Don't all the biggest telescopes on earth use multiple reflectors these days?
  • Dylan16807 8y ago
    The argument is that you can't concentrate two beams to hit the same spot from the same direction. So yes, you can have a bunch of lenses each focusing from a different direction. And if you do a good job of aligning them, each lens will look as bright as the sun/moon from the target. But that's your limit.
• trevyn 8y ago
  Randall might be correct for conventional optics, but what about metamaterial lenses that break the diffraction limit? https://en.m.wikipedia.org/wiki/Superlens https://en.m.wikipedia.org/wiki/Superlens
• hinkley 8y ago
  Isn't the bigger problem that sunlight is nearly parallel and moonlight is reflected off of a spherical surface? How are you violating conservation of energy if you're taking all of the light that would hit a square mile of the earth and concentrating it down to the size of a penny? If you can't concentrate light that way then how do focusing lenses on cutting lasers function? Makes no sense.
  • monochromatic 8y ago
    It’s not parallel though. The sun and the moon subtend almost exactly the same angle in the sky (which is why solar eclipses work).
  • URSpider94 8y ago
    Indeed, you can not focus all of the sunlight that would hit a square mile of Earth down to a penny. You just can’t. When you focus the sun, you’re not making an infinitely small spot, you are making a tiny image of the sun. The bigger your lens, the larger that image. You can’t get any more focused than “in focus”. When you focus a cutting laser, you are imaging the shape of the laser cavity. The emission is coming from a very narrow spatial region, so you can focus it back down to a small spot. However, a laser that is out of alignment, will often not be able to be focused to a small spot.
  • ummonk 8y ago
    It's not because the moon is a spherical surface. If that were the issue, there would be a bright spot on the Moon where you can see the reflection of the Sun (you can see such specular reflection spots on e.g. many cars), and you could use that spot to light a fire. The issue is that the Moon barely reflects any of the light that falls on it. Most of the light is scattered, and most of the rest is absorbed and re-radiated.
    • sopooneo 8y ago
      The scattering seems it should cancel out. As far as total energy reaching a point on earth (or the circular disk of a lens) just as much should get to you only due to scattering as failed to get to you due to scattering. I don't think this changes the ultimate answer to the question.
• Analemma_ 8y ago
  Oh god, can we please get a real physicist in here? This entire thread is a mess of computer programmers “well actually”ing other computer programmers and everyone being wrong.
  • stcredzero 8y ago
    It's just like the old USENET days. It's also how Slashdot used to work. Enough of us programmers would spout enough nonsense to make a real expert angry enough to inform the heck out of us.
  • CydeWeys 8y ago
    The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). A lot of the counter-arguments/speculation here in the comments is wrong. Reading this is is equivalent to reading a thread on a physics forum where with people arguing about an article saying that O(n*lgn) is really the best possible runtime complexity for a comparison-based sorting algorithm, and trying to disprove it. It's also worth pointing out that Randall Munroe is a physicist (to the extent an undergrad degree counts anyway).
    • nkurz 8y ago
      The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). It's interesting what bothers different people. While many of the statements in this thread are probably wrong, not many of them bother me. But I find the lack-of-self-doubt and appeal-to-authority in your message to be genuinely offensive. Where does your certainty come from? With that out of the way, could you give some links to well known arguments that you refer to? Specifically, I feel certain that one can start a fire with sunlight reflected from a room temperature mirror, and don't understand the difference between a mirror and the moon within Munroe's argument. His conclusion might be correct (in practice, you may not be able to concentrate moonlight enough to start a fire) but I don't think the details of his argument can be. I currently don't believe that the temperature of the reflecting surface can be the limiting factor, and I think this is central his argument.
• kayerov 8y ago
  You can with magnifying glass bigger 2.3 million times
• crazygringo 8y ago
  Wow... separately I had no idea the surface of the moon reaches (and goes above) 100°C... that's hot! Literally boiling hot. Turns out at night it goes down to almost –200°C. That's insane. Quick searching of how the astronauts survived this, turns out it seems they timed landings to the lunar dawn for an in-between temperature, that the lunar surface doesn't conduct heat well (all dust?), of course there's no atomsphere to conduct heat, and that their boots were extremely well-insulated.
  • jxcl 8y ago
    Also, since the moon is tidally locked to the earth, it rotates at the same rate it revolves, which means that sunrise and sunset only happen once a month, which makes dawn a pretty long time. There's a pretty good song about how the moon's day/night cycle would affect a lunar mining colony: https://www.youtube.com/watch?v=GDPUdUGJpjc https://www.youtube.com/watch?v=GDPUdUGJpjc
    • crazygringo 8y ago
      Ah, that makes so much more sense then... two weeks of straight unfiltered sunlight is gonna heat things up... and two weeks of total darkness gives it time to freeze, freeze, and freeze some more.
  • InclinedPlane 8y ago
    Yup, lunar landings didn't last long enough to experience the full range of extreme lunar conditions. This is one of the reasons why lunar bases will actually be more difficult than people think. Dealing with extreme environmental conditions is much easier when they are stable, then you can design around them and deal with them. Dealing with conditions that cycle from one extreme to another continuously over extended periods of time is much more challenging. One point of proof of that is the longevity of the Martian and lunar rovers and landers. Many landers and rovers on Mars have lasted for years, some have lasted over a decade. No lunar rover has been able to maintain roving operations longer than a few months. This despite the fact that lunar rovers are in near real-time contact with Earth continuously. The heat/cold cycles and hyper-abrasive clingy dust make the lunar environment particularly harsh on equipment.
• IgorPartola 8y ago
  So thinking about this some more, the argument here is that if you use “just” a magnifying glass of arbitrary size and shape you can’t do this. I can buy that based on the arguments presented. Basically it says that the moon emits (really reflects but that’s immaterial) F photons per second per square meter, and while you can concentrate that into a very small area, all of those photons will not be enough to raise the temperature (that is input enough energy into the system) to a sufficiently high level. This is partially because you can’t make a small enough point with a single lens, and partially because there just aren’t enough photons. The sun doesn’t care because it has such a high flux that the concentration ends up being high enough for the area. The area argument is more important here because the lens cannot increase the number of photons per second, but it can decrease the area. If F = N / (t * A) where N is the number of photons, t is time, and A is area, the lens can change the area, but not to 0. And if you need a sufficiently high F to get to the right temperature, the only way to get there with limited N is to bring A sufficiently close to 0. If you have multiple magnifying glasses and mirrors I am fairly certain that you can. That is the equivalent of using a set of solar panels that power a laser. But that was not what was postulated in the original thought experiment, so it does not apply. I am still fuzzy on the thermodynamic argument, but I was never good at intuiting thermodynamics. The argument presented is that if you have one body at 100 degrees C, and you put another body next to it, you cannot make the second body hotter than the first. That makes sense. But if the first body is constantly generating and transferring heat to the second with at most 100 degrees C temperature, and the second body has some way to store heat energy, then it is possible to heat a local area of the second body to higher than 100 degrees C. The storage of energy here is what I think counts.
  • URSpider94 8y ago
    The thing is, radiative processes (like light) can’t store energy. To do that, you need some kind of engine, either electrical or mechanical. Of course you could put a solar panel connected to a battery in moonlight for a few months and build up enough stored energy to power a laser for long enough to fry something. But that’s not “burning something with moonlight using a magnifying glass.”
    • 8y ago
• aasasd 8y ago
  > Lenses and mirrors work for free; they don't take any energy to operate. Wait a minute. Does that mean that I could get a tiny solar panel and light it up with a lens, instead of getting big panels? Energy output of a panel is proportional to the amount of light that hits it, right? I guess that lenses are ‘free’ only if they have no impurities, but even then, assuming solar panels are costlier than plastic lenses, I could save money. Come to think of it, how is it that I haven't seen or heard about parabolic reflectors with solar panels in the focal point? Right now I've found an article about parabolic troughs that are apparently used to heat old-school fluids instead: https://en.wikipedia.org/wiki/Parabolic_trough https://en.wikipedia.org/wiki/Parabolic_trough
  • detaro 8y ago
    Making solar cells that effectively can handle concentrated sun light is difficult (if they heat up efficiency goes down, ...), so it's easier to just fill an area with solar cells instead of with mirrors pointing at a smaller cell area.
    • aasasd 8y ago
      > if they heat up efficiency goes down So you're saying I should cool them with water and use the steam to move turbines, then I'm golden! \($ ∇ $ )/
      • Klathmon 8y ago
        I'm not sure if you are joking or not, but something like this is still used for solar power generation. https://en.wikipedia.org/wiki/Solar_power_tower https://en.wikipedia.org/wiki/Solar_power_tower
  • wolf550e 8y ago
    Solar panels don't like heat.
  • URSpider94 8y ago
    There are companies that do this. They build tiny solar panels out of materials that can withstand high junction temperatures, and use fresnel lenses to concentrate light onto them.
• bagels 8y ago
  And what of materials that burn at a temperature below the temperature of the Moon? If 100C is the limit, there are materials that burn at much lower temperatures such as Phosphorous (34C).
  • ummonk 8y ago
    I mean you can just take white phosphorous to a warm place and let it spontaneously combust. The use case would be materials that don't burn at the Earth's surface temperature, but do burn at the Moon's peak surface temperature. But you could probably get those hot enough just by rubbing them or something.
• analog31 8y ago
  Here's a thought experiment to consider. Imagine creating a lens to focus the blackbody radiation from a stack of bricks onto a single brick, and heating up the brick. Now focus the light back onto just a small region of the brick pile, and heating up that region, which in turn heats the rest of the bricks by conduction. This in turn increases the amount of heat collected from the pile, ad infinitum or until the brick pile melts. Short of letting the brick pile melt itself, imagine tapping into the excess heat and using it to power an electric motor for a useful purpose.
  • baddox 8y ago
    The problem is that some of the radiation from the moon is not blackbody radiation.
• dekhn 8y ago
  It is a truism: no matter how many times this is explained, some engineer will come up with a complicated system that violates the laws of thermodynamics and refuse to admit that their idea is extremely unlikely. The laws of thermo are some of the best understood and most well-supported physical systems that humans have yet invented. Every time somebody comes up with a perpetual motion machine, it gets shot down because the person who invented it literally ignored all the really well-understood math and physical theory in thermo. it's like there's a brain bug where engineers think they can outsmart 200+ years of scientific progress with a clever arrangement of mirrors.
  • nkurz 8y ago
    It is a truism: no matter how many times this is explained What exactly is the "this" that you refer to? I don't think the issue is that the "engineers" disagree with the physical principles, rather they tend to disagree that the physical principles apply in quite the way that the author claims. Many of the engineers probably believe is that Munroe is correct in claiming that on cannot start a fire with a low temperature blackbody radiation source regardless of the size of one's magnifying glass, but disagree that it is reasonable to consider sunlight reflected by the moon as fitting this model. Presumably, you agree that it's possible to start a fire using a magnifying glass using sunlight on earth. I'd guess you also believe that it's possible reflect the light from small handheld mirror into the magnifying glass, and still start a match, even though the mirror is much lower than the temperature of the sun? While the specular reflection from the mirror is different than the diffuse reflection from the moon, one might note that the words "specular" and "diffuse" don't appear in Munroe's exposition. Would you agree that Munroe's argument would appear to prohibit this behavior? Now assume that the moon was replaced by an equally sized parabolic mirror aimed to be focused on the earth. Would it be possible to light a match using this light if one's magnifying glass was large enough? Which thermodynamical principle am I violating in thinking that it might be possible? And which part of Munroe's argument do I invalidate by making these modifications? Again, my point isn't that Munroe's conclusion is wrong, just that there might be something flawed about the argument he uses to reach that conclusion. This might be a "brain bug", but from the inside it just feels like an attempt to understand truth.
• amluto 8y ago
  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.
  • Dylan16807 8y ago
    > 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. Unconcentrated sunlight is slightly under 1400 watts per square meter. It's equivalent to a temperature of 122C. You can concentrate sunlight coming from the sun, because the sun only fills five millionths of the sky. With a simple lens you can focus hundreds of megawatts per square meter onto a surface. But once you bounce that light off a diffuse surface, whatever concentration you had becomes the new maximum. In your experiment, bathing something in moonlight would max out at 700 watts per square meter. 700 watts per square meter doesn't set things on fire. It can only heat a blackbody to 60 degrees C. Even the full brunt of unaltered sunlight can only bring a blackbody up to 122C. - Treating the moon as a blackbody or not doesn't actually change the equations. The important property is that it diffuses light. It resets your maximum concentration of light, because light that comes evenly from every direction can't be concentrated. (I'm ignoring the part about wavelength filtering because it's confusing and would only make your piece of paper heat up less.)
• skolos 8y ago
  Interesting article. Right in many places. Wrong (possibly) in main conclusion. Entropy argument - correct in the sense that using radiation from black body we cannot use lenses to heat another body to the temperature higher than original. Easy to understand why - the first body has a temperature, radiation has the same temperature, if we apply the radiation to another object it will not heat up more than the radiation's temperature. Also the argument about impossibility of concentrating light into a dot is correct (although even if it were possible we still would not be able to get higher temperature - light would not be energetic enough for that). The important part is - we could concentrate light into a dot only if it consist of parallel rays - i.e. only for an object that is infinitely far away. Moon surface temperature argument is incorrect. A body at 100 degrees Celsius does not radiate in visible spectrum, so the light we see is not produced by Moon's temperature. It is reflected Sun light. So Moon's temperature doesn't matter. Moon surface does absorbs some light, changing spectral composition from about 5.7kK (Sun's surface temperature) to about 4kK. So we should consider moon to be a part of optics not emitter. Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)? I don't have answer for that - need to do calculations. What can be a deal breaker? Remember that Moon is much closer than Sun, so rays come to us even less parallel, so the area into which we can concentrate light reflected from the Moon is even larger than the Sun's, so together with lower intensity of light from Moon we might have trouble achieving the necessary intensity for combustion. However big enough lens probably will work. And yes - I'm a physicist by training.
  • neltnerb 8y ago
    Yeah, this was my immediate thought as well. Yes, you can't concentrate solar energy to heat something hotter than the surface of the sun. But the emission of a 100C blackbody is not visible either, so it's clearly an incorrect hypothesis on it's face that this temperature causes moonlight. It's a bad mirror, not a radiator.
    • tbabb 8y ago
• AlexCoventry 8y ago
  I don't really follow this argument, and I would like to. I think one thing which would help me develop an intuition for it would be to see the calculation of the lens size for heating a one square-centimeter area on the earth to as high a temperature as possible by the light of the moon, and what that optimal temperature is. Anyone reading for whom this is straightforward? Even a description of how to do the calculation would go a long way.
  • Dylan16807 8y ago
    You get the highest temperature by reflecting moonlight from every angle. For a single lens you just want something really big. Make it take up 90+% of the sky from your target. For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get. The calculation way goes as follows: Near earth you get 1400 watts per square meter of sunlight, so if that bounces perfectly off the center of a full moon and gets through the atmosphere with no losses, your target will get 1400 watts per square meter. That's equal to a black body at 122C. After taking into account the spherical shape and atmospheric losses you might get less than half of that, so ambient heat might drown out your results.
    • deleted 8y ago
      [deleted]
    • AlexCoventry 8y ago
      > You get the highest temperature by reflecting moonlight from every angle. I see the thermodynamic principle at work, here, but I don't really understand how it's operating at a mechanistic level. Is it possible to demonstrate that assertion optically?
      • Dylan16807 8y ago
        Take your target and trace a ray in every direction away from every point on its surface. The more of these that hit the energy source, the more energy you're getting. And 100% is obviously the best you can do.
• pontifier 8y ago
  There is a very interesting image in the article. It shows a bunch of light coming into a block and emerging as a beam. It also has a caption saying this is impossible. It struck me as very similar to the setup in this video. https://youtu.be/awADEuv5vWY https://youtu.be/awADEuv5vWY At about 4 minutes in, a nearly identical setup is shown with a beam of light emerging from a block of opaque material using holographic techniques. It seems plausible to me that a specially designed anti-moon hologram could allow reconstruction of the incident light from the sun, thus allowing a fire to be started without violating any law of thermodynamics.
  • pontifier 8y ago
    I think I finally realised what rubs me the wrong way about these xkcd musings, and the related discussions. [Start rant] They take the situation to its absurd conclusion, then quit with a full finality that people take as truth. Further absurd conclusions are ignored, and their word is law. Some more absurd arguments for why you will be able to light a fire with a magnifier and moonlight are as follows: 1: use a pre-magnifier to create a spot on the moon with the same temperature as the sun, then magnify the light from that spot to start your fire. 2: wait a long time... eventually a meteor will hit the moon creating a spot bright enough to focus. 3: wait even longer... Eventually the random mollecular collisions between the wood and the (presumably oxegen rich) air around it will convert it into carbon dioxide and water while serendipitous individual high energy blackbody photons help break it down. 4... insert absurd^4 answer that brings in tunneling effects, or moving mirrors, or some other "impossible" reason that is only impossible because they didn't think of it for you. [End rant]
• proctor 8y ago
  this article seems to refute some aspects of xkcd's answer: https://physics.stackexchange.com/questions/370446/is-randall-munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-th/370600#370600 https://physics.stackexchange.com/questions/370446/is-randal...
  • nkurz 8y ago
    This is indeed a great link (expand the comments on the answer), and shouldn't be languishing at the bottom. Thanks for digging it up!
• michwill 8y ago
  Well, this doesn't look right to me. Imagine that the moon is actually a filter at the path of the sunlight. Sun's temperature is 6000 K. Moon's surface is pretty black: it reflects only 12% of the light. So, effective temperature of the Sun reflected by Moon, considering that thermal radiation is proportional to T^4, is 6000 * .12 ^ (1/4) ~= 3500 K. That's quite enough to light up some fire! Of course, the spectral composition of the light will be not thermal etc, but the estimate should be close enough. Why doesn't the Moon itself heat up like that? Well, the rocks on Earth don't heat up to 6000 K either... I think, it's partly that they are "not surrounded by sun", partly that the Moon is a giant cold heatsink
• empiricus 8y ago
  Reading the comments I found something I don't understand. What is the difference between 1.black body photons and 2.laser photons An object will heat only up to the original temperature of the black body source in the first case, but to an arbitrary high temperature in the laser case...
  • skolos 8y ago
    Black body photons are coming at random times. Laser photons are coherent, so are timed. Think of swing - if you try to push it randomly you'll get it swing as far as hardest push. But if you push periodically, you can swing it very far with small pushes.
    • empiricus 8y ago
      Thanks. So the laser is really special; because it is coherent it can be absorbed and transformed into heat without limit. <br> For random photons, no matter the light intensity, there is a limit temperature of the receiver, which depends on the photon spectrum... <br> From this point of view (energy transfer) are there more than these two kinds of light?
    • MRD85 8y ago
      But won't two hard pushes in a row send you flying? I would have presumed you could find the mean and standard deviation of both the energy of the pushes and the frequency of the pushes and you could build a model to find the expected height.
    • darkmighty 8y ago
      That's not important at all; what's important is that black body radiation has a fixed maximum flux -- its spectrum or lack of coherence isn't why you can't heat another body to a greater temperature. It comes back to etendue, or if you prefer the 2nd law. You could reproduce any fixed black body spectrum (to arbitrary accuracy) from a set of thermal sources and filters (or a set of lasers, LEDs, etc. with random phases) to arbitrary fluxes just like a laser has, and use this light to heat objects to arbitrary temperature. But if the original emission is of black-body type, you cannot -- the flux is given by the quantum mechanical process and a function of local temperature only. From then it follows from etendue conservation you cannot achieve higher temperatures.
• captainsham 8y ago
  You can, if you remember "the scientific principles of the convergence and refraction of light." "The scientific principles of the convergence and refraction of light are very confusing, and quite frankly I can't make head or tail of them, even when my friend Dr. Lorenz explains them to me. But they made perfect sense to Violet." Violet Baudelaire goes on to use the scientific principles of the convergence and refraction of light to set fire to a piece of sail cloth using only moonlight and the lens from a spying glass in "The Wide Window" by Lemony Snicket. It's possible that this book is a work of fiction.
• markbaikal 8y ago
  One has to coat the material to be set on fire with something that has low infrared absorbance (and thus low cooling through heat radiation) but high absorbance for low wavelength (blue/ultraviolet). This is called selective coating. https://en.m.wikipedia.org/wiki/Solar_thermal_collector https://en.m.wikipedia.org/wiki/Solar_thermal_collector Combined with the lens, this might work.
• peterburkimsher 8y ago
  Is it possible to light a fire using sunlight during an eclipse? If not, at what percent totality does it become impossible?
  • BenjiWiebe 8y ago
    Should be possible. Your hot spot under the lens is just as bright as always, it's just smaller, and crescent shaped. As to percent totality, that would depend on how fast your wood/fuel was dissipating heat.
• xupybd 8y ago
  I think They missed something. If the max temp you can get from moon light is 100c that doesn’t mean you can’t start a fire. You just need something that has a very low ignition temp. There must exist something that can start a fire at this temp.
  • e12e 8y ago
    Maybe: https://en.m.wikipedia.org/wiki/Carbon_disulfide https://en.m.wikipedia.org/wiki/Carbon_disulfide Doesn't appear that many substances have auto ignition as low as 100c. If you're hunting for easy to ignite stuff, it might be better to go for low flash point stuff, and strike a spark? Eg gasoline will work in pretty cold environments with a flashpoint of - 43c.
• fpoling 8y ago
  The article reasoning can be shortened to observation that passive optical system does not change the wavelength of photons and to trigger a fire the wavelength has to be short enough. But the conclusion of the article is wrong. The surface temperature of the Moon has very little to do with the wavelength of the reflected photons. Consider a surface covered with ideal tiny mirrors each pointing to random direction. Only a tiny proportion of these mirrors will reflect light from the Sun towards observer. Now consider that 90% of those mirrors are painted black reducing the reflected enrrgy flux by further factor of ten. The Moon is like that. Still the reflected light has original wavelength of the light of the Sun. Collect enough of it and that triggers fire.
  • nshepperd 8y ago
    That's incorrect. It has nothing to do with the wavelength of the light. > Still the reflected light has original wavelength of the light of the Sun. Collect enough of it and that triggers fire. You can't collect enough of it into one place with a passive optical system, because it's been irreversibly scattered by the moon's surface (Read: irreversible increase of https://en.wikipedia.org/wiki/Etendue https://en.wikipedia.org/wiki/Etendue).
    • fpoling 8y ago
      Yes, I stand corrected. Essentially an optical system will bring Moon's surface closer, but even if it brings the surface within 1 cm from the wood, the defused Sun light scattered from that surface is not enough to ignite the fire.
• rdiddly 8y ago
  This explanation does not "click." I can't say whether it's technically wrong; I just note that it lacks some of the features of a successful explanation. First time I've seen a dud from Randall.
• deleted 8y ago
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• kazinator 8y ago
  I'm not buying this. The Moon is only a reflector, not a producer of light. The temperature of the light is that of the reflected source. Moonlight is sunlight, more or less. The fact that the moon reflects poorly is compensated by the size of the gathering lens or mirror. Suppose we build a 100 foot mirror which reflects 10% of sunlight, such that the spectrum remains the same. We could still make a fire with the reflected light, if we just gather 10 times more of it with a larger lens. We could do this in the Arctic, with the mirror's temperature at below zero; the mirror's temperature is irrelevant.
  • username90 8y ago
    What matters is that we have intensity with quadratic falloff. Light from the sun has quadratic falloff based on the distance from the sun and hence can't be bent to become more intense than at the suns surface. Similarly light from the moon has quadratic falloff based on the distance from the moon and hence can't be bent to become more intense than at the moons surface. If you put a mirror on the moon, then the light from it will have quadratic falloff from the reflected sun and not the moon, which is why it can be used to heat to sun level temperatures.
    • kazinator 8y ago
      Yes, since the moon is a scattering reflector, in fact the inverse square dropoff of moonlight is based on the distance from the moon. Note that the article claims that no matter how much moonlight we are able to gather (i.e. we are allowed to overcome the inverse square law however much we want) we cannot create a temperature that will ignite paper.
• tagrun 8y ago
  Physicists here. He's trying to explain everything in terms of a simple blackbody in thermal equilibrium, peacefully radiating its energy away only via thermal photons. That's not the reality of the radiation from the sun or moon. Solar physics is an entire branch of physics, and such simple toy models are not even wrong. Sun doesn't just radiate away its existing energy via thermal photons. First, it keeps burning its fuel via a series of nuclear reactions, which by the way keeps pumping energy into the system, essentially acting like a battery (so there's no perpetual motion here). Second, sun emits photons that are much more energetic than the thermal photons from the surface. Some of the radiation is not thermal, and comes directly from different types of nuclear reactions (which provides signatures regarding the kind of reactions happening in the sun) and various other processes.