
Can you use a magnifying glass and moonlight to light a fire? (2016) - btrask
https://what-if.xkcd.com/145/
======
skolos
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.

~~~
nshepperd
The conclusion is correct. It doesn't rely on the moon being a black body
radiator, but on etendue, which shows that the most you can do with a system
of lenses and mirrors is to create an environment (on earth) as bright as the
environment on the moon. The fact that rocks on the moon's surface only reach
100°C shows that an environment like that is not bright enough to light a
fire.

~~~
skolos
Trying to follow your argument. Suppose we put absolutely reflective mirror on
the orbit. Since its absolutely reflective its temperature will be 0 Kelvin
(in theory, very low in practice - there is a reason solar bound spacecrafts
are covered with reflective surfaces). So we would not be able to use the
light it reflects to heat anything?

~~~
ummonk
Specular reflector preserves the etendue of the light rays that fall on it.
Diffuse reflector (such as the moon) does not; it becomes a new light source
instead, resulting in higher etendue.

In short - perfect reflector preserves etendue, but imperfect does not.

~~~
fpoling
The question is to what extend the moon surface is imperfect reflector. For
example, if the mirror has a hole, it will be an imperfect reflector, but you
can compensate for it using a bigger lense.

~~~
darkmighty
By imperfect it's sufficient to be diffuse. You can verify the Moon is
diffusive to a good approximation because of the approximately uniform
appearance of the full Moon.

The idea is that you can "organize" or "revert" any ray bundle from a system
of non-absorbing lenses and specular reflectors, but if your reflector has
billions of tiny irregularities it's not viable to build such a system (it's
equivalent to an ideal diffuser, in which light is isotropically reflected).
The ideal diffusion process is clearly not reversible by itself: if you shine
a beam onto a diffuser it spreads the light; if you expose it to the same
light (with reversed directions), it again diffuses it instead of reverting to
the original beam. In theory again the physical laws of electromagnetism are
time reversible, but in practice the effort to revert some systems might be
too demanding (you can even do better -- see Maxwell's demon); manipulation of
physical apparatus and information acquisition/manipulation itself has a cost
that surpasses any gains.

------
parliament32
> 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?

~~~
nickparker
You've found the right question to ask.

Your mirror in sunlight works because the reflectivity or albedo of the mirror
is very high relative to whatever target you're lighting on fire.

In a magical closed system where radiative heat transfer was the only factor,
objects of differing reflectivity would eventually reach temperature
equilibrium through black body radiation.

We aren't interested in closed systems though. The moon's temperature is set
by the equilibrium between incident solar radiation and black body emission,
most of which flies off into deep space making the system very open. Just like
your mirror's temperature is the equilibrium of incident radiation, black body
cooling, and convective cooling in Earth's atmosphere.

If the moon had a high reflectivity and/or a powerful cooling mechanism like
convection, its equilibrium temperature would be far lower than the
temperature of its emitted + reflected light. Unfortunately the moon's albedo
is just 0.12 and black body radiation is all it's got, so the modest
difference between its temperature and that of its light isn't enough to start
a fire.

~~~
btilly
Really?

According to the Stefan-Boltzmann Law, the energy put out in blackbody
radiation is proportional to temperature to the 4th power. Therefore something
that is 5000 degrees reflected off of an object with albedo 0.12 is putting
out 0.12 times the energy it originally did, while something that is 2500
degrees only puts out (1/2)^4 = .0625 times as much. So the "temperature of
the Moon's light" should be more than hot enough to light something on fire if
it is focused right.

As a sanity check, compare how much light the moon puts out as a black body in
shadow with what it reflects from the Sun. As another sanity check, compare
how bright the Moon is versus a fire.

What am I missing here?

~~~
tbabb
The angular concentration of the light.

The moon is diffuse, so an incoming ray of sunshine is spread by the optically
rough surface of the moon from an incident solid angle of 6 * 10^-5 steradians
out into 2 pi steradians of the night sky, or a reduction in angular
concentration by a factor of about 100,000 (totaling ~1 million after the
albedo is accounted for).

It is like you are looking at the sun through a mirror so rough that the image
of the sun is blurred over literally half the sky. Because this process does
not create new photons, the blurred image must be far, far dimmer. This
circumstance corresponds to the "most we could do" with lenses and mirrors
focusing the moon, which is to fill the sky with an image of the moon/"blurred
sun".

Unconcentrated moonlight corresponds to the same picture, except we only see a
"cutout" disk of this blurred sun-image which is the size of the moon in the
sky. Our crappy moon-mirror does not fill our vision, it is a porthole letting
through only a tiny fraction of the blurry sun-image.

And of course if you imagine yourself as the ant under the magnifying glass,
with your entire sky filled with moon, there is no way you could spontaneously
become hotter than your moon-y surroundings.

~~~
colanderman
This is the primary reason. You can no more start a fire with sunlight
reflected off the moon at night than with sunlight reflected off a sheet of
paper during the day. (In fact you can do better with the sheet of paper,
because you could in theory surround it with lenses to recapture the diffuse
light.)

------
crazygringo
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
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
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.

------
DenisM
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?

~~~
08-15
> I can cover the entire surface of the earth with solar panels to harvest the
> moon light

Are you sure about that? First off, take an off-the-shelf solar panel, point
it at the moon in the middle of the night. You get a grand total of nothing.
Okay, it was a cheap panel, that might not generalize to anything.

But more importantly, by the argument laid out in the article, your solar
panels _cannot work_ in moon light. (Or maybe they work at horrible
efficiency, because the moon is a bit warmer than the earth.) I'm not sure I
buy that argument; maybe you should run the experiment.

~~~
08-15
I'll repeat for the reflexive downvoters:

Solar panels will work in moonlight _if and only if_ you can make fire from
moonlight with a magnifying glass.

The answer depends on how good a mirror the moon is. It calls for a real
experiment, not a thought experiment. I don't really know which way it will
go.

~~~
DenisM
You're not addressing the essential part of my comment. Let me clarify:

I put in place 1000 solar panels and aim them at the sun thus procuring 200kw
of power. Next I use it to power an arc welder, producing 10,000 C of heat.
Therefore I have used a 6,000C sun to produce 10,000C on earth.

Would a similar setup work with the moon? I don't know. That was not my point,
my point was that it certainly is possible to produce higher temperature at
the target than it was at source.

~~~
URSpider94
Monroe’s explanation only applies to optical systems, ie reflecting and
focusing light, which is a thermodynamically reversible process (yes, even
with dust. Dust doesn’t make the lens thermodynamically irreversible, it just
scatters and absorbs some of the light so it doesn’t reach the lens. The
irreversibility is a property of the lens). Solar panels are converting light
energy into electrical energy, which is not an irreversible process - it
creates entropy, and therefore the rules are different.

------
IgorPartola
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
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.”

~~~
IgorPartola
Right. That’s what I am saying. Using the original setup of the problem, you
cannot use a device to store energy.

------
Analemma_
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.

~~~
CydeWeys
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
_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.

~~~
ummonk
>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.

A mirror does specular reflection and thus conserves the etendue of the
sunlight. You're concentrating the image of the sun in the mirror, not light
from the mirror itself.

The moon in contrast is mostly a diffuse reflector - it scatters most of the
light that falls on it (and absorbs and re-emits most of the rest), so it is
effectively a new light source.

~~~
nkurz
Yes, I can agree with that. The problem for me is fitting it into Munro's
argument. He says (in boldface): "You can't use lenses and mirrors to make
something hotter than the surface of the light source itself".

This is true for a black body, but is it always true for an object being
illuminated by another? I'm don't know that we can consider a diffuse
reflector with a surface temperature of 100C as being equivalent to a black
body with temperature of 100C. I think his conclusion is likely true (the moon
is too dim to start a fire even with a really big magnifying glass) but I
don't think he's right to point to the surface temperature of the moon as
being the evidence of this conclusion.

Assume the sun was much brighter, so that ignition on earth is possible with a
sufficiently large magnifier. Presumably if the moon was the same, this would
mean the surface temperature of the moon was much higher. Now change the moon
to be more heat conductive (causing the surface temperature to drop due to
more heat loss on the dark side), and more reflective (causing the surface
temperature to drop further due to less absorption). I'd guess that if you
tweak the parameters sufficiently, you could end up with a surface temperature
low enough that Munroe's argument would say that ignition is impossible, even
though we've increased the intensity of the moonlight over our baseline.

How does Munroe know that we aren't in that second regime? I don't think
there's enough information in his argument to distinguish. Alternatively
stated, we know that there is some current temperature to which we can heat an
object using concentrated moonlight. We also know that if we can change the
shape and composition of the moon, we can reduce the surface temperature
without reducing the intensity of moonlight. Unless there is some limit to the
effectiveness of the heatsink that we can put on the moon, I think this means
there is some possible arrangement that violates the assumption that the
surface temperature must always exceed the temperature achievable with a
magnifying glass.

(Thanks for helping me to puzzle this out)

~~~
ummonk
An excellent point!

 _> Now change the moon to be more heat conductive (causing the surface
temperature to drop due to more heat loss on the dark side),_

Yeah, although this isn't too big in the case of the moon (unlike the Earth,
it doesn't have an atmosphere and doesn't rotate rapidly, so there isn't too
much redistribution of heat across its surface), it is definitely something
that would confound the calculations. We'd still be able to just look at the
effective temperature of light falling onto the moon and that would limit the
temperature that we could light the object up to. But we wouldn't be able to
use a direct measurement of the temperature of the surface of the moon.

 _> and more reflective (causing the surface temperature to drop further due
to less absorption)._

To the extent that it's a gray body (and most objects are approximately
graybodies), this wouldn't actually lower the temperature. Absorptivity < 1
causes it to absorb less energy from the light, but for a gray body emmisivity
equals absorptivity so it also radiates out less light too, and you actually
end up reaching the same equillibrium temperature as fully absorptive black
body.

~~~
nkurz
Thanks for the kind response. I now see that the final average core
temperature should remain the same, but I'm less sure about the surface
temperature. There are a lot of complex processes involved, and I'm not sure
that one can conclude that everything cancels out to keep the answer constant.

I respond late to offer a link (that's hidden on the second page of this
thread) that I think presents that argument I was trying to make better than I
managed to: [https://physics.stackexchange.com/questions/370446/is-
randal...](https://physics.stackexchange.com/questions/370446/is-randall-
munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-th/370600). I
thought the comments on the answer were a helpful reframing of the problem.

------
tagrun
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.

------
dang
Discussed at the time:
[https://news.ycombinator.com/item?id=11211454](https://news.ycombinator.com/item?id=11211454)

------
zeristor
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
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)

------
amluto
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
> 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.)

~~~
amluto
> 700 watts per square meter doesn't set things on fire. It can only heat a
> blackbody to 60 degrees C.

> (I'm ignoring the part about wavelength filtering because it's confusing and
> would only make your piece of paper heat up less.)

I’m afraid you’re ignoring the interesting bit. You say that 700 W/m^2 doesn’t
set things on fire. This is not true. Sure, 700 W/m^2 applied to some target
that is allowed to radiate its own blackbody light out to the sky won’t get it
very hot, but that’s not what I’m suggesting. I’m suggesting that you insulate
the target very well so that its blackbody emissions don’t escape, but you let
in the short-wavelength moonlight. Thermodynamics requires that you also let
_out_ the short wavelength blackbody emissions, but those are negligible until
the target gets very, very hot.

This effect isn’t science fiction — it’s just the greenhouse effect,
amplified. Greenhouses (the glass ones and the atmospheric ones) exploit the
fact that sunlight doesn’t match the _Earth’s_ blackbody spectrum, so a filter
(glass or gaseous) can allow incoming radiation in but trap most outgoing
radiation.

In effect, I’m suggesting that a very good greenhouse plus some lenses could
get hot enough to start a fire.

~~~
Dylan16807
I see. I couldn't quite follow which parts you were saying to insulate, and
your mention of "half the temperature of the sun" lead me to misinterpret.
That's fine then, I think. Not an expert on wavelength filters.

------
fpoling
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
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
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.

------
michwill
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

------
robertelder
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."

~~~
landryraccoon
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.

~~~
robertelder
If the size of the penny was reduced to just a few atoms, wouldn't changing
the number of atoms in the penny affect the rate at which the penny itself
could emit radiation through blackbody emission? If so, I would expect there
would be a size at which the penny could be small enough that it's temperature
could grow arbitrarily large once the heat flux in = heat flux out from the
penny.

~~~
schoen
I don't know very much thermodynamics, but I'm not sure that thermodynamic
temperature is well-defined for a system consisting of "just a few atoms".

------
hinkley
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.

~~~
ummonk
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
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.

------
IgorPartola
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?

~~~
Dylan16807
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.

------
adammunich
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
You are describing a parabolic concentrator.

------
aasasd
> 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
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
> 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
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)

------
trevyn
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)

------
xupybd
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
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.

------
pontifier
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
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]

------
kazinator
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
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
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.

------
captainsham
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.

------
rdiddly
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.

------
analog31
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
The problem is that some of the radiation from the moon is not blackbody
radiation.

------
bagels
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
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.

------
markbaikal
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.

------
AlexCoventry
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
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.

~~~
AlexCoventry
> 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
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.

~~~
AlexCoventry
From an optical perspective, how does the argument dismissed at the start of
the OP break down? Is there some reason the moon's light, gathered from a lens
covering hundreds of acres, carries insufficient energy to light a fire in
concentrated form?

~~~
Dylan16807
The optical argument goes like this: Measure the brightness (in energy per
square meter) right at the moon's surface. No matter what you do with lenses,
you can't concentrate moonlight beyond this level.

You can make an enormous lens where all the energy coming off a particular
acre of moon goes through it. But you can't focus all of it onto a spot
smaller than an acre, no matter what you do.

~~~
AlexCoventry
Thanks for the explanations.

The current top comment here seems to agree with my original intuition,
though.

[https://news.ycombinator.com/item?id=18739120](https://news.ycombinator.com/item?id=18739120)

~~~
Dylan16807
As a few people pointed out, the reflection off the moon ruins the étendue, so
there's no concentrating it beyond what all the rocks on the surface already
experience.

~~~
AlexCoventry
Thanks again for your help. I think that led me in the right direction.

This is the optical argument for maximum concentration given conservation of
étendue (the same page has an optical argument for the conservation):

[https://en.wikipedia.org/wiki/Etendue#Maximum_concentration](https://en.wikipedia.org/wiki/Etendue#Maximum_concentration)

The angle subtended by the moon is approximately 0.54 degrees, or about 0.01
radians, so the maximum concentration factor is about 10,000. The moon
provides about 0.1 lux of illumination, so the maximum illumination you can
achieve by optics is about 1000 lux. The sun subtends about the same angle,
and we receive 30,000-100,000 lux from it, and the maximum illumination you
can achieve from concentrating it optically is about 1B lux.

I'm willing to believe that you can't light a fire from the moon, if the
intensity's a million times lower.

------
empiricus
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
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.

~~~
darkmighty
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.

~~~
empiricus
Interesting, this makes more sense. Now I have to read more on the
conservation of entendue.

------
ohiovr
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.

------
dekhn
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
_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.

~~~
dekhn
The this I refer to is "the principle of etendue":
[https://en.wikipedia.org/wiki/Etendue](https://en.wikipedia.org/wiki/Etendue)

and I agree it's counterintuitive and it's OK for people to come up with
ideas, but when you hit the point of "hey, the thermo people have a nice
collection of proofs demonstrating this, and it fits very well with the
underlying theory, oh, and if you do manage to violate etendue, you could
probably build a perpetual motion machine", if you willingly continue to argue
and get shot down, it's time to go back and re-read the books.

BTW, what's your obsession with Monroe? What he's referring to is a scientific
phenomenon, Monroe is just a science popularizer, and if he got the details
wrong- well, the point of xkcds like that is more to inspire people with
ideas, than get the exact details right.

~~~
nkurz
_The this I refer to is "the principle of etendue"_

Great principle. I agree with it, and think that most of the people offering
objections here do as well. Our question is whether it's it's being applied
correctly in this case. For me, the sticking point is whether surface
temperature can be used as a proxy for the brightness we care about.

 _BTW, what 's your obsession with Munroe?_

I have none. He's a great explainer, probably occasionally gets details wrong,
and (so far as I can tell) is a positive force for spreading scientific
understanding. My "obsession" seems to be that I have poor tolerance for
overly broad smackdowns of critics. If you are going to tell someone else that
they are wrong (as opposed to Munroe who is trying to explain what he believes
is true) I think you have a higher obligation to get all the details right. I
presume I'm sensitive to it because I'm often on the receiving end.

How about you? Why does it bother you so much that some people say that
Munroe's argument is logically flawed?

~~~
dekhn
I don't have any problem with people complaing about Monroe's argument. I am
fairly certain he wrote what he did in consultation with optics engineers, and
then modified it so it was intellectually comprehensible by a nerd-but-not-
optical engineer audience.

~~~
nkurz
Thanks for the response. If you are still interested in this, there's an
useful link on the second page of this thread to a discussion of the same
question: [https://physics.stackexchange.com/questions/370446/is-
randal...](https://physics.stackexchange.com/questions/370446/is-randall-
munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-th/370600). If
you expand the comments on the answer, the back and forth presents most of the
argument I would make, in a clearer form than I could muster.

------
wefarrell
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.

------
peterburkimsher
Is it possible to light a fire using sunlight during an eclipse?

If not, at what percent totality does it become impossible?

~~~
BenjiWiebe
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.

------
dooglius
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.

------
newnewpdro
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
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, ...

------
proctor
this article seems to refute some aspects of xkcd's answer:

[https://physics.stackexchange.com/questions/370446/is-
randal...](https://physics.stackexchange.com/questions/370446/is-randall-
munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-
th/370600#370600)

~~~
nkurz
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!

------
FiatLuxDave
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_...](http://www.lumec.com/newsletter/architect_06-10/the_sun_the_moon.html)
or [https://physics.stackexchange.com/questions/244922/why-
does-...](https://physics.stackexchange.com/questions/244922/why-does-
moonlight-have-a-lower-color-temperature) . 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
Randall has a response to this argument in the article.

~~~
thatcherc
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...](https://www.lcas-
astronomy.org/articles/display.php?filename=albedo_effects&category=observing)
, also learned this in an astronomy class

------
detaro
EDIT: I'm not sure about this anymore.

~~~
jstanley
I think TFA refutes your argument, but I'm not smart enough to know for sure.
Can you please expand on this?

~~~
randyrand
What if the sun was a bunch of parallel laser beams? Then you could focus them
all into a single point.

~~~
jstanley
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
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
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)

------
dr_orpheus
This is great, I always love the xkcd "What if" explanations. I even have the
book sitting on my desk next to me.

------
kayerov
You can with magnifying glass bigger 2.3 million times

