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Author Topic: Can you start fires with moonlight?  (Read 20403 times)

Offline Bored chemist

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Can you start fires with moonlight?
« Reply #50 on: 06/11/2011 09:48:35 »
"Right, so it's just a matter of magnification. "
No, as was explained here.
http://www.thenakedscientists.com/forum/index.php?topic=41679.msg371880#msg371880
it's not possible.
 

Offline syhprum

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Can you start fires with moonlight?
« Reply #51 on: 06/11/2011 16:16:37 »
I have sent the enclosed email to Prof R Winston and await his reply

"Dear Prof Winston
 
There has been much discussion on the “Nacked Science” forum to which I subscribe as to whether it is possible to produce a temperature on paper (assuming it is a suitable materiel to absorb the radiation) of 451°C reputably sufficient to ignite it from Moonlight.
You have written that with suitable non imaging optics it is possible to produce with Sunlight a temperature greater than the surface of the sun so it may be possible to produce a temperature of approximately twice that of the surface of the Moon with Moonlight.
I write to you as the leading authority in this field and would greatly value your opinion.
 
Regards
 
P Marchese.
"
 
 

Offline Geezer

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Can you start fires with moonlight?
« Reply #52 on: 06/11/2011 17:37:15 »
"Right, so it's just a matter of magnification. "
No, as was explained here.
http://www.thenakedscientists.com/forum/index.php?topic=41679.msg371880#msg371880
it's not possible.

Yes, but the intensity of sunlight on the Moon is even greater than that on Earth, so we know we can create high temperatures on the Moon, and, as you said, the thermal energy must be re-radiated from the moon. It's "simply" a question of building an optical system to capture enough energy.

The system might have to be quite large and be in orbit, but it would still only be capturing Moonlight.

We can't get more energy out than we capture, but if we concentrate any amount of energy, what is it that limits the temperature we produce if we can apply the energy to a sufficiently small area? 

I'm sure there are all sorts of practical limitations that result in significant parasitic losses, but is there something fundamental that limits the energy density?

Presumably each IR photon at a particular frequency delivers the same amount of energy, and the area of the surface that receives it can be made very small. So, how many photons have to bombard that area in time to produce a certain temperature?

Doesn't the answer lie in understanding how small that area is? (Might it be limited by QM :))

 
« Last Edit: 06/11/2011 18:33:37 by Geezer »
 

Offline syhprum

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Can you start fires with moonlight?
« Reply #53 on: 06/11/2011 19:59:43 »
The black body temperature of the Moon at the distance it lies would be 255°K but due to its slow rotation and lack of atmosphere to distribute the heat I think the illuminated surface viewed from the Earth could well be 100°C hotter in would seem that the optical system would need to double this to ignite the paper which seems well within the bounds of possibility.
Let us hope I get a reply from prof winston and see what he thinks. 
 

Offline JP

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Can you start fires with moonlight?
« Reply #54 on: 06/11/2011 20:26:50 »
"Right, so it's just a matter of magnification. "
No, as was explained here.
http://www.thenakedscientists.com/forum/index.php?topic=41679.msg371880#msg371880
it's not possible.

Yes, but the intensity of sunlight on the Moon is even greater than that on Earth, so we know we can create high temperatures on the Moon, and, as you said, the thermal energy must be re-radiated from the moon. It's "simply" a question of building an optical system to capture enough energy.

A lot of the arguments here are against using image forming optics to focus enough energy.  That is a totally different scenario than trying to use non-imaging optics to do the job.  This was pointed out a couple times earlier, but we keep coming back to imaging optics.

So long as you don't violate the 2nd law of thermodynamics, I don't see a limit on non-imaging optics, though I'm sure there are practical limitations.  After all, you could imagine a perfect non-imaging system taking light from many points on the object and putting all that light onto one point on a piece of paper.

Here's a tangentially related video of a guy using multiple mirrors to focus moonlight onto a photovoltaic cell.  He got ~1 volt out.
« Last Edit: 06/11/2011 20:28:32 by JP »
 

Offline Geezer

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Can you start fires with moonlight?
« Reply #55 on: 06/11/2011 21:22:30 »
"Right, so it's just a matter of magnification. "
No, as was explained here.
http://www.thenakedscientists.com/forum/index.php?topic=41679.msg371880#msg371880
it's not possible.

Yes, but the intensity of sunlight on the Moon is even greater than that on Earth, so we know we can create high temperatures on the Moon, and, as you said, the thermal energy must be re-radiated from the moon. It's "simply" a question of building an optical system to capture enough energy.

A lot of the arguments here are against using image forming optics to focus enough energy.  That is a totally different scenario than trying to use non-imaging optics to do the job.  This was pointed out a couple times earlier, but we keep coming back to imaging optics.

So long as you don't violate the 2nd law of thermodynamics, I don't see a limit on non-imaging optics, though I'm sure there are practical limitations.  After all, you could imagine a perfect non-imaging system taking light from many points on the object and putting all that light onto one point on a piece of paper.

Here's a tangentially related video of a guy using multiple mirrors to focus moonlight onto a photovoltaic cell.  He got ~1 volt out.

Why does it make any difference whether it's imaging or not? Is there some theoretical limit to the size of the image formed? If there is no lower limit, the energy density is also unlimited and very high temperatures should be attainable. (Mind you, I'm not saying it won't require some fancy equipment to actually produce ignition.) 

Of course an imaging system will have losses, but are we saying it won't work because of engineering problems, or because of some fundamental constraint?

There are several competing things going on here. As far as we know it really is impossible to amplify energy by any means, but it is not hard to concentrate energy to produce very high temperatures with imaging systems and other means.

Moonlight does have low energy density, but if I can capture enough of it by imaging means, other than losses, what's to stop me using it to ignite a tiny amount of some flammable material?
 

Offline Bored chemist

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Can you start fires with moonlight?
« Reply #56 on: 06/11/2011 21:32:41 »
"Is there some theoretical limit to the size of the image formed?"
Yes.
To get a small image you need a lens with a short focal length, but that requires a small radius of curvature so the lens can't be very big so it cannot gather much energy to focus into that small image so the energy density is still low.

With non imaging optics the answer is, in principle, trivial. A big shiny funnel will do the job.
 

Offline Geezer

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« Reply #57 on: 06/11/2011 23:15:21 »
To get a small image you need a lens with a short focal length, but that requires a small radius of curvature so the lens can't be very big

What's the maximum size?
 

Offline JP

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« Reply #58 on: 06/11/2011 23:39:33 »
Ok, I finally got a rigorous argument.  It's basically burning's, but I had to think about it more.  Recall that the radiance is the spread of light energy over direction and position. 

You can think of the energy from the sun as being evenly spread out over all possible directions through a sphere located around the sun.  This is a consequence of it being a black body, and the even spread is important for what follows.  So the radiance from the sun has some footprint in position (the size of the sun) and takes up all directions.

In terms of the radiance, the intensity at a point in space is just an integral of the radiance function over all directions.  This is rather sensible, since a piece of paper placed at that point is going to intercept all directions of light passing through that point.

Now, if you place an imaging system on the earth, it will capture some subset of the sun's radiance in both position and direction.  Imagine the radiance at the sun was plotted: it would be a constant value over direction and position.  The imaging system just cuts a chunk out of this.  The radiance passing through the imaging system has a constant value over that chunk and is zero for all positions outside the imaging system (obviously) and zero for all directions not captured by the system.

If you ask what an imaging system can physically do while reproducing a clear image: it can squeeze the radiance in position by spreading it in direction or it can squeeze it in direction by allowing it to get wider in position.  All this squeezing does nothing to change the value of the radiance at each point as it's plotted in position-direction (another consequence of imaging optics).  Now, at the most, you can spread the radiance to a hemisphere in direction, the same as you would capture at the surface of the sun.  Since the value of the radiance at each point hasn't changed, the intensity at every point is at most proportional to this angular spread (the hemisphere).  Since this angular spread is the same as at the surface of the sun, the total intensity can't be greater than at the surface of the sun.


--------------------------------------

So all this was a really detailed (and probably confusing) argument that boils down to: for an image of a black body radiator, intensity is proportional to size of the range of directions of light projected onto the piece of paper by your imaging optics.  This is due to the way imaging systems work and the fact that a black body radiates light equally into all directions.  The widest possible range of directions is a hemisphere, but this is the same possible range as was emitted by each point on the sun, so the intensity can't be higher than the sun itself.

For a non-black body, the distribution of the radiance doesn't have to be equal over all directions.  A laser beam can be concentrated over a tiny range of directions.  This is why you can focus a laser beam down to a very hot point--far hotter than at the laser source. 

Imaging optics place very strict requirements on how you can reshape the radiance, even if it isn't a black body.  All this has to do with keeping the radiance constant at each point, while moving those points around in position/direction space.  If you don't care about forming an image, then you're only constrained by conservation of energy which says that the total integral of the radiance over both position and direction remains constant unless you lose light energy somehow.  (You're also limited by uncertainty relationships if you squeeze things down to the scale of wavelengths, or try to have all the light heading in a very narrow range of directions.)
 

Offline Geezer

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Can you start fires with moonlight?
« Reply #59 on: 07/11/2011 02:29:55 »
Ok, I finally got a rigorous argument.  It's basically burning's, but I had to think about it more.  Recall that the radiance is the spread of light energy over direction and position. 

You can think of the energy from the sun as being evenly spread out over all possible directions through a sphere located around the sun.  This is a consequence of it being a black body, and the even spread is important for what follows.  So the radiance from the sun has some footprint in position (the size of the sun) and takes up all directions.

In terms of the radiance, the intensity at a point in space is just an integral of the radiance function over all directions.  This is rather sensible, since a piece of paper placed at that point is going to intercept all directions of light passing through that point.

Now, if you place an imaging system on the earth, it will capture some subset of the sun's radiance in both position and direction.  Imagine the radiance at the sun was plotted: it would be a constant value over direction and position.  The imaging system just cuts a chunk out of this.  The radiance passing through the imaging system has a constant value over that chunk and is zero for all positions outside the imaging system (obviously) and zero for all directions not captured by the system.

If you ask what an imaging system can physically do while reproducing a clear image: it can squeeze the radiance in position by spreading it in direction or it can squeeze it in direction by allowing it to get wider in position.  All this squeezing does nothing to change the value of the radiance at each point as it's plotted in position-direction (another consequence of imaging optics).  Now, at the most, you can spread the radiance to a hemisphere in direction, the same as you would capture at the surface of the sun.  Since the value of the radiance at each point hasn't changed, the intensity at every point is at most proportional to this angular spread (the hemisphere).  Since this angular spread is the same as at the surface of the sun, the total intensity can't be greater than at the surface of the sun.


--------------------------------------

So all this was a really detailed (and probably confusing) argument that boils down to: for an image of a black body radiator, intensity is proportional to size of the range of directions of light projected onto the piece of paper by your imaging optics.  This is due to the way imaging systems work and the fact that a black body radiates light equally into all directions.  The widest possible range of directions is a hemisphere, but this is the same possible range as was emitted by each point on the sun, so the intensity can't be higher than the sun itself.

For a non-black body, the distribution of the radiance doesn't have to be equal over all directions.  A laser beam can be concentrated over a tiny range of directions.  This is why you can focus a laser beam down to a very hot point--far hotter than at the laser source. 

Imaging optics place very strict requirements on how you can reshape the radiance, even if it isn't a black body.  All this has to do with keeping the radiance constant at each point, while moving those points around in position/direction space.  If you don't care about forming an image, then you're only constrained by conservation of energy which says that the total integral of the radiance over both position and direction remains constant unless you lose light energy somehow.  (You're also limited by uncertainty relationships if you squeeze things down to the scale of wavelengths, or try to have all the light heading in a very narrow range of directions.)

Ah! So does this mean that the image can't be any hotter than the object, no matter how small the image is?
« Last Edit: 07/11/2011 02:32:14 by Geezer »
 

Offline JP

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Can you start fires with moonlight?
« Reply #60 on: 07/11/2011 04:09:45 »
That's the net result of it, yes, provided you're using imaging optics on a black body radiator.

Unfortunately, I can't come up with a rigorous argument that's also simple off the top of my head.
 

Offline imatfaal

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Can you start fires with moonlight?
« Reply #61 on: 07/11/2011 10:31:26 »
I liked Dominic's explanation on the podcast.  from the papers point of view the lens is increasing the area of the sky that is filled by sun - the most you could ever fill is a hemisphere.  A hemisphere filled with the sun would approximate the surface of the sun temperature, a hemisphere filled by the moon will approximate the temperature of the surface of the moon.  so even with perfect optics you are limited to the temperature of the surface.
 

Offline syhprum

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« Reply #62 on: 07/11/2011 18:01:33 »
It would be interesting to talk to a designer of hydrogen bombs, they use a big shiny funnel as suggested by Bored Chemist to channel the X radiation generated by a fission bomb onto the Lithium Hydride to produce the fusion explosion.
 

Offline Bored chemist

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« Reply #63 on: 07/11/2011 18:46:00 »
I think it would be interesting to talk to them because they are dead.
It's very difficult to reflect Xrays and so a "shiny " object of any sort (unfundibular or otherwise) is a bit of a non starter. It's more a matter of putting the deuterium in a box with an  atom bomb. When the bomb goes off the deuterium gets very hot very quickly.

As far as I can tell, the biggest lens you can have with a given radius of curvature is a sphere, but I might be talking balls there.
 

Offline syhprum

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« Reply #64 on: 07/11/2011 19:01:31 »
That was Tellers original Idea but it was found to be unsound, compression is what is required and the X radiation must produce sufficient radiation pressure on the fusion materiel to start the fusion process

http://www.progressive.org/images/pdf/1179.pdf

They are still desgning bigger and "better ones" although the origonal inventors are no longer with us.


« Last Edit: 07/11/2011 19:05:24 by syhprum »
 

Offline Bored chemist

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« Reply #65 on: 07/11/2011 21:00:19 »
You might talk me into believing that they made the famed "box where all the sides are mirrors" of which many are discussed on sites like this, but I really don't see a funnel here.
http://en.wikipedia.org/wiki/Teller%E2%80%93Ulam_design
 
 

Offline syhprum

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« Reply #66 on: 07/11/2011 21:40:35 »
The reflective optics that that focus the the X rays onto to the materiel to undergo fusion have to operate at a very shallow grazing in angle to avoid penetration as far as I know they are of multi layer constuction rather like laser mirrors and must be of a funnel like shape.
Detail blueprints of nuclear bomb design are rather hard to come by but that has always been my idea of how they are constructed. 
 

Offline Bored chemist

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« Reply #67 on: 07/11/2011 22:05:33 »
I'm still not convinced.
A funnel has a definite "direction" to it but you want the deuterium compressed fairly symmetrically. A funnel would tend to blast it sideways and you want to blast it from all sides at once. Also there are definite errors (perhaps deliberately) in that report. You can't make a successful fission device as a trigger using just TATNB. you need more than one explosive. Also, they say the shell is a reflector made from uranium. You can't realistically make the equivalent of "dielectric mirrors" with just one material and also Xray mirrors, as you say, only work at grazing incidence. Some of the Xrays in that system have to be turned through 180 degrees.
I think this the the nuclear physics use of the word reflector meaning "if we put enough stuff in the way, it will scatter at least some radiation back the way it came."
Wiki is a bit more polite about it, but says the same thing for neutron reflection.
http://en.wikipedia.org/wiki/Neutron_reflector


Anyway, on the basis that "Detail blueprints of nuclear bomb design are rather hard to come by" and I, for one, would like to keep it that way, perhaps we should get back to the OP.

You could start a fire with the energy reflected from the moon. You could use a solar cell, charge a battery with it ( it would take ages) then discharge the battery into a thin wire so it heated up and started a fire.

 

Offline syhprum

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« Reply #68 on: 07/11/2011 22:48:03 »
Thats how I would do it but thats cheating.
 

Offline Geezer

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« Reply #69 on: 08/11/2011 05:55:17 »
er, a new thread perhaps? Mind you, we might not want to go too far down this particular road in open forum. PM might be better.
 

Offline yor_on

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Can you start fires with moonlight?
« Reply #70 on: 18/11/2011 17:04:51 »
Maybe call it "From Moon light to making your own fire, or if so inclined, hydrogen personnel nuclear activator, In ten easy steps."

NSA will be delighted.
 

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Can you start fires with moonlight?
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