0 Members and 1 Guest are viewing this topic.
"Right, so it's just a matter of magnification. "No, as was explained here.http://www.thenakedscientists.com/forum/index.php?topic=41679.msg371880#msg371880it's not possible.
Quote from: Bored chemist 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#msg371880it'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.
Quote from: Geezer on 06/11/2011 17:37:15Quote from: Bored chemist 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#msg371880it'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.
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
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.)