It looks to me like GusNSpot wants a tube that is free of atmospheric interference, so its top should be in the exospheric layer of the atmosphere, which would, at a minimum, make it over 300 miles long/tall, more likely five or six hundred miles.
See post #34
Tube goes from the surface of the earth to the top of our atmosphere, say 10,000 miles and is perfectly straight and anything except that which can go the full length of the tube without touching the side can get to the bottom. No reflection from the sides of the tube.
If I could see up the tube, how much of the sun would I be seeing?
Looking at a balloon with a flat black 10" straw, I can see just a little more than the whole balloon at xx distance.
Increase to the 10,000 mile tube, how much of the Sun can I see?
The tube needs to be long enough to get above the atmosphere so it won’t bend the light and we can discount it’s possible effects.
So now we have a tube that is 1 mm inside diameter looking 92,500,000 miles ( pick your number to keep the math simple ) so how much of the sun can I see, as a % or actual number, your choice.
Remember the placement of the tube is on the equator so that the Sun can be directly overhead a particular spot & time where we will get the most light down the tube.
Now even with a tube 1mm inside dia. there will be light that goes the total distance from one side of the mouth of the tube to the other side in the 10,000 mile tube. So we will see some part of the Sun that is larger than 1mm IMO. How much?
Which leads to the third question:
If Sun light has a physical component and we make a 10,000 mile tube that is smaller than the size of that particle, can there be a mathematically, open tube, that light can not get in?
If there is no cross section to any part of light, even a tube with an inside dia. the size of an electron can take in infinite numbers of ray/ray’s as postulated above, then why can’t light go through what we perceive as a solid object?
IIRC, some Neutrinos can get through the earth and not hit anything. ( Getting way out of my understanding with this stuff. )
Are these not math problems with known properties ( as we understand them at this point ) of light and particles, etc.?
I am about to fall off the edge of the world here. Try to keep it kinda simple for me please. :dubious:
Cool, I thought it would need to be longer.
Change the math so it is only 600 miles if it makes the math easier. ( to the umpty umpty power does not really compute in my brain so as a percentage I would understand easier)
Thanks
Back to the OP, I think the right way to put it is that using only flat mirrors, you can’t heat something up hotter than the original source (the Sun) – obviously, because if the object was hotter than the Sun, the energy flux back to the Sun from the object would be greater than the flux from the Sun to the object.
But if curved, focusing, mirrors or lenses are allowed, you can heat the object as hot as you want (assuming perfect, arbitrary sized, lenses/mirrors and arbitrarily large Sun). Or have I confused mysefl again?
Your ten thousand miles is more than a little ridiculous. The radius of the Earth is only about 3550, the atmosphere is a very thin layer on top of that, not three times the size of Earth itself. I think satellites and spacecraft do experience some aero drag at 600 miles, but you can count on no appreciable interference to sunlight. What you do have to worry about (yes, I know it is a hypothetical) is getting a perfectly straight tube that has nearly a million to one length-to-diameter aspect ratio and making it so it cannot bend at all over its whole length.
Quercus, the same restriction you stated for flat mirrors also applies to curved mirrors and lenses.
Chronos, can you explain where the line of reasoning that I (and several other people) are offering fails?
That reasoning is that if you concentrate the radiation from a large section of the sun onto a smaller bit of matter, then in order to have the total energy flow equalize, the small thing will have a higher temperature than the large portion of the sun.
I’m not claiming I’m right. I just don’t understand where the error is.
You can’t concentrate it that much. The concentration works by increasing the amount of angular area of sun that your target sees. Once you’ve gotten the angular area up to “all of it”, there’s no further increase available.
There is obviously (to me) that concentrating light from a larger area to a smaller area means more energy per unit area. In radiating away the energy does the wavelength increase to account for the higher amount of incoming energy that is radiated out? (Isn’t higher frequency EM radiation more energetic than than lower frequency radiation?)
But if “all of it” is concentrated on a single square inch, that square inch has to be radiating as much energy as millions of square miles of the sun. In order to radiate that much heat as black body radiation, it has to be at a higher temperature than the sun. (Assume that we’re using some version of unobtanium that allows us to ignore complications like being vaporized.)
OK, I am surprised this has gone on this long. Only one thing to do, go to the math:
Power output from the sun:
Psun=Asunesuns*Tsun^4
where Asun is the area of the sun, esun is emmissivity of the sun, s is the Stefan–Boltzmann constant, and Tsun is the temperature of the sun.
Now we have the same equation for an object. We are going to have this object floating in space around the earth, because then we get to ignore thermal conductivity, atmospheric absorption, etc.
Pobject=Aobjecteobjects*Tobject^4
Lets take a lens perpendicular to rays coming from the sun and concentrate it on the object. For simplicity, lets assume there is no other radiation heating up the object. The percentage of power coming from the sun that goes to the object is now the percent of a sphere surrounding the sun with a radius equal to the distance from the sun that is taken up by our mirror. Lets call this Ratio.
So, the equilibrium equation for the object is:
AobjecteobjectsTobject^4 = RatioAsunesuns*Tsun^4
Lets make the object out of something with the same emissivity as the surface of the sun so the equation is easier and reduce it:
Tobject/Tsun = (Ratio*Asun/Aobject)^(1/4)
This says that the object will be hotter than the surface of the sun when Ratio*Asun/Aobject is great than one.
For a ball in orbit around the earth, I get that you would need a lens with a diameter 430 times that of your object. So if you put a one inch diameter object at the focal point of a lens with a bigger than 36 foot diameter, you get a temperature greater than that of the sun.
QED
And just how do you intend to get the projected image as small as a single inch?
For people thinking of focusing mirrors or lenses, imagine this. You can take a parabolic mirror and focus the rays of the sun on a single point at the focus of the parabola. The object begins to heat up. But we also use parabolic mirrors for the opposite purpose–put a light source at the focus of the parabola, and the mirror will reflect light out in parallel. This is how flashlights or headlights work, the parabolic mirror in back reflects out a parallel beam of light.
So you’ve got the parabola focusing light on the focus, but the focus is also radiating light back out and sending it back out. Assuming we’ve got a perfect mirror here, the system reaches equilibrium when the spot at the focus is radiating as much energy as it absorbs.
Now, it’s a bit tricky because heat and temperature and energy are three different things. But the spot at the focus can’t get hotter than the sun, because if it was hotter than the sun it would be heating up the sun rather than getting heated by the sun. The more perfect mirrors and lenses you add to focus the rays of the sun on the object, the more mirrors and lenses you’ve got that send that energy right back to the sun. Imagine you’ve got a red hot rock in a red hot stove. The rock can’t get any hotter, no matter how big the red hot stove is. To get the rock hotter you have to increase the temperature of the stove.
Replace the parabolic mirror with a lens, and most of the energy will radiate away from the sun. More explanations are needed to get this through our thick skulls.
Looking at this site we see the equations for how small a spot we can make.
We see that
r = theta*f
where r is the radius of our spot, theta is the biggest off-parallel angle of light coming in, and f is the focal length of the lens.
We see that theta is about .00465 radians (light from the edge of the disk of the sun entering the lens) at the distance of the earth from the sun.
So, if we want to have the spot fully hit our 1 inch diameter target, we need a focal length of 107 inches or less, which means a radius of 214 inches. You would have to use a fresnel lens rather than a simple lens, but I don’t see why it couldn’t be done.
As you can see from the equations in my first post, it is all about area ratios. The power exiting the small object (or at least the portion of it facing the lens) gets distributed over the entire disk of the sun. So, you are taking energy from a very big area and concentrating it into a very small area when heating our object. When our object heats the sun it is taking energy from a very small area and distributing it over a very large area.
In fact, this brings up a really interesting thought experiment: think of two objects made of the same material in space: Object A is big and has a surface area of 10000. Object B is small and has a surface area of 1. Now we get a ingenious set of mirrors and lenses so all the radiated energy from A goes to B and B goes to A.
Neither of these objects produces its own heat (like the sun), so they eventually reach an equilibrium where power in equals power out:
AreaAeAsTA^4 = AreaBeBsTB^4
TB/TA = (AreaA/AreaB)^(1/4) = 10
So, even though the energy goes right back and forth between them, due to the difference in surface areas, the small object will be ten times the temperature of the large object.
OK. Imagine that hot rock in that hot stove. Put any number of lenses or mirrors in the stove in any configuration you please, and the hot rock will be at the same temperature as the hot stove, even if the stove and the rock are glowing white hot.
Every lens or mirror you place works in two directions. A mirror that concentrates photons onto one point also takes photons from that point and disperses them. If you had a collection of mirrors and lenses inside the sun, what would happen? Could you use those mirrors and lenses to make one part of the sun hotter than the rest of the sun?
You are dealing with thermal conductivity here. Radiated heat becomes a lot less important when you have conduction (and convection through the air). But, if you remove those, you definitely could make the rock and the stove different temperatures.
But the rock here isn’t inside the sun.
My physics intuition says that you can’t use mirrors/lenses to heat an object hotter than the source. This is an entropy problem, not an energy problem.
If you use a lens to focus the sun onto a rock, what you’re doing is thermodynamically coupling the sun to rock. They will come into thermal equilibrium; that is, they become the same temperature. Using a bigger lens simply couples them more tightly so that they will come into equilibrium faster.
Take the point at which the sun and the rock (the system) are in thermal equilibrium. To move a system away from thermal equilibrium, work must be on the system. Where is the work coming from to make the rock hotter than the sun? I don’t see any.