Divergence and lack of coherence are two very different things (as I fully realized only after I typed up my message, I guess).
Divergence is a result of the angle. If you’re producing light from a local point-source, you have to work very very hard to make sure the angle of the departing rays is as close as you can make it, and you’re still not going to get anywhere even remotely close to 20 feet divided by 149,597,871 km. That’s where all the insane dropoff in the examples you’re talking about is coming from. The rays from the sun, though, are effectively parallel by the time they reach the earth to points 20 feet separated.
The inverse-square law is a result of the power in the beam spreading out over a larger area and spreading out its energy output over a wider area. It’s just a way of expressing that if the beam has spread itself out from hitting 1’x1’ into hitting 10’x10’ at a distance 10 times greater, each square foot of the target will now only get 1/100 of the energy. It won’t get weaker in total, without being absorbed by something along the way; that would violate conservation of energy. In this case the beams are parallel, the target is still 20’x20’ plus some tiny tiny fraction, there is a little bit of absorption by the atmosphere but not enough to make it not bright. The sun’s light goes through the atmosphere and it’s still bright (somewhat brighter if you’re on a mountain or in space, with a lot more UV, but not like night and day.)
I don’t see that coherence fits into this particular part of it in any way; as far as I know, we use lasers for this type of purpose because of their low divergence and the coherence has nothing to do with it. The rays originally from the sun have no coherence and they still manage to make it all the way out here.
Coherence is the key here, I assure you. Incoherent light is subject to the inverse square law in a way that lasers, which demonstrate coherence, are not. Lasers are coherent and collimated, and as such don’t interfere with one another and are parallel contributing to the laser’s ability to remain focused over long distances without spreading out significantly. This collimated nature of laser beams is a direct result of their high degree of spatial coherence, allowing them to maintain intensity over distances where a non-coherent light source would have dispersed according to the inverse square law. You arent reflecting coherent, in-phase, collimated from mirror, even if the suns rays strike the mirror parallel.
Lets assume each of the mirrors reflects 850 watts. The distance to the ISS is 408,000 meters.
The energy reflected by one mirror as received by the ISS is subject to the inverse square law (because it is incoherent).
E = (850 watts) / (4pi408000m)2,, or about 4.06x10 −10 watts/m2
A 5 milliwatt, off the shelf laser pointer with a beam divergence of 1.5 millirads would deliver approximately 4.25x10-9 watts/m2, or about 10x as much energy as the 850 watt mirror.
You can not melt a spy satellite with mirrors. You might be able to with lasers. A laser will be approximately 8.9x106 times as power effecient at getting light from earth to the ISS as a mirror would be. This is directly due to the properties of laser light, specifically coherence and collimation, which make it not subject to the inverse square law.
You’re confused, sir. Light from the sun is collimated, yes, i.e. parallel rays. The correct equation if you’re going to apply the inverse square law is:
Where are you getting this number from? The number you need to be using is the distance from the earth to the ISS, 408,000 meters.
Second, your formulation of the inverse square law is incorrect, in that you are missing the 4pi component, but in the grand scheme of distances we’re looking at, its negligible. It also looks like you may have gotten the order of operations wrong.
Third
149,597,971 km^2 * 149,597,871 km^2
The hell even is it that you think you are representing by these numbers? What is it you think you are saying?
fourth
You can believe or not; I’ve explained it as clearly as I know how.
I will take a bet from you that the energy arriving at the ISS from a laser pointer you or I could purchase off Amazon, so consumer grade, is more than the energy that would arrive at the ISS from a concave mirror that would be used at a solar generating station.
A perfectly flat mirror, exactly like one of the ones in the OP generating station picture. Both are aimed perfectly on target, and the mirror is reflecting light from the sun on a sunny day. With those caveats I’ll bet $1,000. I’m happy with any university physicist or physics professor to be the judge, or Randall Munroe, or make a proposal of some other person if neither of those are acceptable to you. A lower amount of money is also fine if you’re not comfortable with $1,000.
Yeah but what they are not accounting for is that the light actually has to come off the mirror. You can demonstrate this with a hand mirror that the illumination spot gets larger quickly at a distance. Take a mirror and find the sun. Send a reflection to the wall nearest you. Then send the reflection to a wall further away. The reflection on the wall further away is larger and therefore, the energy more spread out. The light coming off the mirror is not perfectly parallel as it had to pass through the atmosphere, then interact with the surface of the mirror. We do use mirrors for calibration in satellite remote sensing, but you will get far far far far more power arriving at something like the ISS coming from a much less powerful laser over such a distance. If we controlled by wattage, a laser will absolute crush a mirror in its ability to transmit energy over a distance.
Send a reflection to the wall nearest you. Then send the reflection to a wall further away. The reflection on the wall further away is larger and therefore, the energy more spread out.
I am very confident that you have not tried this for yourself. I want you to show me pictures of this happening (with sunlight); I think you will find the experience educational.
(Edit: The mirror and the wall must both remain at the same rotational angle – if you angle the mirror to move the spot, and the spot becomes elongated because it’s now coming in at a more acute angle, it doesn’t count. Shining a sunbeam on the edge of a doorframe and then through the door to a faraway wall that’s through the doorway would be a good way to do it.)
Refraction and reflection. Most non specialty (consumer) mirrors have low quality and standards, so are affected by these more than other specialty mirrors.
Yeah but any mirror that isn’t imaginary has some kind of surface coating that’s a different refractive index than the atmosphere, and then the light has to interact with the surface of the mirror, which is not a perfect reflector and has imperfections, not to mention the light just had to pass through 400km of fluid atmosphere of varying density and composition.
Just give up dude , inverse square law doesn’t apply here, you were incorrect and are now making an idiot of yourself trying to incorrectly explain it.
Divergence and lack of coherence are two very different things (as I fully realized only after I typed up my message, I guess).
Divergence is a result of the angle. If you’re producing light from a local point-source, you have to work very very hard to make sure the angle of the departing rays is as close as you can make it, and you’re still not going to get anywhere even remotely close to 20 feet divided by 149,597,871 km. That’s where all the insane dropoff in the examples you’re talking about is coming from. The rays from the sun, though, are effectively parallel by the time they reach the earth to points 20 feet separated.
The inverse-square law is a result of the power in the beam spreading out over a larger area and spreading out its energy output over a wider area. It’s just a way of expressing that if the beam has spread itself out from hitting 1’x1’ into hitting 10’x10’ at a distance 10 times greater, each square foot of the target will now only get 1/100 of the energy. It won’t get weaker in total, without being absorbed by something along the way; that would violate conservation of energy. In this case the beams are parallel, the target is still 20’x20’ plus some tiny tiny fraction, there is a little bit of absorption by the atmosphere but not enough to make it not bright. The sun’s light goes through the atmosphere and it’s still bright (somewhat brighter if you’re on a mountain or in space, with a lot more UV, but not like night and day.)
I don’t see that coherence fits into this particular part of it in any way; as far as I know, we use lasers for this type of purpose because of their low divergence and the coherence has nothing to do with it. The rays originally from the sun have no coherence and they still manage to make it all the way out here.
Coherence is the key here, I assure you. Incoherent light is subject to the inverse square law in a way that lasers, which demonstrate coherence, are not. Lasers are coherent and collimated, and as such don’t interfere with one another and are parallel contributing to the laser’s ability to remain focused over long distances without spreading out significantly. This collimated nature of laser beams is a direct result of their high degree of spatial coherence, allowing them to maintain intensity over distances where a non-coherent light source would have dispersed according to the inverse square law. You arent reflecting coherent, in-phase, collimated from mirror, even if the suns rays strike the mirror parallel.
Lets assume each of the mirrors reflects 850 watts. The distance to the ISS is 408,000 meters.
The energy reflected by one mirror as received by the ISS is subject to the inverse square law (because it is incoherent).
E = (850 watts) / (4pi408000m)2,, or about 4.06x10 −10 watts/m2
A 5 milliwatt, off the shelf laser pointer with a beam divergence of 1.5 millirads would deliver approximately 4.25x10-9 watts/m2, or about 10x as much energy as the 850 watt mirror.
You can not melt a spy satellite with mirrors. You might be able to with lasers. A laser will be approximately 8.9x106 times as power effecient at getting light from earth to the ISS as a mirror would be. This is directly due to the properties of laser light, specifically coherence and collimation, which make it not subject to the inverse square law.
You’re confused, sir. Light from the sun is collimated, yes, i.e. parallel rays. The correct equation if you’re going to apply the inverse square law is:
E = 850 watts / 149,597,971 km^2 * 149,597,871 km^2 = 849.998864 watts
Same reason a signal mirror can reflect a flash as bright as the sun even miles away off a surface a few inches square.
You can believe or not; I’ve explained it as clearly as I know how.
Where are you getting this number from? The number you need to be using is the distance from the earth to the ISS, 408,000 meters.
Second, your formulation of the inverse square law is incorrect, in that you are missing the 4pi component, but in the grand scheme of distances we’re looking at, its negligible. It also looks like you may have gotten the order of operations wrong.
Third
The hell even is it that you think you are representing by these numbers? What is it you think you are saying?
fourth
I can provide sources for all my claims, and I’m pretty sure I got all my math correct, within a ROM. I can’t say the same for your work.
I’m gonna leave the source of that and the other number and where the pi went as an exercise for the reader
So you are like, an actual idiot then. I guess I picked the right meme for you.
Will you bet me money on how signal mirrors work and how bright the flash is at a certain distance from the mirror?
I will take a bet from you that the energy arriving at the ISS from a laser pointer you or I could purchase off Amazon, so consumer grade, is more than the energy that would arrive at the ISS from a concave mirror that would be used at a solar generating station.
A perfectly flat mirror, exactly like one of the ones in the OP generating station picture. Both are aimed perfectly on target, and the mirror is reflecting light from the sun on a sunny day. With those caveats I’ll bet $1,000. I’m happy with any university physicist or physics professor to be the judge, or Randall Munroe, or make a proposal of some other person if neither of those are acceptable to you. A lower amount of money is also fine if you’re not comfortable with $1,000.
The distance you need to account for is Sun to earth plus back to the satellite. Which I think is what they were using.
Yeah but what they are not accounting for is that the light actually has to come off the mirror. You can demonstrate this with a hand mirror that the illumination spot gets larger quickly at a distance. Take a mirror and find the sun. Send a reflection to the wall nearest you. Then send the reflection to a wall further away. The reflection on the wall further away is larger and therefore, the energy more spread out. The light coming off the mirror is not perfectly parallel as it had to pass through the atmosphere, then interact with the surface of the mirror. We do use mirrors for calibration in satellite remote sensing, but you will get far far far far more power arriving at something like the ISS coming from a much less powerful laser over such a distance. If we controlled by wattage, a laser will absolute crush a mirror in its ability to transmit energy over a distance.
I am very confident that you have not tried this for yourself. I want you to show me pictures of this happening (with sunlight); I think you will find the experience educational.
(Edit: The mirror and the wall must both remain at the same rotational angle – if you angle the mirror to move the spot, and the spot becomes elongated because it’s now coming in at a more acute angle, it doesn’t count. Shining a sunbeam on the edge of a doorframe and then through the door to a faraway wall that’s through the doorway would be a good way to do it.)
Refraction and reflection. Most non specialty (consumer) mirrors have low quality and standards, so are affected by these more than other specialty mirrors.
Yeah but any mirror that isn’t imaginary has some kind of surface coating that’s a different refractive index than the atmosphere, and then the light has to interact with the surface of the mirror, which is not a perfect reflector and has imperfections, not to mention the light just had to pass through 400km of fluid atmosphere of varying density and composition.
Its a bet I’m very willing to take.
Just give up dude , inverse square law doesn’t apply here, you were incorrect and are now making an idiot of yourself trying to incorrectly explain it.