It has been more than a year when I did this problem in the Actual Case A, the realistic case. I will go back to proving it one more time just to make sure everything is in its place. I remember I did check my solution couple of times and was pleasantly surprised by the result: A=S.

QuoteOriginally posted by: quantystThe Actual Case A: Radiation from every point of the Sun is projected outward in a uniform fashion.Could you be more precise? If "projected outward in a uniform fashion" means that each point on the surface emits radiation isotropically on the positive hemisphere (i.e. only above the surface) this is quite different from the actual behaviour of the Sun: the edge of the solar disk would appear brighter than the center in that case.

QuoteOriginally posted by: carlitosQuoteOriginally posted by: quantystThe Actual Case A: Radiation from every point of the Sun is projected outward in a uniform fashion.Could you be more precise? If "projected outward in a uniform fashion" means that each point on the surface emits radiation isotropically on the positive hemisphere (i.e. only above the surface) this is quite different from the actual behaviour of the Sun: the edge of the solar disk would appear brighter than the center in that case.The real Sun out there has for sure a very complex behavior, way above my pay scale. I can only tell you my assumption, which is reasonably simple and acceptable enough.All points on the surface of the Sun are assumed equal in their role as point sources of emitted energy. Every point emits energy in all directions (if you will, even back to the Sun itself) in a uniform fashion. I hope this helps.Take a look at my previous comment about getting the same answer: A=S.I am currently working on this in 2D, just to refresh my memory, etc.

QuoteOriginally posted by: quantystTake a look at my previous comment about getting the same answer: A=S.I think that there is a fundamental reason why. In space the equation for the radiation field A is going to be del(A) = 0, curl(A) = 0. If we impose the condition that that the source is symmetric around a point, that gives you enough information to solve for the radiation field. So the result of this is that if the source is spherically symmetrical the radiation field is going to be exactly the same, and so you can replace any spherically symmetric source with a point object and get the same radiation field.

QuoteOriginally posted by: twofishSo the result of this is that if the source is spherically symmetrical the radiation field is going to be exactly the same, and so you can replace any spherically symmetric source with a point object and get the same radiation field.I might be misinterpreting your statement, but are you saying that the Sun (assuming spherical symmetry) will emit light exactly as a punctual source would do?How is it, in that case, that we can see that the Sun is a million kilometers wide and not a point?

(This is off the top of my head, and what I'm doing is writing some radiation hydrodynamics equations and trying to think about what they mean, and some of it doesn't quite make sense so I might be getting something wrong.)QuoteOriginally posted by: carlitosI might be misinterpreting your statement, but are you saying that the Sun (assuming spherical symmetry) will emit light exactly as a punctual source would do?I'm saying that the direction of energy transfer is going to be the same, which is a very different statement. If you are in a room whose walls are 5000 degrees then there is going to be no net energy transfer just like if you are in a room whose walls are 0 degrees, but the rooms are going to look very different.One thing that I think that may change things is how does the earth react to all of this. If the earth is an energy sink that absorbs all of the energy that gets pumped into it, I think you will end up with a different answer than assuming that the system is in steady state and that the earth is in thermal equilibrium with its immediate surroundings.

QuoteOriginally posted by: twofishI'm saying that the direction of energy transfer is going to be the same [...]I understand that the question is about the influx of solar radiation on the Earth surface, and I agree that the net direction is radial from the center of the Sun but I don't think it can be independent of the radius of the Sun.Think of the meatball and sausage planets I draw before (better, think of them as imaginary volumes and not as physical bodies). In fact the argument applies as well, more clearly, if the first planet is just an hemisphere but I can't draw it that way.What is the influx on these surfaces in the following cases?1) Point source (or spherical source emitting only radially)The influx will be the same in both cases, the extra surface is completely useless as there is no extra radiation coming through it.2) Big-ball source (with each point in the surface emitting with a distribution which is non-null for non-perpendicular directions)The influx for the "extended" planet will be bigger, I hope we can all agree on that.I think this shows that case 2 can't be equivalent to case 1.

QuoteOriginally posted by: carlitosThe influx for the "extended" planet will be bigger, I hope we can all agree on that.I think this shows that case 2 can't be equivalent to case 1.That I think is the problem in that the results that I'm getting seem to indicate that case 2 is equivalent to case 1. The basic issue is that if you have two sources of radiation going in opposite directions, do you have a net energy transfer?

QuoteOriginally posted by: twofishThat I think is the problem in that the results that I'm getting seem to indicate that case 2 is equivalent to case 1. The basic issue is that if you have two sources of radiation going in opposite directions, do you have a net energy transfer?If these opposite fluxes are being absorbed by a sink of radiation sitting in the middle, I think you do. At least that was my understanding of the problem, how much energy arrives at the Earth; I don't see why the influx coming from the left and the influx comming from the right would cancel out. Otherwise toasters wouldn't have electrical resistors on both sides :-)And in any case, I don't see why would both cases be equivalent.Think for example of radiation as particles carrying momentum and transferring it to the simplified "planets" consisiting of a) an hemisphere (only the sunny half of the spherical planet in the diagrams)b) an hemisphere + a cilinder (i.e. supress the far end hemisphere from the second planet in the diagrams)I hope we agree that even if you work with vectors the answer to the problem will bea) the integral of some (vector) quantity over the hemispheric surface + the integral over the flat surface = A + B b) the integral over the hemispheric surface + the integral over the surface of the cylindrical part + the integral over the flat surface at the end = A'+ C' + B'The hemispheric surface is the same by construction so in both cases 1 "puntual source" and 2 "extensive source" the integral is the sameA=A' for cases 1,2The flat surface at the end is not exposed to the radiation in any case, so the integral will be always nullB=B' for cases 1,2The surface of the cylindrical part will not receive any radiation in case 1 (or in case 2 if the radius of the source is smaller than the radious of the cylinder).The surface of the cylindrical part will receive radiation in case 2 (if the radius of the source is big enough), and even if you're thinking of a vector quantity and the perpendicular component cancels out you will still have a non-zero value for the integral on the radial direction.C'=0 for case 1C'!=0 for case 2Putting everything together,case 1: point sourceA+B=A'+B'+C' (because A=A', B=B'=0, C'=0)the answer is the same for planets a) and b)case 2: extensive sourceA+B!=A'+B'+C' (because A=A', B=B'=0, but C'!=0)the answer is not the same for planets a) and b)Therefore case 1 and case 2 are not equivalent.Does anyone see any failure in this reasoning?