dI/ds = no(B-I)

1) Is I measured in units of W/sr/m3 ????? B traditionally is given in W/sr/m3, because emission goes out in all directions. Then integrating over a hemisphere gives the flux in W/m2. So the answer to question 1) better be “Yes” or I’ll have to return my academic degrees.

Now let’s imagine we are using a spectrophotometer with a light source that is 3000 K. The emission term is negligible.

dI/ds = -noI

Integrate and get Beer’s Law. Now, I is (normally) measured in W/m2. When I began asking questions, I believed intensity (I) in BOTH equations was measure in W/m2. Now I believe this to be a mistake.

Let’s do a little dimensional analysis, first in W/m2: n*o is (molecules/volume)*(area) or molecules/distance. A beam of area A goes through n*o molecules/m no matter how wide the beam is. When the beam travels an increment ds, it passes through n*o*ds molecules. B and I tell us how many photons/s are emitted or absorbed per molecule. Integrating n*o*ds along a line from point P0 to P1 is simple. [Later I realize this only applies when emission is negligible.]

Now, how do I do the same analysis is spherical coordinates? I start at point P and go to P+ds. I can’t conceive of radiation measured in W/sr/m3 without picturing a sphere with a center somewhere. Do I envision the center of the sphere at point P (the beginning of the increment, ds) or a P0 (the start of my line to P1)? [I find out below, I’m really lost and neither answer is right.]

Let’s go back to W/m2. I’ve cheated. Radiation is actually traveling in all three dimensions, but I’ve ignored the dimensions perpendicular to the line from P0 to P1 and separately focused on the fluxes from P0 to P1 and from P1 to P0. This supposedly is the “two-stream” approach. In the atmosphere, the line P0 to P1 is in the +z direction and I substitute dz for ds.

How do I implement a two-stream approach in spherical coordinates? Do I make the line P0 to P1 have the theta equal to zero and P1 to P0 have theta equal to Pi. Now ds becomes dr. [Nope, this is badly wrong.]

Eureka! ???? The Sch eqn isn’t used to calculate the change in intensity (in W/m2) traveling from along a LINE from P0 to P1. That only works when emission is unimportant. However, for climate we want to calculate how the emission from a SURFACE changes as it passes through the atmosphere to a detector located Z km above the surface. (The math is easier when the detector is located so that the surface can be approximated by an infinite plane rather than a finite sphere.)

So my difficulty arises from how I formulated the problem. Absorption and transmission normally involve beam of radiation (W/m2) traveling along a LINE. The Schwarzschild eqn has nothing to do with line of sight; it has to do with flux from a SURFACE to a detector (or point).

If I had been properly educated (or remembered that education), I would immediately formulate real world emission problems in terms of a surface and a detector. Chemists have the luxury of working with mathematically simpler beams (lines) in a laboratory.

(I’m still working on the questions about the Wiki article on radiative transfer.)

]]>It’s possible I don’t understand your confusion. So maybe this answer doesn’t help at all. Let’s see..

If you take the simple case of a diffuse emitter from a surface and you want to find the flux **perpendicular** to the surface (W/m2) then you have to integrate radiance (W/sr/m2) x cosθ over the hemisphere.

That’s one situation. You end up with a multiple of π

I can email you a scan of the chapter of Incropera and DeWitt which works through the exciting details.

If you work out the radiation through the atmosphere in the Schwarzchild equation it is a different equation. But once again we have to integrate over the solid angle. It’s just a different equation that we are integrating.

]]>dI = nσ.(Bλ(T) – Iλ).ds

In the end, we have a quadruple integration: from all sources (two dimensions), along a straight line path of ds increments and over all wavelengths. I’m beginning to realize that Iλ must be expressed in units of W/sr/m3. I just have difficulty understanding how the intensity of the radiation entering the increment ds can be expressed in units of W/sr/m3.

This is especially true if I claim that the Sch eqn describes the intensity of radiation traveling a straight line from point A to point B. Integration then sums up the contribution of all lines originating at the surface and passing through point B at the edge of space.

]]>You need to integrate the Planck function over a hemisphere to get from watts per square meter per steradian to watts per square meter. See Lambert’s cosine law: https://en.wikipedia.org/wiki/Lambert%27s_cosine_law

That amounts to multiplying by pi.

]]>B(T) = j/a (or j/o ?) is also somewhat unfamiliar, since I’m used to thinking of one cross-section that applies to both emission and absorption. Written as shown belwo, dI/ds is the net result of an emission term using an emission coefficient and an absorption term using an absorption coefficient.

dI/ds = nj – noI = noB – noI

As best I can tell, SOD and Eddington are both using the “two stream approximation), but SOD is using the diffusivity approximation is instead of the Eddington approximation. Is the Eddington approximate needed when scattering is important? If you wish to contact me directly, use frankwhobbs and add icloud.com.

]]>dI/ds = no(B-I)

Clearly there is something wrong if I and dI are in W/m2 and B is W/sr/m3. Assuming I am correct in saying that the Schwarzchild equation is used to calculate changes along a line of sight from point A to point B, this appears(?) to be radiation moving perpendicular to a plane (W/m2). It seems like we really need is

dI/ds = no[ (k)B – I ]

where k has the needed units, sr-m, and perhaps k = 1. If I imagine two infinite planes along a line of sight from A to B separated by a increment of distance ds, perhaps half of the flux towards the forward hemisphere (2*Pi steradians) reaches the second plane, but I’m not sure how to account for the viewing angles. So k may equal 2*Pi.

Now maybe the problem is that the intensity of radiation (I) entering the increment ds is a spectral radiance measured in W/sr/m3 (just like B) and not a spectral flux density (or spectral irradiance) measure in W/m2. Working with climate science makes one think everything is measured in W/m2 (or W/m3 before integrating over all wavelengths) moving perpendicular to a surface.

For non-native speakers, Wikipedia has a handy summary of terminology here. https://en.wikipedia.org/wiki/Radiant_intensity

Also when I look at Figures 2 and 4, they seem “backwards”. If I want to know the TOA flux through 1 m^2 say 100 km above the surface, then I need to integrate the flux from the whole the surface through that 1 m^2. I don’t see why the mathematics for the return direction should be any different or why DLR isn’t the sum of all rays from space to the single point in Figure 2 or 4.

]]>“In the world of practical turbulent flows you have to create a model and get fluid flowing and make measurements. The model doesn’t have to be the same size but there are dimensionality considerations that let you work out how to scale (can’t remember any of them, it’s been so long ago, but you might find that with a 1/10th size model you need 1/(102) speed of fluid flow (or a square root or something else).”

FWIW – Good online source for those interested to follow up:

https://www.rose-hulman.edu/Class/me/ES202/Spring%2006-07/Dimensional%20Analysis.pdf

which is Chapter 7 of

https://www.mheducation.co.uk/9781259921902-emea-fluid-mechanics-fundamentals-and-applications

]]>Wunsch 2007. I don`t think he was invited to be a co-writer with Hansen. ]]>