Thanks.

]]>But still, there’s not a single experiment that shows that co2 can warm anything. Just that it can be warmed. Heat absorption is not warming, when heat is absorbed from a hot surface, it cools the surface. The colder something is, the more heat it absorbs.

]]>Frank,

Yes, you are correct, my mistake – n is a function of height,z, and σ is a function of λ.

n(z)

σ(λ)

In answer to the other questions I would have to sit down for a while and be very precise about the units in each term.

Without thinking about it too hard, reviewing the simpler version, the difference version now corrected:

ΔI_{λ}/Δz = nσ_{λ}.[B_{λ}(T1) – B_{λ}(T2)]

We can see that if we just multiply both sides by Δz we have:

ΔI_{λ} = (nΔz) . σ_{λ}.[B_{λ}(T1) – B_{λ}(T2)]

(nΔz) = number of absorbers in the path (whereas n = per unit volume)

so the LHS = change in monochromatic radiation and RHS = emitted monochromatic radiation – absorbed monochromatic radiation

which is what we expect.

Once again, if we integrate across all wavelengths for a specific location (or difference of heights) we will get the right answer but it won’t be any use. That is, we can’t integrate it after this with respect to height.

There’s also a tidying up requirement in that I_{λ} is spectral intensity and not spectral emissive power, so you either have to integrate over all solid angles, or use the plane parallel approximation.

SOD: Thank you for your help. You are right. There was a lot of hand waving going in when I “integrated” over all wavelengths and replaced B(lamba,T) with oT4.

If I (intensity) is integrated over all wavelengths, the result is power flux (W/m2). Does integrating dI(z,λ)/dz over all wavelengths give dW(z)/dz, where W(z) is the power flux at altitude z? If I express the derivative as a limit, is the integral of the limit, the limit of the integral. That question is beyond my mathematical comfort zone.

On the other side, you wrote n(λ) where I am using n(z) – hopefully correctly – for the density of an GHG at altitude z. For purposes of integrating over all wavelengths, n(z) is a constant. o is a function of wavelength, o(λ). I can move the former outside the integral, but not the latter.

dW(z)/dz = n(z)*o*[Int{B(λ,T2)}dλ – Int{B(λ,T1)}dλ] wrong?

dW(z)/dz = n(z)*o*O*[T2^4 – T1^4] wrong?

dW(z)/dz = n(z)*[Int{o(λ)*[B(λ,T2)dλ} – Int{o(λ)*[B(λ,T1)}dλ] right

This is appears to be a dead end. So I’ll pull a rabbit out of my hat and say that the absorption cross-section of my GHG is a constant (o) everywhere within within the range wavelengths integrated and zero outside. Now the equations labeled “wrong?” could be right. So for this artificial GHG, increasing the amount of the GHG won’t change the outward power flux.

An alternative could be do numerical integrations while increase the mixing ratio for the GHG and see if

Int{o(λ)*[B(λ,T) – I(z)}dλ

begin to vary with the inverse of the mixing ratio when the mixing ratio is high enough. It is my hypothesis that it will for part of the path to space, but every atmosphere gets thin at some altitude.

]]>Frank,

I didn’t realize it was just a notation issue.

Let’s look at it as a difference equation at a specific location instead of a general differential equation, with temperatures T1 & T2:

ΔI_{λ}/Δz = n_{λ}σ.[B_{λ}(T1) – B_{λ}(T2)]

Now you are suggesting to integrate over all wavelengths? You have to integrate outside:

∫ {ΔI_{λ}/Δz} . dλ = ∫ {n_{λ}σ.[B_{λ}(T1) – B_{λ}(T2)]} . dλ

So the problem that I see with what you did originally –

You have integrated only the expression inside the square brackets, which is not possible unless n is a constant. However, it is a function of λ. And also a function of height, although we have fixed this in the revised difference equation.

The whole solution depends on the fact that we can calculate the change in intensity of radiation based on the initial radiation, the number of absorbers as a function of height and the temperature of the gas as a function of height – only when we consider **monochromatic radiation** (for new readers – “one wavelength at a time”).

Once you try integrating across all wavelengths at each height to get total flux you have lost the key information you need to calculate the absorption at the next height.

]]>SOD: Thanks for pointing out a problem in notation. I’ve been mistakenly using I_0 to refer to the radiation entering the path increment ds or dz. If one integrates from z1 to z2 (which I never do personally), it clearly it makes more sense to call I_0 the intensity at z1. Perhaps I can put it more clearly by expressing dependence on altitude explicitly:

dI(z)/dz = n(z)*σ*[B(lamba,T) – I(z)] in general at one wavelength lamba

dI(z2)/dz = n(z)*σ*[B(lamba,T2) – B(lamba,T1)]

where T1 is a composite temperature or blackbody equivalent temperature for the molecules that emitted the average photon traveling upward to reach z2. T1 is the temperature about one mean free path below z2. The greater the density of absorber, the smaller the mean free path and the closer T1 and T2 will be. Things would be simpler if all photons traveled exactly one mean free path.

The rest of the derivation starting with “assuming a graybody” doesn’t change. (Changing to a greybody approximation simplifies the math.)

]]>Frank,

The original equation, e.g. equation 11 in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations is:

dI/ds = nσ.(B_{λ}(T) – I_{λ})

1. You have replaced I_{λ} with I_{0}.

I_{λ} changes along the path s, which is why it is the subject of the differential equation in the first place.

I should probably have written I_{s,λ} but it starts to get clunky, however, I is a function of the wavelength, λ and the path, s.

I_{0} [by which it looks like you mean I_{λ0}, i.e., the monochromatic radiation] – is the intensity at **the surface**, not the intensity of the radiation when it reaches the layer of the atmosphere we are considering.

All the rest of the problems in your derivation appear to come from this original problem.

You can see in equation 16 of the cited article that we have to solve the equation by integration of equation 11 vs height (or optical thickness) to get I_{s,λ} as a function of I_{0,λ}.

dI = n*o*[B(lamba,T2) – I_0]*ds = n*o*[B(lamba,T2) – B(lamba,T1)]*ds

Let’s consider a “composite temperature” T1 for all of the locations that contributed to I_0. (The earth’s blackbody equivalent temperature is a similar composite.)

dI = n*o*[B(lamba,T2) – B(lamba,T1)]*ds

Assuming a graybody, the integral of B(lamba,T) over all wavelengths is eoT^4. Setting T2-T1 = -t, and capitalizing the S-B constant (O)

dI = n*o*e*(OT2^4-OT1^4) = n*o*e*O*(-4T^3*t)*ds

(I’ve dropped terms with lower powers of t and replaced T2 with a generic T.) For a constant lapse rate (r) and a mean free path (d), t = r*d. In general, the mean free path is inversely proportional to the density of GHG (n): d = k/n. making t = r*k/n

dI = -4*o*e*O*k*rT^3)*ds

Now dI appears to not depend upon the density of GHGs. If we call the GHE the difference between surface and TOA flux the GHE and we integrate dI from the surface to space, we have a GHE that doesn’t vary with GHGs (n)! (Unless we add so much GHG that the path to space gets much longer – as it is when we compare Earth and Venus).

I recognize that this result is due to saturation and that most of the GHE on our planet occurs at wavelengths where the mean free path is long enough to reduce the probability of escaping to space (or the surface for DLR) by about 50%. I don’t doubt the output from MODTRAN. However, saying that I_0 is greater than B(lamba,T) and therefore that increasing n must reduce dI appears to be an oversimplification. This argument only applies when B(lamba,T2) – I_0 is not inversely proportional to n!

dI = n*o*[B(lamba,T) – I_0]*ds

Comments (especially corrections) would be appreciated.

PS. If I could devise a situation where the lapse rate decreased with n, we would have a situation where the increasing GHGs would increase the flux to space. That could be caused by GHGs that absorb some SWR, like ozone, and perturb the lapse rate. (:))

]]>which will make billions of dollars from diabetics.

You intend to understand how to cure diabetes. ]]>

A Cool Object Reduces Energy loss from a Hot Object

http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html

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