Alastair McDonald wrote:
> ----- Original Message -----
> From: "Eli Rabett" <[EMAIL PROTECTED]>
>
> > This is the usual confusion.
>
> !
>
> > The saturaters somehow never come to the
> > point that emission also increases.
>
> Oh?
>
> > The base idea is that if
> > atmospheric absorption increases, then the temperature of any packet of
> > air must increase until emission can match absorption.
>
> But emission is independent of temperature. It depends on the number
> of excited molecules which is a function of pressure not temperature. The
> air temperature of the saturated layer will rise until it convects. It won't
> wait around until the temperature rises to a level at which the radiation in
> equals radiation out.
Nope, the temperature determines the proportion of molecules in the
vibrationally excited state which can emit. Technically this follows
the Planck distribution.
>
> > Absorption is
> > measured over a path length, so if it is saturated over 100 meters
> > (reduced by exp(-2) for example), it is not saturated over 10 m. If
> > you double the CO2 mixing ratio, the absorption will be saturated over
> > 50 m, but not 5.
>
> So if you double the concentration you will half the height of the
> layer where the absorption is saturated. i.e. the absorption (greenhouse
> effect) is proportional to the concentration, not the log of the
> concentration.
This can be described as Pielke's falacy. In the limit of small
changes, everything is linear. So speaketh Prov. Taylor For small x,
ln(1+x) ~ x. In the case of absorption Beer's law gives you ln(I/Io)
= -snL (I is intensity, s absorption cross-section, n number density
and L path length. Now let us increase n by some amount x. Then for
the same path we get
ln[(I -i)/Io]= -s(n+x)L,
where i is the extra absorbance due to the increase in number. If i is
small compared to I (which will always be the case if x is small, we
can rearrange this to
ln[(I/Io)(1-i/I)] =ln(I/Io) + ln(1-i/I) = -snl -sxL
for small i/I you get
ln(1-i/I) ~ -i/I = -sxl or
i/I ~ sxL
which is linear. For a larger change in x you get
ln(1-i/I) ~ -sxL
More perhaps anon.
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