## Dealing with multiple greenhouse gases

We now know how to efficiently compute the band-averaged transmission function for a single greenhouse gas acting alone. It is commonly the case, however, that two or more greenhouse gases are simultaneously present - CO2, CH4 and water vapor in Earth's case, for example. How do we compute the averaged transmission function is this situation? The issues are closely related to thosed discussed in Section 4.4.1 in connection with the loss of the multiplicative property in band-averaged transmission functions. Similar reasoning shows that the average transmission function for two greenhouse gases acting together generally differs from the product of the averaged transmission functions of each of the individual gases taken alone. Fortunately, however, the special circumstances under which the multiplicative property holds for multiple gases are expected to be fairly common.

By way of illustration, let's consider an idealized greenhouse gas A with transmission function Ta(v) in a wavenumber band of width A, with the property that Ta = a on a set of wavenumber subintervals with length adding up to r • A, but with Ta = 1 elsewhere; naturally, we require a < 1 and r < 1. Consider a second greenhouse gas B with TB = b on a set of wavenumber subintervals with length adding up to s • A, and with TB = 1 elsewhere. The band-averaged transmission for gas A in isolation is r • a + (1 — r), and for B in isolation is s • b + (1 — s). What is the transmission when both gases act in combination?

The answer depends on how the regions of absorption of gas A - spectral regions where Ta < 1 - line up with those of gas B. We can distinguish three limiting cases. First, the regions of absorption of the two gases may be perfectly correlated, in which case r = s, Ta(v) = a precisely where TB (v) = b, and

Further, a + b — ab = a + (1 — a)b, and so this expression has a value lying between a and b, both of which are less then unity. Hence (a + b — ab — 1) < 0 and r(1 — r)(a + b — ab — 1) < 0, given that 0 < r < 1. We conclude that the mean transmission function for the two gases acting in concert is always greater than or equal to the product of the individual transmission functions, the equality applying only when r = 0 or r = 1, i.e. when there is no absorption or completely uniform absorption. For example, if we take r = 0 and a = b = frac110, the mean transmission function for the two gases acting together is -100, or just a bit over a half, whereas the individual transmission functions are 20 each, multiplying out to 100 or just over a quarter. When the absorption regions of the two gases coincide, the gases acting together transmit considerably more radiation than one would infer by allowing each gas to act independently in sequence. This happens because one of the gases uses up some of the frequencies that the other gas would like to absorb.

Exercise 4.4.1 Derive the final equality in Eq. 4.79. Sketch graphs of transmission functions vs. frequency for the two gases for two different cases illustrating perfectly correlated absorption regions. Evaluate the mismatch between T and TATB for a = b = r = 2. Allowing a,b and r to vary over all possible values, what is the greatest possible mismatch?

At the other extreme, the absorption regions of the two gases may be completely disjoint, so that Ta = 1 wherever TB = b < 1 and TB = 1 wherever Ta = a < 1. For this to be possible, we require r + s < 1. In the disjoint case,

= ra + sb + (1 — (r + s)) = TaTb — rs(1 — a)(1 — b)

In the disjoint case, then, the transmission for the two gases acting together is always less than the compounded transmission of the two gases acting independently.

Exercise 4.4.2 Derive the final equality in Eq. 4.80. Sketch some transmission functions illustrating the disjoint case. Put in a few numerical values for a, b, r and s to show the size of the mismatch between T and TATB. What is the greatest possible mismatch in the disjoint case?

As the final limiting case, suppose that the absorption of the two gases is uncorrelated, so that at any given frequency the probability that Ta = a is r regardless of the value of TB there, and the probability that TB = b is s regardless of the value of Ta there. This situation is also known as the random overlap case. In this case

= ra + sb + (1 — (r + s)) = TaTb — rs(1 — a)(1 — b)

r(1 — s)a + s(1 — r)b + rsab + (1 — r)(1 — s)

TaTb

The reasoning behind the second line is that r(1 — s) is the probability that only the first gas is absorbing, s(1 — r) is the probability that only the first gas is absorbing, rs is the probability that both gases are absorbing, and (1 — r)(1 — s) is the probability that neither gas is absorbing. Multiplying out the terms in the product of Ta and TB we find that in the random overlap case, the mean transmission of the two gases acting together is precisely the same as the compounded transmission of the two gases acting independently.

The properties illustrated by the three cases just discussed can be generalized to an arbitrary set of transmission functions. Let Ta and TB be any two transmission functions, and define the fluctuation

and similarly for TB. Then,

From this we conclude that the transmission of the two gases acting in concert is greater than the product of the individual transmissions if the two transmissions are positively correlated, less than the product of the individual transmissions of the two transmissions are negatively correlated, and equal to the product if the two transmissions are uncorrelated.

In fact, by exercising just a little more mathematical sophistication, it is possible to go further and put an upper bound on the amount by which the mean transmission function for joint action by the two gases deviates from the product of the individual transmission functions. The key is to use a handy and powerful relation known as the Schwarz Inequality, which states that

for any two functions f (x) and g(x), (fg)2 < f g2, where an overbar indicates an average over x. The equality applies only when f (x) is proportional to g(x). Applying the Schwarz Inequality to Eq. 4.83, we find that the deviation satisfies the inequality

In other words, in the worst case the deviation can become as large as the product of the standard deviations of the two individual transmission functions. Since 0 < TAB < 1, the maximum standard deviation is i, occuring when each transmission function is zero for half of the frequencies in the band and unity for the other half; the error in random overlap in this case is +1 if the transmissions are perfectly correlated and — 1 if the transmissions are perfectly anticorrelated. These errors should be compared to the random-overlap value for the limiting case, which is i. The effects of non-random overlap are potentially severe.

Fortunately, the situation is rarely as bad as the worst-case suggests. The positions of spectral lines are a sensitive function of molecular structure, so it is a reasonable guess that the absorption spectra of dissimilar molecules should be fairly uncorrelated. Thus, in most circumstances one can get a reasonable approximation to the joint transmission function by computing the individual transmission functions for each gas and taking the product of the individual transmission functions. It is fairly easy, if computationally expensive, to test the accuracy of the random-overlap assumption in a given band by computing the correlations of the full frequency-dependent transmission functions in the band. However, finding a general characterization of the correction to random-overlap, and the way the correction depends on the concentrations of the individual gases, is an intricate art which we will not pursue here.

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