Adieu Ado About Nothing

Motivated by some claims made by climate scientist Roger Pielke, Sr. (quoted here), I recently considered the question of whether using the global mean temperature instead of the individual temperatures at locations across the earth's surface (hereafter referred to as "base temperatures") introduced an appreciable error into the calculated radiative emission change for the surface as a whole in response to a 1K change in temperature at each location.

After investigating the issue in some detail and actually doing the relevant surface integral for emission increase for a 1K increase in temperature ("odd" as some may find such an approach) , I found that the error associated with using the global mean temperature for the case in question was fairly small (probably a few percent at most). I described why this was so in my last post.

But apart from the variation of base temperature across the globe, shown in this "Temperature VS Latitude" graphic by Robert Rhode, there is a separate issue that also impacts the amount by which radiative emission changes at each location: the amount of temperature change at a given location.

For my previous analysis, in order to isolate the effect of base temperature on total radiative emission change by the earth (the issue originally suggested by climate scientist Roger Pielke, Sr in this post), I made the (admittedly artificial assumption) that temperature went up uniformly by 1K at each location across the earth's surface.

But, in reality a change in the global mean temperature is not the result of a uniform change in temperature across the earth’s surface. In fact, the actual change depends on location. For example, over the past half-century (1955 – 2005), the global mean surface temperature went up by 0.59 deg C, but the changes that were used to calculate that mean depended significantly on the location, as shown in the following NASA graphic:

Figure 3. Temperature change over the past 50 years based on local linear trends

(found on this NASA page)

Note: The black line on the right shows the change in the annual mean for each latitude over the period 1955- 2005 . Though the error bars (standard deviation) for each latitude are not shown, for the analysis below, they were calculated from the 2-D temperature map at the left.

So, the new question I would like to consider is the following:

Does using the change in the global mean temperature rather than the temperature changes as a function of location (shown by the NASA graphic) yield a significantly different answer for "total change in radiative emission by the earth'?

The answer may come as a surprise to some , but “No, it does not."

In fact, if one considers both the details of base temperature as a function of location and the details of temperature change as a function of location in order to calculate the total radiative emission change, one gets a result that is within about 3 % (with a similar associated uncertainty) of the result that one gets if one instead simply uses the global mean temperature and change in the global mean temperature to calculate the total emission change.

Effect of varying temperature change on the surface integral of radiative emission

We can estimate how much radiative emission changes across the entire earth's surface in response to temperature changes across the earth by doing a similar surface integral to the one I did for the original analysis.

But this time, when we do the surface integral, we simultaneously take into account both the base temperature as a function of location and the change in temperature (over some time period) as a function of location (as shown in the above NASA graphic)

For this new integral, we again assume that base temperature depends on surface location according to our previous piecewise linear approximation to the “Temperature VS Latitude” relationship, given on this annotated “Temperature VS Latitude" graphic (original graphic by Robert Rhode), but instead of assuming the 1K change in temperature at all locations across the globe, we now use the temperature change as a function of latitude over the past half-century, shown by the black line on the above NASA "Temperature Change" graph on the right.

As before, we use the “differential of radiative emission” -- proportional to “T^3(dT)” -- to approximate the emission change for a small change in temperature at each location (as indicated in the previous post, for a temperature change on the order of 1 degree K, this approximation introduces very little error into the result). Also as before, the temperature “T” at each location is taken from the central “Temperature vs Latitude” graph (or more precisely, the piecewise linear approximation shown superimposed on the graph).

But this time, instead of assuming that the temperature change “dT’ is 1K for each location, we use the temperature change for each latitude over the past half-century (1955- 2005) shown by the NASA graphic.

We then integrate“T^3(dT)” over the earth’s surface (for latitude-dependent "T" and " dT ") and compare this result to the result obtained by assuming that 1) every square meter of the earth has undergone an increase in temperature of 0.59C (the change in the global mean temperature over the past 50 years, based on linear trend) and 2) every square meter of the earth has emitted at the global mean temperature ( 287.4K -- ie, before the slight increase).

Note: a piecewise linear approximation to the NASA “Temperature Change VS Latitude” curve was used to do the integrals here (admittedly lots of pieces and lots of integrals!). As a check on this approximation, the average over the earth’s surface of temperature change from 1955-2005 was calculated using this piecewise linear approximation. This average was found to be within 0.3% of the 0.59C value that is given by NASA on their graphic (just above the 2-D temperature map on the left above), so the approximation used was indeed quite good.

This new result for "surface integral of T^3(dT) " -- taking into account both the details of base temperature as a function of position on the surface of the earth (represented by the “Temperature vs latitude” curve) and the variation in temperature changes (shown by the “Temperature change Vs Latitude” curve) -- differed by only 3.3% from (was about 3.3% less than) the surface integral obtained by assuming that 1) every square meter of the earth emitted at the global mean temperature and 2) every square meter of the earth underwent an increase in temperature of 0.59C.

Note: Details of the approximations for T and dT (as a function of co-latitude), radiative emission integrals (increase over past 50 years), and global averages of T and dT using the approximations are provided here.

So, in other words, if we assume that both “temperature” and “temperature change” across the surface have the given latitudinal dependencies (given by our two graphics), we get an answer for the “total change in radiative emission across the surface of the earth” that is within 3.3% of the answer obtained using the global average temperature and change in the global average temperature (based on linear trend). The associated uncertainty (see below) is just a few percent.

Uncertainty associated with the above "3.3% difference" result

There is (what turns out to be a relatively small) uncertainty associated with this result. In my previous post, I considered the uncertainty introduced into the surface integral of T^3 by use of my piecewise linear approximation to the “Temperature vs Latitude” graph to represent base temperatures across the earth's surface. I determined that this uncertainty was not likely to be greater than a few percent. I also explained why using the mean temperature value for each latitude rather than the values scattered about the mean introduced a very small error into the integral calculation (of order 0.1%).

For the analysis above, there is an additional uncertainty associated with using the "Temperature change vs Latitude" relationship shown on the NASA graphic (black curve) rather than the individual temperature changes experienced by locations at each latitude. The issue is actually similar to what we encountered before: ie, the use of a mean value rather than the individual values that went into the mean.

In this case, the (black) curve shows the mean temperature change for each latitude. Though the error bands are not shown on the NASA graph at the right, they can be (and were) readily calculated from the temperature map at the left and when this is done and the "scatter" for each latitude was taken into account, it was found to make very little difference.

Why using the NASA "Temperature Change vs latitude" graph introduces very little error into the result

There is a very simple reason why using the mean temperature change for each latitude (shown by the black Temperature vs Latitude curve) yields very nearly the same answer as using the individual temperature change values that went into that mean:

The "differential of emission" -- proportional to T^3(dT)” -- indicates that (for small temperature change) the amount by which radiative emission changes at a given location is directly proportional to the temperature change ("dT") at that location.

As a first approximation, one can assume that all the locations at a given latitude emit at the same base temperature. Though this is not necessary to our final conclusion, it is actually a very good approximation, given the very small base temperature deviation relative to the mean base temperature at each latitude -- and the assumption makes it very clear why using the mean temperature change at each latitude is justified for the purpose at hand.

If one makes the above assumption that all the locations at a given latitude emit at the same base temperature (the mean value for that latitude T_o), one finds that the total amount by which radiative emission at that latitude changes is then proportional to the sum of the individual temperature changes experienced by all locations at that latitude. But the latter sum is just the mean temperature change for that latitude (dT_o) multiplied by the number of locations (N) -- ie, " N (dT_o) "

So, the emission change in that case is equal to

N (T_o)^3 (dT_o)

where " T_o " is the mean base temperature for that latitude and " N " is the number of locations included in the sum (and we have ignored the Stefan Boltzmann constant, emissivity and a factor of 4)

In other words, under the assumption that all locations at a particular latitude emit at the same temperature T_o, we find that the total emission change is precisely what one gets if one assumes that all locations at that latitude have experienced a temperature change equal to the mean change for that latitude (ie, the value given by the black NASA curve on the right above).

In reality, the situation is more involved, since the locations at a given latitude do not all emit at precisely the same base temperature (ie, not all emit at precisely the mean T_o, though the vast majority are not far from it).

But (for a given latitude), if one simultaneously considers the (slight) deviation of the base temperature of each location from the mean " T_o " (for that latitude) and the deviation of the temperature change for each location from the mean temperature change " dT_o " (for that latitude), the result is not much different.

For the mathematically inclined, I provide the details of the latter analysis here (Note: after bringing up the first page, click on the small "tempchange (Set)" graphic to view the next page, of which there are 7).

Final Comment

Climate Scientist Raymond Pierrehumbert made the following comment on Rabett Run regarding the above "issue":

A lot of you guys are getting somewhat led up a garden path. There's no business about "scientists missing an error of 17%" going on no matter how you do the arithmetic. Climate models DO NOT CALCULATE THE ENERGY BUDGET USING A SINGLE GLOBAL MEAN TEMPERATURE,... Climate models do a radiative transfer calculation several times a day at each gridpoint, incorporating the full variation of temperature.

The sort of thing you guys are talking about only tell you how big the errors are in the most primitive blackboard-type zero-dimensional climate calculation, where indeed you do do the energy budget in terms of a global average temperature. The fact that the errors made by doing so are so small is in fact why you can get pretty far with such simple calculations, especially on a planet like Earth or Venus with a thick atmosphere and or ocean to redistribute heat and make temperature more uniform.

I'll have to take Pierrehumbert's word that global climate models do not use global mean temperature to determine the earth's "energy budget". He's the expert and I have no reason to doubt him.

But on the issue of "scientists missing an error of 17%", I would remark that after looking into this issue in some detail, I don't even have to take Pierrehumbert's word for it. I know he is right: There is no business about "scientists missing an error of 17%" (or anything even close) going on no matter how you do the arithmetic. Claims to the contrary simply do not hold up under a careful analysis of the issue.

And you can indeed get pretty far with some relatively simple calculations in this case.

Much (more) ado about nothing?

In a recent post, I investigated an issue suggested by climate scientist Roger Pielke, Sr:

In constructing a global average of T’, its spatial distribution matters since T’ in regions with a baseline colder temperature have a significantly smaller effect on the return of heat energy to space (through infrared emission) than regions with a warmer baseline temperature.”

"Thus, by constructing a global average value of T’ in the equation for dH/dt introduces a significant error in diagnosing the actual global radiative imbalance. A 1C increase in surface temperature in the higher latitudes has less of an effect on the outgoing long wave radiation than a 1C increase in the tropics."

"Thus when the science assessments and policymakers claim we need to keep the global average temperature change below some threshold (e.g. 2K), they should be questioned on how this number is constructed both in the observations and in the models."

The implication seems to be that, if the science assessments and policymakers use the global average temperature (rather than individual temperatures across the earth) to assess the impact of temperature change due to global warming, they could (at least potentially) be making a significant error by doing so.

After doing some analysis, my initial take on this issue was that it did not make much difference whether one used the global mean temperature or the detailed temperatures across the earth's surface to estimate how much the total radiative emission from the earth changes for a small change in temperature. For a 1K change in temperature*, I found the two methods gave an answer within a few percent of one another.

It occurred to me soon after I posted my original piece, however, that I had made a "boo-boo of the first rank" in not considering the uncertainties associated with my answer . My initial (admittedly crude) error approximation then led me to conclude that the uncertainty associated with my answer was somewhere around +- 12.5%, which (if true) meant that I could not claim that the difference was anywhere near as small as my calculation indicated -- not less than about 16%, in fact.

After some additional analysis, I narrowed the probable uncertainty to about +- 6.5%, which meant that the two methods were most probably giving an answer within 10% of one another -- though I could still not be sure whether the two results were actually within just a few % of one another, as my calculation indicated.

But further more detailed analysis of the errors has led me back to my original conclusion that the two methods (calculating total emission change using detailed temperatures or the global mean) yield answers that are very close to one another (probably within a few percent of one another).

Finally, in my investigation of the errors, I discovered just why this is the case -- why the difference between the results from the two methods is relatively small. The reason is actually very simple, and below I link to the details of a very straightforward mathematical argument I came up with to demonstrate that the difference is almost certainly less than 5%.

*************************

*Note: if, instead of assuming a 1K temperature change, one takes into account the actual change in temperature experienced by locations across the earth (over the past 50 years, for example) along with base temperature as a function of location, one finds a very similar result, as discussed in my final post on the subject.

A Brief recap of my investigation of this issue

Above is an annotated version of an original graphic by Robert Rhode (Global Warming Art)

I used the above “Temperature VS Latitude” graphic to estimate the total amount by which radiative emission (proportional to Kelvin temperature raised to the 4^th power, or “T^4”) goes up for a 1K increase in temperature at all locations across the surface of the earth:

First, I approximated the “change in emission for a 1K change in temperature at a particular location” with “ T^3 “ (Kelvin temperature raised to the third power) – basically, the “differential of radiative emission” (ignoring constants).

Then, I integrated “T^3” over the earth’s surface, using a piecewise linear function that closely approximates the “Temperature VS Latitude” relationship (shown superimposed on the above graphic) to represent the temperature “T” at each location on earth.

I found that this result differed by a relatively small percentage from (was about 1.8% less than) the result obtained by assuming that each square meter of the earth radiates at the global mean temperature (about 288K). I estimated (in an update to the initial post) that the uncertainty associated with the surface integral result obtained using the Temperature VS latitude relationship was about +- 6.5%. These results are summarized below:

I used the “2-sigma” curves (above and below the central (black) “temperature VS latitude” curve) to “bracket” the uncertainty associated with the surface integral obtained with the temperature relationship shown with the central Temperature VS latitude curve. First, I did the surface integral of T^3 using the “temperature VS latitude” relationship shown by the upper 2-sigma curve. Then, I did the integral using the relationship shown by the lower curve.

However, this "bracketing" method significantly overestimates the probable uncertainty in the surface integral of T^3 for the central curve because it assumes that all the locations at each latitude either emit at the temperature indicated by the upper curve or emit at the temperature indicated by the lower curve, when, in fact, at each latitude, there will be locations with temperature both above and below the central curve.

A more detailed error analysis narrowed the uncertainty and indicated just why using the mean temperature for each latitude (indicated by the central temperature VS latitude curve) -- instead of the detailed temperatures "scattered about that mean" -- introduces very little error into the "radiative emission increase for a 1K temperature change" result.

Errors in the surface integral calculation due to use of the Temperature VS Latitude Curve?

First, when it comes to calculating total emission increase at each latitude for a 1k increase in temperature, what is the error associated with using the mean value for each latitude (ie, the central "Temperature VS latitude" curve) rather than all the values that went into that mean?

On the “Temperature VS Latitude” graphic above, the annual mean temperatures for a number of different locations at a given latitude are averaged together to give the mean value for that latitude and collectively, these mean values are what is shown by the central black “Temperature vs Latitude” curve.

At each latitude, some of the locations will have annual mean temperature above the central black curve and some below, but the sum of all these deviations from the mean temperature for each latitude is zero.

By virtue of their slightly different temperatures, each of the different locations at a given latitude contributes a slightly different amount to the overall emission increase for the latitude as a whole. In other words, when the temperature goes up by 1K at all locations at that latitude, radiative emission does not go up by precisely the same amount at all locations.

The amount differs very slightly due to the slightly different starting (base) temperature at each location. This is because the amount by which emission goes up at each location is proportional to T^3 (ignoring constants) when temperature increases by 1K (the "differential emission" approximation used in our surface integral).

A location with base temperature (temp before the 1K increase) slightly below the mean for that latitude (dark line) contributes slightly less to the total radiative emission increase for that latitude than does a location with base temperature equal to the mean for the latitude. Similarly, a location with temperature slightly above the mean for that latitude contributes slightly more to the emission increase than does a location with base temperature equal to the mean.

As it turns out, the "slightly lesses" almost balance the "slightly mores", so the net effect is that we are left with a surface integral of T^3 (ie, for radiative emission increase) that is very nearly the same as the result obtained using the mean value for that latitude, which is shown by the central “Temperature vs Latitude” curve. In other words, we get essentially the same result that we get by assuming that all locations at that latitude emit at the mean temp for that latitude.

For illustrative purposes, I will make an approximation (not made in my original surface integral) that greatly simplifies the analysis and makes clear the “balancing” effect. The approximation relies on the fact that at each latitude, every location with temperature lying between the 2-sigma curves has temperature that deviates by a relatively small amount “Delta_T_i” from the temperature value for that latitude (T_o) shown by the (central) "Temperature VS latitude" curve

IMPORTANT NOTE: “Delta_T_i” in this case is not the constant 1K by which temperature was assumed to go up at each location (ie, not the change in temperature associated with the "differential emission" approximation).

Delta_T_i in this case is the deviation of the base temperature at a particular location at a particular latitude from the mean value for that latitude (ie, a point on the central "Temperature VS latitude" curve). For the temperatures lying between the 2-sigma curves on the Temperature VS latitude graphic (which applies to the vast majority of locations on earth), Delta_T_i decreases from a maximum of about 10K at the pole (where T_o = 250K) down to about 4K at the equator (T_o = 298K). So, at the pole, Delta_T_i / T_o < = 0.04 and at the equator, Delta_T_i / T_o <= 0.013

First, at each location at a given latitude, the temperature can be represented as

T = T_o + Delta_T_i

where Delta_T_i is the difference (positive or negative) of the temperature for a particular location at that latitude from the mean value T_o (indicated by the point for that latitude on the central “Temperature Vs Latitude” curve). The radiative emission increase for a 1K temperature increase at each of these locations is proportional to T^3 (to within a small error of about 0.5%, which will be ignored for the moment), where we have ignored “emissivity”, “Stefan Boltzmann” constant and a factor of “4”.

The radiative emission increase for a 1K temperature increase at each location -- (T_o + Delta_T_i)^3 -- can therefore be represented thus:(not showing Stefan Boltzmann constant, emissivity and factor of "4" for the "differential emission" approximation)

(T_o)^3 + 3(T_o)^2 (Delta_T_i) + 3T_o(Delta_T_i)^2 + (Delta_T_i)^3

We can, approximate (T_o + Delta_T_i)^3 with a sum of just the first two terms above

(T_o)^3 + 3(T_o)^2 (Delta_T_i)

where we have dropped the higher order terms " 3T_o(Delta_T_i)^2 " and " (Delta_T_i)^3 " because they are very small compared to the first two terms.

For example, at the pole (T_o ~= 250K), if Delta_T = 10K above the central Temperature VS Latitude curve (on the "2-sigma" curve, where T ~= 260K), the sum of the last two terms

A = 3T_o(Delta_T)^2 + (Delta_T)^3

is only about 0.4% of the sum of the first two terms

B = (T_o)^3 + 3(T_o)^2 (Delta_T)

At the equator (T_o ~= 298K), if Delta_T = 4K above the central curve (on the "2-sigma" curve, where T ~= 302K) , "A" is only about 0.05% of "B". In general, most temperatures will be closer to the central curve than 2-sigma away, which means that A / B fro most cases will have value smaller than the value at 2-sigma away.

Now, to find the integral of T^3 for all the locations at one latitude (ie, “radiative emission increase for a 1K change in temperature”), we add up the contributions made by all the locations at that latitude.

But using the above approximation, the only thing that changes for the individual contributions is “Delta_T_i” So when we add these up, we get the total number of temperatures "N" times " (T_o)^3 "plus a sum of terms involving a constant " (T_o)^2 " times the sum of many different Delta_T_i’s.

But the Delta_T_i’s are both positive and negative and actually sum to zero, because the temperatures for all these locations average to ‘T_o.” In other words, some have temperatures above the average and some below, but the net deviation in temperature from the average (sum of all Delta_T’s) at that latitude is zero.

So, we find that, apart from a very small error involved in the approximation we made for our illustration here, the total contribution to the surface integral of T^3 from all the locations at a given latitude is the same as the result we get by assuming that all these locations are emitting at the mean temperature for that latitude -- ie, the value corresponding to the point on the central "Temperature VS Latitude" curve for that latitude.

The same conclusion holds for every latitude -- ie, the Delta_T_i’s sum to zero, leaving only “ N(T_o)^3 “ -- and the error introduced by dropping higher order terms in Delta_T_i is small, so use of the central “Temperature vs Latitude” curve is indeed an accurate way to calculate the surface integral of T^3 (where T varies). It does not introduce an appreciable error into the surface integral.

The approximation we made here for illustration involves an error (albeit small), because the higher order terms involving (Delta_T_i)^2 and (Delta_T_i)^3 do not cancel out as do the terms involving (Delta_T_i)^1.

For example, the next higher order term in the expansion of (T_o + Delta_T_i)^3 involves (Delta_T_i)^2 , so the contribution from all locations at each latitude is a net positive. But the collective contribution of these terms to the total surface integral of T^3 is less than about 0.1%. -- and the contribution from the (Delta_T_i)^3 term is smaller still.

It turns out (click link below) that the "fractional error' involved in keeping only the first two terms " (T_o)^3 " and " 3(T_o)^2 (Delta_T_i) " in the above sum of (T_o + Delta_T_i)^3 over all the locations at a given latitude can be expressed as

Fractional error ~= 3(sigma)^2 / (T_o)^2

where T_o the mean temperature and "sigma" is the standard deviation of temperatures at each latitude.

We can get the standard deviation at each latitude from the "Temperature VS latitude" graphic, which shows both "sigma" and 2-sigma" curves in addition to the central curve showing the mean temperature for that latitude.

At the pole where sigma = 5K, T_o = 250K, the above error has its maximum value of about 0.001 or 0.1%. The assumption that this is the error at all latitudes then leads to the conclusion that the overall error in the integral of T^3 over the earth's surface due to the source under discussion is not greater than about 0.1%.

The cancellation of positive and negative Delta_T_i's at each latitude means that the error associated with using the mean value for each latitude in the surface integral of T^3 (ie, to calculate total emission change for a 1K change in temperature) rather than considering all the temperatures scattered about that mean value at each latitude is very small .

For details on the above calculations, click here. (Note: after bringing up the first page, click on the small "emission (Set)" graphic to view the next page. 8 total).

There are two other sources of uncertainty in the surface integral of T^3 related to use of the Temperature VS Latitude curve, both of them relatively small:

1) The error associated with my piecewise linear approximation to the Temperature VS Latitude curve:

In my previous post, I found that the surface integral of T^3 calculated using approximations to the upper and lower 2-sigma curves differed from that calculated using my approximation to the central “Temp vs latitude” curve by about +- 6.5%. But it is clear that my piecewise linear approximation is much closer to the central curve than is either of the 2-sigma curves (lower or upper).

Although I can not say precisely what error the use of my piecewise linear approximation introduces into the surface integral of T^3, the above bracketing allows me to gauge the approximate size of that error.

The latter error is certainly significantly less than 6.5%. If I had to guess, I'd say somewhere around 1%, but to find the precise error, I would have to integrate T^3 over the actual Temp VS latitude curve and compare that to the answer obtained with my approximation to that curve. But for the purpose here, I don’t think that is necessary, given that the other errors turn out to be so small. Even if the error associated with my piecewise linear approximation is actually 2%, the overall error (from all sources) would still not be much greater than this.

2) The uncertainty in the “Temperature VS Latitude” curve itself (ie, uncertainty in the mean temperature value for each latitude).

The final error introduced into the surface integral by the uncertainty in the “Temperature VS Latitude” curve itself is very small (0.1% or less). To see this, assume that the error associated with the annual mean temperature for each location at a given latitude is about 0.1C (probably a conservative estimate). Also assume that there are just 10 locations at each latitude going into the mean value for that latitude (again, probably a conservative estimate).

Under those assumptions, the overall 2-sigma uncertainty in the mean for that latitude will then be approximately

2 ( 0.1C / SquareRoot [10] ) = 0.06C.

This uncertainty in temperature will introduce a very small (less than 0.1%) error into the surface integral of T^3 for each latitude and into the total surface integral over all latitudes.

Additional source of uncertainty in the surface integral calculation

There is an error associated with my original assumption that "the change in emission is proportional to 4(T)^3 when temperature changes by only 1K" (ie, for a temperature change that is small compared to T) . In other words, this is the error associated with using the "differential approximation" for emission increase for a 1K increase in temperature in the integral -- ie, integrating 4(T)^3 over the surface of the earth -- rather than integrating the quantity "(T + 1)^4 - T^4 " (which is even worse!).

As it turns out, the error associated with the "differential approximation" is about 0.5% at all latitudes (between 0.6% at the pole and 0.5% at the equator). But in my analysis, I used the differential approximation for all my emission change calculations (including that involving the global mean) and at the end I took a ratio of the "surface integral of T^3 assuming the 'Temperature Vs Latitude' relationship" to the "surface integral of T^3 assuming constant 288K (global mean)". Since the error involved in the T^3 approximation is almost the same (varies from 0.6% to 0.5% ) for all temperatures over the relevant range (250K - 298K), it's overall impact in my ratio is less than the 0.5% quoted above (more like 0.1%).

Total impact of errors on my original conclusion?

Even if one adds the relatively small errors found above to the "1.8% difference" that we obtained previously, we are still left with the same basic conclusion: the calculated radiative emission increase for a small (~1K) temperature increase across the globe”(ie, the surface integral of T^3) is essentially the same whether one takes into account the different individual base temperatures across the earth or uses the global average temperature. The two results most probably do not differ by more than a few percent.

Why the results are indeed very close

By now, the reader should have at least an inkling of just just why the total “radiative emission change for a 1K temperature change” comes out nearly the same (most probably within a few percent) whether one takes into account the specific temperatures across the earth’s surface or simply uses the global average.

From the start, it seemed very unlikely to me that it was mere coincidence that the surface integral of T^3 came out nearly the same whether I used the global average or the detailed Temperature VS latitude relationship for temperatures across the earth.

In the process of revisiting the errors, I discovered just why the answers must come out nearly the same. It essentially boils down to the fact that the “radiative emission increase for a 1K temperature increase” at each location on the earth can be closely approximated by an expression that is comprised of a term that is proportional to the global average temperature (T_av) raised to the third power plus a term involving the deviation (Delta_T_j) of the temperature at that location from the global average:

Emission increase ~= (T_av)^3 + 3(T_av)^2 (Delta_T_j)

where the temperature at each location is expressed as T_av + Delta_T_j

When one sums the contributions from all the different locations across the earth, the (+ and -) terms involving Delta_T_j (raised to the first power) cancel out because of the way the global average is obtained.

The higher order terms involving (Delta_T_j)^2 and (Delta_T_j)^3 do not cancel, but because they are small compared to the other terms, the integral comes out very close to a sum of the the terms involving the global average temperature raised to the third power -- ie, a result that is proportional to N(T_av)^3 , where N is the total number of locations contributing to the sum (and where I have not shown the Stefan-Boltzmann constant, emissivity or the factor of "4" for the "differential approximation" for emission change)

The "fractional error'" involved in keeping only the first two terms " (T_av)^3 " and " 3(T_o)^2 (Delta_T_j) " in the above sum of (T_o + Delta_T_i)^3 over all the locations on earth can be expressed as

Fractional error ~= 3(sigma)^2 / (T_av)^2

where T_av is the global mean temperature and "sigma" is the standard deviation of all surface temperatures across the earth that have gone into the global mean.

In other words, this is the amount by which we would err in our surface integral of T^3 (ie ,in our estimate of "radiative emission increase for a 1K temperature increase") by assuming that all locations on earth emit at the global mean temperature (about 288K) rather than at the individual temperatures they actually emit at.

I honestly can not say what "sigma" is for the earth's surface as a whole (ie, for all temperatures that have gone into the global mean value), but I can say what it is not. It is certainly not 38K or larger -- that being the difference between the annual mean temperature at the pole (250K) and the global mean temperature (288K).

But if sigma were that large -- ie, sigma = 38K -- then the

Fractional error = 3(sigma)^2 / (T_av)^2 ~= 0.05 (about 5%)

We know sigma is smaller than 38K, but even if it were 38K and we used the global average temperature to estimate the emission increase for a 1K temperature increase across the globe, we would err by only 5%. We would still get an answer for total emission increase that was within 5% of (about 5% less than) the answer we would get by considering the details of surface temperature across the globe.

Because sigma is actually less than 38K, however, we know that the answer we get using the global mean has to be even closer to the answer using the detailed temperatures. For example, if sigma were 20K (probably not an unreasonable guess), the above fractional error would be just 0.014, or our result using the global mean would be just 1.4% less than the actual result obtained by using all the individual temperatures to calculate the emission increase.

I give a straightforward derivation of the above result here. (Note: after bringing up the first page, click on the small "emission2 (Set)" graphic to view the next page. 5 total).