## Monday, February 18, 2008

--Maybe...or maybe not!*

*You may want to skip reading what is below and just flip a coin.

It has come to the attention of the “Royal Order of Anonymice” that, just as with real estate, the three most important things with regard to "global average temperature", are (in order of importance) “Location, Location and Location”.

More specifically, as Climate Scientist Roger Pielke, Sr has indicated:

“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. This also means that using the mean daily temperature over land, rather than a diurnal cycle where the long wave emission from the surface is substantially larger at the time of maximum daily temperature, introduces yet another error.

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.

Having had our little mousy curiosity piqued (and being somewhat skeptical of real estate wisdom) we wish to investigate a couple things for ourselves:

1) Whether the diurnal (day/night) variation in temperature at a particular location on the earth's surface has any appreciable effect on the time integral of the radiative emission (total energy radiated per unit area over one day) ; in particular, whether it makes the integral appreciably different from that obtained using the 1-day average temperature at that location.

2) Whether the variation in temperature with latitude has any appreciable effect on the surface integral of the radiative emission (total energy radiated per unit time by the surface of the earth) ; in particular, whether it makes the integral appreciably different from that obtained using the global temperature average.

We will not be debating the statement (above) that “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.” This follows directly from the Stefan-Boltzmann law, which says that "radiative emission" (energy radiated per unit time per unit area) for a body at temperature T is proportional to its temperature to the fourth power (T^4).

But it should be pointed out at the getgo that what is most important with regard to the energy budget of the earth as a whole ("global radiative imbalance", if earth is taking in more energy than is being emitted, for example) is not the energy radiated per square meter per second at any given location on the earth at any given time, but instead the surface integrated radiative emission from the entire globe. So, what we do below will be an effort to address that.

First we will make a reasonable assumption about how temperature (and hence radiative emission) varies at a given location over the period of a full day (ie, including night) and see how that impacts the time integral of the radiative emission over that day at that location. We will be interested in discovering if the result differs significantly from the result we would get by using the (arithmetic) average temperature (T_av) for that same day at that location. In the latter case, the time integrated emission is proportional to T_av raised to the fourth power -- ie, (T_av)^4 -- multiplied by the total time (1 day, in this case).

Next, we will make some reasonable assumptions about how temperature (and hence radiative emission) varies over the globe (as a function of latitude – or more precisely “co-latitude”) and explore the actual impact of that on the surface integral of the radiative emission (ie, the total energy radiated per unit time by the earth's surface) relative to what it would be if every square meter of the globe were assumed to be at the global mean temperature (288K).

Summary

Before we go into details, let me state what I did and give the results:

I first integrated radiative emission over one day at a location having an average temperature for that day of 275K and a total diurnal variation of 20K, assuming that temperature varies sinusoidally over the day. This result deviated by only 0.4% from the result obtained by assuming that the emission for the entire day occurred at the average temp (275K).

I next integrated "change in radiative emission for a 1K change in temperature" over the entire earth's surface, assuming that temperature varies sinusoidally with latitude, from 250K at the pole to 300K at the equator. The answer I got deviated by only 4% from the answer one would get by assuming that every location on the earth's surface was radiating at the global mean temperature (288K). [the percent deviation might even be less: see "update" below]

Update2: I goofed!! I did not include the uncertainty estimate in my answer, which turns out to be on the order of +- 12.5% (or perhaps bigger)! The difference may indeed be significant (if one considers 10-15% significant, at least). It is really not possible to say (at least not to any great degree of certainty) based on the temperature graph I used to estimate temperature across the earth's surface. The errors are just too big.

Note: this does not mean that the difference between the two results is now 12.5%. It means that it is impossible to say one way of the other whether the 4% difference that I calculated is accurate. The actual difference might be 4%, or it might be less, or as big as 16.5%. But that does not help at all in deciding whether the two methods give the same result to within 10% or so (if that is what one considers significant).

Update 3: Using a better approximation to the temperature vs latitude relationship across the globe, the difference between the two results (obtained assuming a varying temperature and a constant global average temperature) was further narrowed from about 4% to about 2% and the uncertainty was also narrowed from the above-stated 12.5% to about 7%. The new result allows one to say with some confidence that the difference is most probably less than 10%. See details of "Update 3" below.

So, in order to settle the question, one would need actual temperature data at locations across the globe, which the global climate models already use to do the emission calculation, at any rate. (**see Pierrehumbert comment at end).

If nothing else, this illustrates the limitations of very simple models. They are useful for order of magnitude estimates but of rather limited use when one is trying to determine whether two quantities are within about 10% of one another.

Warning: If you want definitive conclusions, you may want to stop here!

Effect of diurnal Temperature variation on emission integrated over 1 day?

First, let's consider the effect of the diurnal (day/night) variation in temperature at a particular location on the earth's surface on the integrated emission for one full day.

We know the temperature variation over the day/night cycle leads to a change in the radiative emission from one time of the day to another, but we want to know if it also affects the integral of the radiative emission over 1 day (ie, the total energy emitted by unit area in one day at that location).

Specifically, instead of taking into account these daily variations when we are trying to determine how much energy is radiated back to space per unit area over the course of 1 day (and night) at a particular location, can we safely assume that the time integral of the radiative emission is essentially (within some small error) equal (ignoring constants) to the average temperature at that location on that day raised to the fourth power? That’s the question we would like to answer.

The temperature at a given location on a given day can be fairly accurately represented by a sinusoidal oscillation about the arithmetic average for that (full) day as shown in this graphic.

For example, let us represent T as a function of time “t” at a particular location on a particular day in the following way:

T_t = T_av – 10cos[(Pi)(t)/12]

where T_av is the arithmetic average of the temperature at that location for that (full 24 hour) day.

The above equation assumes temperature starts (at time t=0) at a minimum value of T_av – 10, and then ramps up co-sinusoidally to a maximum value of T_av + 10, twelve hours later.

The total temperature change from min to max is 20K degrees, though the actual amount is not critical to our argument below. This is a reasonable daily change for one location on the surface of the earth, at any rate.

For calculation purposes here, we will also assume that T_av = 275K, which corresponds to a realistic value for the surface of the earth at some location .

As you can see in this graphic, temperature values for different locations on the earth vary between about 250K and 300K.

By Stefan-Boltzmann, the radiative emission (E_t) as a function of time at the given location is proportional to the temperature T_t raised to the fourth power

(T_av – 10cos[(Pi)(t)/12] )^4

We get the total energy emitted per unit area at that given location for the given day by integrating E_t over a full day.

If we perform that integral, we find that E_total per unit area is equal (leaving out constants and showing only the temperature dependence, which is the only thing that might change) to

1.004 (T_av)^4

In other words, the time integral of the radiative emission for that day differs by only about ½ of one percent from the result we would get if we had used the average temp T_av for that day raised to the fourth power.

It’s fairly easy to see why the two answers do not differ appreciably. For example, consider the emission for the two times separated by twelve hours: time t=0, with temperature T_0 = 265 and time t=12, with temperature T_12 = 285.

If we add together the energy radiated per unit area for each of these two hours, we find that the total emission (total energy radiated per unit area ) is equal to the sum (265)^4 + (285)^4. (Note: We are making a simplifying approximation in assuming that the temperature remains at one value for the full hour, of course. We made no such approximation above when we actually did the integral.)

This sum turns out to be 2.016 times the result for 275K. In other words, SUM = 2.016(T_av)^4 where T_av = 275 in this case. If we were instead using T_av to calculate the combined emission over the two hours, we would have found a number equal to 2(275)^4 for the summed emission. But instead we find 2.016(275)^4, and since 2.016/2 = 1.008, the error is 0.8% (less than 1%)

We can pair up other temperatures throughout the day in a similar manner (ones that when added together give approximately 2X (275)^4 + some small error . For every temperature above the average, there is a corresponding one below it that yields a similar offset from the emission at 275K, but with an opposite sign (so the offsets tend to cancel when we add them together). But note that the particular two we chose to pair up (265K and 285K) give the maximum error because they lie at the extremes of the temperature variation about the average value (275K in this case).

In fact, adding the 1-hour integrated radiative emission corresponding to a temperature just 1K degree above T_av to 1-hour integrated radiative emission corresponding to a temperature 1K degree below T_av gives a value for the summed (two hour) emission that deviates almost not at all from 2(275)^4. The error for each of the pairs that we can form for the hours between 0 and 12 varies between 0 and about 0.8%, so the average error is about half this, 0.4%, which is just what we determined by actually doing the full integral above.

So, let us reiterate what we have found here:at a particular location on the surface of the earth, the radiative emission integrated over one day comes out the same (to within ½ percent) whether one takes into consideration the actual temperature (and therefore emission) of the surface at each given time during the day, or whether one assumes that all the emission occurred at a temperature of T_av (275K in the above case).

Finally let us note that the specific T_av (275K) and diurnal change (20K) that were used in the analysis above are in no way critical to the conclusion. As long as one uses a reasonable T_av for one location on the earth’s surface (between about 250K and 300K) and a plausible diurnal temperature variation (about 20-30K), the main conclusion is the same: It just does not make much difference to speak of.

Next, we will consider just what effect the variation in temperature with latitude has on the integral of radiative emission over the earth's surface.

Effect of variation in temperature with latitude has on the integral of the radiative emission over the earth's surface?

Most everyone who is not comatose knows that temperature varies over the globe. We will ignore variation with altitude for now and just consider variation with latitude. The graphic shown here indicates how temperature varies with latitude. It is basically a sinusoidal variation with a constant added to it.

In fact, the relationship can be represented very well by the equation

T_phi = 275 - 25 cos (2phi)

where T_phi is the temperature in degrees Kelvin as a function of the co-latitude "phi".

phi = 0 corresponds to the poles (with a mean temperature of about 250K)

phi = pi/2 corresponds to the equator (with a mean temperature of about 300K)

The good match between the above sinusoidal function (shown in light blue) and temperature as a function of latitude (black line) is shown in the graphic below

Original graph is by Robert Rhode (Global Warming Art)

2) Total radiative emission (total energy radiated per unit time per unit area) is proportional to T^4, but we will be concerned here primarily with the change (eg, increase) in radiative emission that accompanies a small change in the temperature at a given location on the earth’s surface.

If one considers a "small" enough change in temperature (“small compared to the starting value”), the change in radiative emission is essentially equal to the differential of sigmaT^4, -- ie, proportional to the third power of the starting temperature (T^3) and also proportional to the temperature change dT, where dT is in Kelvin degrees.

For our purposes here, we will assume dT is 1K, which certainly is very small compared to the temperatures we will consider here (the temperature of the earth's surface varies between about 250K and 300K from the poles to the equator, and 1 is certainly small compared to 250 or 300. So this approximation makes very little difference to our answer.)

delta_E is proportional to (T^3)dT = T^3 (for 1K change)

Note: Though the earth is not a black body, the T^3 dependence for differential (small temperature change) radiative emission still holds. The normal constants out front (emissivity and Stefan Boltzmann constant) are not important to the analysis here, so we have omitted them.

3) Plugging the temperature T_phi (temperature as a function co-latitude) into the differential radiative emission equation (ie, the one with the T^3 dependence), we find that the radiative emission change for a 1K temperature change varies across the surface of the earth approximately as

(T_phi)^3 = [275 - 25cos(2phi)]^3

= (275)^3 [ 1 - (25/275)cos(2phi) ]^3

= (275)^3 [ 1 - (1/11)cos(2phi) ]^3

4) Note that the change in radiative emission (delta_E_phi) for a 1K change in temperature near the pole does differ significantly from the change in radiative emission for the same 1K change near the equator. At the equator, delta_E_phi is about 73% larger than it is at the pole.

But what we really need to estimate is the change in the radiative emission for the entire surface of the earth associated with the 1 K temperature change (the change in the total energy radiated per unit time by the earth's surface). One gets this by integrating the differential radiative emission over each square meter of the earth's surface, taking into account the temperature at each location (using the above formula for delta_E_phi) Note: I am not going to worry about the diurnal temp variation here because we found above that using the average value at a particular location is quite legitimate – ie, does not introduce an appreciable error.

Since we wish to know the change in the energy radiated per unit time (due to a 1K change in temp at all locations on the surface of the earth), we need to integrate delta_E_phi over the earth's surface, where T_phi is given as the above function of the co-latitude.

5) In spherical coordinates, an element of surface area is given by

R^2 sin(phi) dPhi dTheta

where “phi” is co-latitude, “theta” is longitude, R is the earth's radius and “dPhi” and “dTheta” are the respective differentials of the two angles.

So, the increase in energy radiated per unit time by the earth's surface due to the 1K temperature increase is given by

SURFACE INTEGRAL of delta_E_phi R^2 sin(phi) dPhi dTheta

If we do this integral (it’s pretty straightforward, so I won’t bore you with the details) with the delta_E_phi (ie, temperature dependence) given above, we find that the change in emission from the entire surface of the earth due to the one degree K temperature change is equal (again ignoring constants and showing only the temperature dependence, which is the only thing that might change) to

1.10(275)^3

So, how does this compare with the result that we would get by using the global mean temperature in our emission equation?

It is very important to note here that 275K is not the global mean temperature, so it is not simply a matter of reading the "1.10" in front of (275)^3 and concluding "the result is 10% greater than the result obtained by assuming that the whole earth surface radiates at the average".

If the entire surface of the earth were at the actual global mean temperature (288K), the emission change due to the 1 K change in temperature would be equal (ignoring constants) to (288)^3 so this is what we must compare the former result -- 1.10(275)^3 -- to

The former result as a fraction of the latter result gives

1.10(275)^3 / (288)^3 = 0.96

Again, this compares the change in the energy radiated per unit time from the entire surface of the earth (due to a 1K change in temperature) obtained with two different methods:

Method 1) takes into account the variation in temperature with latitude across the earth's surface (yielding 1.10(275)^3 )
Method 2) assumes every square meter of the earth radiates at the average temperature 288K (yielding (288)^3 )

Update: The above comparison (proportion) uses the global mean temperature determined experimentally (288K). If we average our assumed temperature variation [275 - 25cos(2phi)] over the globe, we get a value of 283.3K. If we then make the same comparison as before, but instead use 283.3 for the global mean, we get for the proportion of the two results

1.10 (275)^3 / (283.3)^3 = 1.006

which makes the two emission results even closer (within 0.6%)

The fact that we get a value of 283.3K (instead of 288) when we average over the surface means that the approximate function we assumed for temperature variation over the surface is not perfect (!). No surprise there. But given it's simplicity (crudeness?) and given the error margin on the temperature vs latitude graph from which it was devised, it ain't bad at all. After all, it gives a value within 2% of the actual global mean temperature.

Of course, I could adjust my temperature function so that it yields a global (surface) average that matches the experimental value even better and then go back and recalculate things. But I'm not going to do that. My hunch is that it won't make much difference as far as the conclusion is concerned. ( I did just that: See "Update 3" below).

Update2: I goofed!! I did not include the uncertainty estimate in my answer, which turns out to be on the order of +-12.5% (or perhaps more)!

Gray regions on the above temperature vs latitude graph indicate the first and second standard deviations for the temperature.

To get an idea of the error involved using the method I used, assume the error in temp over the entire globe is +- 10K, which corresponds to the 2-sigma error shown on the above graph for the pole (where the error is greatest)

Since the change in emission for a 1K change in temperature is proportional to T^3, we can estimate the 2-sigma error associated with that emission change with this proportion for the pole (assuming a temp 250 +- 10K)

[260^3 – 250^3]/250^3 ~= 12.4%

This changes my conclusion ! I can no longer claim that the results using the two different methods are within 4% o f one another, since the 2-sigma error is 3X as large as that.

Note: this does not mean that the difference between the two results is now 12.5%. It means that it is impossible to say whether the 4% difference that I calculated is accurate. It might be 4% or it might be 16.5% (or perhaps a little bigger). But that does not help at all in deciding whether the two methods give the same result to within 10% or so.

I should also note that I did not include temperature change due to altitude, which, though most of the earth’s surface is at sea level, must be added into the uncertainty budget. This might add another few degrees to the 2-sigma temperature uncertainty, making the integrated emission uncertainty even greater.

Update 3:

Using a better approximation to the temperature vs latitude relationship across the globe, the difference between the two results (obtained assuming a varying temperature and a constant global average temperature) was further narrowed from about 4% to about 2% and the uncertainty was also narrowed from the above-stated 12.5% to about 7%. The new result allows one to say with some confidence that the difference is most probably less than 10%.

The analysis of the original post above used a sinusoidal function to approximate the variation of surface temperature with latitude.

That function is not a bad approximation to the temperature vs latitude curve, yielding as it does a global average temperature of 283K, which is within 2% of the generally accepted value of 288K.

But it is nonetheless possible to do better by approximating the temperature vs latitude relationship shown on the graphic with a piecewise linear function.

On the graphic, the latitude (and hence co-latitude) from 0 to 90 degrees is divided into ranges and a different linear function is used to represent the temperature as a function of co-latitude over each range yielding the results given below:

Click for full view

The original analysis was repeated. The new approximation yielded a calculated value for the global average temperature of 285.5K, which is yet closer (within 1%) to the accepted value of 288K than the global average calculated in the original analysis using the sinusoidal approximation to temperature vs latitude.

The global surface integral of “radiative emission change for a 1K change in temperature” was also repeated using the new (piecewise linear) temperature vs latitude approximation. The result for the surface integral was largely consistent with the original result obtained using the sinusoidal approximation.

With the piecewise linear approximation, the difference between the integral (of radiative emission increase for a 1K change in temperature) obtained using the globally varying (with latitude) temperature and a constant temperature (the global average) across the globe was even smaller (within about 1.8 %) than what was found above using the sinusoidal temperature function (within about 4%). This is to be expected, since the sinusoidal temperature function actually falls slightly below the actual temperature vs latitude line for much of the graphic.

In addition to recalculating the relevant surface integral for the new piecewise linear temperature approximation, the uncertainty associated with the result can also be narrowed.

The original estimate for the 2-sigma (95% confidence level) uncertainty , +-12.5%, was somewhat crude, assuming as it did that everywhere over the temperature vs latitude graph, the 2-sigma uncertainty in temperature was equal to 10K.

But the actual uncertainty is less than this. This was actually a maximum, based as it was on the uncertainty associated with temperature at the pole. The 2-sigma uncertainty band on the temperature vs latitude graphic actually decreases from about 9K at the pole down to about 4K at the equator (ie, it is not a constant 10K across the entire latitude range).

A better estimate of the 2-sigma uncertainty on the integral of Delta_E was obtained by approximating the upper and lower 2-sigma (gray) curves on the temperature vs latitude graph with the piecewise linear functions given below. The surface integral of Delta_E was then re-calculated using those approximations and the results compared to the value of the surface integral obtained using the central (black) temperature vs latitude curve (or, more precisely, the piecewise linear approximation to it).

Click for full view

Note: As previously, the "constants" (Stefan Boltzmann constant and emissivity, which is also assumed constant) have been left off the surface integrals and since temperature was assumed to depend only on latitude (not longitude and not altitude), only the "Phi" (co-latitude) part of the surface integral was considered (because the "theta" and "r" parts come out the same for the two cases and drop out when one takes a proportion in the end).

The narrowed estimates for uncertainty in temperature lead to a smaller uncertainty in the surface integral of the radiative emission increase associated with a 1K temperature increase at each location on the globe. The uncertainty in the surface integral is thereby narrowed from about 12.5% (the estimate obtained previously) to within about 7 % (+ 6.7% and - 6.4%, for the upper and lower temperature curves, respectively. see "Summary" graphic)

Combined with the more accurate estimate for the surface integral (using the piecewise linear temperature approximation), this narrowed uncertainty means that we can now say with some confidence that the surface integrals obtained with the two different methods

-- 1) assuming varying temperature over earth’s surface and

-- 2) assuming constant (global average) temperature across the earth’s surface

are within about 9 % (7% + 2%) of one another (perhaps less, but most probably not greater).

The graphic below summarizes the new results:

Click for full view

The line labeled "T=T_c(phi)" on the graphic above represents the global surface integral of Delta_E for the central temperature vs latitude curve (black), for which temperature depends on co-latitude, Phi. The line labeled "T=285.5K" represents the surface integral of Delta_E for T assumed to have the constant value 285.5K over the earth's surface. The line labeled "T=288K" represents the global surface integral for T assumed to have the constant value 288K (ie, the accepted value for global average) over the earth's surface. The other two lines "T_LB(phi)" and 'T_UB(phi)" are the corresponding surface integrals of Delta_E for the lower and upper 2-sigma temperature boundary (gray) curves, resp.

Though we can not claim that the actual difference between the surface integral for the case where T varies across the surface as a function of Phi -- ie, T=T_c(phi) -- and the T=288K case is only 1.8% (the calculated difference using the new piecewise linear approximation to the temp vs latitude curve), we have narrowed the range for the difference to about 9%, which allows us to say with some degree of confidence (even based on this simple model, as it were) that “the difference is most probably not 15% or bigger (most probably less than 10%, in fact).

As I indicated previously, the actual difference must be determined by calculating the surface integral of ‘radiative emission increase for a 1 K increase in temperature” at each location using the temperature at each location (not simply the temp vs latitude relationship that was derived from that data).

And what is the conclusion?

The surface integral of the radiative emission change (ie, total change in energy radiated per unit time by the entire earth surface due to the 1 K temperature change) that we get by taking into account the latitude-related temperature variations across the earth's surface is very close (within 4%) to the result obtained assuming every location on the earth's surface emits at the global mean temperature.

Update2: The difference may be significant, but it is really not possible to say based on the temperature vs latitude information on the graph I used. One would need temperature data from across the globe to settle the issue.

Is such a small difference "significant"?

Update 2: The difference may not be as small as I calculated -- and it might be significant (10% or better). It is impossible to say based on the analysis above, at least not to any great degree of certainty.

Update 3: The difference may not be as small as I calculated but is nonetheless probably less than about 10%. To narrow the uncertainty more, the surface integral would have to be repeated with temperatures at each location across the globe rather than the temperature vs latitude curve.

I suppose that depends.

For modeling certain aspects of the the earth's climate, it is undoubtedly important to know what is happening at locations across the earth as a function of time (ie, not just the overall picture) and even small errors might make a difference. The global climate models already do take such things into account, at any rate.

But when it comes to assessing whether "we need to keep the global average temperature change below some threshold (e.g. 2K)", would concern about relatively small errors really be warranted?

I'll let you decide.

**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.