## A Flanner in the Works for Snow and Ice calculations

#### Posted on 23 January 2011 by MarkR

**Endnote and calculations (warning, maths!)**

I was always told to show my working so that everyone else could see where I'd gone wrong. So here's some maths for those unfamiliar with climate feedbacks. It's an endnote to my examination of a new paper, those who want to read something interesting should go there.

The change in temperature of the climate in response to a forcing F can be written:

We then define a feedback factor, Y:

Which turns the first equation into:

where Y has been decomposed into a sum of N Y_{i} values, each representing an individual feedback like snow, clouds or water vapour.

We can further decompose each feedback into a product of differentials:

Where φ_{i} is the i^{th} feedback and the differentials explain how it is related to temperature, and how the radiation response is related to it. So far all we've done is use the properties of differentials in an entirely mathematical sense and then made 2 assumptions: that we see an approximately linear response and that the feedback parameter can be split into some linear superposition of individual elements that can be measured.

In the case of snow and ice we need to know its change in albedo for a change in temperature and the change in flux for a change in albedo to calculate Y using the above equation. The change in albedo requires some more information, all of which can be measured by satellite and this is how Flanner *et al* calculated their feedback parameter.

The first feedback calculated is usually the 'blackbody feedback' - as Earth warms up it radiates more and this has a value of about 3.3 W m^{-2} K^{-1}. By definition, positive feedbacks are subtracted from this value, whilst negative ones are added (blackbody is positive since it dumps heat from the Earth as temperatures rise). You can use this to do calculations by knowing the initial radiative forcing: a common one is the 3.7 W m^{-2} you get from doubling CO_{2}. Therefore the 'climate sensitivity' or temperature change from a doubling of CO_{2}, assuming approximate linearity in the range we're looking at so that ΔT and ΔF can be used is:

where the minus sign comes from the convention of now taking positive feedbacks to be a positive value of Y, and the summation starts at i=2 because I gave the blackbody feedback the index of 1 and it is included as the 3.3 in the denominator.

A climate sensitivity of 3 °C is the classic middle value given by the IPCC. This implies a Y value of about 1.2 using the model results. If Flanner *et al*'s values are correct and are maintained, then global feedback is increased to 0.39 from the current estimate of 0.2, assuming that Southern Hemisphere changes match model predictions. This new value of Y boosts the sensitivity to about 3.5-3.6 °C, which is a 20% increase.

This is, of course, an approximation and I haven't bothered to apply this modified Y to the entire probability distribution function of climate sensitivities, but now you have the tools to DIY!

**Further edit**

Ken Lambert was unhappy with 'jumping' straight to equilibrium, but wanted to see what happened when you looked at the change moment by moment. So here we go, I'll try the *transient* climate change case and we will find exactly the same answer as above!

Assume that there is some 'forcing' applied to the climate, dF, and that the Earth changes in response to this. It will warm up or cool down and this will change the rate of heat escape from Earth. e.g. dF causes a warming so the Earth radiates more. I'll call the feedback df.

So the net heat flow dQ is:

However, the feedbacks are a function of temperature, so we can re-write the df using the basic properties of calculus:

And by defining the differential as the letter 'Y' we have our feedback parameter again. By measuring between two finite temperatures, we can take the mean of the path integral of the Y from T1 to T2 as a value of Y that is valid in the equation:

The climate has heat capacity C, and the change in temperature must be the total change in energy in the system (dΔQ.dt) divided by the heat capacity, which you can rearrange using the properties of calculus to:

and then substitute this value for ΔQ into the original heat flow equation to get a first order differential equation:

This can be solved by integrating factor, and we take t = 0 when ΔF = 0. The solution is:

The solution for a general t at constant ΔF is:

i.e. as t becomes very large, the exponential tends to 0 so you are left with:

which is the same result as before!

Steve Bloomat 08:28 AM on 24 January, 2011Pete Dunkelbergat 18:15 PM on 24 January, 2011MarkRat 20:40 PM on 24 January, 2011Pete Dunkelbergat 01:45 AM on 25 January, 2011MarkRat 02:56 AM on 25 January, 2011ifthe current measurements are representative of what we ultimately get then this means a big boost to climate sensitivity. But I also tried to make it clear that perhaps it's a case of things just melting more quickly than we expected to begin with and maybe albedo feedback will 'slow down' and head back to model values. All the paper shows is that models underestimate what we expect to see so far, we can probably be confident about that (but ofc we need to see further work for confirmation!) - but more work is needed to have greater confidence that we've underestimated the long term feedback.roccoat 08:33 AM on 25 January, 2011MarkRat 10:51 AM on 25 January, 2011Ken Lambertat 11:47 AM on 25 January, 2011MarkRat 19:23 PM on 25 January, 2011entirelyin flux-space, and if you know each element contributing to flux (radiative forcing + feedbacks) then you would have some function to explain it. From the properties of calculus, I get dT = (dT/dF)dF. I usedchange in flux arriving at the surfaceand you have usednet radiative forcingat the top of the atmosphere - which are very different! When you assume a constant forcing just after (1), you're saying the difference between heat in and heat out is constant all the time. In a world with no feedbacks the world would warm up under a constant forcing and your value of F would actually be an F(t). You made some function of F=F(in)+F(out), which sounds sensible but can actually get very confusing. Whereas I am using some function of F(in). (which =F(out) because I'm considering equilibrium)roccoat 21:38 PM on 25 January, 2011Ken Lambertat 23:55 PM on 25 January, 2011Riccardoat 00:44 AM on 26 January, 2011MartinSat 00:47 AM on 26 January, 2011MarkRat 01:48 AM on 26 January, 2011at equilibrium, they are the result of the integral over time. Your simplistic model is not complete: a more complete model is ΔQ = ΔF - YΔT Where Q is the net flux (which you originally labelled F), F is the radiative forcing and Y is the feedback parameter. For transient calculations, you should use this equation - your original calculation assumed forcing increased at exactly the rate as feedbacks (includingheat dumped by the warming Earth's surface). My calculation was for equilibrium, i.e. where ΔQ = 0 and the equation becomes ΔF / Y = ΔT, which is what I put in this post! The time varying nature is included because you take an average Y from the path integral of the feedbacks through temperature space. Also, you don't have to integrate F(t) wrt t because the total energy in the system in the past doesn't necessarily correspond to the temperature today. Analogy: take 2 kettles, boil one and leave the other one cold. Wait a day, then measure their temperatures. Both are the same temperature, even though the integrated energy history of the one that was boiled is higher! Equilibrium is achieved whenpowerin equalspowerout.MarkRat 02:05 AM on 26 January, 2011Albatrossat 02:23 AM on 26 January, 2011Eric (skeptic)at 04:46 AM on 26 January, 2011muoncounterat 05:11 AM on 26 January, 2011Ken Lambertat 19:55 PM on 26 January, 2011Riccardoat 22:13 PM on 26 January, 2011Ken Lambertat 22:32 PM on 26 January, 2011Riccardoat 00:58 AM on 27 January, 2011MarkRat 21:00 PM on 27 January, 2011Ken Lambertat 00:09 AM on 28 January, 2011MarkRat 01:49 AM on 28 January, 2011notyour imbalance, ΔQ is. ΔF is radiative forcing; heat flow that doesn't react to the climate state on the timescales we're considering. Changes in CO2, volcanism and solar activity are examples. YΔT is the feedback sum that responds to the state of the climate. We have estimates of sensitivity (and therefore net Y in the past) from palaeoclimate. We have direct measurements of Y from the past few decades and climate models implicitly calculate it So far climate models have generally been in agreement, or underestimated observational values.Riccardoat 05:02 AM on 28 January, 2011^{-1}is the climate sensitivity. Howver, this model is a bit crude. Indeed, there are several response times relative to the various parts of the climate system. As a grossly aproximated single "effective" response time you can take something like 40-50 years. To have the time dependence of forcings and feedbacks you need to run a climate model, better if many runs of many models; something I have never seen either.Ken Lambertat 01:10 AM on 30 January, 2011MarkRat 07:44 AM on 31 January, 2011^{-1}) so every time you read a paper you have to double check all this. So where I have: ΔQ = ΔF - YΔT that means that to restore equilibrium (i.e. ΔQ = 0), if Y is smaller then you need a bigger temperature change to restore equilibrium.Ken Lambertat 01:13 AM on 2 February, 2011Riccardoat 01:59 AM on 2 February, 2011^{4}−T_{o}^{4}) for ΔT=T−T_{o}small compared to T_{o}you can write it as ΔF=4εσT_{o}^{3}ΔT=YΔT with Y independent on temperature. Similar approximations apply to any other feedback, you can always linearise something if changes are small.MarkRat 03:45 AM on 2 February, 2011^{-2}from an average of over 400 W m^{-2}. So we are looking at mathematically 'small' changes, and the approximation is good.MarkRat 20:31 PM on 2 February, 2011_{2}^{3}/ T_{1}^{3}gives you the fractional change in feedback factor from T_{1}to T_{2}. For 288 K to 291 K it's a change of ~3% so linear feedback isn't a bad approximation here.Ken Lambertat 22:33 PM on 2 February, 2011MarkRat 01:42 AM on 3 February, 2011^{-2}. The 'exact' model calculates 2.939 W m^{-2}. The fractional error in changes seen so far from linearisation is 0.46% or 0.013 W m^{-2}. Absolutely tiny compared to the full fluxes, and effectively impossible to measure to that precision in the climate anyway. Full models include the 'exact' version, this simple linear model appears to be a good approximation according to them. For the blackbody feedback it's very good, for the others it seems to be reasonable.Ken Lambertat 00:23 AM on 7 February, 2011Riccardoat 00:38 AM on 7 February, 2011muoncounterat 02:16 AM on 7 February, 2011MarkRat 05:38 AM on 8 February, 2011faster than exponentiallywith time. Even if it was 'just' exponential then you would expect accelerating global warming because the time taken to return to equilibrium doesn't increase at the same rate as the final expected warming does.Ken Lambertat 23:33 PM on 8 February, 2011Riccardoat 03:33 AM on 9 February, 2011Ken Lambertat 00:14 AM on 11 February, 2011