A Flanner in the Works for Snow and Ice calculations

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:

$\Delta T = \Delta F / \sum_{i=1}^NY$

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

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

Where ?i is the ith 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 CO2. Therefore the 'climate sensitivity' or temperature change from a doubling of CO2, assuming approximate linearity in the range we're looking at so that ?T and ?F 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:

$dQ = dF - df$

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

$df = \frac{df}{dT}dT$

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:

$\Delta Q = \Delta F - Y \Delta T$

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

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

$\frac{d\Delta T}{dt} + \frac{1}{C}[Y\Delta T - \Delta F] = 0$

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

$\Delta T(t) = e^{-Yt/C} \int_0^t \frac{\Delta F}{C}e^{Yt'/C} dt'$

The solution for a general t at constant ?F is:

$\Delta T(t) = \frac{\Delta F}{Y}[1-e^{Yt/C}]$

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

$\Delta T_{eq} = \frac{\Delta F}{Y}$

which is the same result as before!

Posted by MarkR on Sunday, 23 January, 2011