Brian E. J. Rose, University at Albany
You really should be looking at The Climate Laboratory book by Brian Rose, where all the same content (and more!) is kept up to date.
Here you are likely to find broken links and broken code.
This document uses the interactive Jupyter notebook format. The notes can be accessed in several different ways:
github at https://github.com/brian-rose/ClimateModeling_coursewareAlso here is a legacy version from 2015.
Many of these notes make use of the climlab package, available at https://github.com/brian-rose/climlab
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# Ensure compatibility with Python 2 and 3
from __future__ import print_function, division
We will call this abrupt decrease in OLR the radiative forcing, a positive number in W m$^{-2}$
$$ \Delta R = OLR_i - OLR_{2xCO2} $$$\Delta R$ is a measure of the rate at which energy begins to accumulate in the climate system after an abrupt increase in greenhouse gases, but before any change in climate (i.e. temperature).
What happens next?
First note that at equilibrium we must have
$$ASR = OLR$$and in our very simple model, there is no change in ASR, so $$ ASR_f = ASR_f $$ (with standing for final.)
From this we infer that $$ OLR_f = OLR_i $$
The new equilibrium will have exactly the same OLR as the old equilibrium, but a different surface temperature.
The climate system must warm up by a certain amount to get the OLR back up to its original value! The question is, how much warming is necessary? In other words, what is the new equilibrium temperature?
We now define the Equilibrium Climate Sensitivity (denoted ECS or $\Delta T_{2xCO2}$):
The global mean surface warming necessary to balance the planetary energy budget after a doubling of atmospheric CO$_2$.
The temperature must increase so that the increase in OLR is exactly equal to the radiative forcing:
$$ OLR_f - OLR_{2xCO2} = \Delta R $$From last lecture, we have linearized our model for OLR with a slope $\lambda_0 = 3.3$ W m$^{-2}$ K$^{-1}$. This means that a global warming of 1 degree causes a 3.3 W m$^{-2}$ increase in the OLR. So we can write:
$$OLR_f \approx OLR_{2xCO2} + \lambda_0 \Delta T_0 $$where we are writing the change in temperature as
$$ \Delta T_0 = T_f - T_i $$(and the subscript zero will remind us that this is the response in the simplest model, in the absence of any feedbacks)
To achieve energy balance, the planet must warm up by
$$ \Delta T_0 = \frac{\Delta R}{\lambda_0} $$As we will see later, the actual radiative forcing due CO$_2$ doubling is about 4 W m$^{-2}$.
So our model without feedback gives a prediction for climate sensitivity:
In [2]:
# Repeating code from Lecture 2
sigma = 5.67E-8 # Stefan-Boltzmann constant in W/m2/K4
Q = 341.3 # global mean insolation in W/m2
alpha = 101.9 / Q # observed planetary albedo
Te = ((1-alpha)*Q/sigma)**0.25 # Emission temperature (definition)
Tsbar = 288. # global mean surface temperature in K
beta = Te / Tsbar # Calculate value of beta from observations
lambda_0 = 4 * sigma * beta**4 * Tsbar**3
In [3]:
DeltaR = 4. # Radiative forcing in W/m2
DeltaT0 = DeltaR / lambda_0
print( 'The Equilibrium Climate Sensitivity in the absence of feedback is {:.1f} K.'.format(DeltaT0))
Question: what are the current best estimates for the actual warming (including all feedbacks) in response to a doubling of CO$_2$?
We’ll now look at the feedback concept. Climate feedbacks tend to amplify the response to increased CO$_2$. But $\Delta T_0$ is a meaningful climate sensitivity in the absence of feedback.
$\Delta T_0 = 1.2$ K is the warming that we would have if the Earth radiated the excess energy away to space as a blackbody, and with no change in the planetary albedo.
The result of a loop system can either be amplification or dampening of the process, depending on the sign of the gain in the loop, which we will denote $f$.
We will call amplifying feedbacks positive ($f>0$) and damping feedbacks negative ($f<0$).
We can think of the “process” here as the entire climate system, which contains many examples of both positive and negative feedback.
The capacity of the atmosphere to hold water vapor (saturation specific humidity) increases exponentially with temperature. Warming is thus accompanied by moistening (more water vapor), which leads to more warming due to the enhanced water vapor greenhouse effect.
Positive or negative feedback?
Colder temperatures lead to expansion of the areas covered by ice and snow, which tend to be more reflective than water and vegetation. This causes a reduction in the absorbed solar radiation, which leads to more cooling.
Positive or negative feedback?
Make sure it’s clear that the sign of the feedback is the same whether we are talking about warming or cooling.
Now consider what happens in the presence of a feedback process. For a concrete example, let’s take the water vapor feedback. For every degree of warming, there is an additional increase in the greenhouse effect, and thus additional energy added to the system.
Let’s denote this extra energy as $$ f \lambda_0 \Delta T_0 $$
where $f$ is the feedback amount, a number that represents what fraction of the output gets added back to the input. $f$ must be between $-\infty$ and +1.
For the example of the water vapor feedback, $f$ is positive (between 0 and +1) – the process adds extra energy to the original radiative forcing.
The amount of energy in the full "input" is now $$ \Delta R + f \lambda_0 \Delta T_0 $$ or $$ (1+f) \lambda_0 \Delta T_0 $$
But now we need to consider the next loop. A fraction $f$ of the additional energy is also added to the input, giving us $$ (1+f+f^2) \lambda_0 \Delta T_0 $$
and we can go round and round, leading to the infinite series $$ (1+f+f^2+f^3+ ...) \lambda_0 \Delta T_0 = \lambda_0 \Delta T_0 \sum_{n=0}^{\infty} f^n $$
Question: what happens if $f=1$?
It so happens that this infinite series has an exact solution
$$ \sum_{n=0}^{\infty} f^n = \frac{1}{1-f} $$So the full response including all the effects of the feedback is actually
$$ \Delta T = \frac{1}{1-f} \Delta T_0 $$This is also sometimes written as $$ \Delta T = g \Delta T_0 $$
where
$$ g = \frac{1}{1-f} = \frac{\Delta T}{\Delta T_0} $$is called the system gain -- the ratio of the actual warming (including all feedbacks) to the warming we would have in the absence of feedbacks.
So if the overall feedback is positive, then $f>0$ and $g>1$.
And if the overall feedback is negative?
ECS is an important number. A major goal of climate modeling is to provide better estimates of ECS and its uncertainty.
Latest IPCC report AR5 gives a likely range of 1.5 to 4.5 K. (There is lots of uncertainty in these numbers – we will definitely come back to this question)
So our simple estimate of the no-feedback change $\Delta T_0$ is apparently underestimating climate sensitivity.
Saying the same thing another way: the overall net climate feedback is positive, amplifying the response, and the system gain $g>1$.
Let’s assume that the true value is $\Delta T_{2xCO2} = 3$ K (middle of the range). This implies that the gain is
$$ g = \frac{\Delta T_{2xCO2}}{\Delta T_0} = \frac{3}{1.2} = 2.5 $$The actual warming is substantially amplified!
There are lots of reasons for this, but the water vapor feedback is probably the most important.
Question: if $g=2.5$, what is the feedback amount $f$?
$$ g = \frac{1}{1-f} $$or rearranging,
$$ f = 1 - 1/g = 0.6 $$The overall feedback (due to water vapor, clouds, etc.) is positive.
Now what if we have several individual feedback processes occurring simultaneously?
We can think of individual feedback amounts $f_1, f_2, f_3, ...$, with each representing a physically distinct mechanism, e.g. water vapor, surface snow and ice, cloud changes, etc.
Each individual process takes a fraction $f_i$ of the output and adds to the input. So the feedback amounts are additive,
$$ f = f_1 + f_2 + f_3 + ... = \sum_{i=0}^N f_i $$This gives us a way to compare the importance of individual feedback processes!
The climate sensitivity is now
$$ \Delta T_{2xCO2} = \frac{1}{1- \sum_{i=0}^N f_i } \Delta T_0 $$The climate sensitivity is thus increased by positive feedback processes, and decreased by negative feedback processes.
We can also write this in terms of the original radiative forcing as
$$ \Delta T_{2xCO2} = \frac{\Delta R}{\lambda_0 - \sum_{i=1}^{N} \lambda_i} $$where
$$ \lambda_i = \lambda_0 f_i $$known as climate feedback parameters, in units of W m$^{-2}$ K$^{-1}$.
With this choice of sign conventions, $\lambda_i > 0$ for a positive feedback process.
Individual feedback parameters $\lambda_i$ are then additive, and can be compared to the no-feedback parameter $\lambda_0$.
Based on our earlier numbers, the net feedback necessary to get a climate sensitivity of 3 K is
$$ \sum_{i=1}^N \lambda_i = \lambda_0 \sum_{i=1}^N f_i = (3.3 \text{ W m}^{-2} \text{ K}^{-1}) (0.6) = 2 \text{ W m}^{-2} \text{ K}^{-1} $$We might decompose this net climate feedback into, for example
These individual feedback processes may be positive or negative. This is very powerful, because we can measure the relative importance of different feedback processes simply by comparing their $\lambda_i$ values.
The "Planck feedback" represented by our reference parameter $\lambda_0$ is not really a feedback at all.
It is the most basic and universal climate process, and is present in every climate model. It is simply an expression of the fact that a warm planet radiates more to space than a cold planet.
As we will see, our estimate of $\lambda_0 = -3.3 ~\text{W} ~\text{m}^{-2} ~\text{K}^{-1} $ is essentially the same as the Planck feedback diagnosed from complex GCMs. Unlike our simple zero-dimensional EBM, however, most other climate models (and the real climate system) have other radiative feedback processes, such that
$$\lambda = \lambda_0 - \sum_{i=1}^{N} \lambda_i \ne \lambda_0 $$This figure is reproduced from the recent IPCC AR5 report. It shows the feedbacks diagnosed from the various models that contributed to the assessment.
(Later in the term we will discuss how the feedback diagnosis is actually done)
See below for complete citation information.
Figure 9.43 | (a) Strengths of individual feedbacks for CMIP3 and CMIP5 models (left and right columns of symbols) for Planck (P), water vapour (WV), clouds (C), albedo (A), lapse rate (LR), combination of water vapour and lapse rate (WV+LR) and sum of all feedbacks except Planck (ALL), from Soden and Held (2006) and Vial et al. (2013), following Soden et al. (2008). CMIP5 feedbacks are derived from CMIP5 simulations for abrupt fourfold increases in CO2 concentrations (4 × CO2). (b) ECS obtained using regression techniques by Andrews et al. (2012) against ECS estimated from the ratio of CO2 ERF to the sum of all feedbacks. The CO2 ERF is one-half the 4 × CO2 forcings from Andrews et al. (2012), and the total feedback (ALL + Planck) is from Vial et al. (2013).
Figure caption reproduced from the AR5 WG1 report
Legend:
Things to note:
This is Figure 9.43 from Chapter 9 of the IPCC AR5 Working Group 1 report.
The report and images can be found online at http://www.climatechange2013.org/report/full-report/
The full citation is:
Flato, G., J. Marotzke, B. Abiodun, P. Braconnot, S.C. Chou, W. Collins, P. Cox, F. Driouech, S. Emori, V. Eyring, C. Forest, P. Gleckler, E. Guilyardi, C. Jakob, V. Kattsov, C. Reason and M. Rummukainen, 2013: Evaluation of Climate Models. In: Climate Change 2013: The Physical Science Basis. Contribution of Working Group I to the Fifth Assessment Report of the Intergovernmental Panel on Climate Change [Stocker, T.F., D. Qin, G.-K. Plattner, M. Tignor, S.K. Allen, J. Boschung, A. Nauels, Y. Xia, V. Bex and P.M. Midgley (eds.)]. Cambridge University Press, Cambridge, United Kingdom and New York, NY, USA, pp. 741–866, doi:10.1017/CBO9781107415324.020
In homework you will be asked to include a new process in the zero-dimensional EBM: a temperature-dependent albedo.
We use the following formula:
$$ \alpha(T) = \left\{ \begin{array}{ccc} \alpha_i & & T \le T_i \\ \alpha_o + (\alpha_i-\alpha_o) \frac{(T-T_o)^2}{(T_i-T_o)^2} & & T_i < T < T_o \\ \alpha_o & & T \ge T_o \end{array} \right\}$$with parameter values:
For intermediate temperature, this formula gives a smooth variation in albedo with global mean temperature. It is tuned to reproduce the observed albedo $\alpha = 0.299$ for $T = 288$ K.
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%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
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def albedo(T, alpha_o = 0.289, alpha_i = 0.7, To = 293., Ti = 260.):
alb1 = alpha_o + (alpha_i-alpha_o)*(T-To)**2 / (Ti - To)**2
alb2 = np.where(T>Ti, alb1, alpha_i)
alb3 = np.where(T<To, alb2, alpha_o)
return alb3
In [6]:
def ASR(T, Q=341.3):
alpha = albedo(T)
return Q * (1-alpha)
def OLR(T, sigma=5.67E-8, tau=0.57):
return tau * sigma * T**4
def Ftoa(T):
return ASR(T) - OLR(T)
In [7]:
T = np.linspace(220., 300., 100)
plt.plot(T, albedo(T))
plt.xlabel('Temperature (K)')
plt.ylabel('albedo')
plt.ylim(0,1)
plt.title('Albedo as a function of global mean temperature')
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In [8]:
plt.plot(T, OLR(T), label='OLR')
plt.plot(T, ASR(T), label='ASR')
plt.plot(T, Ftoa(T), label='Ftoa')
plt.xlabel('Surface temperature (K)')
plt.ylabel('TOA flux (W m$^{-2}$)')
plt.grid()
plt.legend(loc='upper left')
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The graphs meet at three different points! That means there are actually three different possible equilibrium temperatures in this model.
In [9]:
# Use numerical root-finding to get the equilibria
from scipy.optimize import brentq
# brentq is a root-finding function
# Need to give it a function and two end-points
# It will look for a zero of the function between those end-points
Teq1 = brentq(Ftoa, 280., 300.)
Teq2 = brentq(Ftoa, 260., 280.)
Teq3 = brentq(Ftoa, 200., 260.)
print( Teq1, Teq2, Teq3)
In [10]:
%load_ext version_information
%version_information numpy, scipy, matplotlib
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The author of this notebook is Brian E. J. Rose, University at Albany.
It was developed in support of ATM 623: Climate Modeling, a graduate-level course in the Department of Atmospheric and Envionmental Sciences
Development of these notes and the climlab software is partially supported by the National Science Foundation under award AGS-1455071 to Brian Rose. Any opinions, findings, conclusions or recommendations expressed here are mine and do not necessarily reflect the views of the National Science Foundation.
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