(absorptivity) * incident - emission
$$ R_s = F_s - E_s $$$$ R_1 = \epsilon_1 F_1 - 2 E_1 $$$$ R_2 = \epsilon_2 F_2 - 2 E_2 $$Need to add the solar energy source. We assume atm is transparent, solar is all absorbed at the surface.
$$ R_1 = R_2 = 0$$$$ R_s = - (1-\alpha) Q $$Introduce useful notation shorthand:
$$ (1-\alpha) Q = \sigma T_e^4 $$This gives a 3x3 system which is linear in $T^4$ (divide through by $\sigma$)
$$ - T_s^4 + \epsilon_1 T_1^4 + (1-\epsilon_1) \epsilon_2 T_2^4 + T_e^4 = 0 $$$$ \epsilon_1 T_s^4 - 2 \epsilon_1 T_1^4 + \epsilon_1 \epsilon_2 T_2^4 = 0$$$$ \epsilon_2 (1-\epsilon_1) T_s^4 + \epsilon_1 \epsilon_2 T_1^4 - 2 \epsilon_2 T_2^4 = 0$$Here we use the sympy module to solve the algebraic system symbolically.
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import sympy
sympy.init_printing()
T_s, T_1, T_2, T_e, e_1, e_2 = sympy.symbols('T_s, T_1, T_2, T_e, e_1, e_2', positive=True )
system = [-T_s**4 + e_1*T_1**4 + e_2*(1-e_1)*T_2**4 + T_e**4,
e_1*T_s**4 - 2*e_1*T_1**4 + e_1*e_2*T_2**4,
e_2*(1-e_1)*T_s**4 + e_1*e_2*T_1**4 - 2*e_2*T_2**4]
out1 = sympy.solve( system, [T_s**4, T_1**4, T_2**4])
out1
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quarter = sympy.Rational(1,4)
out2 = {}
for var4, formula in out1.iteritems():
var = (var4)**quarter
out2[var] = sympy.simplify(formula**quarter)
out2
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# The special case of equal absorptivities
e = sympy.symbols('e')
out3 = {}
for var4, formula in out1.iteritems():
var = (var4)**quarter
simple_formula = sympy.cancel(formula.subs([(e_2, e),(e_1, e)]))
out3[var] = sympy.simplify( simple_formula**quarter )
out3
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The solution is
\begin{align} T_s^4 &= T_e^4 \frac{4 - \epsilon_1 \epsilon_2}{4 + \epsilon_1 \epsilon_2 - 2 \epsilon_1 - 2 \epsilon_2} \\ T_1^4 &= T_e^4 \frac{2 -\epsilon_1 \epsilon_2 + \epsilon_2}{4 + \epsilon_1 \epsilon_2 - 2 \epsilon_1 - 2 \epsilon_2} \\ T_2^4 &= T_e^4 \frac{ 1}{2 - \epsilon_2} \end{align}In the special case $\epsilon_1 = \epsilon_2$ this reduces to
\begin{align} T_s^4 &= T_e^4 \frac{2+\epsilon}{2-\epsilon} \\ T_1^4 &= T_e^4 \frac{1+\epsilon}{2-\epsilon} \\ T_2^4 &= T_e^4 \frac{ 1}{2 - \epsilon} \end{align}
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out2[T_s].subs([(T_e, 255), (e_1, 0.4), (e_2, 0.4)])
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for var, formula in out2.iteritems():
print formula.subs([(T_e, 255), (e_1, 0.4), (e_2, 0.4)])
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# Coding up the analytical solutions for radiative equilibrium
# These use the analytical results returned by sympy and wrap them in callable functions
def Ts(Te, e1, e2):
#return Te*((4-e1*e2)/(4+e1*e2-2*(e1+e2)))**0.25
return out2[T_s].subs([(T_e, Te), (e_1, e1), (e_2, e2)])
def T1(Te, e1, e2):
#return Te*((2+e2-e1*e2)/(4+e1*e2-2*(e1+e2)))**0.25
return out2[T_1].subs([(T_e, Te), (e_1, e1), (e_2, e2)])
def T2(Te, e1, e2):
#return Te*(1/(2-e2))**0.25
return out2[T_2].subs([(T_e, Te), (e_1, e1), (e_2, e2)])
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%matplotlib inline
import numpy as np
from climlab import constants as const
from climlab.model import column
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mycolumn = column.SingleColumnModel( num_levels=2 )
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print mycolumn
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mycolumn.integrate_years(10.)
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print mycolumn.Ts
print mycolumn.Tatm
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(e1, e2)= mycolumn.subprocess['LW'].absorptivity
print e1, e2
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ASR = (1-mycolumn.param['albedo_sfc'])*mycolumn.param['Q']
Te = (ASR/const.sigma)**0.25
print Te
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tol = 0.01
def test_2level(col):
(e1, e2)= col.subprocess['LW'].absorptivity
ASR = (1-col.param['albedo_sfc'])*col.param['Q']
Te = (ASR/const.sigma)**0.25
print 'Surface:'
num = col.Ts
anal = Ts(Te,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
print 'Level 1'
num = col.Tatm[0]
anal = T1(Te,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
print 'Level 2'
num = col.Tatm[1]
anal = T2(Te,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
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test_2level(mycolumn)
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e1 = 0.3
e2 = 0.4
mycolumn.subprocess['LW'].absorptivity = np.array([e1,e2])
mycolumn.integrate_years(10.)
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test_2level(mycolumn)
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e1 = 0.6
e2 = 0.6
mycolumn.subprocess['LW'].absorptivity = np.array([e1,e2])
mycolumn.integrate_years(10.)
test_2level(mycolumn)
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col1 = column.SingleColumnModel(num_levels=2, abs_coeff=1.9E-4)
(e1, e2) = col1.subprocess['LW'].absorptivity
print e1, e2
col1.integrate_years(10.)
test_2level(col1)
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col1 = column.SingleColumnModel(num_levels=2, abs_coeff=1.9E-4)
lw = col1.subprocess['LW']
(e1, e2) = lw.absorptivity
e1 *= 1.2
lw.absorptivity = np.array([e1, e2])
col1.integrate_years(10.)
test_2level(col1)
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T_s, T_0, T_1, T_2, T_e, epsilon_0, epsilon_1, epsilon_2, sigma = \
sympy.symbols('T_s, T_0, T_1, T_2, T_e, epsilon_0, epsilon_1, epsilon_2, sigma', positive=True )
Define the transmissivities $\tau_i$ for layers $i=0, 1, \dots, N-1$
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tau_0 = (1-epsilon_0)
tau_1 = (1-epsilon_1)
tau_2 = (1-epsilon_2)
tau_0, tau_1, tau_2
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Note that if the atmosphere has $N$ layers then $\epsilon_N = 0$
Define the emissions for each layer:
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E_s = sigma*T_s**4
E_0 = epsilon_0*sigma*T_0**4
E_1 = epsilon_1*sigma*T_1**4
E_2 = epsilon_2*sigma*T_2**4
E_s, E_0, E_1, E_2
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Define the longwave fluxes incident on each layer, $F_{i}$
Note that if the atmosphere has $N$ layers then $F_{N}$ is the OLR (emission to space)
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F_s = E_0 + tau_0*E_1 + tau_0*tau_1*E_2
F_0 = E_s + E_1 + tau_1*E_2
F_1 = tau_0*E_s + E_0 + E_2
F_2 = tau_1*tau_0*E_s + tau_1*E_0 + E_1
F_3 = tau_2*tau_1*tau_0*E_s + tau_2*tau_1*E_0 + tau_2*E_1 + E_2
F_s, F_0, F_1, F_2, F_3
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Now define the net absorbed longwave radiation (flux divergence) in each layer.
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R_s = F_s - E_s
R_0 = epsilon_0*F_0 - 2*E_0
R_1 = epsilon_1*F_1 - 2*E_1
R_2 = epsilon_2*F_2 - 2*E_2
R_s, R_0, R_1, R_2
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out1 = sympy.solve([R_s + sigma*T_e**4, R_0, R_1, R_2],
[T_s**4, T_0**4, T_1**4, T_2**4])
out1
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quarter = sympy.Rational(1,4)
out2 = {}
for var4, formula in out1.iteritems():
var = (var4)**quarter
out2[var] = sympy.simplify(formula**quarter)
out2
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Now wrap these analytical radiative equilibrium solutions in callable functions:
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def Ts(Te, e0, e1, e2):
return out2[T_s].subs([(T_e, Te), (epsilon_0, e0), (epsilon_1, e1), (epsilon_2, e2)])
def T0(Te, e0, e1, e2):
return out2[T_0].subs([(T_e, Te), (epsilon_0, e0), (epsilon_1, e1), (epsilon_2, e2)])
def T1(Te, e0, e1, e2):
return out2[T_1].subs([(T_e, Te), (epsilon_0, e0), (epsilon_1, e1), (epsilon_2, e2)])
def T2(Te, e0, e1, e2):
return out2[T_2].subs([(T_e, Te), (epsilon_0, e0), (epsilon_1, e1), (epsilon_2, e2)])
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tol = 0.01
def test_3level(col):
(e0, e1, e2)= col.subprocess['LW'].absorptivity
ASR = (1-col.param['albedo_sfc'])*col.param['Q']
Te = (ASR/const.sigma)**0.25
print 'Surface:'
num = col.Ts
anal = Ts(Te,e0,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
print 'Level 0'
num = col.Tatm[0]
anal = T0(Te,e0,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
print 'Level 1'
num = col.Tatm[1]
anal = T1(Te,e0,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
print 'Level 2'
num = col.Tatm[2]
anal = T2(Te,e0,e1,e2)
print ' Numerical: %.2f Analytical: %.2f Same:' %(num, anal) , abs(num - anal)<tol
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col = column.SingleColumnModel( num_levels=3 )
col.integrate_years(10.)
test_3level(col)
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e0 = 0.3
e1 = 0.6
e2 = 0.2
col.subprocess['LW'].absorptivity = np.array([e0,e1,e2])
col.integrate_years(10.)
test_3level(col)
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# The 3-layer solution reduces to two layer solution if we set e_2 = 0
out3 = {}
for var, formula in out2.iteritems():
if var is not T_2:
out3[var] = sympy.simplify(formula.subs(epsilon_2,0))
out3
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Now substitute $n \rightarrow -n$ in the second summation.
\begin{align} F_i &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{n-1} \tau_{i+j} \bigg) E_{i+n} + \sum_{-n=1}^{i-1} \bigg( \prod_{j=0}^{-n-1} \tau_{i-j} \bigg) E_{i+n} \right\} \\ &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{n-1} \tau_{i+j} \bigg) E_{i+n} + \sum_{n=1-i}^{-1} \bigg( \prod_{j=0}^{-n-1} \tau_{i-j} \bigg) E_{i+n} \right\} \end{align}And substitude $j \rightarrow -j$ in the second product.
\begin{align} F_i &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{n-1} \tau_{i+j} \bigg) E_{i+n} + \sum_{n=1-i}^{-1} \bigg( \prod_{-j=0}^{-n-1} \tau_{i+j} \bigg) E_{i+n} \right\} \\ &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{n-1} \tau_{i+j} \bigg) E_{i+n} + \sum_{n=1-i}^{-1} \bigg( \prod_{j=0}^{1+n} \tau_{i+j} \bigg) E_{i+n} \right\} \\ &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{|n|-1} \tau_{i+j} \bigg) E_{i+n} + \sum_{n=1-i}^{-1} \bigg( \prod_{j=0}^{1- |n|} \tau_{i+j} \bigg) E_{i+n} \right\} \\ &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{|n|-1} \tau_{i+j} \bigg) E_{i+n} + \sum_{n=1-i}^{-1} \bigg( \prod_{j=0}^{sign(n) (|n|-1)} \tau_{i+j} \bigg) E_{i+n} \right\} \end{align}Now combine both sums together
\begin{align} F_i &= \frac{1}{\tau_i} \left\{ \sum_{n=1}^{N-i} \bigg( \prod_{j=0}^{|n|-1} \tau_{i+j} \bigg) E_{i+n} + E_i + \sum_{n=1-i}^{-1} \bigg( \prod_{j=0}^{sign(n) (|n|-1)} \tau_{i+j} \bigg) E_{i+n} \right\} - \frac{E_i}{\tau_i} \\ &= \frac{1}{\tau_i} \left\{ \sum_{n=1-i}^{N-i} \bigg( \prod_{j=0}^{sign(n) \big(|n|-1\big)} \tau_{i+j} \bigg) E_{i+n} - E_i \right\} \end{align}with the convention that $\prod_{j=0}^{-1} = 1$.
Alternatively if we set $\prod_{j=0}^{-1} = 0$ then
$$ F_i = \frac{1}{\tau_i} \left\{ \sum_{n=1-i}^{N-i} \bigg( \prod_{j=0}^{sign(n) \big(|n|-1\big)} \tau_{i+j} \bigg) E_{i+n} \right\} $$Let's use this notation since it simplifies our expressions.
(absorptivity) * incident - emission
\begin{align} R_i &= \epsilon_i F_i - 2 E_i \\ &= \frac{\epsilon_i}{\tau_i} \left\{ \sum_{n=1-i}^{N-i} \bigg( \prod_{j=0}^{sign(n) \big(|n|-1\big)} \tau_{i+j} \bigg) E_{i+n} \right\} - 2 E_i \end{align}We will define the transmissivity between layer i and layer i+n
$$ T_{in} = \left\{ \begin{array}{cc} \prod_{j=1}^{n-1} \tau_{i+j} & n > 1 \\ 1 & n = 1 \\ 0 & n = 0 \\ 1 & n = -1 \\ \prod_{j=1}^{-n-1} \tau_{i-j} & n < -1 \end{array} \right\} $$Then the incident flux follows directly
$$ F_i = \sum_{n=1-i}^{N-i} T_{in} E_{i+n} $$and the net radiation is
\begin{align} R_i &= \epsilon_i F_i - 2 E_i \\ &= \epsilon_i \sum_{n=1-i}^{N-i} T_{in} E_{i+n} - 2 E_i \end{align}Now make the substitution $i+n \rightarrow m$
\begin{align} F_i &= \sum_{m=1}^{N} T_{im} E_{m} \\ T_{im} &= \left\{ \begin{array}{cc} \prod_{j=1}^{m-i-1} \tau_{i+j} & m > 1+i \\ 1 & m = i+1 \\ 0 & m = i \\ 1 & m = i-1 \\ \prod_{j=1}^{i-m-1} \tau_{i-j} & m < i-1 \end{array} \right\} \end{align}Or using the Einstein summation notation, since $m$ is a repeated index, we can just write
$$ F_i = T_{im} E_{m} $$and the net radiation is
$$ R_i = \epsilon_i F_i - 2 E_i $$
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