Stefan-Boltzman answer

Calculate the definite integral of $\int_0^5 a x^3 dx$ for a given value of a and compare it with the analytic answer: $y=\frac{a}{4} 5^4$

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import matplotlib.pyplot as plt
import numpy as np

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xvec=np.arange(0,5,0.01) #vector from 0 to 5 in 0.01 step sizes

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xvec[:10] #print the first 10 values

Now make a vector of intervals $dx$ using the diff function

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dx=np.diff(xvec) #note that dx is one item shorter than xvec

integration is just the sum over rectangles of width $dx$ and height $ax^3$

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print "analytic: %8.4f, approximate: %8.4f" % (analytic,integral)

Apply this to the Planck function (Stull Chapter 2, Equation 2.13, p. 36):

$E_\lambda^* = \frac{c1}{\lambda^5 \left ( exp(c_2/(\lambda T)) - 1 \right )}$

or Planck's law

$E_\lambda* =\pi \frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}}-1}$


$h$=Planck's constant ($6.63 \times 10^{-34}$ Joule seconds})

$c$= Speed of light in a vacuum ($3.00 \times 10^{8}\ \mathrm{meters/second}$)

$k_b$ =Boltzman's constant ($1.38 \times 10^{-23}\ \mathrm{Joules/Kelvin}$)

(note that $E_\lambda*$ is the blackbody flux, or irradiance, with units of $W/m^2 /\mu m$) Wikipedia (and Wallace and Hobbs) give the blackbody radiance which differs by by a factor of $\pi$ and has units of $W/m^2/\mu m /sr$. We will talk a lot about this in the next few weeks, don't worry about it for now.

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c=2.99792458e+08  #m/s -- speed of light in vacumn
h=6.62606876e-34  #J s  -- Planck's constant
kb=1.3806503e-23  # J/K  -- Boltzman's constant

def planckwavelen(wavel,Temp):
    """input wavelength in microns and Temp in K, output
    bbr in W/m^2/micron
    wavel=wavel*1.e-6  #convert to meters
    Elambda=1.e-6*np.pi*c1/(wavel**5.*(np.exp(c2/(wavel*Temp)) -1))
    return Elambda

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the_wavelengths=np.linspace(0.1,30,5000) #microns
the_temp=300 #K

Assignment: Use sum and diff along with the function planckwavelen to integrate the Planck function over wavelength and verify the Stefan-Boltzman equation on Stull chapter 2 p. 37:

$E^* = \int_0^\infty E_\lambda^* d\lambda = \sigma T^4$

where $\sigma=5.67 \times 10^-8$ $W/m^2/K^{4}$

Answer (I extended the maximum wavelength and the number of points until I got agreement to a percent or so -- but it wasn't necessary for the spirit of this exercise).

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the_wavelengths=np.linspace(0.1,200,50000) #microns
the_temp=300 #K
sigma=5.67e-8  #W/m^2/K^4
print "approx integral: %9.4f, exact stefan boltzman %9.4f" % (the_int,stefan)