Application: Blackbody Radiation

The spectrum of a star is well approximated as a blackbody. The Planck function describes the intensity of a blackbody:

\[B_\lambda = \frac{2 h c^2}{\lambda^5} \frac{1}{\exp\left (\frac{hc}{\lambda k_B T}\right) - 1}\]

This has units of energy / area / wavelength / steradian.

Integrating over wavelength

Let’s integrate out wavelength:

\[B = \int_0^\infty B_\lambda d\lambda\]

our integral is then:

\[B = 2 h c^2 \int_0^\infty \frac{1}{\lambda^5} \frac{1}{\exp\left (\frac{hc}{\lambda k_B T}\right) - 1} d\lambda\]

defining a dimensional quantity, \(x\):

\[x = \frac{hc}{\lambda k_B T}\]

we have

\[dx = - \frac{hc}{\lambda^2 k_B T} d\lambda = - \frac{x^2k_BT}{hc} d\lambda\]

finally, we have:

\[B = \frac{2(k_B T)^4}{h^3 c^2} \int_0^\infty \frac{x^3}{e^x - 1} dx\]

Integrating to infinity

Now the challenge becomes — how do we do this integral, when the integration limits extend to infinity!

Note: it turns out that this integral has an analytic solution:

\[\int_0^\infty \frac{x^3}{e^x - 1} dx = \frac{\pi^4}{15}\]

and the original integral is

\[B = \frac{\sigma T^4}{\pi}\]

with the Stephan-Boltzmann constant defined as:

\[\sigma \equiv \frac{2 k_B^4 \pi^5}{15 h^3 c^2}\]

so we can check our answer.

Tip

It’s always a good idea to start by plotting the integrand.

import numpy as np
import matplotlib.pyplot as plt

def f(x):
    # we need to take the limit x -> 0 if the input x is 0
    return np.where(x == 0, 0, x**3 / (np.exp(x) - 1))

fig = plt.figure()
ax = fig.add_subplot()

x = np.linspace(0, 10, 100)
ax.plot(x, f(x))

/tmp/ipykernel_2491/922372595.py:3: RuntimeWarning: invalid value encountered in divide
  return np.where(x == 0, 0, x**3 / (np.exp(x) - 1))

[<matplotlib.lines.Line2D at 0x7f29d0443510>]

From the plot, we see that it peaks near x = 3 and then drops off, but it will only asymptotically approach 0, so we need to still be careful about the integration limits.

Let’s consider the transform:

\[z = \frac{x}{c + x}\]

where \(c\) is chosen to be close to the maximum of the integrand. This transformation maps the interval \(x \in [0, \infty)\) into \(z \in [0, 1)\) — that makes the integral finite, and we can use our methods (like trapezoid) to do it.

The inverse transform is:

\[x = c \frac{z}{1 - z}\]

SMALL = 1.e-30

def zv(x, c):
    """ transform the variable x -> z """
    return x/(c + x) 

def xv(z, c):
    """ transform back from z -> x """
    return c*z/(1.0 - z + SMALL)

Let’s plot it again, but in terms of \(z\)

z = np.linspace(0, 1, 100)
c = 3

ax.clear()
ax.plot(z, f(xv(z, c)))
fig

/tmp/ipykernel_2491/922372595.py:3: RuntimeWarning: overflow encountered in exp
  return np.where(x == 0, 0, x**3 / (np.exp(x) - 1))
/tmp/ipykernel_2491/922372595.py:3: RuntimeWarning: invalid value encountered in divide
  return np.where(x == 0, 0, x**3 / (np.exp(x) - 1))

Notice that the transform puts the peak near the center of the interval, and the function nicely goes to zero as \(z\rightarrow 1\). This is much better behaved to integrate.

Consider a general integral:

\[I = \int_0^\infty f(x) dx\]

Let’s rewrite this in terms of \(z\). Since

\[x = c \frac{z}{1 - z}\]

we have

\[dx = \frac{c dz}{(1 - z)^2}\]

so

\[I = c \int_0^1 f\left(c \frac{z}{1 - z} \right ) (1 -z)^{-2} dz\]

We need to include these extra factors in our quadrature for computing the integral.

def I_t(func, N=10, c=5):
    """composite trapezoid rule for integrating from [0, oo].
    Here N is the number of intervals"""
    
    # there are N+1 points corresponding to N intervals
    z = np.linspace(0.0, 1.0, N+1, endpoint=True)
    
    I = 0.0
    for n in range(N):
        I += 0.5 * (z[n+1] - z[n]) * (func(xv(z[n], c)) / (1.0 - z[n] + SMALL)**2 +
                                      func(xv(z[n+1], c)) / (1.0 - z[n+1] + SMALL)**2)
    I *= c
    return I

Now we can check our result. First, let’s compute the analytic solution:

I_analytic = np.pi**4/15

for nint in [2, 4, 8, 16, 32, 64]:
    I_trap = I_t(f, N=nint, c=3)
    print(f"{nint:5} {I_trap:10.5f} {np.abs(I_trap - I_analytic):20.10g}")

    2    8.48810          1.994163429
    4    6.09974         0.3941968295
    8    6.49205       0.001888841238
   16    6.49392      1.669019097e-05
   32    6.49394      5.396286928e-07
   64    6.49394      3.352448719e-08

/tmp/ipykernel_2491/922372595.py:3: RuntimeWarning: invalid value encountered in scalar divide
  return np.where(x == 0, 0, x**3 / (np.exp(x) - 1))
/tmp/ipykernel_2491/922372595.py:3: RuntimeWarning: overflow encountered in exp
  return np.where(x == 0, 0, x**3 / (np.exp(x) - 1))

We see that our answer matches the analytic solution very well.

C++ implementation

A C++ implementation is available here: zingale/computational_astrophysics