This page discusses some of the issues related to getting the correct plots.

The primary cause of confusion is due to the way that electromagnetic radiation is treated.

- When working with visible light, the frequency (f) is indicated by the wavelength (λ in either nm or angstroms)
- When working with infrared (IR) radiation, the frequency is indicated by the wavenumber (wn = 1/λ typically scaled to cm
^{-1}) - When working with x-rays and higher energy radiation, the frequency is indicated by the energy in electronvolts (eV)
- In some cases, the frequency is simply stated in hertz (Hz)

λf = c The speed of light E = hf Energy in a photon of frequency 'f' h - Planck's constant |

λν = c E = hν |

ν = c/λ E = hc/λ E = h c wn |

(SpectralCalc.com uses σ and ν for wavenumber and frequency, respectively.)

p = 2pi hc^2 / lambda^5 (exp(hc/lambda kt) - 1) p = 2πh c |

Δp = { 2πh c |

Replacing the wavelength with the wavenumber is simple enough, however

λ = 1/wn Δλ = 1/wn2 - 1/wn1 = (wn1-wn2)/(wn1 * wn2) = -Δwn/wn |

Δp = [2πh c |

The minus sign is generally ignored because it indicates a flipping of the x-axis. In both cases, plots are generally made with zero on the left. Shorter wavelengths have more energy. When plotting verses wavelength, then energy per photon increases toward the left. However, when plotting verses wavenumber, the shorter (higher energy) wavelengths are on the right.

As shown, the wavelengths are expressed in meters and wavenumbers are expressed in
inverse meters (m^{-1}). When other units are used (such as cm^{-1}),
adjustments need to be made.

The red curve (peak near 580 cm-1) uses the wavenumber to the 3rd power, the purple curve (peak on the right, near 1031 cm-1) uses the wavelength (lambda) to the 5th power. The graphs are scaled so that when they are integrated over wavenumber they both have the same area. (Without this scaling, it would not be possible to show both curves on the same graph.)

Notice that the 2 peaks are not even close, neither in maximum power nor in the frequency of the peak. This is because the integral over small changes in frequency (wavenumber) is not the same as an integral over small changes in wavelength.

The y-axis values go from 0.0 to 0.5 W/m2/cm-1. As a result, the area of a 1 x 500 square should be

0.1 W/m2/cm-1 * 500 cm-1 = 50 W/m2 |

When convolving a blackbody envelope with the HITRAN data, the red curve provides the correct result (I think).

Another way to indicate the number of degrees in an angle is to divide the length
of the associated arc by the radius.
In this way,
360° = 2π **radians**
and
90° = 2π/4 = π/2 **radians**.

By analogy, when working with 3-dimentional objects,
a **solid angle** is defined as the ratio of the area of a part of a sphere divided
by the radius squared. Even though the number is actually dimensionless,
this ratio is usually expressed in **steradians** (sr).
Since the area of a sphere is 4πr^2,
there are 4π steradians in a sphere.

Some sources express the blackbody equation as

w / m |

2π sr * 1/2 m |

(I am no longer sure where the `1/2 m ^{2}` came from.
I believe it is the result of integrating the area over all angles, but
I am still working on this.)

I am not sure what to do when integrating over a 1.0 m^{3} emitting sphere
in the atmosphere.
For the equations I've seen, the only way to get the dimensions to work out
is to use the energy emitted from the imagined surface of the bounding sphere.
Assuming that this is a valid approach, this means that a unit volume of atmosphere
will emit 4 times the amount of energy as a unit area on the surface of the Earth
- half toward the surface and half toward space.

Of course, for a gas the amount of energy emitted is actually the temperature, pressure, and doppler broadened spectra, scaled by the number of molecules present, and convolved with the blackbody distribution for that temperature.

When computing the number of photons (amount of energy) absorbed by the atmosphere, the thickness of the atmosphere between the surface and free space is also dependent on the angle.

p = 2pi h c^2 / lambda^5 (exp(hc/lambda kt) - 1) (no dimensions provided) p = 2π h c |

p = 2 h c^2 / lambda^5 (exp(hc/lambda kt) - 1) W / m2 /sr / m p = 2 h c |

At any rate, this is the value I was trying to explain in the previous section.

SpectralCalc.com provides complete derivations of the different blackbody equations. The associated Blackbody Calculator provides a simply way to verify that your program is working correctly.

This is a Flash based blackbody simulation. It works pretty good, but the x-axis is always wavelength.

Solid Angle provides a good definition, equations, and a couple of non-conic projections as examples.

**The Warm Earth: Thermal Remote Sensing**
section 9.1 - Planck Radiation (Blackbody) Law
provides the equation used in the program.

Author: Robert Clemenzi