Science Facts - Blackbody Equations

When studying Global Warming there are a lot of graphs showing the blackbody thermal emissions of the Earth and Sun. When I tried to reproduce those graphs myself, I had a lot of trouble - my plots did not match some of those others were using.

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

Background

When I started plotting blackbody spectra, I used an equation from a NASA site. When I compared the results to other sites, the peak wavelengths did not match. I eventually found 4 different equations. Hyperphysics was able to explain how 3 of them were related ... it appears that the forth equation was simply wrong (actually a transcription error).

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

Of course, just to keep everyone confused, many references use the Greek letter nu (ν) to represent the frequency.

λν = c
E  = hν

Using substitution

ν = c/λ
E = hc/λ
E = h c wn

Some references also use the lower case sigma (σ) to represent the wavenumber (wn). However, I have not done that here because Stefan's constant (which can be derived from Planck's equation) is usually represented by σ.

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

Planck's function

According to many sites, this is the basic blackbody equation (from GSFC).

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

p = 2πh c² / [ λ⁵ (e^{(hc/λ kt)}-1)]

However, that is not quite correct. (It would be obvious if I included the units.) That equation actually provides the energy emitted at the specified frequency (in this case, specified by lambda [λ]). Since the total energy emitted is the integral of that equation, it is necessary to multiply it by Δλ and then perform the integral.

Δp = { 2πh c² / [ λ⁵ (e^{(hc/λ kt)}-1)] } Δλ  w/m²

It is the final Δλ that causes the problem. When working with some data (such as that from HITRAN) integrations must be made over wavenumber, not wavelength.

Replacing the wavelength with the wavenumber is simple enough, however

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

Substituting back into the main equation gives

Δp = [2πh c² wn⁵/ (e^{(hc wn / kt)}-1)] (-Δwn/wn²)

Δp = [-2πh c² wn³/ (e^{(hc wn / kt)}-1)] Δwn

Notice that this changes a function related to the 5th power of the wavelength to a function related to the 3rd power of the wavenumber. This is extremely important .. it completely changes the results when convolving the HITRAN absorption spectra with the blackbody envelope.

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.

Example

This is a plot of the two equations showing the difference. Temperature = 300K, Integral = 459 W/m2.

Graph showing the difference between the 2 blackbody equations

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).

Solid Angle

By definition, a circle is divided into 360°. One forth of the circle is 360/4 = 90°.

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-dimensional 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 / m² / sr / cm^-1  or   w m^-2 sr^-1 (cm^-1)^-1

From this, it is obvious that to determine the energy emitted into half a sphere, you just multiply by 2π sr. While that gives the correct result for a point source (or a small spherical volume in the atmosphere), it gives the wrong result in some cases. The problem is due to the fact that a unit area on the surface of a sphere does not "appear" to have the same area in all directions. The apparent, or effective, area is the actual area times the cosine of the angle with respect to the normal. Assuming a circle, when viewed from an angle, it will appear as an ellipse. When integrated over half a sphere, and assuming a circular area of one square meter, you get

2π sr * 1/2 m² = π sr m²

(I am no longer sure where the 1/2 m² came from. I believe it is the result of integrating the area over all angles, but I am still working on this.)

NOTE - from spectralcalc's notes on Integrating the Planck Equation

A common mistake in deriving this result is to assume the factor is 2π rather than π, because there are 2π steradians in the hemisphere, but this neglects the cosθ reduction from Lambert’s cosine law.

I am not sure what to do when integrating over a 1.0 m³ 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.

Confusion

The blackbody equation used in my program (from GSFC)

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

p = 2π h c² / [ λ⁵ (e^{(hc/λ kt)}-1)]

differs significantly from the spectralcalc equation

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

p = 2 h c² / [ λ⁵ (e^{(hc/λ kt)}-1)]

Notice? the pi (π) is missing. Supposedly, this is just because one equation is the total radiation, and the other is per steradian. However, there are 4π steradians in a sphere .. and the equations differ only by a factor of π. (There are also a few powers-of-ten that I only partly understand .. presumably, these are simply for unit conversion.)

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

References

HyperPhysics is one of the great sites for understanding physics. Their Blackbody Radiation page shows the equations for frequency and wavelength, but not for wavenumber. This is the page that made it obvious that more research was necessary.

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.

The Ultraviolet Catastrophe (2016 video by Dianna Cowern - aka Physics Girl) - an entertaining overview of the development of blackbody theory and related issues, but without any technical details

Author: Robert Clemenzi