Black Body Radiation: Wien's Displacement Law

Main Author: David Vartanyan
Co-authors: John Pharo and Mee Chatarain

Since problem 2 was relatively long compared to 1, we split it between John and I, while Mee did 1.

2)
Abstract: We apply the results of problem 1) (the relationships between energy density, intensity, and flux) to arrive at several general results for blackbody emission spectra. Recall that intensity is energy times multiplied by the speed of light and flux is intensity times surface area of object through which flux passes. This is consistent with the conservation of flux over an area, as intensity is conserved regardless of where the emitter is observed (it is an intrinsic property).

a) We will show: B(T)=2*c^2/[Lambda]^5*1/(e^(h*c/(T*k*[Lambda])-1)
[Lambda]=c/(v) where v is frequency.

Initially we thought to substitute in v=c/(Lambda), but this didn't give us the textbook answer, and reading the hint we realized that energy per dv didn't equal energy per d[Lambda].

We know that the integral of B(T) where the intensity is defined for a wavelength has to equal the integral of B(T) where the intensity is defined for a frequency, since these integrals are proportional to the conserved energy density.

Intensity as a function of frequency:



so d/d([Lambda] Integral[2*h*v^3/c^2*1/(e^(h*v/(k*T)-1)*dv]
=d/d([Lambda]) Integral[2*h*c/[Lambda]^3*1/(e^(h*v/(k*T)-1)*c/[Lambda]^2*d[Lambda]]
and by the fundamental theorem of calculus we get
B(T) as defined for wavelengths
=2c^2/[Lambda]^5*1/(e^(h*c/(T*k*[Lambda])-1)
where we have dv/d[Lambda]~c/[Lambda]^2 (where we have implicitly taken the absolute value to maintain the positive sign).

b) [Lambda] max is defined as the wavelength where the intensity function (in terms of wavelength) B(T)=2*c^2/[Lambda]^5*1/(e^(h*c/(T*k*[Lambda])-1) has a maximum.

so, d/d[Lambda] (2*c^2/[Lambda]^5*1/(e^(h*c/(T*k*[Lambda])-1))=0
= 2*c^2*-5/[Lambda]^6*1/(e^(h*c/(T*k*[Lambda])-1))
+2*c^2/[Lambda]^5*1*(e^(h*c/(T*K*Lambda])-1)^-1*c*h/(k*T*[Lambda]^2)

Equating this and simplifying, we get
h*c/(k*T*[Lambda]*e^(h*c/(k*T*[Lambda]))/(e^(h*c/(k*T*[Lambda])-1)=5
Substituting x=h*c/(k*T*[Lambda], we solve to get x=4 using first order approximation e^x=1+x
So h*c/(k*T*[Lambda]=4 and T*[Lambda]=.003 mole Kelvins, close to 3.15 in C&O.

c) In Rayleigh Jean's limit, thermal energy dominates photon energy so k*T>>h*v. The statistical term 1/(e^(k*T*v])-1) becomes 1/(e^x-1) where x=h*v/k*T. For small x, as in our case, e^x=1+x to first order so B~k*T/(h*v). The brightness temperature indicates that the brightness of low energy stars is proportional its temperature. For a blackbody intensity plot vs wavelength, this forms the right tail where photon energy is low. The other extreme is when photon energy>>thermal energy, and this forms the left tail of the blackbody intensity plot.

Conclusion: Having understood the math behind perfect emitters, we were introduced to some actual blackbodies. Interestingly, the perfect blackbody is the universe, if we plot CMB radiation intensity vs frequency. Other notable but non-ideal blackbodies are the sun, and to a lesser degree, even the human body.

The sun as a blackbody:



Human Body as a blackbody: