Overview In this chapter, we explore one transformative experiment that helped pave the way to a new quantum theory: blackbody radiation and the ultraviolet catastrophe.   This experiment revealed a new quantum paradigm: the quantization of energy in matter.

In this section, we explore blackbody radiation.   A blackbody is a theoretical object that absorbs all incident frequencies of radiation, hence the name.  A blackbody that is in thermal equilibrium also emits all frequencies of radiation, though not equally.   The peak wavelength and energy density of emitted radiation depend on the blackbody's temperature.   Early classical attempts to explain blackbody radiation imagined that electromagnetic radiation emitted by the walls of the blackbody was caused by the oscillation of the electrons in constituent atoms. Implicit in this assumption was the idea that the electronic oscillators could have any arbitrary energy and that all oscillators contribute equally to the energy.   These ideas led to the following expression for the emitted energy density, known as the Rayleigh-Jeans Law:

(1)   Rayleigh-Jeans Law

Plot Rayleigh-Jeans Law for a blackbody at room temperature using the Maple input below:

 > $\mathrm{with}\left(\mathrm{plots}\right):\mathrm{with}\left(\mathrm{ScientificConstants}\right):$
 > $\mathrm{ρRJ}\left(T,\mathrm{nu}\right)≔\frac{8\cdot \mathrm{Pi}\cdot \mathrm{evalf}\left(\mathrm{Constant}\left(k\right)\right)\cdot T}{\mathrm{evalf}{\left(\mathrm{Constant}\left(c\right)\right)}^{3}}\cdot {\mathrm{\nu }}^{2};$
 ${\mathrm{ρRJ}}{≔}\left({T}{,}{\mathrm{\nu }}\right){↦}\frac{{8}{\cdot }{\mathrm{\pi }}{\cdot }{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({k}\right)\right){\cdot }{T}{\cdot }{{\mathrm{\nu }}}^{{2}}}{{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({c}\right)\right)}^{{3}}}$ (2.1)
 > $T≔300;$
 ${T}{≔}{300}$ (2.2)
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What is wrong with the Rayleigh-Jeans law?

 Answer The energy density emitted increases as frequency increases, diverging to infinity!  So even at room temperature, a blackbody would be emitting UV-radiation and even higher energy radiation!

The 'ultraviolet catastrophe' was resolved when Max Planck proposed the quantization of energy, the revolutionary idea that the energy of each electric oscillator is limited to discrete values and cannot be varied arbitrarily. Furthermore, the permitted energies of an oscillator of a given frequency are integer multiples of a constant, h:

where n = 0, 1, 2, 3, ...        (2)  Quantization of energy

where the constant h was an unknown proportionality constant to give the proper units of energy. According to Planck's hypothesis, oscillators are only excited if they can acquire an energy of at least hn\027. Therefore, for the higher frequency oscillators, this energy is too large for the walls to supply, and they remain unexcited. Accounting for the quantization of energy leads to the Planck distribution:

(3)   Planck's Distribution

Planck's distribution contained the unknown parameter, h.   A fit to experimental curves revealed h = 6.626 ×${10}^{-34}$, now known as Planck's constant! Using the Maple input below, plot Planck's distribution and vary T to see how the distribution depends on temperature.

 > $T≔'T':$
 > $\mathrm{ρ}\left(T,\mathrm{ν}\right)≔\frac{8\cdot \mathrm{Pi}\cdot \mathrm{evalf}\left(\mathrm{Constant}\left(h\right)\right)\cdot {\mathrm{\nu }}^{3}}{\mathrm{evalf}{\left(\mathrm{Constant}\left(c\right)\right)}^{3}}\cdot \frac{1}{\mathrm{exp}\left(\frac{\mathrm{evalf}\left(\mathrm{Constant}\left(h\right)\right)\cdot \mathrm{nu}}{\mathrm{evalf}\left(\mathrm{Constant}\left(k\right)\right)\cdot T}\right)-1};$
 ${\mathrm{\rho }}{≔}\left({T}{,}{\mathrm{\nu }}\right){↦}\frac{{8}{\cdot }{\mathrm{\pi }}{\cdot }{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({h}\right)\right){\cdot }{{\mathrm{\nu }}}^{{3}}}{{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({c}\right)\right)}^{{3}}{\cdot }\left({{ⅇ}}^{\frac{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({h}\right)\right){\cdot }{\mathrm{\nu }}}{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({k}\right)\right){\cdot }{T}}}{-}{1}\right)}$ (2.3)
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$\mathbf{T}$

What do you notice about r as T increases?

 Answer As T increases, the energy density increases and becomes more broad, and the peak wavelength shifts to the blue, as expected!



Applications in Physics

Treating the Universe as a Blackbody!

Now that you have a foundation in blackbody radiation, let's see how knowledge of blackbody radiation can be used.   Although there is no true blackbody, many objects behave like blackbodies, such as iron rods in a fire, a stovetop, or even stars! Astronomers estimate the temperature of stars based on their peak wavelength and Planck's distribution!

What if we were to treat the universe as a blackbody???  Could we estimate its temperature?   The universe has cooled since the Big Bang, and we can use the 'cosmic microwave background' (CMB), the remnant of the Big Bang serendipitously discovered and measured in 1964 by Arno Penzias and Robert Wilson (earning them the Nobel Prize in Physics in 1978), to estimate the temperature of the universe!

Using the Maple input below, plot the experimentally measured CMB:

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 > $\mathrm{plot}\left(\mathrm{CMB}\right);$

Notice the plot looks like Planck's distribution!  Using Maple's interactive plot feature, you will fit the CMB (red curve) to Planck's distribution (blue curve).  Use the slide bar to vary T until the curves overlay each other!

 > $A≔\mathrm{Spline}\left(\mathrm{CMB},\mathrm{nu}\right):$
 > $\mathrm{rho}\left(\mathrm{nu},T\right)≔\frac{8\cdot \mathrm{evalf}\left(\mathrm{Constant}\left(h\right)\right)\cdot \mathrm{Pi}\cdot {\mathrm{\nu }}^{3}}{\mathrm{evalf}{\left(\mathrm{Constant}\left(c\right)\right)}^{3}\cdot \left(\mathrm{exp}\left(\frac{\mathrm{evalf}\left(\mathrm{Constant}\left(h\right)\right)\cdot \mathrm{nu}}{\mathrm{evalf}\left(\mathrm{Constant}\left(k\right)\right)\cdot T}\right)-1\right)};$
 ${\mathrm{\rho }}{≔}\left({\mathrm{\nu }}{,}{T}\right){↦}\frac{{8}{\cdot }{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({h}\right)\right){\cdot }{\mathrm{\pi }}{\cdot }{{\mathrm{\nu }}}^{{3}}}{{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({c}\right)\right)}^{{3}}{\cdot }\left({{ⅇ}}^{\frac{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({h}\right)\right){\cdot }{\mathrm{\nu }}}{{\mathrm{evalf}}{}\left({\mathrm{Constant}}{}\left({k}\right)\right){\cdot }{T}}}{-}{1}\right)}$ (3.1.1)

You will use the interactive plot function in Maple to determine the temperature that best fits the Planck distribution to the CMB.  When the interactive window appears, set the plot type to Interactive Plot with 1 parameter.   Select Plot = 2D Plot.   x-axis = nu, min = 1e11, max = 6e11.   Slidable parameter 1 = T,  min = 0, max = 5.   Then choose Plot.  Adjust the slide bar until the two curves match to determine the temperature of the universe!

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What is the temperature of the Universe?

 Answer According to the fit, the temperature of the universe is approximately 2.72 K.

 References https://lambda.gsfc.nasa.gov/product/