# 1.1: Blackbody Radiation cannot Be Explained Classically

Overview

One experimental phenomenon that could not be adequately explained by classical physics was black-body radiation.

What is meant by the phrase “black body” radiation? The point is that the radiation from a heated body depends to some extent on the body being heated. To see this most easily, let’s back up momentarily and consider how different materials absorb radiation. Some, like glass, seem to absorb light hardly at all—the light goes right through. For a shiny metallic surface, the light isn’t absorbed either, it gets reflected. For a black material like soot, light and heat are almost completely absorbed, and the material gets warm.  How can we understand these different behaviors in terms of light as an electromagnetic wave interacting with charges in the material, causing these charges to oscillate and absorb energy from the radiation? In the case of glass, evidently this does not happen, at least not much.  Why not?  A full understanding of why needs quantum mechanics, but the general idea is as follows:  there are charges—electrons—in glass that are able to oscillate in response to an applied external oscillating electric field, but these charges are tightly bound to atoms, and can only oscillate at certain frequencies. It happens that for ordinary glass none of these frequencies corresponds to visible light, so there is no resonance with a light wave, and hence little energy absorbed. That’s why glass is perfect for windows! But glass is opaque at some frequencies outside the visible range (in general, both in the infrared and the ultraviolet). These are the frequencies at which the electrical charge distributions in the atoms or bonds can naturally oscillate.

How can we understand the reflection of light by a metal surface? A piece of metal has electrons free to move through the entire solid. This is what makes a metal a metal: it conducts both electricity and heat easily, both are actually carried by currents of these freely moving electrons.  (Well, a little of the heat is carried by vibrations.) But metals are recognizable because they’re shiny—why’s that? Again, it’s those free electrons: they’re driven into large (relative to the atoms) oscillations by the electrical field of the incoming light wave, and this induced oscillating current radiates electromagnetically, just like a current in a transmitting antenna. This radiation is the reflected light.  For a shiny metal surface, little of the incoming radiant energy is absorbed as heat, it’s just reradiated, that is, reflected.

Now let’s consider a substance that absorbs light: no transmission and no reflection. We come very close to perfect absorption with soot.  Like a metal, it will conduct an electric current, but nowhere near as efficiently.  There are unattached electrons, which can move through the whole solid, but they constantly bump into things—they have a short mean free path.  When they bump, they cause vibration, like balls hitting bumpers in a pinball machine, so they give up kinetic energy into heat.  Although the electrons in soot have a short mean free path compared to those in a good metal, they move very freely compared with electrons bound to atoms (as in glass), so they can accelerate and pick up energy from the electric field in the light wave. They are therefore very effective intermediaries in transferring energy from the light wave into heat.

### Relating Absorption and Emission

Having seen how soot can absorb radiation and transfer the energy into heat, what about the reverse? Why does it radiate when heated? The pinball machine analogy is still good: imagine now a pinball machine where the barriers, etc., vibrate vigorously because they are being fed energy. The balls (the electrons) bouncing off them will be suddenly accelerated at each collision, and these accelerating charges emit electromagnetic waves.  On the other hand, the electrons in a metal have very long mean free paths, the lattice vibrations affect them much less, so they are less effective in gathering and radiating away heat energy.  It is evident from considerations like this that good absorbers of radiation are also good emitters.

In fact, we  can be much more precise: a body emits radiation at a given temperature and frequency exactly as well as it absorbs the same radiation. This was proved by Kirchhoff: the essential point is that if we suppose a particular body can absorb better than it emits, then in a room full of objects all at the same temperature, it will absorb radiation from the other bodies better than it radiates energy back to them. This means it will get hotter, and the rest of the room will grow colder, contradicting the second law of thermodynamics.However, a metal glows when it’s heated up enough (Figure $$\PageIndex{1}$$): why is that? As the temperature is raised, the lattice of atoms vibrates more and more, these vibrations scatter and accelerate the electrons.  Even glass glows at high enough temperatures, as the electrons are loosened and vibrate.

Figure $$\PageIndex{1}$$: Hot metalwork. The yellow-orange glow is the visible part of the thermal radiation emitted due to the high temperature. Everything else in the picture is glowing with thermal radiation as well, but less brightly and at longer wavelengths than the human eye can detect. A far-infrared camera can observe this radiation.

A body emits radiation at a given temperature and frequency exactly as well as it absorbs the same radiation

Any body at any temperature above absolute zero will radiate to some extent, the intensity and frequency distribution of the radiation depending on the detailed structure of the body. To begin analyzing heat radiation, we need to be specific about the body doing the radiating:  the simplest possible case is an idealized body which is a perfect absorber, and therefore also (from the above argument) a perfect emitter. For obvious reasons, this is called a “black body.

Figure $$\PageIndex{2}$$: Blackbody radiator is any object that is a perfect emitter and a perfect absorber of radiation.

So how do we construct a perfect absorber in the laboratory? OK, nothing’s perfect, but in 1859 Kirchhoff had a good idea: a small hole in the side of a large box is an excellent absorber, since any radiation that goes through the hole bounces around inside, a lot getting absorbed on each bounce, and has little chance of ever getting out again. So, we can do this in reverse: have an oven with a tiny hole in the side, and presumably the radiation coming out the hole is as good a representation of a perfect emitter as we’re going to find.

By the 1890’s, experimental techniques had improved sufficiently that it was possible to make fairly precise measurements of the energy distribution in this cavity radiation, or as we shall call it black body radiation. In 1895, at the University of Berlin, Wien and Lummer punched a small hole in the side of an otherwise completely closed oven, and began to measure the radiation coming out. The beam coming out of the hole was passed through a diffraction grating, which sent the different wavelengths/frequencies in different directions, all towards a screen. A detector was moved up and down along the screen to find how much radiant energy was being emitted in each frequency range. They found a radiation intensity/frequency curve close to the distributions in Figure $$\PageIndex{2}$$.

Figure $$\PageIndex{3}$$: Graphic representation of spectral distribution of blackbody radiation at different temperatures. Image used with permission (CC-SA-BY 3.0; 4C).

By measuring the black body curves at different temperatures (Figure $$\PageIndex{3}$$), they were also able to construct two Laws on importance: Stefan-Boltmann’s Law and Wien’s Displacement Law.

### The Stefan-Boltzmann Law

The first quantitative conjecture based on experimental observations was the Stefan-Boltzmann Law (1879) which states the total power P radiated from one square meter of black surface at temperature $$T$$ goes as the fourth power of the absolute temperature (Figure $$\PageIndex{4}$$):

$P = \sigma T^4 \label{Eq1}$

where

• P is the total amount of radiation emitted by an object per square meter ($$Watts\; m^{-2}$$)
• $$\sigma$$ is a constant called the Stefan-Boltzman constant ($$5.67 \times 10^{-8} Watts\; m^{-2} K^{-4}$$)
• $$T$$ is the absolute temperature of the object (in K\)

Figure $$\PageIndex{4}$$: Graph of a function of total emitted energy of a black body proportional to the fourth power of its thermodynamic temperature $$T$$ according to the Stefan–Boltzmann law. Image used with permission (CC -SA-BY 3.0; xJaM).

The Stefan-Boltzmann Law is easily observed by comparing the integrated value (i.e., under the curves) of the experimental black-body radiation distribution in Figure $$\PageIndex{3}$$ at different temperatures. In 1884, Boltzmann derived this $$T^4$$ behavior from theory by applying classical thermodynamic reasoning to a box filled with electromagnetic radiation, using Maxwell’s equations to relate pressure to energy density. (The tiny amount of energy coming out of the hole would of course have the same temperature dependence as the radiation intensity inside.)

Exercise $$\PageIndex{1}$$

The sun’s surface temperature is 5700 K. How much power is radiated by one square meter of the sun’s surface? Given that the distance to earth is about 200 sun radii, what is the maximum power possible from a one square kilometer solar energy installation?

Exercise $$\PageIndex{2}$$

If surface body temperature is 90 °F. How much radiant energy in $$W\, m^{-2}$$ would your body emit? What is the total radiant energy emitted by your body in Watts?

### Wien’s Displacement Law

The second finding was Wien’s Displacement Law. Wien's law identifies the dominant wavelength, or color, of light coming from a body at a given temperature. As the oven temperature varies, so does the frequency at which the emitted radiation is most intense (Figure $$\PageIndex{3}$$). In fact, that frequency is directly proportional to the absolute temperature:

$\nu_{max} \propto T \label{Eq2}$

Wien himself deduced this law theoretically in 1893, following Boltzmann’s thermodynamic reasoning. It had previously been observed, at least semi-quantitatively, by an American astronomer, Langley. This upward shift in $$\nu_{max}$$ with $$T$$ is familiar to everyone—when an iron is heated in a fire (Figure $$\PageIndex{1}$$), the first visible radiation (at around 900 K) is deep red, the lowest frequency visible light. Further increase in T causes the color to change to orange then yellow, and finally blue at very high temperatures (10,000 K or more) for which the peak in radiation intensity has moved beyond the visible into the ultraviolet.

Another representation of Wien's Law (Equation $$\ref{Eq2}$$) in terms of the peak wavelength of light is

$\lambda_{max} = \dfrac{b}{T} \label{Eq2a}$

where $$T$$ is the absolute temperature in kelvin and $$b$$ is a constant of proportionality called Wien's displacement constant, equal to $$2.89 \times 10^{−3} m\, K$$, or more conveniently to obtain wavelength in micrometers, $$b≈2900\; μm \cdot K$$. This is an inverse relationship between wavelength and temperature. So the higher the temperature, the shorter or smaller the wavelength of the thermal radiation. The lower the temperature, the longer or larger the wavelength of the thermal radiation. For visible radiation, hot objects emit bluer light than cool objects.

Example $$\PageIndex{1}$$: The temperature of the Sun

For example, if the Sun has a surface temperature of 5700 K, what is the wavelength of maximum intensity of solar radiation?

Solution

If we substitute 5700 K for T in Equation $$\ref{Eq2a}$$, we have

$λ_{max} = \dfrac{0.0029}{5700} = 5.1 \times 10^{-7} \, m$

Knowing that violet light has a wavelength of about 4.0 × 10-7 meters, yellow about 5.6 × 10-7 meters, and red about 6.6 × 10-7 meters, what can we say about the color of the Sun's peak radiation? The peak wavelength of the Sun's radiation is at a slightly shorter wavelength than the color yellow, so it is a slightly greenish yellow. To see this greenish tinge to the Sun, you would have to look at it from space. It turns out that the Earth's atmosphere scatters some of the shorter waves of sunlight, which shifts its peak wavelength to pure yellow.

Remember that thermal radiation always spans a wide range of wavelengths (Figure $$\PageIndex{2}$$); the mathematical form given above specifies the single wavelength that is the peak of the spectrum. So although the Sun appears yellowish-white, when you disperse sunlight with a prism you see radiation with all the colors of the rainbow. Yellow just represents the average wavelength of the emission.

Exercise $$\PageIndex{3}$$

1. At what wavelength does the sun emit most of its radiation if it has a temperature of 5,778 K?
2. At what wavelength does the earth emit most of its radiation if it has a temperature of 288 K?

1. 500 nm
2. 10.0 microns

### The Ultraviolet Catastrophe

Lord Rayleigh and J. H. Jeans developed an equation which explained blackbody radiation at low frequencies.The equation which seemed to express blackbody radiation was built upon all the known assumptions of physics at the time. The big assumption which Rayleigh and Jean implied was that infinitesimal amounts of energy were continuously added to the system when the frequency was increased. Classical physics assumed that energy emitted by atomic oscillations could have any continuous value.This was true for anything that had been studied up until that point, including things like acceleration, position, or energy. The resulting Rayleigh-Jeans law  was

$d\rho \left( \nu ,T \right) = \rho_{\nu} \left( T \right) d\nu = \dfrac{8 \pi k_B T}{c^3} \nu^2 d\nu \label{Eq3}$

Experimental data performed on the black box showed slightly different results than what was expected by the Rayleigh-Jeans law. The law had been studied and widely accepted by many physicists of the day, but the experimental results did not lie, something was different between what was theorized and what actually happens. The experimental results showed a bell type of curve but according to the Rayleigh-Jeans law the frequency diverged as it neared the ultraviolet region (Equation $$\ref{Eq3}$$). Ehrenfest later dubbed this the ultraviolet catastrophe.

It’s worth emphasizing that Equation $$\ref{Eq3}$$ is a classical result: the only inputs are classical dynamics, and Maxwell’s electromagnetic theory. Notice that the charge e of the oscillator does not appear: the result is independent of the coupling strength between the oscillator and the radiation, the coupling only has to be strong enough to ensure thermal equilibrium.

.

Figure $$\PageIndex{5}$$: Relationship between the Temperature of an Object and the Spectrum of Blackbody Radiation It Emits. At relatively low temperatures, most radiation is emitted at wavelengths longer than 700 nm, which is in the infrared portion of the spectrum. The dull red glow of the hot metalwork in Figure $$\PageIndex{5}$$ is due to the small amount of radiation emitted at wavelengths less than 700 nm, which the eye can detect. As the temperature of the object increases, the maximum intensity shifts to shorter wavelengths, successively resulting in orange, yellow, and finally white light. At high temperatures, all wavelengths of visible light are emitted with approximately equal intensities.

Experimental data performed on the black box showed slightly different results than what was expected by the Rayleigh-Jeans law.  The law had been studied and widely accepted by many physicists of the day, but the experimental results did not lie, something was different between what was theorized and what actually happens.The experimental results showed a bell type of curve but according to the Rayleigh-Jeans law the frequency diverged as it neared the ultraviolet region (Equation $$\ref{Eq3}$$). This inconsistency was termed the ultraviolet catastrophe.