The Physics of Energy, page 91
Box 22.1 More on Black (and Not so Black) Bodies
The interaction of an object with electromagnetic radiation is characterized by its absorptivity , transmittance ),emissivity , and reflectivity , all of which can depend on the frequency ω of the radiation. All aredimensionless numbers describing the fraction of radiation of frequency ω that is absorbed, etc. If anobject is opaque (it does not transmit any radiation) then light incident on it must be either reflected orabsorbed, so and must sum to one when . German physicist Gustave Kirchhoff, who first introduced theconcept of blackbody radiation, used the existence of thermal equilibrium to prove that the absorptivity andemissivity must be equal, since otherwise the object could not reach a thermal equilibrium in the presence ofincoming thermal radiation without violating the second law of thermodynamics. Thus, any opaque object has
(22.15)
A black body is defined as a perfect absorber – and therefore perfect emitter – of radiation, with for all ω. Blackbody radiation is the radiation emitted by such an object when it is in thermal equilibrium. A close approximation to a black body is given by a small opening in a wall enclosing a cavity held at constant temperature. A glass blower’s furnace provides a good example (Figure 6.2). The hole allows all radiation that impinges on it to enter the cavity, where it bounces around until it is absorbed by the walls. Thus, the hole has to a very good approximation. The radiation that is emitted from the walls and finds its way out of the hole is characteristic of the temperature T of the cavity. Quantum mechanics allows us to view radiation as a gas of photons, which is in thermal equilibrium within the cavity, and can be observed when it leaks out of the hole.
If an object has an emissivity less than one but independent of frequency, it is termed a gray body. The emissivity of some common materials was mentioned in §6.3.3.
The discovery of Planck’s law (22.13) was a key step in the development of quantum mechanics. In the early 1900s, English physicists John Strutt (Lord Rayleigh) and James Jeans used classical reasoning to argue that the rate of energy radiation in the frequency range should go as . This Rayleigh–Jeans law agrees with eq. (22.13) at small ω (including the overall constant). This hypothesis matched experimental data at the time for low frequencies, but suffered from two problems. First, experimental data showed that the Rayleigh–Jeans law is violated at high frequency. Second, Rayleigh–Jeans predicted an infinite amount of radiated energy when integrated over all frequencies, a problem known as the ultraviolet catastrophe. Planck solved these problems by postulating that excitations of the electromagnetic field in a cavity are quantized in units of , leading to eq. (22.13), five years before Einstein postulated the quantization of electromagnetic energy in photons to explain how light can knock electrons off the surface of metals.
Example 22.1 Blackbody Radiation From Toaster Heating Coils
A household toaster oven is heated by two coils, each composed of a metal cylinder of diameter 0.6 cmand length 1.2 m. If the toaster oven converts 1500 W of electric power into thermal radiation, what is thetemperature of the coils and the frequency at which the radiation power spectrum is peaked?
The total surface area of the coils is m. The radiated power is related to the temperature through
Solving for temperature, we get K ℃. At this temperature, the frequency associated with maximum power is . This corresponds to infrared radiation of wavelength m. The tail in the frequency distribution of thermal radiation at this temperature, however, contains almost a milliwatt of radiation in the visible spectrum, so the coil glows red when radiating at this temperature and power (Problem 22.2).
22.2.3 The Solar Blackbody Spectrum
The distribution of solar energy across the electromagnetic spectrum can be estimated from eq. (22.13), where we approximate the Sun as a black body at a temperature of 5780 K. Figure 22.3 shows the distribution of blackbody radiation at 5780 K, both as a function of frequency/photon energy and as a function of wavelength. Solar radiation is strongly peaked in the visible range of 400–750 nm. Because the Sun is not a perfect black body, the spectrum in the figure is only approximate. In fact, though the total luminosity corresponds to a temperature of roughly 5780 K, the shape and location of the peak of solar energy is closer to that of a blackbody distribution at 6000 K. For most purposes, it is sufficient to consider the solar spectrum as a blackbody spectrum at 5780 K. We discuss some further details of the solar radiation spectrum and its absorption in the atmosphere in §23.5.
Figure 22.3 The spectrum of solar radiation approximated as a perfect black body at 5780 K, graphed (a) as a function of photon energy, and (b) as a function of wavelength. The band of visible photons is shaded in both graphs.
Thermal Radiation
Any object at temperature T radiates energy in the form of electromagnetic waves. The power per unit area in the radiated energy is distributed among photons of frequency ω according to the Planck radiation law
where characterizes the emissivity of the object. If ε is constant, then the total power is given by the Stefan–Boltzmann law
where σ is the Stefan–Boltzmann constant
22.3Derivation of the Blackbody Radiation Formula
The spectrum of thermal radiation plays a central role in determining Earth’s climate and the solar energy available for human use, and warrants a derivation for the interested reader. Furthermore, the Planck radiation law (22.13) can be derived from first principles using the methods of quantum and statistical physics that have been developed in §7 and §8, and provides an excellent example of the power of those methods.
We begin with a brief summary of the argument, which may serve as a road map for the analysis that follows. As described in Box 22.1, we can regard blackbody radiation as the spectrum of radiation emitted from a small opening in a cavity at temperature T. The first step is to enumerate the standing wave solutions of Maxwell’s equations in the cavity. These are modes that oscillate sinusoidally in time with certain allowed frequencies. Quantum mechanically, each mode of oscillation with frequency ω behaves like an independent harmonic oscillator with energy levels , with m a nonnegative integer. Next we turn to statistical mechanics, and in particular to the Boltzmann distribution, which determines the probability of finding each of these field oscillators in the energy level when the system is in equilibrium at temperature T. The internal energy of the radiation field is obtained by multiplying the probability for a given mode to be in a given state by the energy of that state and summing over all states and over all modes of the radiation field. Finally, we imagine that a hole on the side of the box allows electromagnetic radiation to escape, and we compute the power spectrum and total rate of energy loss through the hole. This leads to the Planck radiation law, including an expression for the Stefan–Boltzmann constant in terms of , , and c. The analysis presented here can be found in many texts; see for example [21].
We begin by considering a cubic cavity with sides of length L. We choose this shape to simplify the calculation, though the final result turns out to be an integral over the volume that is independent of the shape of the cavity. We take coordinates inside the box to run from 0 to L. We take the box itself to be a conductor, although the final result is independent of the nature of the material as long as it interacts with the electromagnetic field. We wish to consider the quantum electromagnetic field inside the box (see Figure 22.4).
Figure 22.4 To find the blackbody radiation spectrum we quantize the electromagnetic field in a conducting box.
The assumption that the box is made of a conducting material implies that the electric field must be perpendicular to all the boundaries of the box, so no charges are accelerated. For example, at , the electric field can only have a nonzero component . The modes of the electromagnetic field are then standing wave solutions of Maxwell’s equations in a vacuum that satisfy this boundary condition. Finding these modes is a problem similar to solving the Schrödinger equation for a particle in a 3D box, which we considered in §7. The spatial dependence of each component of the electric field can be written as a product of sine and cosine modes. The boundary conditions restrict the possible combinations for the E field, so that the set of mode solutions is given by
(22.16)
where the modes are parameterized by a 3-tuple (triplet) of positive integers , with . E must satisfy Gauss’s law , which reduces to a single linear condition on the three amplitudes . There are two independent solutions to this equation for each choice of k, corresponding to the two independent polarizations of light waves. Each standing wave mode can be written as a linear combination of plane waves. This decomposition uniquely fixes the fields of the standing waves, though we will not need the explicit form here. The frequency ω associated with each mode follows from substituting (22.16) into (22.12), giving
(22.17)
Each mode is essentially an independent harmonic oscillator of frequency . We can therefore think of the quantum electromagnetic field in a conducting box as a system of many independent harmonic oscillators. The oscillators are parameterized by a triplet of positive integers n, and there are two oscillators (polarizations) for each allowed triplet. We know that a quantum harmonic oscillator of frequency ω has a spectrum of states with energy levels . Since the oscillators are independent, the total energy of the system is simply given by the sum of the energies in all the oscillators.
In §8.6, we determined that in thermal equilibrium the probability of finding a system in a state with energy E is given by the Boltzmann distribution (8.32). As an application, we computed the internal energy of a crystal modeled as 3N independent oscillators (Einstein’s model of the specific heat). That result, eq. (8.54), can be adapted to the present situation by removing the factor of 3N. The average energy in the mode with frequency is then
(22.18)
The ground state energy of the oscillators is simply an additive constant, known as the zero-point energy, which does not affect the physics, so we drop the factor of henceforth.3
Given the average energy in each mode of the electromagnetic field, it remains only to sum over all the modes,
(22.19)
where the factor of two comes from the two polarizations of the electromagnetic field. If we assume that the size of the box L is sufficiently large,4 we can approximate the sums by integrals,
(22.20)
The last step in this equation makes use of a change from cartesian coordinates to spherical polar coordinates (Appendix B.1.2).
We can now use eq. (22.17) to replace by , giving an expression for the average value of the total energy as an integral over the spectrum parameterized by frequency ω,
(22.21)
where we have replaced by the volume V and where is the spectral density of energy,
(22.22)
describes the density of energy per unit volume contained in modes of the electromagnetic field with frequency between ω and .
We have thus computed the energy distribution in the electromagnetic field inside a volume V at temperature T. Finally, we must relate this to the energy that is emitted from a hole of area A in the surface of the box surrounding the cavity. Those photons headed towards the hole can continue out of the box. If all photons in the nearby volume were headed directly towards the hole, the rate at which energy associated with photons at frequency ω would leave the box would be (density speed area). If we put the hole, say, on the face at , then a mode n will be moving towards the hole if . The angle of incidence between the mode and the face with the hole is given by (see Figure 22.5). Integrating over all angles of incidence on the hole, the total energy leaving the hole is reduced by a factor of . Thus, the flux of energy (energy per unit area per unit time) per unit frequency leaving the black body is given by
(22.23)
which is precisely the Planck radiation law (22.13) for a perfect black body with emissivity for all ω.
Figure 22.5 A fraction of the photons passing through points in an element of surface area are heading toward the hole A and will pass out of the cavity.
To compute the total radiation, we simply integrate eq. (22.23); replacing , the integral of the term in parentheses in eq. (22.23) is
(22.24)
This gives us the Stefan–Boltzmann law
(22.25)
where (including the constant factor from eq. (22.23))
(22.26)
Thus, we have derived the Planck and Stefan–Boltzmann laws for the electromagnetic field in a conducting cavity. While we have assumed a cubic cavity, note that the energy density scales as the volume of the cavity. Though we have not proven this explicitly, the energy does not in fact depend upon the shape of the cavity walls; the electromagnetic field in any region has an energy density in thermal equilibrium at temperature T that takes the form of eq. (22.22).
Discussion/Investigation Questions
22.1 The interior of a glass blower’s furnace may contain shelves and other objects of various shapes and colors. As the furnace is heated, objects in the interior become more and more difficult to discern through a small opening, until finally, when the furnace is hot enough, the interior objects are all but lost in a uniform glow (see Figure 6.2). Explain what is going on here.
22.2 As a source of blackbody radiation becomes hotter, the peak in its radiation spectrum moves from the visible to the ultraviolet and beyond. Does this imply that the object can no longer be seen by the unaided human eye?
Problems
22.1 [C] Compute or estimate the fraction of radiated thermal energy that is in the visible range (wavelength 400–750 nm) for the following two radiation spectra: (a) solar radiation, assumed to be perfect blackbody radiation at 5780 K; (b) EM radiation from a pot of boiling water, again assume a perfect blackbody distribution.
22.2 [C] Compute the power radiated in the visible range from the toaster oven coil radiating with a total power of 1500 W, as described in Example 22.1.
22.3 Estimate the rate of thermal radiation from a household hot-water radiator with a surface area of 1 m and a temperature of 80℃.
22.4 The incandescent light bulb is a notoriously inefficient way to convert electric power into visible light. The tungsten filament emits blackbody radiation at a temperature that is limited by its melting point. Define the lighting efficiency as the ratio of the power emitted in the visible range (400–750 nm) to the total power emitted. What is the theoretical maximum efficiency of a tungsten light bulb, given that tungsten melts at 3422℃?
22.5 Compute the frequency corresponding to the maximum power of radiation for a blackbody at temperature K. This is roughly the average temperature of Earth’s surface. Compute the wavelength corresponding to this frequency.
22.6 [T] Estimate the gravitational energy in the Sun, using eqs. (22.1) and (22.2), assuming that the mass is distributed uniformly. Can you confirm the statement in the text that gravitational potential energy could only power the Sun at its present luminosity for y? At what rate would the radius of the Sun have to be decreasing (in, say, mm/s) if its luminosity came from gravitational potential energy?
22.7 [T] Compute the energy released in each step of the solar PPI fusion chain eq. (22.5), and confirm that the total energy released matches eq. (22.6).
22.8 [T] Show that the combination of the Boltzmann factor and the tunneling probability give a probability for fusion that is maximized at , as stated in §22.1.1.
22.9 [T] Show that the peak of the blackbody spectrum as a function of ω is given by eq. (22.14).
22.10 [T] Rewrite the blackbody power spectrum as a function of and compute the value where the power density is maximized. Find the frequency ω corresponding to . Why is ω not equal to , where is given by eq. (22.14)? Compare to for a black body at temperature 5780 K.
22.11 [T] Use dimensional analysis to show that the wavelength scale of blackbody radiation is given by . The average radiation energy in a cavity depends only on its volume and not on its shape when , where L is a typical length scale characterizing the cavity. How low must the temperature be for the shape to matter if cm?
22.12 [T] Show that the classical Rayleigh–Jeans law for radiation follows from eq. (22.13) in the limit as . Show that the total power radiated diverges in this limit (the ultraviolet catastrophe that helped lead to the discovery of quantum mechanics).
22.13 [T] In §7, we showed that the energy of a particle of mass M in an box in the state labeled is . The probability of finding a particle in the state at temperature T is given by the Boltzmann distribution . In a hot gas the energy levels are very close together, so E can be regarded as a continuous variable. Show that the probability of finding the particle with energy between E and at temperature T is given by the Maxwell–Boltzmann distribution, . [Hint: Employ the methods used to compute the spectrum of blackbody radiation, including eq. (22.20).]
* * *
1 Recall (from §18) that we use to denote fully ionized nuclei, such as those involved in fusion processes.
2 Neutrinos only carry away about 3% of the energy from the Sun, though there are a lot of them: a flux of roughly solar neutrinos/m s pass through Earth and everything on it. The luminosity quoted in eq. (22.3) includes only EM radiation and not neutrinos.
3 This zero-point energy relates to the discussion in §21.1.5; if we include modes with wavelengths up to the Planck length, this gives a ground state energy that is over 100 orders of magnitude larger than the observed cosmological constant.
