# Planck distribution

Figure 1: Planck blackbody spectral emissive power from eq. (1)
Figure 2: Plot of generalized spectral blackbody emissive power.
Figure 3: Relation of angle and radius for (a): Planar angle and (b) Solid angle
Figure 4: Projected area and intensity
Figure 5: Solid angle in terms of angles dθ and dφ
Figure 6: Hemisphere for integration of intensity over $0 \le \theta \le \pi /2$ and $0 \le \phi \le 2\pi$

The historical development of the expression for the spectral distribution of blackbody thermal radiation is quite interesting and important, as the search for the correct relation led directly to the development of quantum theory, which must be invoked to explain how the derivation of the blackbody relation can be reconciled with experimental measurements. Max Planck derived the correct relation in 1901, and was forced to invoke quantum arguments to explain the form that agrees with experiment. Planck's fundamental relation for the rate of energy emission (into all directions) from an ideal blackbody is

${E_{\lambda b}}\left( {\lambda ,T} \right) = \frac{{2\pi {C_1}}}{{{\lambda ^5}\left( {{e^{{C_2}/\lambda T}} - 1} \right)}} \qquad \qquad(1)$

where Eλb is the spectral emissive power of the blackbody, which is the rate of thermal radiation per unit area and per unit wavelength interval emitted by a blackbody at temperature T and at the wavelength λ. The Eλb has units in SI of (W / m2 − μm). The subscript b indicates that this expression applies to a blackbody, and the λ subscript indicates that it is wavelength dependent. The constants C1 and C2 are combinations of more fundamental physical constants (Planck's constant h, the Boltzmann constant k, and the speed of light in a vacuum, co), and have values of

${C_1} = hc_o^2 = 0.59552 \times {10^8}({\rm{W - \mu }}{{\rm{m}}^{\rm{4}}}{\rm{/}}{{\rm{m}}^{\rm{2}}})$
C2 = hco / k = 14,388(μmK)

Observe that eq. (1) is the rate of energy leaving the surface; the blackbody generally will also be absorbing energy from other radiating sources in its surroundings, and these will have to be considered to find the net radiative heat transfer rate to/from the blackbody element.

Blackbody spectral emissive power Eλb clearly depends on both temperature and wavelength in a complex way. A plot of Eλb at three temperatures is shown in Fig. 1. Some observations from this plot include that as temperature increases, the spectral emissive power at each wavelength increases so that the curves never cross; the wavelength for peak emission decreases with increasing temperature; and the total emission (area under the curves) increases rapidly with temperature. Each of these observations has important consequences in analyzing heat transfer by thermal radiation.

A more general relation for Eλb can be formed by dividing eq. (1) by T5, resulting in

$\frac{{{E_{\lambda b}}}}{{{T^5}}} = \frac{{2\pi {C_1}}}{{{{\left( {\lambda T} \right)}^5}\left( {{e^{{C_2}/\lambda T}} - 1} \right)}}\qquad \qquad(2)$

It is clear that the ratio Eλb / T5 is now a function of only the product λT rather than λ and T separately. Equation (2) can then be plotted as a single curve as in Fig 2. From this plot, it is seen that the maximum in the blackbody function occurs at a particular value of the λT product, given by

${\left( {\lambda T} \right)_{\max }} = {C_3} = 2897.8(\mu m \cdot K)\qquad \qquad(3)$

This result can be derived analytically by taking the derivative of eq. (2), and setting it equal to zero to find the location of the maximum of the function. Equation (3) is called Wien's Displacement Law, and provides a quick way to determine the wavelength where most blackbody thermal energy is emitted by a blackbody at temperature T.

For engineering calculation, the total energy emitted by a blackbody surface at all wavelengths is generally of interest, and can be found by integrating eq. (1) over all wavelengths. This can be done through a change of variables using ξ = C2 / λT and results in

${E_b} = \int_{\lambda = 0}^\infty {\frac{{2\pi {C_1}}}{{{\lambda ^5}\left( {{e^{{C_2}/\lambda T}} - 1} \right)}}d\lambda } = \frac{{2\pi {C_1}{T^4}}}{{C_2^4}}\int_{\xi = 0}^\infty {\frac{{{\xi ^3}}}{{{e^\xi } - 1}}} d\xi = \sigma {T^4}\qquad \qquad(4)$

Where σ is the Stefan-Boltzmann constant, again made up of a combination of the more fundamental constants, $\sigma \equiv 2{C_1}{\pi ^5}/(15C_2^4)$. It has a numerical value in SI units of

$\sigma {\rm{ = 5}}{\rm{.6704}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{ W/}}{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^{\rm{4}}}{\rm{.}}$

The Stefan-Boltzmann Law, eq. (4), is one of the fundamental relations of thermal radiation heat transfer, showing that total energy emission from a blackbody depends on the fourth power of the absolute temperature of a material. (The word total in thermal radiation is used to indicate integration over all wavelengths.) Again, energy absorbed by the blackbody surface that originates from other radiating materials must be accounted for before radiation heat transfer can be calculated. The fourth power dependence of total blackbody emissive power indicates that radiation will rapidly become important or dominating over other heat transfer modes at higher absolute temperatures. This is certainly the case in power plant boilers, where radiation can contribute up to 90 percent of the energy transfer from combustion gases to the steam tubes.

The small numerical value of the Stefan-Boltzmann constant indicates that thermal radiation may be unimportant at lower absolute temperatures; however, this conclusion can be misleading. If the other heat transfer modes are small or suppressed, then radiation may be the important mode even at low temperatures. This is the case, for example, in spacecraft thermal analysis, heat loss in evacuated thermos bottles, and even for energy loss from surfaces at relatively low temperatures where radiative transfer may be comparable to free convection losses.

When two materials exchange radiation energy, they will be oriented with each other in a particular way. The energy leaving surface A that reaches surface B will depend on this orientation, so an additional quantity is needed to prescribe the radiant energy going into a particular direction (remember that Eb includes the radiation emitted into all directions). It is necessary now to look at the three-dimensional analog of the relation between the angle dθ subtended by an arc length dS swept out by a radius arm of length R. In two dimensions, this relationship is the familiar dθ = dS/R where θ is in radians (Fig. 3a). In a three-dimensional rendering (Fig. 3b), the analogous solid angle dΩ is subtended by the area dAs that is normal to the radius R. The relation among these factors is dΩ dAs / R2. The solid angle has units of steradians, sr. Now consider a hemisphere of radius R, with a blackbody element at temperature T centered at its base (Figure 4). If an observer on the surface of the hemisphere looks at the element, he/she will detect energy reaching the hemisphere surface. As the observer moves so that the angle θ is increased, the apparent size of the element to the observer will vary as dA cosθ, so that at θ = 0 the full element size is seen, but as the observer nears an angle of 90°, an edge-on view is approached and the apparent element size approaches zero. This apparent area dAp = dA cosθ is called the projected area of the element. The energy reaching the observer from the element dA thus varies as cos θ, and this effect is known as Lambert's cosine law. It is valid for a blackbody, but may not be followed for real surfaces.

We can now define a quantity that will quantify the radiation emitted from a blackbody in a particular direction. This quantity is called the intensity; it is also wavelength and temperature dependent, and for a blackbody is given the symbol Iλb. The spectral blackbody intensity is defined as the rate of energy deλ emitted per unit solid angle subtended by dAs, per unit projected area dA cosθ, per unit wavelength interval, or

${I_{\lambda b}} = \frac{{d{e_\lambda }}}{{dA\cos \theta d\omega d\lambda }} \qquad \qquad(5)$

Suppose the elements dA and dAs in Fig. 4 are both blackbodies at the same temperature T. The energy leaving dA in the direction of dAs is, from eq. (5)

$d{e_{\lambda ,dA - d{A_s}}} = {I_{\lambda b}}(\theta )dA\cos \theta d\Omega d\lambda = {I_{\lambda b}}\left( \theta \right)dA\cos \theta \left( {\frac{{d{A_s}}}{{{R^2}}}} \right)d\lambda \qquad \qquad(6)$

Similarly, the energy from dAs that is incident on dA is

$d{e_{\lambda ,d{A_s} - d{A_{}}}} = {I_{\lambda b}}({\theta _s} = 0)d{A_s}d\Omega d\lambda = {I_{\lambda b}}({\theta _s} = 0)d{A_s}\left( {\frac{{dA \cdot \cos \theta }}{{{R^2}}}} \right)d\lambda \qquad \qquad(7)$

The projected area of dA has been used in evaluating the subtended solid angle of dA when viewed from dAs. Also, dAs is normal to R so that cosθs = 1. To avoid violating the second law, the energy exchange between the two elements at the same temperature must be the same, and equating (6) and (7) results in

${I_{\lambda b}}(\theta ) = {I_{\lambda b}}({\theta _s} = 0) = {I_{\lambda b}} \qquad \qquad(8)$

so that the blackbody intensity is independent of angle of emission. The intensity is what the eye interprets as the brightness of an emitter.

Examination of Fig. 5 allows an alternate definition of the solid angle dΩ subtended by the area dAs on the surface of the hemisphere in terms of the angles θ and φ as

$d\Omega = \frac{{d{A_s}}}{{{R^2}}} = \frac{{d{S_1} \times d{S_2}}}{{{R^2}}} = \frac{{\left( {R\sin \theta d\phi } \right) \times \left( {Rd\theta } \right)}}{{{R^2}}} = \sin \theta d\theta d\phi \qquad \qquad(9)$

where dS1 and dS2 are the arc lengths of the sides of dAs. This allows finding the relation between Iλb and Eλb by integrating the energy deλ that leaves through a particular solid angle over all solid angles in the hemisphere (Figure 6).

Using eq. (6) gives

$\begin{array}{l} {E_{\lambda b}} = \int_{{A_s}}^{} {d{e_{\lambda ,dA - d{A_s}}}} = \int_{\Omega = 2\pi }^{} {{I_{\lambda b}}\cos \theta d\Omega } \\ = {I_{\lambda b}}\int_{\phi = 0}^{2\pi } {\int_{\theta = 0}^{\pi /2} {\cos \theta \sin \theta d\theta d\phi } } = 2\pi {I_{\lambda b}}\int_{\theta = 0}^{\pi /2} {\cos \theta \sin \theta d\theta } = \pi {I_{\lambda b}} \\ \end{array} \qquad \qquad(10)$

resulting in a remarkably simple relation between the blackbody spectral emissive power Eλb and the blackbody spectral intensity Iλb of Eλb = πIλb.

## References

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.