Fundamentals of Blackbody Radiation

Prof. Elias N. Glytsis November 8, 2018

School of Electrical & Computer Engineering National Technical Univerity of Athens This page was intentionally left blank...... BLACKBODY RADIATION†

A blackbody absorbs and emits radiation perfectly, i.e. it does not favor any particular range of radiation frequencies over another. Therefore the intensity of the emitted radiation is related to the amount of energy in the body at thermal equilibrium. The history of the development of the theory of the blackbody radiation is very interesting since it led to the discovery of the quantum theory [1]. Early experimental studies established that the emissivity of a blackbody is a function of frequency and temperature. A measure of the emissivity can be the term ρ(ν, T ) which is the density of radiation energy per unit volume per unit frequency (J/m3Hz) at an absolute temperature T and at frequency ν. The ﬁrst theoretical studies used the very successful at that point theory of Maxwell equations for the determination of the density of electromagnetic modes and from that the determination of ρ(ν, T ). For example, Wilhelm Wien in 1896 used a simple model to derive the expression

ρ(ν, T )= αν3 exp(−βν/T ) (1) where α, β were constants. However, the above equation failed in the low frequency range of the experimental data. In June 1900 Lord Rayleigh published a model based on the modes of electromagnetic waves in a cavity. Each mode possessed a particular frequency and could give away and take up energy in a continuous manner. Using the standard electromagnetic theory of a cavity resonator (see Fig. 1) with perfect conductor walls the following dispersion equation can be easily obtained [3, 4, 5] (see Appendix A) :

mπ 2 pπ 2 qπ 2 2πν 2 + + = n2, (2) a b d c where n is the index of refraction of the medium, and a, b, and d are the dimensions of the cavity resonator in the x, y, and z directions, and m, p, q are positive integers.

† c 2018 Prof. Elias N. Glytsis, Last Update: November 8, 2018

Figure 1: Cavity box for the determination of the density of electromagnetic modes.

If for simplicity it is assumed that the cavity is a cube, a = b = d then the previous equation can be written as 2ν 2 2νna 2 m2 + p2 + q2 = a2n2 = . (3) c c In order to count the electromagnetic modes up to frequency ν it is necessary to evaluate the number of modes that ﬁt in the one eighth of the sphere that is shown in Fig. 2. Thus, the total number of electromagnetic modes N(ν) can be determined as follows (1/8) cavity volume (1/8)(4/3)π(2νan/c)3 4 ν3n3a3 N(ν)= = = π . (4) volume of a mode 1 × 1 × 1 3 c3 Due to TE and TM mode degeneracy the above number should be multiplied by a factor of 2. Therefore, the total number of electromagnetic modes per volume, N (ν), is N(ν) 8 ν3n3 N (ν)= = π . (5) V olume = a3 3 c3 Then the density of electromagnetic modes per frequency is dN (ν) 8πν2n3 = . (6) dν c3 In the last equation it is assumed that the refractive index n is independent of frequency (or freespace wavelength). Usually for all materials there is dispersion, i.e. dependence of the

2 refractive index on the frequency (or wavelength) of the electromagnetic radiation. In the latter case n = n(ν) and in the above derivative over frequency this dependence must be taken into account. Then the previous equation can be written as follows [6] dN (ν) 8πν2n2(n + ν dn ) 8πν2n2n = dν = g . (7) dν c3 c3

dn dn where ng = n + ν dν = n − λ dλ is the group refractive index and is important in materials such semiconductors and ﬁbers where the refractive index dependence on frequency (or wavelength) can be signiﬁcant. For the remainder of this section it will be assumed that the refractive index

is independent of frequency (or wavelength) for the sake of simplicity. §

Figure 2: The eighth of the sphere in the mpq space for the determination of the number of electromagnetic modes up to frequency ν.

Rayleigh assigned an energy kBT/2 to each electromagnetic mode (kBT/2 for the electric

−23 ◦ ﬁeld oscillation and kBT/2 for the magnetic ﬁeld oscillation, where kB = 1.38066×10 J/ K). Therefore, the electromagnetic energy density per unit frequency ρ(ν, T ) becomes dN (ν) 8πν2n3 ρ(ν, T )= k T = k T. (8) dν B c3 B The last equation is known as the Rayleigh-Jeans distribution of a blackbody radiation and fails dramatically in the ultraviolet part of the spectrum (historically referred as the “ultraviolet catastrophe”). This can be seen in the Rayleigh-Jeans curve of Fig. 3.

3 It has been suggested that Planck discovered his radiation formula in the evening of October 7, 1900 [1]. Planck had taken into account some additional experimental data by Heinrich Reubens and Ferdinand Kurlbaum as well as Wien’s formula and he deduced an expression that “ﬁtted” all the available experimental data. His formula was the now known as the blackbody radiation formula given by 8πν2n3 hν ρ(ν, T )= 3 , (9) c exp(hν/kBT ) − 1 where h = 6.626×10−34Joule·sec is known as Planck’s constant. The above expression reduces to Wien’s formula for high frequencies (i.e. hν/kB T 1) and to Rayleigh-Jeans formula for low frequencies (i.e. hν/kBT 1). An example of Planck’s radiation formula is shown in Fig. 3 along with Rayleigh-Jeans and Wien’s approximations for a blackbody of absolute temperature T = 6000◦K. Having obtained his formula Planck was concerned to discover its physical basis. It was hard to argue about the density of electromagnetic modes determination. Therefore, he focused on the average energy per electromagnetic mode. Planck made the hypothesis that electromagnetic energy at frequency ν could only appear as a multiple of the step size hν which was a quantum of energy (later it was called photon). Energies between hν and 2hν do not occur. Then he used Boltzmann’s statistics to compute the average energy of an electromagnetic mode. If

E1, E2, E3, ... , are the allowed energies then according to Boltzmann’s statistics the relative probability that Ej can occur is exp(−Ej/kBT ). Then using the quantum hypothesis Ej = jhν the average energy of an electromagnetic mode can be determined as follows 1hνe−hν/kB T + 2hνe−2hν/kB T + ··· ∞ nhνe−nhν/kB T hν hEi = = n=1 = . (10) −hν/kB T −2hν/kBT ∞ −nhν/kB T 1+ e + e + ··· P n=0 e exp(hν/kB T ) − 1 Using the above calculation of the average energyP of an electromagnetic mode the Planck’s formula can be rewritten with the physical meaning of each of its terms photon energy 8πν2n3 1 ρ(ν, T )= 3 hν . (11) c z }| { exp(hν/kBT ) − 1 Number of em modes Number of photons/mode | {z } | {z } The density of electromagnetic modes can also be expressed per wavelength and is given by dN (λ) 8πn3 = − , (12) dλ λ4

4 ×10-15 Blackbody Radiation, T = 6000°K

Planck Rayleigh-Jeans Hz) 3 1.5 Wien (J/m ν ρ

0.5 Spectral Energy Density, 0 0 2 4 6 8 10 Frequency, ν (Hz) ×1014

Figure 3: Blackbody radiation for T = 6000◦ K. The initial theories by Rayleigh-Jeans and Wien are also shown for comparison. and the corresponding density of electromagnetic radiation of a blackbody per wavelength is

8πn3 hc/λ ρ(λ, T )= 4 . (13) λ exp(hc/λkBT ) − 1

Frequently in the literature the blackbody radiation formula is expressed in terms of the radiant exitance (or radiant emittance) of the blackbody (power/unit area = W/m2). The radiant exitance expresses the total power emitted by a source in a hemisphere (towards the direction of emission) per unit area of the source. The Poynting vector expresses the power per unit area of the electromagnetic radiation. Therefore, the Poynting vector is given by 2 Pavg = (1/2η)|E| where E is the electric ﬁeld amplitude of the electromagnetic wave, and η = 2 µ0/n 0 is the intrinsic impedance of the nonmagnetic medium in which the electromagnetic p radiation propagates. The energy density of the electromagnetic radiation is given by wem = 2 2 (1/2)n 0|E| . Therefore, Pavg = (c/n)wem. However, wem = ρ(ν, T )dν = ρ(λ, T )dλ and then

Pavg can be determined as follows

8πn2ν2 hν Pavg = 2 dν, c exp(hν/kBT ) − 1

8πn2c hc/λ Pavg = 4 dλ. (14) λ exp(hc/λkB T ) − 1

5 The radiance L (in W/m2sr where sr = steradian) of a radiant source (that could be a 2 2 blackbody radiator) is defined as L = d P/dA⊥/dΩ where d P is the differential electromagnetic power that is emitted by the source in a specified direction, dA⊥ is the differential source area element perpendicular to the specified direction of propagation, and dΩ is the differential solid angle inside which the differential power is propagated in the specified direction [2]. A blackbody emits radiation equally in all directions and consequently it seems similarly bright from any direction observed. This means that its radiance L is constant and independent of the observation angle. Such a source is called Lambertian [2]. Therefore, a blackbody is always a Lambertian source. Integrating the radiance all over the solid angles it can be easily shown 2 that Ω LdΩ = 4πL = (d P/dA⊥) = Pavg. Then the radiance of blackbody radiation can be R R expressed as follows

2n2ν2 hν L = 2 dν, c exp(hν/kBT ) − 1

2n2c hc/λ L = 4 dλ. (15) λ exp(hc/λkBT ) − 1

From radiometry [2] it can be easily determined that the radiance L and the radiant exitance (emittance) M (W/m2) of a blackbody (or a Lambertian source in general) can be related from 2 the equation M = Lπ. This is straightforward to show since M = Ω[d P/(dAsdΩ)]dΩ = π/2 2π R Ω L cos θdΩ = θ=0 φ=0 L cos θ sin θdθdφ = Lπ (where dAs = dA⊥/cosθ and dΩ = sinθdθdφ). R R R Therefore, the radiant exitance per unit frequency (power/unit area/frequency = W/m2Hz) of a blackbody radiator can be determined to be

2πn2ν2 hν Mν = 2 , (16) c exp(hν/kB T ) − 1 while the same exitance expressed per wavelength interval (power/unit area/wavelength = W/m2m) is given by 2πn2c hc/λ Mλ = 4 . (17) λ exp(hc/λkB T ) − 1 Integrating the above equations over all frequencies (or wavelengths) the radiant exitance M of a blackbody radiator at temperature T can be determined. This is known as Stefan’s law and

6 is expressed by the following equation

∞ 5 4 2π kB 2 4 2 4 M = Mλdλ = 3 2 n T = σn T (18) Z0 15h c where σ = 5.67 × 10−8 W/m2 ◦K−4 = Stefan-Boltzmann constant (usually the refractive index is considered that of vacuum or air, i.e. n ' 1). The maxima of the blackbody radiator curve can be found from the solution of the equation

dMλ(λm) hc ◦ = 0 ⇒ = 4.96511423 ⇒ λmT = 2897.821 µm K, (19) dλ λmkBT where the last part of the above equation described how the peak of the blackbody radiation shifts with the temperature and it is known as Wien’s displacement law. An example of Mλ for T = 6000 ◦K and Wien’s displacement law are shown in Fig. 4. A similar Wien’s displacement law can be deﬁned for Mν . The maxima for Mν can be found by dM (ν ) hν ν ν m = 0 ⇒ m = 2.82143937 ⇒ m = 5.878924 × 1010 Hz/◦K. (20) dν kBT T

◦ An example of Mν for T = 6000 K and Wien’s displacement law are shown in Fig. 5. It is mentioned that the peak of Mλ, λm, and the peak of Mν, νm, are not related by λmνm = c since the corresponding spectral exitances are per unit wavelength and per unit frequency respectively. The blackbody radiation represents the upper limit to the amount of radiation that a real body may emit at a given temperature. At any given wavelength λ, emissivity ε(λ), is deﬁned as the ratio of the actual emitted radiant exitance M˜ λ over the emitted radiant exitance of a blackbody Mλ. M˜ λ ελ = . (21) Mλ Emissivity is a measure of how strongly a body radiates at a given wavelength. Emissivity ranges between zero and one for all real substances (0 ≤ ελ ≤ 1). A gray body is deﬁned as a substance whose emissivity is independent of wavelength, i.e. ελ = ε. In the atmosphere, clouds and gases have emissivities that vary rapidly with wavelength. The ocean surface has near unit emissivity in the visible regions. For a body in local thermodynamic equilibrium the amount of thermal energy emitted must be equal to the energy absorbed. Otherwise the body would heat up or cool down in time,

7 4 107 Blackbody Radiation and Wien Displacement 10 12 3 T = 6000°K

10 2.5 K) ° m)

8 2 2

(W/m 6 1.5

4 1 Exitance M Absolute Temperature T, ( 2 0.5

0 0 0 0.5 1 1.5 2 2.5 3 Freespace Wavelength ( m)

◦ Figure 4: Blackbody radiation spectral exitance (emittance), Mλ(λ), for T = 6000 K as a function of freespace wavelength. The Wien’s displacement law is also shown for the same wavelength range. The maximum of Mλ occurs for λm =0.483 µm.

4 109 Blackbody Radiation and Wien Displacement 10 12 6 T = 6000°K

10 5 K) ° Hz)

2 8 4 (W/m 6 3

4 2 Exitance M 2 1 Absolute Temperature T, (

0 0 0 0.5 1 1.5 2 2.5 3 Frequency (Hz) 1015

◦ Figure 5: Blackbody radiation spectral exitance (emittance), Mν (ν), for T = 6000 K as a function of frequency. The Wien’s displacement law is also shown for the same frequency range. The maximum of Mν 14 occurs for νm =3.52 × 10 Hz.

8 contrary to the assumption of equilibrium. As a result of this it can be said that materials that are strong absorbers at a given wavelength are also strong emitters at that wavelength. Similarly weak absorbers are weak emitters. Blackbody radiation is also used to establish a color scale as a function of the absolute temperature. The color temperature of a light specimen is the temperature of a blackbody with the closest spectral distribution. For example, the sun has a typical color temperature of 5500◦ K.

9 APPENDIX A: Determination of Electromagnetic Modes in Rectangular Metallic Cavities

The purpose of this Appendix is to review the determination of modes in a rectangular-shaped cavity which is considered to have perfectly conducting walls while the material filling the cavity is homogeneous and isotropic [3, 4, 5]. The approach that will be presented here is rather independent from the knowledge of the solutions of rectangular metallic waveguides solutions which is normally the traditional manner in determining the cavity modes. The rectangular cavity with the corresponding coordinate system is shown in Fig. 1. It is assumed that the determination of the T Empq modes is seeked, i.e., it is assumed that Ez = 0 while all other field components Ex, Ey, Hx, Hy, Hz are in general nonzero. Every field component satisfies the Helmholtz equation

2 2 2 2 2 2 ∂ ∂ ∂ 2 2 ∇ S + k0n S = 2 + 2 + 2 S + k0n S = 0, (A.1) ∂x ∂y ∂z where S = Ex, Ey, Hx, Hy, Hz , k0 = ω/c = 2π/λ0 is the freespace wavenumber, and n is the refractive index of the material inside the cavity. Because of the rectangular geometry it is reasonable to seek solutions based on the method of separation of variables, i.e., S(x,y,z) =

X(x)Y (y)Z(z) where X(x) = A cos(kxx)+ B sin(kxx), Y (y) = C cos(ky y)+ D sin(kyy), and 2 2 2 2 2 Z(z)= E cos(kzz)+F sin(kzz), with kx +ky +kz = k0n . From the two curl Maxwell’s equations 2 ∇× E~ = −jωµ0H~ , and ∇× H~ =+jω0n E~ , for the T Empq modes the following equations are derived:

1 ∂Hz ∂Hy Ex = 2 − (A.2) jω0n ∂y ∂z 1 ∂Hx ∂Hz Ey = 2 − (A.3) jω0n ∂z ∂x 1 ∂Hy ∂Hx Ez = 2 − =0 (A.4) jω0n ∂x ∂y 1 ∂Ey Hx = + (A.5) jωµ0 ∂z 1 ∂Ex Hy = − (A.6) jωµ0 ∂z 1 ∂Ey ∂Ex Hz = − − (A.7) jωµ0 ∂x ∂y

10 Since the cavity is surrounded by perfect conducting walls the boundary conditions on the various field components are that the normal to the wall boundary magnetic field components are zero as well as the tangential to the boundaries electric field components. These conditions can be expressed by the following equations:

Hx(x = 0,y,z) = Hx(x = a,y,z)=0 (A.8)

Hy(x,y = 0,z) = Hy(x,y = b,z)=0 (A.9)

Hz (x,y,z =0) = Hz (x,y,z = d)=0 (A.10)

Ex(x,y = 0,z)= Ex(x,y = b,z) = Ex(x,y,z =0)= Ex(x,y,z = d)=0 (A.11)

Ey(x = 0,y,z)= Ey(x = a,y,z) = Ey(x,y,z =0)= Ey(x,y,z = d)=0 (A.12)

where it is reminded that for the T Empq modes Ez = 0, ∀ x,y,z. In order to satisfy the boundary conditions for the Hx component the X(x)= sin(kxmx) where kxm =(mπ/a) and m = 0, 1, ··· .

Similarly, for Hy to satisfy the boundary conditions the Y (y)= sin(kypy) where kyp = (pπ/b) and p = 0, 1, ··· . Therefore, the solutions for Hx and Hy take the following form

2 mπ mπ 2 2 2 2 H (x,y,z) = sin x Y1(y)Z1(z), with + k + k = k0n , (A.13) x a a y z 2 pπ 2 pπ 2 2 2 Hy(x,y,z) = X2(x) sin y Z2(z), with kx + + kz = k0n . (A.14) b b

In order to force the Ez ﬁeld component to be zero from Eq. (A.4) the following should hold ∀ x,y,z,

1 dX2 pπ 1 mπ dY1 2 sin y Z2(z) = 2 sin x Z1(z) ∀x,y,z. (A.15) jω0n n dx b o jω0n n a dy o

Using X2(x) = A2 cos(kxx)+ B2 sin(kxx) and Y1(y) = C1 cos(kyy)+ D1 sin(kyy) it is straightforward to show that B2 =0= D1, and kx =(mπ/a), ky =(pπ/b), and the coeﬃcients of the

Z1(z) and Z2(z) are related in such a way that the ﬁeld components Hx and Hy are expressed by the following equations:

mπ pπ Hx(x,y,z) = sin x cos y [E1 cos(kzz)+ F1 sin(kz z)], (A.16) a b mπ pπ pπ/b Hy(x,y,z) = cos x sin y [E1 cos(kzz)+ F1 sin(kz z)], (A.17) a b mπ/a

11 2 2 2 2 2 where, of course (mπ/a) +(pπ/b) +kz = k0n . Now in order to satisfy the boundary condition for the Hz ﬁeld component the following solution is valid qπ Hz(x,y,z)= H0zX3(x)Y3(y) sin z . (A.18) d

From the z-dependence of Hz it is implied that the Ex and Ey ﬁeld components have the following form due to Eq. (A.7) qπ E (x,y,z) = E0 X1(x)Y1(y) sin z , (A.19) x x d qπ Ey(x,y,z) = E0yX2(x)Y2(y) sin z , (A.20) d where E0x, E0y are amplitude constants. Then applying Eqs. (A.5) and (A.6) for the Hx and

Hy components respectively, in conjunction with Eqs. (A.16). (A.17), (A.19), and (A.20), the following conditions must be satisﬁed ∀ x,y,z,

mπ pπ 1 qπ qπ sin x cos y [E1 cos(kz z)+ F1 sin(kzz)] = + E0y cos z X2(x)Y2(y) , a b jωµ0 h d d i mπ pπ pπ/b 1 qπ qπ cos x sin y [E1 cos(kz z)+ F1 sin(kzz)] = − E0x cos z X1(x)Y1(y) . a b mπ/a jωµ0 h d d i

From the last two equations the following solutions for the Ex, Ey, Hx, Hy ﬁelds can be obtained

E0y qπ mπ pπ qπ Hx = sin x cos y cos z , (A.21) jωµ0 d a b d E0y pπ/b qπ mπ pπ qπ Hy = cos x sin y cos z , (A.22) jωµ0 mπ/a d a b d pπ/b mπ pπ qπ Ex = −E0y cos x sin y sin z , (A.23) mπ/a a b d mπ pπ qπ Ey = E0y sin x cos y sin z . (A.24) a b d

The last component to be determined is the Hz. Using the solutions for Ex and Ey as well as

Eq. (A.7) and Eq. (A.18) the following solution for Hz is obtained

2 2 E0y a mπ pπ mπ pπ qπ Hz = − + cos x cos y sin z . (A.25) jωµ0 mπh a b i a b d In order to write the equations in the usual format [4, 5] found in the literature the coeﬃcient of 2 2 2 2 the Hz component can be deﬁned as C = −(E0y/jωµ0)(a/mπ)kc where kc =(mπ/a) +(pπ/b) .

Using C as the free parameter in the expressions of the ﬁelds of the T Empq mode the ﬁelds are summarized in the form.

12 T Empq Modes :

jωµ0 pπ mπ pπ qπ Ex = C 2 cos x sin y sin z , (A.26) kc b a b d jωµ0 mπ mπ pπ qπ Ey = −C 2 sin x cos y sin z , (A.27) kc a a b d Ez = 0, (A.28) 1 mπ qπ mπ pπ qπ Hx = −C 2 sin x cos y cos z , (A.29) kc a d a b d 1 pπ qπ mπ pπ qπ Hy = −C 2 cos x sin y cos z , (A.30) kc b d a b d mπ pπ qπ Hz = C cos x cos y sin z . (A.31) a b d

In exactly similar manner the solutions of the T Mmpq modes can be calculated where the

Hz = 0. These solutions are summarized next for completeness.

T Mmpq Modes :

1 mπ qπ mπ pπ qπ Ex = −D 2 cos x sin y sin z , (A.32) kc a d a b d 1 pπ qπ mπ pπ qπ Ey = −D 2 sin x cos y sin z , (A.33) kc b d a b d mπ pπ qπ E = D sin x sin y cos z , (A.34) z a b d 2 jω0n pπ mπ pπ qπ Hx = D 2 sin x cos y cos z , (A.35) kc b a b d 2 jω0n mπ mπ pπ qπ Hy = −D 2 cos x sin y cos z , (A.36) kc a a b d Hz = 0. (A.37)

where now D has been selected as the free parameter coeﬃcient. For both T Empq and T Mmpq modes the dispersion relation and the corresponding resonance frequencies are given by the following equations

2 2 2 2 2 mπ pπ qπ k0n = + + , (A.38) a b d c mπ 2 pπ 2 qπ 2 ωmnq = + + . (A.39) nr a b d

13 REFERENCES

1. J. Baggott, The Meaning of Quantum Theory, (Oxford University Press, New York, 1992).

2. W. L. Wolfe, Introduction to Radiometry, (Tutorial Texts in Optical Engineering, vo. TT29, SPIE Optical Engineering Press, 1998).

3. C. A. Balanis, Advanced Engineering Electromagnetics, (John Wiley & Sons, New York, 1989).

4. D. K. Cheng, Fields and Waves in Electromagnetics, 2nd Ed., (Addison Wesley,New York, 1989).

5. S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communications Elec- tronics, 3rd Ed., (John Wiley and Sons, Inc., New York, 1993).

6. J. T. Verdeyen, Laser Electronics, 3rd Ed., ch. 7, (Prentice Hall, New Jersey, 1995).