Fundamentals of Blackbody Radiation

Fundamentals of Blackbody Radiation Prof. Elias N. Glytsis Nov. 08, 206 School of Electrical & Computer Engineering National Technical Univerity of Athens

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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 ce it led to the discovery of the quantum theory []. 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/m 3 Hz 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 896 used a simple model to derive the expression ρ(ν, T = αν 3 exp( βν/t ( where α, β were constants. However, the above equation failed in the low frequency range of the experimental data. In June 900 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. Ug the standard electromagnetic theory of a cavity resonator (see Fig. with perfect conductor walls the following dispersion equation can be easily obtained [3, 4, 5] (see Appendix A : 2 + a b 2 + ( qπ d 2 ( 2πν 2n = 2, (2 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 206 Prof. Elias N. Glytsis, Last Update: Nov. 08, 206

Figure : 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 2a ( 2νna 2. m 2 + p 2 + q 2 = 2 n 2 = (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 N(ν = (/8 cavity volume volume of a mode = (/8(4/3π(2νan/c3 = 4 3 πν3 n 3 a 3 c 3. (4 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 = Then the density of electromagnetic modes per frequency is N(ν V olume = a 3 = 8 3 πν3 n 3 c 3. (5 dn(ν dν = 8πν2 n 3 c 3. (6 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 dν = 8πν2 n 2 (n + ν dn dν c 3 = 8πν2 n 2 n g c 3. (7 where n g = n + ν dn = n dν λdn 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 k B T/2 to each electromagnetic mode (k B T/2 for the electric field oscillation and k B T/2 for the magnetic field oscillation, where k B =.38066 0 23 J/ K. Therefore, the electromagnetic energy density per unit frequency ρ(ν, T becomes ρ(ν, T = dn(ν k B T = 8πν2 n 3 k dν c 3 B T. (8 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, 900 []. 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 ρ(ν, T = 8πν2 n 3 c 3 hν exp(hν/k B T, (9 where h = 6.626 0 34 Joule sec is known as Planck s constant. The above expression reduces to Wien s formula for high frequencies (i.e. hν/k B T and to Rayleigh-Jeans formula for low frequencies (i.e. hν/k B T. 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 E, E 2, E 3,..., are the allowed energies then according to Boltzmann s statistics the relative probability that E j can occur is exp( E j /k B T. Then ug the quantum hypothesis E j = jhν the average energy of an electromagnetic mode can be determined as follows n= = nhνe nhν/k BT = n=0 e nhν/k BT E = hνe hν/k BT + 2hνe 2hν/k BT + + e hν/k BT + e 2hν/k BT + hν exp(hν/k B T. (0 Ug the above calculation of the average energy of an electromagnetic mode the Planck s formula can be rewritten with the physical meaning of each of its terms ρ(ν, T = } 8πν 2 n 3 {{ c 3 } Number of em modes photon energy {}}{ hν exp(hν/k B T }{{}. ( Number of photons/mode The density of electromagnetic modes can also be expressed per wavelength and is given by dn(λ dλ = 8πn3 λ 4, (2 4

0-5 Blackbody Radiation, T = 6000 K Spectral Energy Density, ρ ν (J/m 3 Hz.5 0.5 Planck Rayleigh-Jeans Wien 0 0 2 4 6 8 0 Frequency, ν (Hz 0 4 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 ρ(λ, T = 8πn3 λ 4 hc/λ exp(hc/λk B T. (3 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/m 2. 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 P avg = (/2η E 2 where E is the electric field amplitude of the electromagnetic wave, and η = µ0 /n 2 ɛ 0 is the intrinsic impedance of the nonmagnetic medium in which the electromagnetic radiation propagates. The energy density of the electromagnetic radiation is given by w em = (/2n 2 ɛ 0 E 2. Therefore, P avg = (c/nw em. However, w em = ρ(ν, Tdν = ρ(λ, Tdλ and then P avg can be determined as follows P avg = 8πn2 ν 2 c 2 hν exp(hν/k B T dν, P avg = 8πn2 c λ 4 hc/λ dλ. (4 exp(hc/λk B T 5

The radiance L (in W/m 2 sr where sr = steradian of a radiant source (that could be a blackbody radiator is defined as L = d 2 P/dA /dω where d 2 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 that Ω LdΩ = 4πL = (d 2 P/dA = P avg. Then the radiance of blackbody radiation can be expressed as follows L = 2n2 ν 2 c 2 L = 2n2 c λ 4 hν exp(hν/k B T dν, hc/λ dλ. (5 exp(hc/λk B T From radiometry [2] it can be easily determined that the radiance L and the radiant exitance (emittance M (W/m 2 of a blackbody (or a Lambertian source in general can be related from the equation M = Lπ. This is straightforward to show ce M = Ω [d2 P/(dA s dω]dω = Ω L θdω = π/2 2π L θ θdθdφ = Lπ (where da θ=0 φ=0 s = da /θ and dω = θdθdφ. Therefore, the radiant exitance per unit frequency (power/unit area/frequency = W/m 2 Hz of a blackbody radiator can be determined to be M ν = 2πn2 ν 2 c 2 hν exp(hν/k B T, (6 while the same exitance expressed per wavelength interval (power/unit area/wavelength = W/m 2 m is given by M λ = 2πn2 c λ 4 hc/λ exp(hc/λk B T. (7 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 M = 0 M λ dλ = ( 2π 5 kb 4 n 2 T 4 = σn 2 T 4 (8 5h 3 c 2 where σ = 5.67 0 8 W/m 2 K 4 = Stefan-Boltzmann constant (usually the refractive index is considered that of vacuum or air, i.e. n. The maxima of the blackbody radiator curve can be found from the solution of the equation dm λ (λ m dλ = 0 hc λ m k B T = 4.965423 λ mt = 2897.82µm K, (9 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 M λ for T = 6000 K and Wien s displacement law are shown in Fig. 4. A similar Wien s displacement law can be defined for M ν. The maxima for M ν can be found by dm ν (ν m dν = 0 hν m k B T = 2.8243937 ν m T = 5.878924 00 Hz/ K. (20 An example of M ν for T = 6000 K and Wien s displacement law are shown in Fig. 5. It is mentioned that the peak of M λ, λ m, and the peak of M ν, ν m, are not related by λ m ν m = c ce 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 M λ. ε λ = M λ M λ. (2 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 ε λ. A gray body is defined as a substance whose emissivity is independent of wavelength, i.e. ε λ = 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

0 7 Blackbody Radiation and Wien Displacement T = 6000 K 2 0 4 3 Exitance M λ (W/m 2 µm 0 8 6 4 2 2.5 2.5 0.5 Absolute Temperature T, ( K 0 0 0 0.5.5 2 2.5 3 Freespace Wavelength λ (µm Figure 4: Blackbody radiation spectral exitance (emittance, M λ (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 M λ occurs for λ m = 0.483 µm. 0 9 2 Blackbody Radiation and Wien Displacement 0 4 6 Exitance M ν (W/m 2 Hz 0 8 6 4 2 5 4 3 2 Absolute Temperature T, ( K 0 0 0.5.5 2 2.5 0 3 Frequency ν (Hz 0 5 Figure 5: Blackbody radiation spectral exitance (emittance, M ν (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 M ν occurs for ν m = 3.52 0 4 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.. It is assumed that the determination of the TE mpq modes is seeked, i.e., it is assumed that E z = 0 while all other field components E x, E y, H x, H y, H z are in general nonzero. Every field component satisfies the Helmholtz equation ( 2 S + k0 2 2 n2 S = x + 2 2 y + 2 S + k 2 2 z 2 0 n2 S = 0, (A. where S = E x, E y, H x, H y, H z, k 0 = ω/c = 2π/λ 0 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(xY (yz(z where X(x = A (k x x + B (k x x, Y (y = C (k y y + D (k y y, and Z(z = E (k z z+f (k z z, with k 2 x +k2 y +k2 z = k2 0 n2. From the two curl Maxwell s equations E = jωµ 0 H, and H = +jωɛ 0 n 2 E, for the TE mpq modes the following equations are derived: E x = E y = E z = jωɛ 0 n 2 ( Hz jωɛ 0 n 2 ( Hx y H y (A.2 z z H z (A.3 x = 0 (A.4 ( Hy jωɛ 0 n 2 x H x y H x = + E y jωµ 0 z H y = E x jωµ 0 z H z = ( Ey jωµ 0 x E x y 0 (A.5 (A.6 (A.7

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: H x (x = 0, y, z = H x (x = a, y, z = 0 H y (x, y = 0, z = H y (x, y = b, z = 0 H z (x, y, z = 0 = H z (x, y, z = d = 0 E x (x, y = 0, z = E x (x, y = b, z = E x (x, y, z = 0 = E x (x, y, z = d = 0 E y (x = 0, y, z = E y (x = a, y, z = E y (x, y, z = 0 = E y (x, y, z = d = 0 (A.8 (A.9 (A.0 (A. (A.2 where it is reminded that for the TE mpq modes E z = 0, x, y, z. In order to satisfy the boundary conditions for the H x component the X(x = (k xm x where k xm = (mπ/a and m = 0,,. Similarly, for H y to satisfy the boundary conditions the Y (y = (k yp y where k yp = (pπ/b and p = 0,,. Therefore, the solutions for H x and H y take the following form H x (x, y, z = H y (x, y, z = X 2 (x a x Y (yz (z, with b y Z 2 (z, with kx 2 + 2 + k 2 a b y + kz 2 = k0n 2 2, 2 + k 2 z = k0n 2 2. (A.3 (A.4 In order to force the E z field component to be zero from Eq. (A.4 the following should hold x, y, z, { dx2 } jωɛ 0 n 2 dx b y Z 2 (z = { jωɛ 0 n 2 a x dy } dy Z (z x, y, z. (A.5 Ug X 2 (x = A 2 (k x x + B 2 (k x x and Y (y = C (k y y + D (k y y it is straightforward to show that B 2 = 0 = D, and k x = (mπ/a, k y = (pπ/b, and the coefficients of the Z (z and Z 2 (z are related in such a way that the field components H x and H y are expressed by the following equations: H x (x, y, z = H y (x, y, z = a x a x b y [E (k z z + F (k z z], (A.6 b y pπ/b mπ/a [E (k z z + F (k z z], (A.7

where, of course (mπ/a 2 +(pπ/b 2 +k 2 z = k2 0 n2. Now in order to satisfy the boundary condition for the H z field component the following solution is valid ( qπ H z (x, y, z = H 0z X 3 (xy 3 (y d z. (A.8 From the z-dependence of H z it is implied that the E x and E y field components have the following form due to Eq. (A.7 ( qπ E x (x, y, z = E 0x X (xy (y d z, (A.9 ( qπ E y (x, y, z = E 0y X 2 (xy 2 (y d z, (A.20 where E 0x, E 0y are amplitude constants. Then applying Eqs. (A.5 and (A.6 for the H x and H y components respectively, in conjunction with Eqs. (A.6. (A.7, (A.9, and (A.20, the following conditions must be satisfied x, y, z, a x a x b y [E (k z z + F (k z z] = + [ qπ ( qπ ] E 0y jωµ 0 d d z X 2 (xy 2 (y, pπ/b b y mπ/a [E (k z z + F (k z z] = [ qπ ( qπ ] E 0x jωµ 0 d d z X (xy (y. From the last two equations the following solutions for the E x, E y, H x, H y fields can be obtained H x = E ( 0y qπ jωµ 0 d a x H y = E 0y pπ/b qπ jωµ 0 mπ/a( d pπ/b = E 0y E x E y = E 0y mπ/a a x a x b y ( qπ b y d z, (A.2 ( qπ a x b y d z, (A.22 b y ( qπ d z, (A.23 ( qπ d z. (A.24 The last component to be determined is the H z. Ug the solutions for E x and E y as well as Eq. (A.7 and Eq. (A.8 the following solution for H z is obtained H z = E 0y jωµ 0 a mπ [ 2 + a b 2 ] ( qπ a x b y d z. (A.25 In order to write the equations in the usual format [4, 5] found in the literature the coefficient of the H z component can be defined as C = (E 0y /jωµ 0 (a/mπk 2 c where k 2 c = (mπ/a 2 +(pπ/b 2. Ug C as the free parameter in the expressions of the fields of the TE mpq mode the fields are summarized in the form. 2

TE mpq Modes : E x = C jωµ ( 0 pπ kc 2 b E y = C jωµ 0 kc 2 E z = 0, H x H y H z a a x a x ( qπ b y d z, (A.26 ( qπ b y d z, (A.27 (A.28 = C ( qπ ( qπ kc 2 a d a x b y d z, (A.29 = C ( qπ ( qπ kc 2 b d a x b y d z, (A.30 ( qπ = C a x b y d z. (A.3 In exactly similar manner the solutions of the TM mpq modes can be calculated where the H z = 0. These solutions are summarized next for completeness. TM mpq Modes : E x = D ( qπ ( qπ kc 2 a d a x b y d z, (A.32 E y = D ( qπ ( qπ kc 2 b d a x b y d z, (A.33 ( qπ E z = D a x b y d z, (A.34 H x = D jωɛ 0n 2 ( qπ kc 2 b a x b y d z, (A.35 H y = D jωɛ 0n 2 ( qπ a a x b y d z, (A.36 H z = 0. k 2 c (A.37 where now D has been selected as the free parameter coefficient. For both TE mpq and TM mpq modes the dispersion relation and the corresponding resonance frequencies are given by the following equations 2 2 ( qπ 2, k0n 2 2 = + + (A.38 a b d ω mnq = c (mπ 2 2 ( qπ 2. + + (A.39 n a b d 3

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