Radiation

check all the nomenclature/definition

Planck’s law

Definition 4 (Black-body)

Spectral energy density for black-body radiation

\[u_{f}(f, T) = \frac{8 \pi h f^3}{c^3} \frac{1}{e^{\frac{hf}{k_B T}} - 1}\]

Spectral energy density has the physical dimension \(\frac{\text{Energy}}{\text{Volume} \times \text{Frequency}}\)

Spectral radiance reads1 \(B_f = \frac{1}{4 \pi} u_f(f,T) c\) and thus for a black-body

\[B_{f}(f, T) = \frac{2 h f^3}{c^2}\frac{1}{e^{\frac{hf}{k_B T}} - 1} \ .\]

It has physical dimension \(\frac{\text{Energy}}{\text{Volume} \times \text{Frequency}} \times \text{Velocity} = \frac{\text{Power}}{\text{Area} \times \text{Frequency}}\).

Wien’s law

Wien’s law states that the maximum of the spectral energy is obtained at a frequency \(f^*\), whose value is proportional with the temperature of the body.

\[f^* = x^* \frac{k_B}{h} T \simeq 2.82 \frac{k_B}{h} T \ ,\]

where \(x^* \simeq 2.82\) is the solution of the algebraic equation \(3(1-e^x)-x=0\).

Proof of Wien’s law

From direct evaluation of the derivative of the spectral radiance as a function of \(f\),

\[\begin{split}\begin{aligned} \partial_f B_{f}(f,T) & = \frac{2 h}{c^2} \left[ 3 f^2 \frac{1}{e^{\frac{hf}{k_B T}}-1} + f^3 \left(-\frac{\frac{h}{k_B T} e^{\frac{hf}{k_B T}}}{\left( e^{\frac{hf}{k_B T}} - 1 \right)^2} \right) \right] = \\ & = \frac{2 h f^2 e^{\frac{hf}{k_B T}}}{c^2 \left( e^{\frac{hf}{k_B T}} - 1 \right)^2} \left[ 3 \left( 1 - e^{-\frac{hf}{k_B T}} \right) - \frac{h f}{k_B T} \right] \ . \end{aligned}\end{split}\]

Now, if \(\partial_f B_{f}(f,T) = 0\) the frequency is either \(f = 0\), or the solution of the nonlinear algebraic equation

\[0 = 3 \left(1 - e^{-\frac{h f}{k_B T}} \right) - \frac{hf}{k_B T} \ .\]

Defining \(x := \frac{h f}{k_B T}\), this equation becomes

\[0 = 3 (1 - e^x) - x \ ,\]

whose solution \(x^* \approx 2.82\) can be easily evaluated with an iterative method (or expressed in term of the Lambert’s function \(W\), so loved at Stanford and on Youtube: they’d probaly like to look at tabulated values, or pose). Once the solution \(x^*\) of this non-dimensional equation is found, the frequency where maximum energy density occurs reads

\[f^* = \frac{k_B T}{h} x^* \simeq 2.82 \frac{k_B}{h} T \ .\]

Energy transfer by radiation

A list of physical quantities used to describe a radiation process follows.

• Radiant flux, \(\Phi_e\). Physical dimension: \(\text{W}\)

• Radiant intensity \(I_{e, \Omega}\) is the radiant flux per unit solid angle \(\Omega\). Physical dimension: \(\frac{\text{W}}{\text{sr}}\)

• Spectral radiant intensity \(I_{e, \Omega, f}\) is the radiant flux per unit solid angle per frequency of the radiation: \(\frac{\text{W}}{\text{sr} \, \text{Hz}}\)

• Radiant exitance, \(M_e\) is the flux across an emispherical receiver surface, per unit surface of the source. Physical dimension \(\frac{\text{W}}{\text{Area}}\)

Definition 5 (Lambert medium)

A Lambert medium is defined as a medium that follows Lambert cosine law,

\[I_{1}(\vec{r}_{12}, \hat{n}_1) = I_{1,0} \, \hat{r}_{12} \cdot \hat{n}_1 \ ,\]

with \(\vec{r}_{12} = \vec{r}_2 - \vec{r}_1\).

Elementary power flux of the radiation emitted by a source in \(1\) through a solid angle \(d \Omega_{21}\), as seen from \(1\), reads

\[\begin{split}\begin{aligned} d P_{1 \rightarrow 2} & = I_{1}(\vec{r}_{12}, \hat{n}_1) \, d \Omega_{2/1} \\ \end{aligned}\end{split}\]

The solid angle \(d \Omega_{21}\) of a surface \(d S_2\) as seen from a point in \(\vec{r}_1\) is \(d \Omega_{21} = \frac{\hat{n}_2 \cdot \vec{r}_{12}}{|\vec{r}_{12}|^3} \, dS_2\). From the expression of the elementary power flux,

\[ d P_{1 \rightarrow 2} = I_{1}(\vec{r}_{12}, \hat{n}_1) \, \frac{\vec{r}_{12} }{|\vec{r}_{12}|^3} \cdot \hat{n}_2 \, dS_2 = \vec{s}_{12} \cdot \hat{n}_2 \, dS_2 \ , \]

it’s possible to find the expression of the power flux density vector

\[\vec{s}_{12} = I_{12} \frac{\vec{r}_{12}}{|\vec{r}_{12}|^3} \ .\]

For a Lambert medium then

(6)\[\begin{split}\begin{aligned} d P_{1 \rightarrow 2} & = I_{1}(\vec{r}_{12}, \hat{n}_1) \, \frac{\vec{r}_{12} }{|\vec{r}_{12}|^3} \cdot \hat{n}_2 \, dS_2 = \\ & = I_{0,1} \frac{\vec{r}_{12}}{|\vec{r}_{12}|} \cdot \hat{n}_1 \, \frac{\vec{r}_{12} }{|\vec{r}_{12}|^3} \cdot \hat{n}_2 \, dS_2 \ . \end{aligned}\end{split}\]

Stefan-Boltzmann law

The radiant exitance, \(M_e^\circ\), - i.e. the total power through an emisphere emitted by a source, per unit area of the source - of a black body at temperature \(T_1\) is

\[M_e^\circ = \sigma T_1^4 \ ,\]

with the Stefan-Boltzmann constant, \(\sigma = 5.67 \cdot 10^{-8} \frac{\text{W}}{\text{m}^2 \text{K}^4}\).

Total radiation and Stefan-Boltzmann constant \(\ \sigma\)

Integration over frequency \(f \in [0, +\infty]\), and over an emisphere receiver surface (or better, over the equivalent solid angle) for a Lambert medium provide the radiant exitance

\[\begin{split}\begin{aligned} M_{e}^\circ = M_{1,2^{emisphere}} & = \frac{d P_{1 \rightarrow 2}}{d S_1} = \\ & = \int_{f=0}^{+\infty} \int_{\Omega_{emisphere}} I_{1,0}(f) \cos \theta_{12} \, df \, d\Omega_{2/1} = \\ & = \int_{f=0}^{+\infty}I_{1,0}(f) \, df \, \int_{\phi = 0}^{2\pi} \int_{\theta=0}^{\pi_2} \cos \theta \sin \theta \, d \theta \, d\phi \ , \end{aligned}\end{split}\]

with the solid angle with spherical coordinates, \(d\Omega_{2/1} = \sin \theta \, d \theta \, d\phi\). The geometric integral gives

\[ \int_{\phi = 0}^{2\pi} \int_{\theta=0}^{\pi_2} \cos \theta_{} \, d \theta \, d\phi = \pi \ . \]

For a black-body the integral over frequency gives

\[\begin{split}\begin{aligned} \int_{f = 0}^{+\infty} I_{1,0}(f) \, df & = \int_{f=0}^{+\infty} \frac{2 h f^3}{c^2} \frac{1}{e^{\frac{hf}{k_B T}}-1} \, df \\ & = T^4 \frac{2 h}{c^2} \left(\frac{k_B}{h}\right)^4 \int_{u=0}^{+\infty} \frac{u^3}{e^u-1} \, du =: T^4 \frac{\sigma}{\pi} \ , \end{aligned}\end{split}\]

i.e. it’s proportional to \(T^4\) with a constant of proportionality built on constants of nature, speed of light \(c\), Boltzmann constant \(k_B\), and Planck constant \(h\). The integral is non-dimensional and its value2 is \(\frac{\pi^4}{15}\). Stefan-Boltzmann constant \(\sigma\) is then defined as

\[\sigma := \pi \cdot \frac{2 h}{c^2} \left( \frac{k_B}{h} \right)^4 \int_{u=0}^{+\infty} \frac{u^3}{e^{u}-1} du = \frac{2 \pi^5}{15}\frac{k_B^4}{c^2 h^3} = 5.67 \cdot 10^{-8} \frac{\text{W}}{\text{m}^2 \text{K}^4} \ .\]

Integration over a generic surface \(S_2\) of the power flux emitted by a source \(1\) per unit surface of the source gives

\[\begin{aligned} M_{12} = \frac{d P_{1 \rightarrow 2} }{d S_1} = T_1^4 \, \frac{\sigma}{\pi} \int_{S_2} \hat{n}_1 \cdot \frac{\vec{r}_{12}}{|\vec{r}_{12}|} \, \frac{\vec{r}_{12}}{|\vec{r}_{12}|^3} \cdot \hat{n}_2 \, dS_2 \end{aligned}\]

Thus the elementary power per unit surface of the source \(1\) and the receiver \(2\) is

\[dP_{1 \rightarrow 2} = T_1^4 \, \frac{\sigma}{\pi} \, \hat{n}_1 \cdot \frac{\vec{r}_{12}}{|\vec{r}_{12}|} \, \frac{\vec{r}_{12}}{|\vec{r}_{12}|^3} \cdot \hat{n}_2 \, dS_1 \, dS_2 \]

Comparing this expression of the power flux from source \(1\) to the elementary surface \(d S_2\) with the expression (6) using the radiant intensity \(I_{0,1}\), it immediately follows that the elementary radiant intensity \(d I_{0,1}\) of the elementary surface \(d S_1\) of the source reads

\[d I_{0,1} = T_1^4 \, \frac{\sigma}{\pi} \, d S_1 \ .\]

Geometry effects on radiation

The elementary power is the product of \(\frac{\sigma T_1^4}{\pi}\) and a factor depending only on the geometry of the problem,

\[d G_{12} := \hat{n}_1 \cdot \frac{\vec{r}_{12}}{|\vec{r}_{12}|} \, \frac{\vec{r}_{12}}{|\vec{r}_{12}|^3} \cdot \hat{n}_2 \, dS_1 \, dS_2 \ .\]

This factor is symmetric, i.e. inverting the roles of the source and the receiver is irrelevant. Letting \(G_{12} = \int_{S_1} \int_{S_2} dG_{12}\), if the source surface \(1\) has uniform temperature \(T_1\) and the receiver surface has uniform temperature \(T_2\), (and assuming emissvity \(\varepsilon = 1\), for black-bodies) thus the power fluxes from \(1\) to \(2\) and from \(2\) to \(1\) read

\[\begin{split}\begin{aligned} P_{1 \rightarrow 2} & = T_1^4 \frac{\sigma}{\pi} G_{12} \\ P_{2 \rightarrow 1} & = T_2^4 \frac{\sigma}{\pi} G_{12} \ . \end{aligned}\end{split}\]

Thus the net power into surfaces \(1\) and \(2\) are

\[\begin{split}\begin{aligned} P_{\rightarrow 1} & = P_{2 \rightarrow 1} - P_{1 \rightarrow 2} \\ P_{\rightarrow 2} & = P_{1 \rightarrow 2} - P_{2 \rightarrow 1} \ , \end{aligned}\end{split}\]

so that \(P_{\rightarrow 1} + P_{\rightarrow 2} = 0\).

In the limit of small surfaces compared with their distance, an approximate expression of the integral \(\int_{S_1} \int_{S_2} d G\) allows to write the power transmitted by radiation from surface \(1\) to surface \(2\) as

\[\begin{aligned} P_{1 \rightarrow 2} = \frac{\sigma T_1^4}{\pi} G_{12} = \frac{\sigma T_1^4}{\pi} A_{1, \perp \vec{r}_{12}} \Omega_{2/1} = \frac{\sigma T_1^4}{\pi} A_{2, \perp \vec{r}_{12}} \Omega_{1/2} \end{aligned}\]

i.e. as the product of the radiant … todo find the right term \(\frac{\sigma T_1^4}{\pi}\), the projection of the surface \(S_{1}\) in a plane orthogonal to the distance vector \(\vec{r}_{12}\) and the solid angle \(\Omega_{2/1}\) of the surface \(2\) as seen by surface \(1\), or viceversa.

Proof of the small surface limit

Emissivity and coefficient of absorption

The radiation emission of a generic body differs from the radiation of a black-body. Its emittance can be written comparing its emission with that of of a black body. In general, the emission could differ as a function of the frequency of the radiation and the direction of the radiation.

…

If there’s no dependence on \(f\) and \(\hat{r}\) - or taking the average values, neglecting the dependence on \(f\) and \(\hat{r}\) if using the average coefficient - the radiant exitance is

\[M_1 = \varepsilon_1 \sigma T_1^4 \ .\]

Reflection coefficient \(r\) and absorbtion (or transmission) coefficient \(a\) are then defined as the reflected and absorbed ratios of the incident radiation. Given incident radiation with power \(P_i\), power of reflected and absorbed radiation reads

\[P_r = r \, P_i \quad , \quad P_a = a \, P_i \ .\]

and for “power balance”, the sum of absorbed and reflected power must be equal to the incident power

\[P_i = P_r + P_a \ ,\]

and thus the coefficients must satistfy the following relation check check the definition of the coefficients, if they should take into account the emission…

\[1 = r + a \ .\]

Example 19 (Energy exchange in isolated systems and relations between coefficients - todo check)

Radiation transmission is assumed to be instantaneous here: this assumption holds for bodies close enough for the time \(\Delta t = \frac{d_{12}}{c}\) to be small enough to be treated as zero.

A system composed of two bodies. Energy balance of a body immediately follows the first principle of thermodynamics: time derivative of energy of a system equals the net power done on the system. Here the net power equals the difference between the absorbed incoming power and the emitted power,

\[\begin{split}\begin{cases} \dot{E}_1 = P^a_{\rightarrow 1} - P_1^e \\ \dot{E}_2 = P^a_{\rightarrow 2} - P_2^e \end{cases}\end{split}\]

Incoming power. The incoming power is due to the power emitted and reflected by the other body,

\[\begin{split}\begin{cases} P_{\rightarrow 1} = P_{2}^e + r_2 P_{\rightarrow 2} \\ P_{\rightarrow 2} = P_{1}^e + r_1 P_{\rightarrow 1} \end{cases}\end{split}\]

From these two relations, the incoming powers can be explitly written as function of the emitted powers,

\[\begin{split}\begin{cases} P_{\rightarrow 1} = \dfrac{P_2^e + r_2 P_1^e}{1 - r_1 r_2} \\ P_{\rightarrow 2} = \dfrac{P_1^e + r_1 P_2^e}{1 - r_1 r_2} \end{cases}\end{split}\]

Energy balance of the system. Energy of an isolated system is constant and thus the dime derivative of the energy is zero,

\[\begin{split}\begin{aligned} 0 & = \dot{E} = \dot{E}_1 + \dot{E}_2 = \\ & = a_1 \dfrac{P_2^e + r_2 P_1^e}{1 - r_1 r_2} - P_1^e + a_2 \dfrac{P_1^e + r_1 P_2^e}{1 - r_1 r_2} - P_2^e = \\ & = P_1 \frac{a_2 + a_1 r_2 - 1 + r_1 r_2}{1 - r_1 r_2} + P_2 \frac{a_1 + a_2 r_1 - 1 + r_1 r_2}{1 - r_1 r_2} \\ & = T_1^4 \sigma \varepsilon_1 \frac{a_2 + a_1 r_2 - 1 + r_1 r_2}{1 - r_1 r_2} + T_2^4 \sigma \varepsilon_2 \frac{a_1 + a_2 r_1 - 1 + r_1 r_2}{1 - r_1 r_2} \ . \end{aligned}\end{split}\]

If this relation must hold in any condition, for any value of the emitted powers and for any independent choice of the materials of the bodies, the coefficients must be zero

\[\begin{split}\begin{cases} a_2 + a_1 r_2 - 1 + r_1 r_2 = 0 \\ a_1 + a_2 r_1 - 1 + r_1 r_2 = 0 \\ \end{cases}\end{split}\]

Subtracting,

\[a_2 (1 - r_1) = a_1 (1 - r_2) \ ,\]

and, since the relation must hold for any choice of pair of materials, the relation \(a_1 = 1 - r_1\) is found again.

Bodies in thermal equlibrium. If each body is in thermal equilibrium,

\[\begin{split}\begin{cases} 0 = \dfrac{P_2^e + r_2 P^{e}_1}{ 1 - r_1 r_2} - P_1^e = \sigma G_{12} \dfrac{ \varepsilon_2 T_2^4 + ( r_2 - 1 + r_1 r_2 ) \varepsilon_1 T_1^4 }{ 1 - r_1 r_2 } \\ 0 = \dfrac{P_1^e + r_1 P^{e}_2}{ 1 - r_1 r_2} - P_2^e = \sigma G_{12} \dfrac{ \varepsilon_1 T_1^4 + ( r_1 - 1 + r_1 r_2 ) \varepsilon_2 T_2^4 }{ 1 - r_1 r_2 } \\ \end{cases}\end{split}\]

Kirchhoff’s law

For systems in thermodynamic equilibrium,…

• constant ratio of…

• \(\alpha = \varepsilon\)? for a system in thermal equilibrium only

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1

Show how to get this expression from the spectral energy density.

2

todo Evaluate it