Radiative Transfer

A Simple Macroscopic Model for the Interaction of Radiation with Matter

In free space, the specific intensity $I_\nu$ of radiation is conserved along a ray:
$$ {d I_\nu \over d s} = 0  ,$$
where $s$ is the coordinate along the ray between the source at $s = 0$ and the detector at $s = s_{\rm o}$. What happens when there is an intervening medium between $s_{\rm in}$ and $s_{\rm out}$?



Absorption between a source and a detector.  The coordinate along the ray increases from $s_{\rm in}$ at the input end of the absorber to $s_{\rm out}$ at the output end of the absorber.  The optical depth $\tau$ is measured in the opposite direction, starting from $\tau = 0$ at $s_{\rm out}$ and increasing as $s$ decreases.

Absorption

Think of the ray as a beam of photons, or light particles, some of which may be absorbed by the medium and vanish. The infinitesimal probability $dp_\nu$ of a photon with frequency $\nu$ being absorbed (e.g., by hitting an absorbing particle) in a thin slab of thickness $ds$ is directly proportional to $ds$: $dp_\nu= \kappa_\nu ds$.  This frequency-dependent constant of proportionality
$$\bbox[border:3px blue solid,7pt]{\kappa_\nu \equiv {dp_\nu \over ds}}\rlap{\quad \rm {(2B1)}}$$
is called the linear absorption coefficient, and its dimension is inverse length. A value $\kappa_\nu = 1 {\rm ~m}^{-1}$ means that over some small distance $\Delta s \ll \kappa_\nu^{-1}$ along the ray, $\Delta s = 10^{-3}$ m for example, the small fraction $\kappa_\nu \Delta s = 1 {\rm ~m}^{-1} \times 10^{-3} {\rm ~m} = 10^{-3}$ of the photons will be absorbed.  Why consider such thin layers rather than the whole absorber at once? Only if the layer is thin enough that $dp_\nu$ is infinitesimal does the probability of absorption increase linearly with thickness $ds$ so that $\kappa_\nu$ is a constant.  In a layer thick enough to absorb a significant fraction of the photons, fewer photons are absorbed near the far end of the layer simply because fewer photons have survived to be absorbed there, so the probability of absorption increases nonlinearly with thickness.

This simple macroscopic model doesn't depend on the microscopic physical processes by which photons are absorbed.  It is like the ideal gas model, which describes the macroscopic properties of a gas without going into the microscopic details about how individual gas molecules behave.

The fraction of the specific intensity lost to absorption in the infinitesimal distance $ds$ along the ray is
$$ {d I_\nu \over I_\nu} = -dp_\nu = -\kappa_\nu ds \rlap{\rm\quad ~(absorption~only)}$$
Integrating both sides of this equation along the absorbing path gives the output specific intensity as a fraction of the input specific intensity:
$$\int_{s_{\rm in}}^{s_{\rm out}} {d I_\nu \over I_\nu} = -\int_{s_{\rm in}}^{s_{\rm out}} \kappa_\nu(s')ds' = \ln I_\nu \big|_{s_{\rm in}}^{s_{\rm out}}$$
$$\ln[I_\nu(s_{\rm out})] - \ln[I_\nu(s_{\rm in})] = - \int_{s_{\rm in}}^{s_{\rm out}} \kappa_\nu (s') ds'$$ $$I_\nu(s_{\rm out}) = I_\nu(s_{\rm in}) \times \exp\biggl[ -\int_{s_{\rm in}}^{s_{\rm out}} \kappa_\nu(s')ds'\biggl]$$
The (dimensionless) quantity
$$\bbox[border:3px blue solid,7pt]{\tau_\nu \equiv \int_{s_{\rm out}}^{s_{\rm in}} -\kappa_\nu(s') ds'}\rlap{\quad \rm {(2B2)}}$$
is called the optical depth or opacity of the absorber.  Note that $d\tau = -\kappa_\nu\ ds$.  If $\tau \ll 1$ the absorber is said to be optically thin; if $\tau \gg 1$ it is optically thick. The "backwards" direction of the path integration from the source to the observer was chosen in Equation 2B2 to make $\tau_\nu > 0$ for an absorber. Thus
$$I_\nu(s_{\rm out}) = I_\nu( s_{\rm in}) \times \exp (-\tau_\nu) {\rm ~~(absorption~only)}$$

Emission

The intervening medium may also generate, or emit, photons, again by some unspecified microscopic process. In any small volume ($ds d\sigma$) of space, the probability per unit time that a photon having frequency $\nu$ will be emitted into the solid angle $d \Omega$ is directly proportional to that volume and solid angle:
$$P_{\rm em} \propto ds d\sigma d\Omega$$
The emission coefficient $\epsilon_\nu$ is defined so that
$$\bbox[border:3px blue solid,7pt]{\epsilon_\nu \equiv {dI_\nu \over ds}}\rlap{\quad \rm {(2B3)}}$$
if we ignore absorption.  The dimensions of $\epsilon_\nu$ follow from its definition; they are power per unit volume per unit frequency per unit solid angle, and the corresponding mks units are W m$^{-3}$ Hz$^{-1}$ sr$^{-1}$.

Considering both emission and absorption, we get the equation of radiative transfer
$$\bbox[border:3px blue solid,7pt]{{d I_\nu \over ds} = -\kappa_\nu I_\nu + \epsilon_\nu}\rlap{\quad \rm {(2B4)}}$$

Since we have not yet specified anything about the absorption and emission processes, $\kappa_\nu$ and $\epsilon_\nu$ might seem to be independent. However, they are not independent in thermodynamic equilibrium (TE). In TE, matter and radiation are in equilibrium at the same temperature $T$. Equilibrium radiation at temperature $T$ is isotropic and has a blackbody spectrum. One source of blackbody radiation is a cavity defined by opaque walls pierced by a small hole. The intensity and spectrum of the radiation emerging from the hole is independent of the wall material (e.g., wood with green paint, shiny copper, gray concrete, etc.) and any absorbing material (e.g., gas, dust, fog, etc.) that may be inside the cavity.



Thermal radiation from a cavity depends only on its temperature and is independent of both the material used to make the cavity walls and the contents of the cavity.


Kirchoff proved this statement by a thought experiment illustrated in the figure below.



Kirchoff's thought experiment invokes two cavities in thermodynamic equilibrium connected through a filter that passes radiation in the narrow frequency range $\nu$ to $\nu+d\nu$.  The cavities may be made of different materials and contain different emitting/absorbing particles.

Consider two cavities at the same temperature $T$ whose windows are connected by a narrow-band passive (meaning, no energy is needed for it to work) filter transparent only between frequencies $\nu$ and $\nu + d\nu$. Radiation can carry no net power from one cavity to the other, lest one cavity cool down and the other heat up. Cooling one and heating the other would violate the second law of thermodynamics because the two cavities at different temperatures could be used to drive a heat engine.

In (full) thermodynamic equilibrium (TE) at temperature $T$,
$${d I_\nu \over ds} = 0 {\rm ~and~} I_\nu = B_\nu(T)$$
so
$${d I_\nu \over ds} = 0 = -\kappa_\nu B_\nu (T) + \epsilon_\nu$$
yielding Kirchoff's law for  a sytem in TE:
$$\bbox[border:3px blue solid,7pt]{{\epsilon_\nu (T) \over \kappa_\nu (T )} = B_\nu(T)}\rlap{\quad \rm {(2B5)}}$$

This argument is clearly independent of the shapes of the cavities, and consequently blackbody radiation must be isotropic.

Although Kirchoff's law was derived for a system in thermodyamic equilibrium, its applicability is not limited to blackbody radiation in full thermodynamic equilibrium with its material environment. Kirchoff's law applies whenever the radiating/absorbing material is in thermal equilibrium, for any radiation field. If the emitting/absorbing material is in thermal equilibrium (has a well-defined temperature $T$), even if it is not in equilibrium with the radiation field, it is said to be in local thermodyamic equilibrium (LTE). For example, most of the Earth's atmosphere is in LTE . It has a well-defined gas temperature $T \sim 300$ K measureable with an ordinary thermometer, but it is in a distinctly nonequilibrium radiation field (anisotropic $T \sim 5800$ K sunlight during the day, the cold dark sky at night, plus anisotropic emission from the ground). 

Kirchoff's law (Eq. 2B5) applies in LTE as well as TE.  To show this, note that $B_\nu (T)$ is a function of the radiation temperature $T$ only and does not depend on the properties of the radiating/absorbing material.  In contrast, both $\epsilon_\nu( T)$ and $\kappa_\nu (T)$ depend only on the material properties of the cavity (e.g., whether the walls are made of copper or concrete) and on the temperature of that material; they do not depend on the radiation field or its spectrum. 

This generalized version of Kirchoff's law is a valuable tool for calculating the emission coefficient from the absorption coefficient or vice-versa.  For example, in Section 4B we will calculate $\epsilon_\nu$ for free-free emission by an HII region (a cloud of ionized hydrogen) in LTE and use Kirchoff's law to obtain $\kappa_\nu$ immediately.

Kirchoff's law is not intuitively obvious to us because any room-temperature ($T \sim 300$ K) object in our environment is much too cold ($h \nu / k T \gg 1$) to emit a detectable amount of visible light. Thus a glass of water might absorb 10% of the sunlight passing through it, but we will not see any emitted light because 10% of the light emitted by a room-temperature blackbody is nearly zero.  One familiar example of Kirchoff's law is a charcoal fire with flames and glowing coals. The infrared (heat) radiation from barely glowing black coals in LTE is much more intense than that from the hotter but nearly transparent visible flames, as you can verify by using a shield to cover either the coals or the flames.

Kirchoff's law is always significant at radio wavelengths, though, because $h \nu / k T < 1$. This point is illustrated by radio emission and absorption in the Earth's atmosphere.

Emission and absorption of radio waves in the Earth's atmosphere

At high ($\nu > 10$ GHz) radio frequencies, absorption by the Earth's atmosphere can be large enough to affect the accuracy of flux-density measurements.  Radio astronomers often determine the amount of atmospheric absorption by measuring the change in atmospheric emission at different angles from the zenith (the zenith is the direction straight up) and using Kirchoff's law.  They use Kirchoff's law to calculate the zenith opacity of the (roughly isothermal) atmosphere at frequency $\nu$ from its kinetic (thermometer) temperature $T_{\rm A}$ and its radio brightness.  (The celestial sky above the atmosphere is relatively cold, so the "background'' emission above the atmosphere can usually be ignored.)



Most atmospheric emission and absorption occurs in a layer whose height $h$ is only a few km, much smaller than the radius of the Earth, so the plane-parallel approximation for the spherical Earth shown here is accurate.  


Procedure: Tip the radio telescope, measure $I_{\nu}$ as a function of the zenith angle $z$, the angle between the zenith (vertical) and the incoming ray. Practical complications include:

Since $B_\nu$ is directly proportional to $T$ in the Rayleigh-Jeans low-frequency approximation: $$ B_\nu \approx {2 k T \nu^2 \over c^2}~,$$ radio astronomers often find it convenient to specify the spectral brightness $I_\nu$, even if $I_\nu \ne B_\nu$, in terms of the equivalent blackbody brightness temperature $T_{\rm b}$ defined by
$$I_\nu = {2 k T_{\rm b}(\nu) \nu^2 \over c^2}$$
Thus for
any $I_\nu$, blackbody or not,
$$\bbox[border:3px blue solid,7pt]{T_{\rm b}(\nu) \equiv {I_\nu c^2 \over 2 k \nu^2}}\rlap{\quad \rm {(2B6)}}$$

Brightness temperature is a practical way to specify brightness because radio telescopes are often calibrated by absorbers or "loads" of known temperature and because the temperature of a radio source is frequently a quantity of physical interest. However, be careful not to confuse brightness temperature with "real" physical temperature. They might be equal for an opaque ($\tau \gg 1$) source, but the brightness temperature of a semitransparent thermal source (like the Earth's atmosphere) will always be less than its kinetic temperature.

The observed output from the radio telescope might vary with $z$ like this:



The receiver output from a tipping scan shows the increase in sky brightness with zenith angle, but the zero point of the sky brightness temperature is not well determined. 


To understand and interpret these observations, we start with the radiative transfer equation
$${d I_\nu \over d S} = -\kappa_\nu I_\nu + \epsilon_\nu$$
Since the atmosphere is in LTE, we can eliminate the unknown $\epsilon_\nu$ using Kirchoff's law: $$\epsilon_\nu = \kappa_\nu B_\nu(T_{\rm A})~,$$
where $T_{\rm A}$ is the kinetic temperature of the atmosphere as measured by an ordinary thermometer. Dividing by $\kappa_\nu$ yields
$${ 1 \over \kappa_\nu} {d I_\nu \over ds} = {-d I_\nu \over d \tau} = -I_\nu + B_\nu (T_{\rm A})$$
Multiply both sides of this differential equation by $\exp(-\tau)$ and integrate along the ray in the telescope beam from the top of the atmosphere to the ground. Let $\tau_{\rm A}(z)$ be the total optical depth of the atmosphere along the ray at zenith angle $z$. Then
$$\int_0^{\tau_{\rm A}} e^{-\tau} {d I_\nu \over d \tau} d \tau = \int_0^{\tau_{\rm A}} [I_\nu - B_\nu(T_{\rm A})] e^{-\tau} d\tau$$
Next integrate the left side by parts and take $B_\nu(T_{\rm A})$ outside the integral (that is, make the approximation that the atmosphere below altitude $h$ is nearly isothermal).
$$e^{-\tau} I_\nu \vert_0^{\tau_{\rm A}} - \int_0^{\tau_{\rm A}} -e^{-\tau} I_\nu\ d \tau = \int_0^{\tau_{\rm A}} I_\nu\ e^{-\tau} d\tau - B_\nu(T_{\rm A}) \int_0^{\tau_{\rm A}} e^{-\tau} d \tau$$
$$I_\nu(\tau = \tau_{\rm A}) e^{-\tau_{\rm A}} - I_\nu(\tau = 0) = B_\nu(T_{\rm A}) (e^{-\tau_{\rm A}} - 1)$$
We can neglect $I_\nu(\tau = \tau_{\rm A}) \approx 0$ because the brightness of emission above the atmosphere is relatively low at high radio frequencies. 
$$I_\nu(\tau = 0) = ( 1 - e^{-\tau_{\rm A}}) B_\nu(T_{\rm A})$$
The path length through a plane-parallel atmosphere is proportional to ${\rm sec}\, z$ so at any zenith angle $z$,
$$\tau_{\rm A} = \tau_{\rm Z}\, {\rm sec}\, z$$
where $\tau_{\rm Z} \equiv \tau_{\rm A} ( z = 0)$ is the zenith opacity of the atmosphere above the radio telescope.
$$I_\nu = [1 - \exp(-\tau_{\rm Z}\, {\rm sec} \, z)]{2 k T_{\rm A} \nu^2 \over c^2}$$
Thus we obtain an equation for the brightness temperature of the atmospheric emission as a function of zenith angle:
$$T_{\rm b} = {I_\nu c^2 \over 2 k \nu^2} = T_{\rm A} \biggl[1 - \exp (-\tau_{\rm Z}\,{\rm sec}\,z)\biggr]$$


Ratios of atmospheric brightness temperature $T_{\rm b}$ to atmospheric kinetic temperature $T_{\rm A}$ for different zenith opacities $\tau_{\rm z}$ as functions of zenith angle $z$.


Example: In Green Bank, in freezing weather ($T_{\rm A} = 273$ K), the zenith opacity $\tau_{\rm z} \approx 0.1$ at $\nu = 90$ GHz. Thus we expect that the brightness temperature of the atmosphere will vary with $z$ according to
 $$T_{\rm b} \approx 273\,{\rm K}\, \biggl[ 1 - \exp ( -\tau_{\rm Z}\, {\rm sec}\,z)\biggr]$$
By fitting the observed brightness temperatures to this equation, we can deduce $\tau_{\rm z}$ and hence the opacity at any zenith angle.

Green Bank $T_{\rm A} = 273$ K, $\nu = 90$ GHz, $\tau_{\rm z} \approx 0.1$:

$z$(deg)   sec($z$)       $T_{\rm b}$ (K)

  0
1.00
26.0
30
1.15
29.8
45
1.41
36.0
60
2.00
49.5



Emission, Absorption, and Reflection From an Opaque Body

A photon striking an opaque ($\tau \gg 1$) body is either absorbed or reflected. The absorption coefficient $a_\nu$ is the probability that an incident photon of frequency $\nu$ will be absorbed. The reflection coefficient $r_\nu$ is the probability that an incident photon of frequency $\nu$ will be reflected. Thus $a_\nu = 1$ for a black body and $r_\nu = 1$ for a perfect mirror. Since each photon is either absorbed or reflected by any opaque body, $a_\nu + r_\nu = 1$. The emission and absorption coefficiencts of an opaque body in local thermodynamic equilibrium are not independent; they are related by another form of Kirchoff's law.

Imagine an opaque body surrounded by a filter passing frequencies in the range $\nu$ to $\nu + d \nu$ inside a cavity in thermodynamic equilibrium at some temperature $T$.



Thought experiment involving an opaque body surrounded by a filter passing frequencies $\nu$ to $\nu + d\nu$ inside a cavity in thermodynamic equilibrium.



In equilibrium, the spectral power absorbed per unit area, $P_{\rm a}$, must equal the power emitted per unit area. 
$$P_{\rm a} = a_\nu \int B_\nu \cos \theta d \Omega$$
$$P_{\rm a} = a_\nu \int_{\phi = 0}^{2\pi} \int_{\theta=0}^{\pi/2}
B_\nu \cos\theta \sin\theta d \theta d \phi$$
$$P_{\rm a} = a_\nu 2 \pi B_\nu \int_0^{\pi/2} \cos \theta \sin \theta d \theta$$
$$P_{\rm a} = a_\nu \times \pi B_\nu$$
The spectral emissivity $e_\nu$ is defined as the spectral power emitted per unit area by the opaque body divided by the spectral power emitted per unit area by a blackbody of the same temperature: $P_{\rm e} = \pi B_\nu \times e_\nu$. Then $P_{\rm a} = P_{\rm e}$ implies
$$\bbox[border:3px blue solid,7pt]{e_\nu = a_\nu = 1 - r_\nu}\rlap{\quad \rm {(2B7)}}$$
This is just another form of Kirchoff's law. It implies that the brightness temperature $T_{\rm b}$ of an opaque body in LTE at physical temperature $T$ is given by
$$T_{\rm b}(\nu) = a_\nu T = (1 - r_\nu) T$$
Since $a \le 1$, $T_{\rm b} \leq T$. A perfect reflector ($r = 1$) will always have $T_{\rm b} = 0$; that is, it does not emit.


Example: The special paint used on the surface of the Green Bank Telescope (GBT) exploits this form of Kirchoff's law to perform three separate functions simultaneously.  It is white in the visible portion of the spectrum to reflect sunlight ($T \approx 5800$ K) because differential solar heating would expand and deform the reflector.  It is black in the mid-infrared so that the GBT ($T \approx 300$ K) can cool itself efficiently by reradiation.  It is transparent at radio wavelengths so that it neither absorbs incoming radio waves nor emits thermal noise at radio wavelengths.

Surface of GBT in partial sun

Temperature differences across the GBT reflector caused by differential solar heating would deform the surface and degrade its performance.  A special paint that is white at visible wavelengths, black in the mid-infrared, and transparent at radio wavelengths keeps the surface cool and does not harm performance at radio wavelengths by absorbing incoming radio waves or emitting radio noise. Image credit