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A blackbody is the simplest source: it absorbs and re-emits radiation with 100\% efficiency. The frequency content of blackbody radiation is given by the {\it Planck Function}: \begin{equation} B_\nu={2h\nu^3\over c^2(e^{\frac{h\nu}{kT}}-1)} \end{equation} \begin{figure}[h!]

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Since radio photons typically have very low energies compared with the kinetic energy of most emitters, so that $h\nu\ll kT$, it is usually safe in radio astronomy to take the {\it Raleigh-Jeans} approximation to the blackbody spectrum: \begin{equation} B_\nu\approx\frac{2kT}{\lambda^2} \end{equation} Even though we've written $\lambda$ in place of $c/\nu$ above, remember that we are talking about $B_\nu$, which is a per-bandwidth (i.e. per Hz) quantity.

The Raleigh-Jeans approximation is so common in radio astronomy that it is used to define a {\it brightness temperature}, $T_b$, given by: \begin{equation} \frac{2kT_b}{\lambda^2}\equiv I_\nu. \end{equation} The brightness temperature is the temperature a blackbody would have to be at to produce match the intensity $I_\nu$ at the frequency in question. Note that this is not saying that $I_\nu$ is a blackbody spectrum, or is at all thermal in nature. It's just matching the emission at a particular frequency to an equivalent temperature. Brightness temperatures are generally a function of frequency for non-thermal sources. \begin{figure}[h!]

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\subsection*{Derivation}

If you take Bose-Einstein statistics as a given, deriving the Planck function is pretty straightforward; it boils down to counting degenerate states for a given energy.

To begin with, we need to relate the specific intensity of radiation from a blackbody, $B_\nu$, to an energy density that we can then count up states for: \begin{equation} \frac{4\pi}{c}B_\nu=D_\nu\cdot f(\nu)\cdot E, \end{equation} where $D_\nu$ counts up the specific degenerate states per volume for a given energy, $f(\nu)$ is the probability of occupying that energy, and $E$ is the energy of that state. Hence, the right side of this equation expresses specific energy density. On the left side, since $B_\nu$ has units of ${\rm ergs}/{\rm s}\cdot{\rm Hz}\cdot{\rm cm^2}\cdot{\rm sr}$, dividing by $c$ converts the specific flux to a specific energy density, with units of ${\rm ergs}/{\rm Hz}\cdot{\rm cm^3}\cdot{\rm sr}$ (recall "specific" means "per frequency"). The factor of $4\pi$ comes from assuming isotropic emission, and cancels out the factor of ${\rm sr}$.

For energy, we can use $E=h\nu$, and assuming Bose-Einstein statistics (photons are bosons, after all), we have that $f(\nu)= 1/(e^\frac{h\nu}{kt}-1)$. All that remains is to figure out $D_\nu$, the number of degenerate states per volume per frequency for a given energy. To do this, let's examine a cube of space with length $L$ on a side. For a finite region of space, the energy states a photon can take are quantized. This is because, to be occupied in this volume, a photon go through an integral number of wavelengths in distance $L$ along each of the $\hat x, \hat y$, and $\hat z$ axes. An equivalent way of saying this is that frequency comes in units of $c/L$.

So if we want to count the number of degenerate states that all have the same energy for some differential frequency interval $d\nu$, we can calculate that by calculating the volume these states occupy in phase space. Since $E=h\nu$, the surface area of a shell of constant energy is $4\pi\nu^2$, and since each quantized state occupies a volume $(c/L)^3$ is phase space, the total number of states is given by: \begin{equation} D_\nu\cdot V\cdot d\nu= \frac{4\pi\nu^2d\nu}{(c/L)^3}\cdot g, \end{equation} where the volume $V=L^3$, and $g=2$ is a final degeneracy factor that accounts for the fact that photons can have 2 different polarizations.

Working out the algebra and plugging everything into our original equation for $B_\nu$, we end up with: \begin{equation} B_\nu = \frac{2h\nu^3}{c^2}\frac1{e^\frac{h\nu}{kT}-1} \end{equation}

\section*{ Kirchoff's Law}

Kirchoff's Law is a statement that may alternately be regarded as obvious or deeply insightful. Simply put, it states that, for an object in a (let's say, Planckian) radiation bath in thermal equilibrium, the energy absorbed equals the energy radiated. As obvious as this may seem, Kirchoff's Law can have non-obvious implications. For example, let's take the equation for Radiative Transport: \begin{equation} \frac{dI_\nu}{ds}=j_\nu - \alpha_\nu I_\nu. \end{equation} In the optically thick limit for blackbody radiation, $I_\nu=B_\nu$ and $dI_\nu/ds=0$. Hence, \begin{equation} j_\nu=\alpha_\nu B_\nu(T). \end{equation} On one level, this is simply saying that for a medium in local thermodynamic equilibrium, $S_\nu\equiv j_\nu/\alpha_\nu = B_\nu(T)$. But on another level, since the values of $j_\nu$ and $\alpha_\nu$ are inherent properties of a medium and don't depend on thermal equilibrium, this is a fundamental relationship between $j_\nu$ and $\alpha_\nu$. Kirchoff's law says that good emitters are good absorbers, and vice versa.

This fundamental relationship between $j_\nu$ and $\alpha_\nu$ ensures that, when things are brought into thermal equilibrium, the resultant spectrum is always Planckian, and does not depend on the emissivity or absorptivity of the medium.

\section*{Blackbody Sources}

There are not many examples of true blackbody sources of radiation, but what few there are tend to be quite important.

\subsection*{Cosmic Microwave Background}

The mother of all blackbodies, this is the relic 2.7K radiation left over from the hot plasma that was generated by the Big Bang. Photons were produced and absorbed via the thermal radiation of charged particles (mostly electrons) scattering off one another, converging quickly on a blackbody spectrum. Once the universe cooled to the point that protons and electrons could bind to form neutral hydrogen without being immediately ionized, these photons stopped scattering, and have been free-streaming ever since. As the universe expands, the wavelength of these photons stretch with it, causing photon energies (and hence, the characteristic temperature) to gradually decline with time. Note that, at the current temperature of 2.7K, the CMB peaks at 160.2 GHz.

\subsection*{Sun}

Stars are generally blackbody-like, although most of them are so faint in the radio band as to be nearly unobservable. Just by sheer proximity, our Sun is the brightest radio source in the sky, but, particularly at lower frequencies, not by a huge margin. Emission from the quiet Sun is dominated at radio frequencies by the photosphere (6000K) around 100 GHz, the chromosphere (10,000K) at 1 GHz, and the corona (1,000,000K) at 100 MHz. As a result, the spectrum of the quiet Sun within a relatively narrow band is blackbody-like, but broadly departs from the characteristic blackbody spectrum.

It should also be noted that during sunspot activity, which varies on an 11-year solar cycle, solar emission departs dramatically from that of a blackbody, and becomes dominated by intense emission from electrons trapped in the magnetic fields around sunspots. The perturbed sun can be orders of magnitude brighter than the quiet Sun, and with emission dominated by particular hot spots that rotate with the Sun.

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