Units

Here are some terms pertaining to telescope observations:

aperture area (${\displaystyle \Delta A}$), solid angle on sky (${\displaystyle \Delta \Omega }$), exposure time (${\displaystyle \Delta t}$), collects energy (${\displaystyle \Delta E}$), over waveband (${\displaystyle \Delta \nu }$), but ${\displaystyle \Delta \lambda \neq {c \over \Delta \nu }}$.

${\displaystyle {\Delta E \over \Delta t\Delta \nu \Delta A\Delta \Omega }\iff {dE \over dtd\nu dAd\Omega }\equiv I_{\nu }\,\!}$

${\displaystyle I_{\nu }}$ is the specific intensity per unit frequency.

Flux density is power per unit frequency passing through a differential area whose normal is ${\displaystyle {\hat {n}}}$. Thus, flux density is:

${\displaystyle {F_{\nu }\equiv \int I_{\nu }\cos \theta d\Omega }\,\!}$

Proof that Specific Intensity is conserved along a ray

The power received by the telescope is:

${\displaystyle P_{rec}=I_{\nu }d\Omega dA\,\!}$

where ${\displaystyle I_{\nu }(\alpha ,\delta )}$ is the intensity as a function of right-ascension (${\displaystyle \alpha }$) and declination (${\displaystyle \delta }$). Say that ${\displaystyle \Sigma _{\nu }(\alpha ,\delta )}$ is the surface luminosity of a patch of sky (that is, the emitted intensity). Then power emitted by patch of sky is:

${\displaystyle P_{emit}=\Sigma _{\nu }{dA \over r^{2}}d{\tilde {A}}\,\!}$

Recognizing that ${\displaystyle d{\tilde {A}}=d\Omega r^{2}}$:

${\displaystyle \Sigma _{\nu }=I_{\nu }\,\!}$

This derivation assumes that we are in a vacuum and that the frequencies of photons are constant. If frequencies change, then though specific intensity ${\displaystyle I_{\nu }}$ is not conserved, ${\displaystyle {I_{\nu } \over \nu ^{3}}}$ is. Also, for redshift ${\displaystyle z}$,

${\displaystyle I_{\nu }\propto \nu ^{3}\propto {\frac {1}{(1+z)^{3}}}\,\!}$

so intensity decreases with redshift. Finally:

${\displaystyle {I_{\nu } \over \eta ^{2}}\,\!}$

is conserved along a ray, where ${\displaystyle \eta }$ is the index of refraction.

The Blackbody

A blackbody is the simplest source: it absorbs and reemits radiation with 100% efficiency. The frequency content of blackbody radiation is given by the Planck Function:

${\displaystyle B_{\nu }={h\nu \over \lambda ^{2}}{2 \over (e^{h\nu \over kT}-1)}\,\!}$
${\displaystyle {B_{\nu }={2h\nu ^{3} \over c^{2}(e^{h\nu \over kT}-1)}\neq B_{\lambda }}\,\!}$

(The Planck Function for Black Body Radiation)

Derivation:

The # density of photons having frequency between ${\displaystyle \nu }$ and ${\displaystyle \nu +d\nu }$ has to equal the # density of phase-space cells in that region, multiplied by the occupation # per cell. Thus:

${\displaystyle n_{\nu }d\nu ={4\pi \nu ^{2}d\nu \over c^{3}}{2 \over e^{h\nu \over kT}-1}\,\!}$

However,

${\displaystyle h\nu {n_{\nu }c \over 4\pi }=I_{\nu }=B_{\nu }\,\!}$

so we have it. In the limit that ${\displaystyle h\nu \gg kT}$:

${\displaystyle B_{\nu }\approx {2h\nu ^{3} \over c^{2}}e^{-{h\nu \over kT}}\,\!}$

Wein tail

If ${\displaystyle h\nu \ll kT}$:

${\displaystyle B_{\nu }\approx {2kT \over \lambda ^{2}}\,\!}$

Rayleigh-Jeans tail Note that this tail peaks at ${\displaystyle \sim {3kT \over h}}$. Also,

${\displaystyle \nu B_{\nu }=\lambda B_{\lambda }\,\!}$