Units

Here are some terms pertaining to telescope observations:

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

${\Delta E \over \Delta t\Delta \nu \Delta A\Delta \Omega }\iff {dE \over dtd\nu dAd\Omega }\equiv I_{\nu }\,\!$ $I_{\nu }$ is the specific intensity per unit frequency.

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

${F_{\nu }\equiv \int I_{\nu }\cos \theta d\Omega }\,\!$ Proof that Specific Intensity is conserved along a ray

The power received by the telescope is:

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

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

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

$I_{\nu }\propto \nu ^{3}\propto {\frac {1}{(1+z)^{3}}}\,\!$ so intensity decreases with redshift. Finally:

${I_{\nu } \over \eta ^{2}}\,\!$ is conserved along a ray, where $\eta$ 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:

$B_{\nu }={h\nu \over \lambda ^{2}}{2 \over (e^{h\nu \over kT}-1)}\,\!$ ${B_{\nu }={2h\nu ^{3} \over c^{2}(e^{h\nu \over kT}-1)}\neq B_{\lambda }}\,\!$ (The Planck Function for Black Body Radiation)

Derivation:

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

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

$h\nu {n_{\nu }c \over 4\pi }=I_{\nu }=B_{\nu }\,\!$ so we have it. In the limit that $h\nu \gg kT$ :

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

If $h\nu \ll kT$ :

$B_{\nu }\approx {2kT \over \lambda ^{2}}\,\!$ Rayleigh-Jeans tail Note that this tail peaks at $\sim {3kT \over h}$ . Also,

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