## Lecture 1

### 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 ($\Delta \nu [Hz^{-1}]$ ), but $\Delta \lambda \neq {\tfrac {c}{\Delta \nu }}$ .

${\Delta E \over \Delta t\Delta \nu \Delta A\Delta \Omega }\iff {dE \over dtd\nu dAd\Omega }\equiv I_{\nu }$ $I_{\nu }$ 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, ${\tfrac {I_{\nu }}{\nu ^{3}}}$ is. Also, for redshift $z$ , $I_{\nu }\propto \nu ^{3}\propto {\tfrac {1}{(1+z)^{3}}}$ so intensity decreases with redshift. Finally: ${\tfrac {I_{\nu }}{\eta ^{2}}}$ is conserved along a ray, where $\eta$ is the index of refraction.

### The Black Body

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:

$B_{\nu }={\frac {h\nu }{\lambda ^{2}}}{2 \over (e^{\frac {h\nu }{kT}}-1)}$ $B_{\nu }={\frac {2h\nu ^{3}}{c^{2}(e^{\frac {h\nu }{kT}}-1)}}\neq B_{\lambda }$ ### 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:

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

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

$B_{\nu }\approx {\frac {2h\nu ^{3}}{c^{2}}}e^{-{\frac {h\nu }{kT}}}$ If $h\nu \ll kT$ :

$B_{\nu }\approx {\frac {2kT}{\lambda ^{2}}}$ Note that this tail peaks at $\sim {\tfrac {3kT}{h}}$ . Also, $\nu B_{\nu }=\lambda B_{\lambda }$ 