# Difference between revisions of "Radiative Processes in Astrophysics"

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## Lecture 1

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

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

${\displaystyle I_{\nu }}$ 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, ${\displaystyle {\tfrac {I_{\nu }}{\nu ^{3}}}}$ is. Also, for redshift ${\displaystyle z}$, ${\displaystyle I_{\nu }\propto \nu ^{3}\propto {\tfrac {1}{(1+z)^{3}}}}$ so intensity decreases with redshift. Finally: ${\displaystyle {\tfrac {I_{\nu }}{\eta ^{2}}}}$ is conserved along a ray, where ${\displaystyle \eta }$ 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:

${\displaystyle B_{\nu }={\frac {h\nu }{\lambda ^{2}}}{2 \over (e^{\frac {h\nu }{kT}}-1)}}$

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

### Derivation

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

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

However,

${\displaystyle h\nu {\frac {n_{\nu }c}{4\pi }}=I_{\nu }=B_{\nu }}$

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

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

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

${\displaystyle B_{\nu }\approx {\frac {2kT}{\lambda ^{2}}}}$

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