# Difference between revisions of "Radiative Processes in Astrophysics"

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− | == The Black Body == | + | == Lecture 1 == |

+ | |||

+ | === Units === | ||

+ | |||

+ | Here are some terms pertaining to telescope observations: aperture area (<math>\Delta A</math>), solid angle on sky (<math>\Delta \Omega | ||

+ | </math>), exposure time (<math>\Delta t</math>), collects energy (<math>\Delta E</math>), over waveband (<math>\Delta \nu [Hz^{-1}]</math>), but <math>\Delta\lambda\ne\tfrac{c}{\Delta\nu}</math>. | ||

+ | |||

+ | <math>{\Delta E\over\Delta t\Delta\nu\Delta A\Delta\Omega} \iff | ||

+ | {dE \over dt d\nu dA d\Omega} \equiv I_\nu</math> | ||

+ | |||

+ | <math>I_\nu</math> is the specific intensity per unit frequency. | ||

+ | Flux density is power per unit frequency passing through a differential area whose normal is <math>\hat n</math>. Thus, flux density is: <math>{F_\nu\equiv\int I_\nu\cos\theta d\Omega}</math> | ||

+ | |||

+ | === Proof that Specific Intensity is conserved along a ray === | ||

+ | |||

+ | The power received by the telescope is: | ||

+ | |||

+ | <math>P_{rec}=I_\nu d\Omega dA</math> | ||

+ | |||

+ | where <math>I_\nu(\alpha,\delta )</math> is the intensity as a function of right-ascension (<math>\alpha</math>) and declination (<math>\delta</math>). Say that <math>\Sigma_\nu(\alpha,\delta)</math> is the surface luminosity of a patch of sky (that is, the emitted intensity). Then power emitted by patch of sky is: | ||

+ | |||

+ | <math>P_{emit}=\Sigma_\nu{dA\over r^2}d\tilde A</math> | ||

+ | |||

+ | Recognizing that <math>d\tilde A=d\Omega r^2</math>: <math>\Sigma_\nu=I_\nu</math> This derivation assumes that we are in a vacuum and that the frequencies of photons are constant. If frequencies change, then though specific intensity <math>I_\nu</math> is not conserved, <math>\tfrac{I_\nu}{\nu^3}</math> is. Also, for redshift <math>z</math>, <math>I_\nu\propto\nu^3 \propto\tfrac{1}{(1+z)^3}</math> so intensity decreases with redshift. Finally: <math>\tfrac{I_\nu}{\eta^2}</math> is conserved along a ray, where <math>\eta</math> is the index of refraction. | ||

+ | |||

+ | |||

+ | === The Black Body === | ||

A blackbody is the simplest source: it absorbs and reemits radiation with | A blackbody is the simplest source: it absorbs and reemits radiation with |

## Revision as of 20:39, 9 February 2010

## Lecture 1

### Units

Here are some terms pertaining to telescope observations: aperture area (), solid angle on sky (), exposure time (), collects energy (), over waveband (), but .

is the specific intensity per unit frequency. Flux density is power per unit frequency passing through a differential area whose normal is . Thus, flux density is:

### Proof that Specific Intensity is conserved along a ray

The power received by the telescope is:

where is the intensity as a function of right-ascension () and declination (). Say that is the surface luminosity of a patch of sky (that is, the emitted intensity). Then power emitted by patch of sky is:

Recognizing that : This derivation assumes that we are in a vacuum and that the frequencies of photons are constant. If frequencies change, then though specific intensity is not conserved, is. Also, for redshift , so intensity decreases with redshift. Finally: is conserved along a ray, where 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*:

### Derivation

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

However,

so we have it. In the limit that :

If :

Note that this tail peaks at . Also,