# 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 (<math>\Delta A</math>), solid angle on sky (<math>\Delta \Omega | + | 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 </math>), but <math>\Delta \lambda \ne {c\over \Delta \nu }</math>. |

− | </math>), exposure time (<math>\Delta t</math>), collects energy (<math>\Delta E</math>), over waveband (<math>\Delta \nu | ||

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

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

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

− | 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> | |

− | <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>{I_\nu \over \nu ^3}</math> is. Also, for redshift <math>z</math>, | |

− | |||

− | the '' | ||

− | <math> | + | <math> I_\nu \propto \nu ^3 \propto {1 \over (1+z)^3} \,\!</math> |

− | |||

− | </math> | ||

− | + | so intensity decreases with redshift. Finally: | |

− | |||

− | |||

− | + | <math> {I_\nu \over \eta ^2} \,\!</math> | |

− | |||

− | |||

− | |||

− | <math>n_\nu d\nu= | + | is conserved along a ray, where <math>\eta </math> 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'': | ||

+ | |||

+ | <math> B_\nu ={h\nu \over \lambda ^2}{2\over (e^{{h\nu \over kT}}-1)} \,\!</math> | ||

+ | |||

+ | <math> {B_\nu ={2h\nu ^3\over c^2(e^{{h\nu \over kT}}-1)}\ne B_\lambda } \,\!</math> | ||

+ | |||

+ | (The Planck Function for Black Body Radiation)Derivation:The # density of photons having frequency between <math>\nu </math> and <math>\nu +d\nu </math> has to equal the # density of phase-space cells in that region, multiplied by the occupation # per cell. Thus: | ||

+ | |||

+ | <math> n_\nu d\nu ={4\pi \nu ^2d\nu \over c^3}{2\over e^{{h\nu \over kT}}-1} \,\!</math> | ||

However, | However, | ||

− | <math>h\nu\ | + | <math> h\nu {n_\nu c\over 4\pi }=I_\nu =B_\nu \,\!</math> |

+ | |||

+ | so we have it. In the limit that <math>h\nu \gg kT</math>: | ||

+ | |||

+ | <math> B_\nu \approx {2h\nu ^3\over c^2}e^{-{h\nu \over kT}} \,\!</math> | ||

+ | |||

+ | Wein tailIf <math>h\nu \ll kT</math>: | ||

+ | |||

+ | <math> B_\nu \approx {2kT\over \lambda ^2} \,\!</math> | ||

+ | |||

+ | Rayleigh-Jeans tail Note that this tail peaks at <math>\sim {3kT\over h}</math>. Also, | ||

+ | |||

+ | <math> \nu B_\nu =\lambda B_\lambda \,\!</math> | ||

+ | |||

+ | = Lecture 2 = | ||

+ | |||

+ | == Reiteration: Conservation of specific intensity == | ||

+ | |||

+ | Conservation of specific intensity told us the intensity collected by your telescope is absolutely equal to the surface intensity emitted into the angle leftrightarrowonding to your pixel on the sky. ''Specific intensity is distance independent''. <math>1 m^2</math> of sky emitted into angle of one square degree is equal to the specific intensity measured by a <math>1m^2</math> telescope pointed at a square-degree pixel of sky. | ||

+ | |||

+ | == Units == | ||

+ | |||

+ | We have flux, flux density, surface brightness, and spectral energy distribution to describe luminosity. When in doubt, look at the units. Surface brightness, like specific intensity, is distance independent. It is simply specific intensity integrated over frequency. | ||

+ | |||

+ | == Lenses == | ||

+ | |||

+ | If you look at something through a lens, the specific intensity of a point is conserved. In the lens, the specific intensity may be greater, because <math>I_\nu \over \eta ^2</math> is conserved (remember that <math>\eta </math> is the index of refraction of the medium), but once the light goes out again, even if the object looks bigger, the specific intensity is the same. ''This does not violate conservation of energy because the light is being focused to a smaller area.''Now if something is ''inside'' the lens (say, a bowl of water), then the lens bends light rays closer to each other, increasing the specific intensity. Thus, the object is emitting/scattering higher density photons, and appears more luminous, and the object also appears larger. This means that there must be some places where the object, omitting the lens, would have been visible, but now isn’t. | ||

+ | |||

+ | == Blackbodies == | ||

+ | |||

+ | The CMB is the perfect blackbody because in the past, the surface of the black body (the edge of the universe) was not allowed to leak out energy. This begs the question: does it matter what the boundary of the black body is made of? The answer is: No. In the short term, the boundary may imperfectly reflect a photon, but then the resulting change in the temperature of boundary dictates that it will tend to reemit the energy into the photon gas. Overall: | ||

+ | |||

+ | <math> I_\nu I_{emitted}=I_\nu I_{absorbed}=B_\nu f_{abs} \,\!</math> | ||

+ | |||

+ | However, <math>f_{emission}B_\nu \equiv I_\nu I_{emitted}</math>, so: | ||

+ | |||

+ | <math> {f_{emission}=f_{abs}} \,\!</math> | ||

+ | |||

+ | (Kirchoff’s Law)Note that Kirchoff’s Law is true regardless of equilibrium. | ||

+ | |||

+ | == Fundamental equation of transfer == | ||

+ | |||

+ | The fundamental equation of transfer is governed by emission and extinction. Extinction is brought about by absorption (which changes photon energy) or by scattering (which does not). Examples of scattering are Thomson scattering off of cold electrons, Rayleigh scattering in the atmosphere, and Line scattering (reemission in a different direction). An example of absorption is photoionization (where a photon ionizes an atom, say by knocking off an electron). | ||

+ | * Absorption:Let’s say radiation <math>I_\nu </math> passes through a region <math>ds</math> of absorption/scattering on its way to us. Then: | ||

+ | |||

+ | <math> dI_\nu =-\alpha _\nu I_\nu ds \,\!</math> | ||

+ | |||

+ | where <math>\alpha _\nu </math> is the ''extinction'' coefficient (units of <math>cm^{-1}</math>). We may compute <math>\alpha _\nu </math> a couple different ways: | ||

+ | |||

+ | <math> \alpha _\nu =\overbrace{n}^{\# \ density} \overbrace{\sigma _\nu }^{cross\ section} =\overbrace{\rho }^{mass\ density}\overbrace{\kappa _\nu }^{opacity} \,\!</math> | ||

+ | |||

+ | Solving for intensity: | ||

+ | |||

+ | <math> \begin{align} I_\nu (s)& =I_\nu (0)e^{-n\sigma _\nu s}\\ & =I_\nu (0)e^{-\tau _\nu }\\ \end{align} \,\!</math> | ||

+ | |||

+ | where <math>\tau _\nu </math> is the ''optical depth'' at <math>\nu </math>. Optical depth is often computed as: | ||

+ | |||

+ | <math> \tau _\nu =n\sigma _\nu s=N\sigma _\nu \,\!</math> | ||

+ | |||

+ | where <math>N</math>, the ''column density'', is in <math>cm^{-2}</math> and is the # of extinguishers per unit area. Similarly, | ||

+ | |||

+ | <math> \tau _\nu =\rho \kappa _\nu s=\Sigma \kappa _\nu \,\!</math> | ||

+ | |||

+ | where <math>\Sigma </math> is the mass surface density. | ||

+ | |||

+ | <math> \tau _\nu \begin{cases} \ll 1 & optically\ thin\\ \gg 1 & optically\ thick\end{cases} \,\!</math> | ||

+ | |||

+ | The ''Mean Free Path'' is given by: <math>\lambda _{mfp,\nu }=\alpha _\nu ^{-1}={1 \over n\sigma _ nu} ={1 \over \rho K_\nu }</math>. Thus: | ||

+ | |||

+ | <math> \tau _\nu ={s\over \lambda _{mfp,\nu }} \,\!</math> | ||

+ | |||

+ | That is, the optical depth is the number of mean-free-paths deep a medium is. For Poisson processes, the probability of absorption is given by: | ||

+ | |||

+ | <math> P(n)={e^{-{s\over \lambda _{mfp,\nu }}}\left({s\over \lambda _{mfp,\nu }}\right)^ n\over n!} \,\!</math> | ||

+ | |||

+ | Therefore: | ||

+ | |||

+ | <math> I_\nu (s)=I_\nu (0)e^{-\alpha _\nu s} \,\!</math> | ||

+ | |||

+ | * Emission:If <math>j_\nu </math> is the emissivity, then the contribution of the emissivity of a medium to the flux is: | ||

+ | |||

+ | <math> dI_\nu =j_\nu ds \,\!</math> | ||

+ | |||

+ | * Emission and Extinction together: | ||

+ | |||

+ | <math> {{dI_\nu \over ds}=j_\nu -\alpha _\nu I_\nu } \,\!</math> | ||

+ | |||

+ | (Fundamental Equation of Transfer)It is often convenient to express this in terms of optical depth. Dividing by <math>\alpha _\nu </math> and recognizing <math>d\tau _\nu = ds \alpha _\nu </math>: | ||

+ | |||

+ | <math> \begin{align} {dI\nu \over d\tau _\nu }& ={j_\nu \over \alpha _\nu }-I_\nu \\ & =S_\nu -I_\nu \\ \end{align} \,\!</math> | ||

+ | |||

+ | where <math>S_\nu </math> is a “source function”. In general, | ||

+ | |||

+ | <math> S_\nu \big |_{scattering}\propto \int {I_\nu d\Omega } \,\!</math> | ||

+ | |||

+ | There is a formal solution for <math>I_\nu </math>. Let’s define <math>\tilde I\equiv Ie^{\tau \nu }</math> and <math>\tilde S\equiv Se^{\tau \nu }</math>. Then: | ||

+ | |||

+ | <math> {d\tilde I\over d\tau _\nu }=\tilde S \,\!</math> | ||

+ | |||

+ | <math> \tilde I(\tau _\nu )=\tilde I(0)+\int _0^{\tau _\nu }{\tilde Sd\tilde\tau _\nu } \,\!</math> | ||

+ | |||

+ | <math> {I_\nu (\tau _\nu )=\overbrace{I_\nu (0)e^{-\tau _\nu }}^{atten\ bg\ light} +\overbrace{\int _0^{\tau _\nu }{S_\nu (\tau _\nu ^\prime ) \underbrace{e^{-(\tau _\nu -\tau _\nu ^\prime )}}_{self-absorption} d\tau _\nu ^\prime }}^{glowing\ medium}} \,\!</math> | ||

+ | |||

+ | If <math>S_\nu </math> is constant with <math>\tau _\nu </math>, then: | ||

+ | |||

+ | <math> I_\nu (\tau _\nu )=I_\nu (0)e^{-\tau _\nu }+S_\nu (1-e^{-\tau _\nu }) \,\!</math> | ||

+ | |||

+ | That second term on the righthand side can be approximated as <math>S_\nu \tau _\nu </math> for <math>\tau _\nu \ll 1</math>, since self-absorption is negligible. Similarly, for <math>\tau _\nu \gg 1</math>, it may be approximated as <math>S_\nu </math>. The source function <math>S_\nu </math> is everything. It has both the absorption and emission coefficients embedded in it. | ||

+ | |||

+ | = Lecture 3 = | ||

+ | |||

+ | For a sphere to be a black body, it requires that <math>f_{abs} = f_{emis} = 1</math>. However it can still be Planckean if <math>0<f_{abs}=f_{emis}\le 1</math>. | ||

+ | |||

+ | == Local Thermodynamic Equilibrium (LTE) == | ||

+ | |||

+ | Local Thermodynamic Equilibrium means | ||

+ | |||

+ | <math> S_\nu =B_\nu (T) \,\!</math> | ||

+ | |||

+ | where <math>T</math> is ''local''. Take a sphere at surface temp <math>T</math>, with some gas (matter) inside it. Allow to come to thermal equilibrium. Inside the sphere, you see a Planckean spectrum (<math>I_\nu = B_\nu </math>). Now suppose you take a box of gas (it is both emissive and absorbent) and shoot rays of photons through that box suffer some absorption: | ||

+ | |||

+ | <math> dI_\nu = -\alpha _\nu ds I_\nu \,\!</math> | ||

+ | |||

+ | But those photons also pick up some intensity: | ||

+ | |||

+ | <math> dI_\nu = +j_\nu ds \,\!</math> | ||

+ | |||

+ | But since the photons should not pick up energy going through the box (everything is at the same temperature), | ||

+ | |||

+ | <math> S_\nu \equiv {j_\nu \over \alpha _\nu }=I_\nu =B_\nu \,\!</math> | ||

+ | |||

+ | Now suppose you let all the photons out of this sphere. Those photons will no longer be in thermal equilibrium with the gas. Suppose we magically fix the temperature of the gas at this point. The source function will remain the same (it will still be the case that <math>S_\nu = B_\nu </math>) because the source function ''is a property of the matter alone'' (the absorptive and emissive properties of it). However, <math>I_\nu \ne B_\nu </math>.Suppose we look along a column of this gas. At each point, the source function will be: <math>S_\nu =B_\nu (T)</math>. Therefore, <math>I_\nu </math> along that column will be: | ||

+ | |||

+ | <math> I_\nu =\int _0^{T_\nu }{S_\nu (T_{\nu ^\prime })e^{-(T_{\nu ^\prime }-T_\nu )}dT_{\nu ^\prime }} \,\!</math> | ||

+ | |||

+ | == Einstein Coefficients == | ||

+ | |||

+ | Derivation identical to Rybicki’s. We should memorize these.These coefficients govern the interaction of radiation with discrete energy levels. Say we have 2 energy levels with a difference <math>\Delta E=h\nu _0</math>. There is some uncertainty associated with <math>\nu </math>, but we’ll say it’s small for now.There are 3 coefficients: | ||

+ | * <math>{A_{21}}</math> governs decay from 2 to 1, and is the transition probability per unit time. The probability of spontaneous de-excitation and release of photon is Poisson-distributed with mean rate <math>{A_{21}}</math>. So <math>{A_{21}}^{-1}</math> is the mean lifetime of the excited state. e.g. For <math>H_\alpha </math> (n=3 to n=2): <math>{A_{21}}\approx 10^9 s^{-1}</math>. | ||

+ | * <math>{B_{12}}</math> governs absorptions causing transitions <math>1\to 2</math>. The transition probability per unit time is <math>{B_{12}}J_\nu </math>, where <math>{B_{12}}</math> is the probability constant, and <math>J_\nu </math> is: | ||

+ | |||

+ | <math> J_\nu \equiv {\int {I_\nu d\Omega } \over 4\pi } \,\!</math> | ||

+ | |||

+ | It depends on <math>I_\nu </math> (the intensity), but it does not depend on direction, so we integrate over all angles. The <math>4\pi </math> is a normalization constant which makes <math>J_\nu </math> the mean intensity, instead of the total intensity. However, we have to remember that there are uncertainties in the energy-level separations. <math>\phi (\nu )\equiv </math> is called the line profile function. It describes some (maybe gaussian) distribution of absorption around <math>\nu _0</math> (the absorption frequency), and is subject to the requirement that: | ||

+ | |||

+ | <math> \int _0^\infty {\phi (\nu )d\nu }=1 \,\!</math> | ||

+ | |||

+ | Say that <math>\Delta \nu </math> is the width of the distribution around <math>\nu _0</math>. <math>\Delta \nu </math> is affected by many factors: <math>{A_{21}}</math> (the natural, uncertainty-based broadening of at atom in isolation), <math>\nu _0{V_ T\over c}</math> (the thermal, Doppler-based broadening), and <math>n_{coll}\sigma _{coll}v_{rel}</math> (collisional broadening, a.k.a. pressure broadening). So really, the transition probability per unit time is: | ||

+ | |||

+ | <math> R_{ex}^{-1}={B_{12}}\int _0^\infty {J_\nu \phi (\nu )d\nu }\approx {B_{12}}\bar J \,\!</math> | ||

+ | |||

+ | * <math>{B_{21}}</math> governs stimulated emission. In this example, we are in energy state 2, and an incoming photon causes a transition to energy level 1 and the emission of 2 photons. The transition per unit time is <math>{B_{21}}\bar J</math>. | ||

+ | |||

+ | == Einstein Relations among coefficients == | ||

+ | |||

+ | Assume we have many atoms with 2 energy states, and <math>n_1</math> is the # density in state 1, ditto for <math>n_2</math>. Assume we are in thermal, steady-state equilibrium, so: | ||

+ | |||

+ | <math> n_1{B_{12}}\bar J=n_2{A_{21}}+n_2{B_{21}}\bar J \,\!</math> | ||

+ | |||

+ | This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: <math>\bar J = {n_2{A_{21}}\over n_1{B_{12}}-n_2{B_{21}}}</math>. Using that <math>{n_2\over n_1}={g_2\over g_1}e^{{-h\nu _0 \over kT}}</math>: | ||

+ | |||

+ | <math> \bar J={{{A_{21}}\over {B_{21}}}\over {g_1{B_{12}}\over g_2{B_{21}}}e^{{-h\nu _0 \over kT}}-1} \,\!</math> | ||

+ | |||

+ | In thermal equilibrium <math>J_\nu =B_\nu </math>: | ||

+ | |||

+ | <math> \begin{align} \bar J& \equiv \int _0^\infty {J_\nu \phi (\nu )d\nu }\\ & =\int _0^\infty {B_\nu \phi (\nu )d\nu }\\ & \approx B_\nu (\nu _0)\\ & ={2h\nu _0^3\over c^2(e^{{-h\nu _0 \over kT}}-1)}\\ \end{align} \,\!</math> | ||

+ | |||

+ | Combining this with <math>\bar J</math> earlier, we get: | ||

+ | |||

+ | <math> {g_1{B_{12}}=g_2{B_{21}}} \,\!</math> | ||

+ | |||

+ | and | ||

+ | |||

+ | <math> {{A_{21}}\over {B_{21}}}={2h\nu ^3\over c^2} \,\!</math> | ||

+ | |||

+ | == Rewriting <math>j_\nu , \alpha _\nu </math> in terms of Einstein coeffs == | ||

+ | |||

+ | In a small volume <math>dV</math>: | ||

+ | |||

+ | <math> \begin{align} j_\nu & \equiv {dE\over dt\, dV\, d\nu \, d\Omega }\\ & ={h\nu _0{A_{21}}n_2\phi (\nu )\over 4\pi }\\ \end{align} \,\!</math> | ||

+ | |||

+ | We can express <math>\alpha _\nu </math> in terms of the Einstein coefficients. The excitation probability per time is <math>n_1B_{12}\bar J</math>, and the energy lost in crossing the small volume <math>\propto n_1{B_{12}}{I_\nu d\Omega \over 4\pi }\phi (\nu )d\nu </math> (it is the probability per time per volume of going <math>1\to 2</math> by absorbing <math>I_\nu </math> from a cone of solid angle <math>d\Omega </math> and frequency range <math>[\nu ,\nu +d\nu ]</math>). Thus, the energy is given by: | ||

− | + | <math> \begin{align} E& =n_1{B_{12}}{I_\nu d\Omega \over 4\pi }\phi (\nu )d\nu h\nu dt\, dV\\ & =\alpha _\nu I_\nu ds\, dt\, d\Omega \, dA\, d\nu \\ \end{align} \,\!</math> | |

− | <math> | + | Recognizing that <math>dV=dA\, ds</math>: |

− | + | <math> \alpha _\nu ={n_1{B_{12}}\phi (\nu )\over 4\pi }h\nu \,\!</math> | |

− | + | Correcting for stimulated emission, we get: | |

− | + | <math> {\alpha _\nu ={(n_1{B_{12}}-n_2{B_{21}})\phi (\nu )h\nu \over 4\pi }} \,\!</math> | |

− |

## Revision as of 15:53, 10 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 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*:

(The Planck Function for Black Body Radiation)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 :

Wein tailIf :

Rayleigh-Jeans tail Note that this tail peaks at . Also,

# Lecture 2

## Reiteration: Conservation of specific intensity

Conservation of specific intensity told us the intensity collected by your telescope is absolutely equal to the surface intensity emitted into the angle leftrightarrowonding to your pixel on the sky. *Specific intensity is distance independent*. of sky emitted into angle of one square degree is equal to the specific intensity measured by a telescope pointed at a square-degree pixel of sky.

## Units

We have flux, flux density, surface brightness, and spectral energy distribution to describe luminosity. When in doubt, look at the units. Surface brightness, like specific intensity, is distance independent. It is simply specific intensity integrated over frequency.

## Lenses

If you look at something through a lens, the specific intensity of a point is conserved. In the lens, the specific intensity may be greater, because is conserved (remember that is the index of refraction of the medium), but once the light goes out again, even if the object looks bigger, the specific intensity is the same. *This does not violate conservation of energy because the light is being focused to a smaller area.*Now if something is *inside* the lens (say, a bowl of water), then the lens bends light rays closer to each other, increasing the specific intensity. Thus, the object is emitting/scattering higher density photons, and appears more luminous, and the object also appears larger. This means that there must be some places where the object, omitting the lens, would have been visible, but now isn’t.

## Blackbodies

The CMB is the perfect blackbody because in the past, the surface of the black body (the edge of the universe) was not allowed to leak out energy. This begs the question: does it matter what the boundary of the black body is made of? The answer is: No. In the short term, the boundary may imperfectly reflect a photon, but then the resulting change in the temperature of boundary dictates that it will tend to reemit the energy into the photon gas. Overall:

However, , so:

(Kirchoff’s Law)Note that Kirchoff’s Law is true regardless of equilibrium.

## Fundamental equation of transfer

The fundamental equation of transfer is governed by emission and extinction. Extinction is brought about by absorption (which changes photon energy) or by scattering (which does not). Examples of scattering are Thomson scattering off of cold electrons, Rayleigh scattering in the atmosphere, and Line scattering (reemission in a different direction). An example of absorption is photoionization (where a photon ionizes an atom, say by knocking off an electron).

- Absorption:Let’s say radiation passes through a region of absorption/scattering on its way to us. Then:

where is the *extinction* coefficient (units of ). We may compute a couple different ways:

Solving for intensity:

where is the *optical depth* at . Optical depth is often computed as:

where , the *column density*, is in and is the # of extinguishers per unit area. Similarly,

where is the mass surface density.

The *Mean Free Path* is given by: . Thus:

That is, the optical depth is the number of mean-free-paths deep a medium is. For Poisson processes, the probability of absorption is given by:

Therefore:

- Emission:If is the emissivity, then the contribution of the emissivity of a medium to the flux is:

- Emission and Extinction together:

(Fundamental Equation of Transfer)It is often convenient to express this in terms of optical depth. Dividing by and recognizing :

where is a “source function”. In general,

There is a formal solution for . Let’s define and . Then:

If is constant with , then:

That second term on the righthand side can be approximated as for , since self-absorption is negligible. Similarly, for , it may be approximated as . The source function is everything. It has both the absorption and emission coefficients embedded in it.

# Lecture 3

For a sphere to be a black body, it requires that . However it can still be Planckean if .

## Local Thermodynamic Equilibrium (LTE)

Local Thermodynamic Equilibrium means

where is *local*. Take a sphere at surface temp , with some gas (matter) inside it. Allow to come to thermal equilibrium. Inside the sphere, you see a Planckean spectrum (). Now suppose you take a box of gas (it is both emissive and absorbent) and shoot rays of photons through that box suffer some absorption:

But those photons also pick up some intensity:

But since the photons should not pick up energy going through the box (everything is at the same temperature),

Now suppose you let all the photons out of this sphere. Those photons will no longer be in thermal equilibrium with the gas. Suppose we magically fix the temperature of the gas at this point. The source function will remain the same (it will still be the case that ) because the source function *is a property of the matter alone* (the absorptive and emissive properties of it). However, .Suppose we look along a column of this gas. At each point, the source function will be: . Therefore, along that column will be:

## Einstein Coefficients

Derivation identical to Rybicki’s. We should memorize these.These coefficients govern the interaction of radiation with discrete energy levels. Say we have 2 energy levels with a difference . There is some uncertainty associated with , but we’ll say it’s small for now.There are 3 coefficients:

- governs decay from 2 to 1, and is the transition probability per unit time. The probability of spontaneous de-excitation and release of photon is Poisson-distributed with mean rate . So is the mean lifetime of the excited state. e.g. For (n=3 to n=2): .
- governs absorptions causing transitions . The transition probability per unit time is , where is the probability constant, and is:

It depends on (the intensity), but it does not depend on direction, so we integrate over all angles. The is a normalization constant which makes the mean intensity, instead of the total intensity. However, we have to remember that there are uncertainties in the energy-level separations. is called the line profile function. It describes some (maybe gaussian) distribution of absorption around (the absorption frequency), and is subject to the requirement that:

Say that is the width of the distribution around . is affected by many factors: (the natural, uncertainty-based broadening of at atom in isolation), (the thermal, Doppler-based broadening), and (collisional broadening, a.k.a. pressure broadening). So really, the transition probability per unit time is:

- governs stimulated emission. In this example, we are in energy state 2, and an incoming photon causes a transition to energy level 1 and the emission of 2 photons. The transition per unit time is .

## Einstein Relations among coefficients

Assume we have many atoms with 2 energy states, and is the # density in state 1, ditto for . Assume we are in thermal, steady-state equilibrium, so:

This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: . Using that :

In thermal equilibrium :

Combining this with earlier, we get:

and

## Rewriting in terms of Einstein coeffs

In a small volume :

We can express in terms of the Einstein coefficients. The excitation probability per time is , and the energy lost in crossing the small volume (it is the probability per time per volume of going by absorbing from a cone of solid angle and frequency range ). Thus, the energy is given by:

Recognizing that :

Correcting for stimulated emission, we get: