# 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 (${\displaystyle \Delta \nu }$), but ${\displaystyle \Delta \lambda \neq {c \over \Delta \nu }}$.

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

${\displaystyle I_{\nu }}$ 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, ${\displaystyle {I_{\nu } \over \nu ^{3}}}$ is. Also, for redshift ${\displaystyle z}$,

${\displaystyle I_{\nu }\propto \nu ^{3}\propto {1 \over (1+z)^{3}}\,\!}$

so intensity decreases with redshift. Finally:

${\displaystyle {I_{\nu } \over \eta ^{2}}\,\!}$

is conserved along a ray, where ${\displaystyle \eta }$ 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:

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

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

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

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

However,

${\displaystyle h\nu {n_{\nu }c \over 4\pi }=I_{\nu }=B_{\nu }\,\!}$

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

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

Wein tailIf ${\displaystyle h\nu \ll kT}$:

${\displaystyle B_{\nu }\approx {2kT \over \lambda ^{2}}\,\!}$

Rayleigh-Jeans tail Note that this tail peaks at ${\displaystyle \sim {3kT \over h}}$. Also,

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

# 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. ${\displaystyle 1m^{2}}$ of sky emitted into angle of one square degree is equal to the specific intensity measured by a ${\displaystyle 1m^{2}}$ 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 ${\displaystyle I_{\nu } \over \eta ^{2}}$ is conserved (remember that ${\displaystyle \eta }$ 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:

${\displaystyle I_{\nu }I_{emitted}=I_{\nu }I_{absorbed}=B_{\nu }f_{abs}\,\!}$

However, ${\displaystyle f_{emission}B_{\nu }\equiv I_{\nu }I_{emitted}}$, so:

${\displaystyle {f_{emission}=f_{abs}}\,\!}$

(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 ${\displaystyle I_{\nu }}$ passes through a region ${\displaystyle ds}$ of absorption/scattering on its way to us. Then:

${\displaystyle dI_{\nu }=-\alpha _{\nu }I_{\nu }ds\,\!}$

where ${\displaystyle \alpha _{\nu }}$ is the extinction coefficient (units of ${\displaystyle cm^{-1}}$). We may compute ${\displaystyle \alpha _{\nu }}$ a couple different ways:

${\displaystyle \alpha _{\nu }=\overbrace {n} ^{\#\ density}\overbrace {\sigma _{\nu }} ^{cross\ section}=\overbrace {\rho } ^{mass\ density}\overbrace {\kappa _{\nu }} ^{opacity}\,\!}$

Solving for intensity:

{\displaystyle {\begin{aligned}I_{\nu }(s)&=I_{\nu }(0)e^{-n\sigma _{\nu }s}\\&=I_{\nu }(0)e^{-\tau _{\nu }}\\\end{aligned}}\,\!}

where ${\displaystyle \tau _{\nu }}$ is the optical depth at ${\displaystyle \nu }$. Optical depth is often computed as:

${\displaystyle \tau _{\nu }=n\sigma _{\nu }s=N\sigma _{\nu }\,\!}$

where ${\displaystyle N}$, the column density, is in ${\displaystyle cm^{-2}}$ and is the # of extinguishers per unit area. Similarly,

${\displaystyle \tau _{\nu }=\rho \kappa _{\nu }s=\Sigma \kappa _{\nu }\,\!}$

where ${\displaystyle \Sigma }$ is the mass surface density.

${\displaystyle \tau _{\nu }{\begin{cases}\ll 1&optically\ thin\\\gg 1&optically\ thick\end{cases}}\,\!}$

The Mean Free Path is given by: ${\displaystyle \lambda _{mfp,\nu }=\alpha _{\nu }^{-1}={1 \over n\sigma _{n}u}={1 \over \rho K_{\nu }}}$. Thus:

${\displaystyle \tau _{\nu }={s \over \lambda _{mfp,\nu }}\,\!}$

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:

${\displaystyle P(n)={e^{-{s \over \lambda _{mfp,\nu }}}\left({s \over \lambda _{mfp,\nu }}\right)^{n} \over n!}\,\!}$

Therefore:

${\displaystyle I_{\nu }(s)=I_{\nu }(0)e^{-\alpha _{\nu }s}\,\!}$

• Emission:If ${\displaystyle j_{\nu }}$ is the emissivity, then the contribution of the emissivity of a medium to the flux is:

${\displaystyle dI_{\nu }=j_{\nu }ds\,\!}$

• Emission and Extinction together:

${\displaystyle {{dI_{\nu } \over ds}=j_{\nu }-\alpha _{\nu }I_{\nu }}\,\!}$

(Fundamental Equation of Transfer)It is often convenient to express this in terms of optical depth. Dividing by ${\displaystyle \alpha _{\nu }}$ and recognizing ${\displaystyle d\tau _{\nu }=ds\alpha _{\nu }}$:

{\displaystyle {\begin{aligned}{dI\nu \over d\tau _{\nu }}&={j_{\nu } \over \alpha _{\nu }}-I_{\nu }\\&=S_{\nu }-I_{\nu }\\\end{aligned}}\,\!}

where ${\displaystyle S_{\nu }}$ is a “source function”. In general,

${\displaystyle S_{\nu }{\big |}_{scattering}\propto \int {I_{\nu }d\Omega }\,\!}$

There is a formal solution for ${\displaystyle I_{\nu }}$. Let’s define ${\displaystyle {\tilde {I}}\equiv Ie^{\tau \nu }}$ and ${\displaystyle {\tilde {S}}\equiv Se^{\tau \nu }}$. Then:

${\displaystyle {d{\tilde {I}} \over d\tau _{\nu }}={\tilde {S}}\,\!}$

${\displaystyle {\tilde {I}}(\tau _{\nu })={\tilde {I}}(0)+\int _{0}^{\tau _{\nu }}{{\tilde {S}}d{\tilde {\tau }}_{\nu }}\,\!}$

${\displaystyle {I_{\nu }(\tau _{\nu })=\overbrace {I_{\nu }(0)e^{-\tau _{\nu }}} ^{atten\ bg\ light}+\overbrace {\int _{0}^{\tau _{\nu }}{S_{\nu }(\tau _{\nu }^{\prime })\underbrace {e^{-(\tau _{\nu }-\tau _{\nu }^{\prime })}} _{self-absorption}d\tau _{\nu }^{\prime }}} ^{glowing\ medium}}\,\!}$

If ${\displaystyle S_{\nu }}$ is constant with ${\displaystyle \tau _{\nu }}$, then:

${\displaystyle I_{\nu }(\tau _{\nu })=I_{\nu }(0)e^{-\tau _{\nu }}+S_{\nu }(1-e^{-\tau _{\nu }})\,\!}$

That second term on the righthand side can be approximated as ${\displaystyle S_{\nu }\tau _{\nu }}$ for ${\displaystyle \tau _{\nu }\ll 1}$, since self-absorption is negligible. Similarly, for ${\displaystyle \tau _{\nu }\gg 1}$, it may be approximated as ${\displaystyle S_{\nu }}$. The source function ${\displaystyle S_{\nu }}$ 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 ${\displaystyle f_{abs}=f_{emis}=1}$. However it can still be Planckean if ${\displaystyle 0.

## Local Thermodynamic Equilibrium (LTE)

Local Thermodynamic Equilibrium means

${\displaystyle S_{\nu }=B_{\nu }(T)\,\!}$

where ${\displaystyle T}$ is local. Take a sphere at surface temp ${\displaystyle T}$, with some gas (matter) inside it. Allow to come to thermal equilibrium. Inside the sphere, you see a Planckean spectrum (${\displaystyle I_{\nu }=B_{\nu }}$). 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:

${\displaystyle dI_{\nu }=-\alpha _{\nu }dsI_{\nu }\,\!}$

But those photons also pick up some intensity:

${\displaystyle dI_{\nu }=+j_{\nu }ds\,\!}$

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

${\displaystyle S_{\nu }\equiv {j_{\nu } \over \alpha _{\nu }}=I_{\nu }=B_{\nu }\,\!}$

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 ${\displaystyle S_{\nu }=B_{\nu }}$) because the source function is a property of the matter alone (the absorptive and emissive properties of it). However, ${\displaystyle I_{\nu }\neq B_{\nu }}$.Suppose we look along a column of this gas. At each point, the source function will be: ${\displaystyle S_{\nu }=B_{\nu }(T)}$. Therefore, ${\displaystyle I_{\nu }}$ along that column will be:

${\displaystyle I_{\nu }=\int _{0}^{T_{\nu }}{S_{\nu }(T_{\nu ^{\prime }})e^{-(T_{\nu ^{\prime }}-T_{\nu })}dT_{\nu ^{\prime }}}\,\!}$

## 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 ${\displaystyle \Delta E=h\nu _{0}}$. There is some uncertainty associated with ${\displaystyle \nu }$, but we’ll say it’s small for now.There are 3 coefficients:

• ${\displaystyle {A_{21}}}$ 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 ${\displaystyle {A_{21}}}$. So ${\displaystyle {A_{21}}^{-1}}$ is the mean lifetime of the excited state. e.g. For ${\displaystyle H_{\alpha }}$ (n=3 to n=2): ${\displaystyle {A_{21}}\approx 10^{9}s^{-1}}$.
• ${\displaystyle {B_{12}}}$ governs absorptions causing transitions ${\displaystyle 1\to 2}$. The transition probability per unit time is ${\displaystyle {B_{12}}J_{\nu }}$, where ${\displaystyle {B_{12}}}$ is the probability constant, and ${\displaystyle J_{\nu }}$ is:

${\displaystyle J_{\nu }\equiv {\int {I_{\nu }d\Omega } \over 4\pi }\,\!}$

It depends on ${\displaystyle I_{\nu }}$ (the intensity), but it does not depend on direction, so we integrate over all angles. The ${\displaystyle 4\pi }$ is a normalization constant which makes ${\displaystyle J_{\nu }}$ the mean intensity, instead of the total intensity. However, we have to remember that there are uncertainties in the energy-level separations. ${\displaystyle \phi (\nu )\equiv }$ is called the line profile function. It describes some (maybe gaussian) distribution of absorption around ${\displaystyle \nu _{0}}$ (the absorption frequency), and is subject to the requirement that:

${\displaystyle \int _{0}^{\infty }{\phi (\nu )d\nu }=1\,\!}$

Say that ${\displaystyle \Delta \nu }$ is the width of the distribution around ${\displaystyle \nu _{0}}$. ${\displaystyle \Delta \nu }$ is affected by many factors: ${\displaystyle {A_{21}}}$ (the natural, uncertainty-based broadening of at atom in isolation), ${\displaystyle \nu _{0}{V_{T} \over c}}$ (the thermal, Doppler-based broadening), and ${\displaystyle n_{coll}\sigma _{coll}v_{rel}}$ (collisional broadening, a.k.a. pressure broadening). So really, the transition probability per unit time is:

${\displaystyle R_{ex}^{-1}={B_{12}}\int _{0}^{\infty }{J_{\nu }\phi (\nu )d\nu }\approx {B_{12}}{\bar {J}}\,\!}$

• ${\displaystyle {B_{21}}}$ 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 ${\displaystyle {B_{21}}{\bar {J}}}$.

## Einstein Relations among coefficients

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

${\displaystyle n_{1}{B_{12}}{\bar {J}}=n_{2}{A_{21}}+n_{2}{B_{21}}{\bar {J}}\,\!}$

This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: ${\displaystyle {\bar {J}}={n_{2}{A_{21}} \over n_{1}{B_{12}}-n_{2}{B_{21}}}}$. Using that ${\displaystyle {n_{2} \over n_{1}}={g_{2} \over g_{1}}e^{-h\nu _{0} \over kT}}$:

${\displaystyle {\bar {J}}={{{A_{21}} \over {B_{21}}} \over {g_{1}{B_{12}} \over g_{2}{B_{21}}}e^{-h\nu _{0} \over kT}-1}\,\!}$

In thermal equilibrium ${\displaystyle J_{\nu }=B_{\nu }}$:

{\displaystyle {\begin{aligned}{\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{aligned}}\,\!}

Combining this with ${\displaystyle {\bar {J}}}$ earlier, we get:

${\displaystyle {g_{1}{B_{12}}=g_{2}{B_{21}}}\,\!}$

and

${\displaystyle {{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}\,\!}$

## Rewriting ${\displaystyle j_{\nu },\alpha _{\nu }}$ in terms of Einstein coeffs

In a small volume ${\displaystyle dV}$:

{\displaystyle {\begin{aligned}j_{\nu }&\equiv {dE \over dt\,dV\,d\nu \,d\Omega }\\&={h\nu _{0}{A_{21}}n_{2}\phi (\nu ) \over 4\pi }\\\end{aligned}}\,\!}

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

{\displaystyle {\begin{aligned}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{aligned}}\,\!}

Recognizing that ${\displaystyle dV=dA\,ds}$:

${\displaystyle \alpha _{\nu }={n_{1}{B_{12}}\phi (\nu ) \over 4\pi }h\nu \,\!}$

Correcting for stimulated emission, we get:

${\displaystyle {\alpha _{\nu }={(n_{1}{B_{12}}-n_{2}{B_{21}})\phi (\nu )h\nu \over 4\pi }}\,\!}$