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

$I_{\nu }I_{emitted}=I_{\nu }I_{absorbed}=B_{\nu }f_{abs}\,\!$ However, $f_{emission}B_{\nu }\equiv I_{\nu }I_{emitted}$ , so:

${f_{emission}=f_{abs}}\,\!$ (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 $I_{\nu }$ passes through a region $ds$ of absorption/scattering on its way to us. Then:

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

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

{\begin{aligned}I_{\nu }(s)&=I_{\nu }(0)e^{-n\sigma _{\nu }s}\\&=I_{\nu }(0)e^{-\tau _{\nu }}\\\end{aligned}}\,\! where $\tau _{\nu }$ is the optical depth at $\nu$ . Optical depth is often computed as:

$\tau _{\nu }=n\sigma _{\nu }s=N\sigma _{\nu }\,\!$ where $N$ , the column density, is in $cm^{-2}$ and is the # of extinguishers per unit area. Similarly,

$\tau _{\nu }=\rho \kappa _{\nu }s=\Sigma \kappa _{\nu }\,\!$ where $\Sigma$ is the mass surface density.

$\tau _{\nu }{\begin{cases}\ll 1&optically\ thin\\\gg 1&optically\ thick\end{cases}}\,\!$ The Mean Free Path is given by: $\lambda _{mfp,\nu }=\alpha _{\nu }^{-1}={\frac {1}{n\sigma _{n}u}}={\frac {1}{\rho K_{\nu }}}$ . Thus:

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

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

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

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

$dI_{\nu }=j_{\nu }ds\,\!$ • Emission and Extinction together:
${{dI_{\nu } \over ds}=j_{\nu }-\alpha _{\nu }I_{\nu }}\,\!$ (Fundamental Equation of Transfer)

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

{\begin{aligned}{dI\nu \over d\tau _{\nu }}&={j_{\nu } \over \alpha _{\nu }}-I_{\nu }\\&=S_{\nu }-I_{\nu }\\\end{aligned}}\,\! where $S_{\nu }$ is a “source function”. In general,

$S_{\nu }{\big |}_{scattering}\propto \int {I_{\nu }d\Omega }\,\!$ There is a formal solution for $I_{\nu }$ . Let’s define ${\tilde {I}}\equiv Ie^{\tau \nu }$ and ${\tilde {S}}\equiv Se^{\tau \nu }$ . Then:

${d{\tilde {I}} \over d\tau _{\nu }}={\tilde {S}}\,\!$ ${\tilde {I}}(\tau _{\nu })={\tilde {I}}(0)+\int _{0}^{\tau _{\nu }}{{\tilde {S}}d{\tilde {\tau }}_{\nu }}\,\!$ ${I_{\nu }(\tau _{\nu })=\overbrace {I_{\nu }(0)e^{-\tau _{\nu }}} ^{atten\ bg\ light}+\overbrace {\int _{0}^{\tau _{\nu }}{S_{\nu }(\tau _{\nu }^{\prime })\underbrace {e^{-(\tau _{\nu }-\tau _{\nu }^{\prime })}} _{self-absorption}d\tau _{\nu }^{\prime }}} ^{glowing\ medium}}\,\!$ If $S_{\nu }$ is constant with $\tau _{\nu }$ , then:

$I_{\nu }(\tau _{\nu })=I_{\nu }(0)e^{-\tau _{\nu }}+S_{\nu }(1-e^{-\tau _{\nu }})\,\!$ That second term on the righthand side can be approximated as $S_{\nu }\tau _{\nu }$ for $\tau _{\nu }\ll 1$ , since self-absorption is negligible. Similarly, for $\tau _{\nu }\gg 1$ , it may be approximated as $S_{\nu }$ . The source function $S_{\nu }$ is everything. It has both the absorption and emission coefficients embedded in it.