Astronomical Techniques

Cheat sheet by Lukas Wenzl

Based on an Astronomical Techniques lecture by Prof. Gordon Stacey and Prof. Jim Cordes in 2019. All mistakes were added by me.


$c = \lambda \nu$; $E_{\gamma} = h \nu = \hbar \omega$; $p_\gamma = h/\lambda$


$$I_\nu \displaystyle \begin{bmatrix} \frac{\text{Energy}}{\text{Time Area }\Delta\Omega \, \Delta \nu} \end{bmatrix} $$

Planck Spectrum

$$\displaystyle B_\nu = 2 \frac{h \nu^3}{c^2} \frac{1}{\mathrm{exp}{\frac{h\nu} {k_B T}} -1}$$

Intensity, 2 degrees of freedom for polarization, Quantum State Density, Photon Energy, Bose-Einstein Statistics

For low frequencies one gets the Rayleigh-Jeans Limit (Taylor expansion): $B_\nu \approx \frac{2 \nu^2} {c^2}kT$, For high frequencies you get the Wien Limit $B_\nu \approx \frac{2 \nu^2} {c^2} e^{-h\nu \beta}$, Maximum: Wiensches Verschiebungsgesetz: $\nu_{max} \approx 3 \frac {kT} h$; In wavelength: $B_\lambda = \frac{2hc^2}{\lambda^5}\frac{1}{exp\frac{hc}{\lambda k_B T}-1} $

from astropy.modeling import blackbody
T_CMB = 2.735#K
blackbody.blackbody_nu(1e12, T_dust)#frequ in Hz


$F_\nu \displaystyle \begin{bmatrix} \frac{\text{Energy}}{\text{Time Area }\Delta \nu} \end{bmatrix} $, unit: $ 1 \text{ Jy} \equiv 10^{-23} \frac{erg}{s\, cm^2 \, Hz} $

$F_\nu = \Omega I_\nu $

$F = \int F_\lambda d\lambda = \int F_\nu d\nu$, but $F_\lambda \neq F_\nu$, only $d\lambda \, F_\lambda = d\nu\, F_\nu$

On the surface of a light source with Temperature $T_*$

$$F = \sigma_B T^4 $$

Observed Flux from a distance d of an Object with Radius $R$ and Surface Temperature $T$

$$\displaystyle F_{obs} = \Big(\frac{R}{d}\Big)^2 \sigma T^4 $$


$$P_\nu \displaystyle \begin{bmatrix} \frac{\text{Energy}}{\text{Time }\Delta \nu } \end{bmatrix} = \int F_\nu dA $$

$P \approx I_\nu \, \Delta \nu \, \Omega \, A $


The Luminosity of an Object with Radius $R$ and Surface Temperature $T$ is

$$L = 4\pi R^2F = 4\pi R^2 \sigma_B T^4$$


kind of insane, but convention. basically a log of the flux. Definition ”apparent Magnitude” in Band x

$$m = -2.5 log_{10} \frac{F_{x,obs}}{F_{x,ref}}$$

with $F_{bla} = \int d\lambda f_{bla}(\lambda) g_x(\lambda)$; f flux per wavelength, g band function (fraction of ENERGY throughput, sometimes fraction of photons)

different References:

AB (most common)
just assume: $f_{ref}(\nu) = \text{const} = 3.631e-23 \frac{J}{s m^2 Hz}$

measure vega flux in Band x for $f_{ref}$. Gets funny in southern sky

Spectral Index

Dependence of Flux on frequency, denoted $\alpha$
$F_\nu \propto \nu^\alpha$

-0.1 to 2 in radio -> thermal source
<< 0 in radio -> synchrotron emission


Solid Angles

The whole sky is $\Omega = 4\pi$. For small circles with radius $\theta$ we have $\Omega = \pi \theta^2$. Everything is in sr, which I imagine as rad$^2$. An object with diameter d and distance R appears under the angle $\theta \approx \frac{d}{R}$ and therefore has solid angle $\Omega = \frac{\pi d^2}{R^2} = \frac{A}{R^2}$.


Described mainly by total focal length f and diameter D of main mirror.

F number $f^{\textrm{#}} \equiv \frac{f}{D}$ describes the opening angle in the beam, it is really the only info you need about light reaching an element you are investigating

Diffraction limit $\theta_{diff} \equiv \frac{1.22 \, \lambda}{D}$, this is for first null, if you want FWHM use 1.03, plate scale $x \equiv \theta f$, Nyquist sampling gives factor 1/2

Magnification $M \equiv f_{prim}/f_{eye/eyepiece}$

Thin lens formula $\frac{1}{q}-\frac{1}{p}=\frac{1}{f}$, q distance to object, p distance to picture, f>0 for convex lens (normal, ball shaped), other: convex; (maybe add more details)

Etendue is constant: $\Omega A = \text{ const}$ inside telescope; plate scale $p \equiv \frac{206265}{f(mm)} [\frac{”}{mm}] = \frac{1}{f(mm)} [\frac{rad}{mm}]$


Fabry Perot

Finesse: $F = \frac{\Delta\lambda}{\delta\lambda_{FWHM}}$ transmission peak separation over width. Get rid of everything but light in $\delta \lambda_{FWHM}$, tune so that this is your resolving power ($RP = \frac{\lambda}{\delta \lambda_{FWHM}}$). Can put them back to back, useful since F is usually only 50.

Fourier Transformation


Resolution $R=Nn$, $N\equiv G/d$ Grating size/grove separation. This is for diffraction. If limited by slit we get a factor $\theta_{diffraction}/\theta_{slit}$, Notice that the height difference of peaks needs to be $h =0.5\lambda n$ and $d=h/sin(\alpha)$ where $\alpha$ is the blazing angle.

We can rewrite these equations as:

Chromatic resolution (slit <= diffraction spot, i.e. diffraction limited): $ R = \frac{2 G_{proj} tan(\alpha)}{\lambda}$, with $G_{proj} \equiv G cos(\alpha)$ and $\delta \lambda = \frac{\lambda d}{G n} $

Spectral purity (slit >= diffraction spot): $R = \frac{2 G_{proj} tan(\alpha)}{\theta_{beam}D_{tel}}$, with $\Delta \lambda = – cos(\alpha) \frac{W’ d}{f_{coll}n}$


Resolution $R = \lambda/W_{\lambda}$ with $W_{\lambda}=$ limiting angular resolution ( e.g. $\lambda/$beam size, or seeing ) * $d\lambda/d\theta$. Typically in optical 15000, strong wavelength dependence.

Radio Astronomy

$T_B (\nu) \approx \frac{B_\nu \lambda^2}{2k_B}$ note alternative forms: $=\frac{F_\nu \lambda^2}{2k_B \Omega} = \frac{B_\lambda \lambda^4}{2k_B c}$ only works if the source is extended over diffraction spot i think

Everything is a temperature in the Rayleigh-Jeans limit. The different temperatures from the different sources add.

$P_\nu = kT \Delta \nu$

Radiometer Equation

$\frac{S}{N} = \frac{T_{src}}{T_{noise}}\sqrt{\tau \Delta \nu}$

lecture version: $\frac{\sigma_T}{T_{sys}} = \frac{1}{\sqrt{\Delta \nu \tau}}$, it is the same thing, the signal here is the temperature of the source given by $T_{source} = G F$ , with Gain $G\equiv \frac{A_e}{2k}$

Radiative Transfer

For a cloud with optical depth $\tau$ and Temperature $T_{cloud}$ the observed Temperature $T_{obs}$ is

$$T_{obs} = T_{em} e^{-\tau} + T_{cloud} (1-e^{-\tau})$$

Where we also included a background source with $T_{em}$


Wiener-Khinchen Theorem

V(t) & \overset{F.T.}{\longleftrightarrow} & V'(\nu)\\
\downarrow & & \downarrow |.|^2\\
R(\tau) = \langle V(t) V^*(t+\tau)\rangle & \overset{F.T.}{\longleftrightarrow} & S(\nu) \end{matrix}$$

Where the bottom left is the autocorelation, V is the voltage, i.e. the electric field and S is the final spectrum.


Van Cittert–Zernike theorem

E(\vec x, t) & \overset{F.T.}{\longleftrightarrow} & E'(\vec k, \omega) = \int dx \int dt E(\vec x, t) e^{i(\vec k \vec x – \omega t)}\\
\downarrow & & \downarrow |.|^2\\
\Gamma(\delta \vec x,\tau) = \langle E(\vec x, t) E^*(\vec x + \delta \vec x, t+\tau)\rangle & \overset{F.T.}{\longleftrightarrow} & I(\vec \theta) \end{matrix}$$

Same as spectrum, just now also including spacial separation between dishes. Define baseline $b \equiv \delta \vec x$ then:

$$ I (\vec \theta)$ = \int d\vec b \Gamma(b) e^{i \vec k \cdot \vec b}$$

For N antennas, # pairs $\frac{N(N-1)}{2} $

use earth rotation, in the end still artifacts: use deconvolution (clean algorithm)