# Making Spectral Energy Distributions (SEDs)

SEDs are energy plotted against some measure of the photon -- frequency or wavelength. The reason astronomers do this is to see how much energy is produced by the object as a function of frequency or wavelength. Now it's really going to get a little hairy! Steel your nerves and plunge onwards... it really all comes down to unit conversion.

The units that are used in Spitzer and Herschel data are Janskys. 1 Jy = ${\displaystyle 10^{-23}}$ erg/s/cm^2/Hz (in cgs units rather than mks units, sorry -- and just to be clear, I mean erg / (s*cm^2*Hz), it's just easier to read in the way I wrote it above). A Jansky is technically a unit of "flux density," represented by ${\displaystyle F_{\nu }}$. We want to plot "energy density", so one way to do this (DON'T DO THIS TO YOUR DATA YET) is to get rid of the "per Hz", e.g., multiply the Jy by the frequency (${\displaystyle \nu }$) of the bandpass center, or ${\displaystyle \nu F_{\nu }}$. You could just stop here, and plot ${\displaystyle \nu F_{\nu }}$ vs. ${\displaystyle \nu }$ to get something that is technically a spectral energy distribution. BUT, now it starts to get a little hairy, because there are some cultural influences here. Do you know off the top of your head what the frequency of the IRAC-1 band is? I don't either, but I do know its wavelength -- 3.6 microns. Astronomers coming from the longer wavelengths (mm, radio, etc.), because they think in units of frequencies, will tend to plot up ${\displaystyle \nu F_{\nu }}$ against ${\displaystyle \nu }$ (nu). The units of ${\displaystyle F_{\nu }}$ really are Janskys. BUT, astronomers coming from the shorter wavelengths (optical, etc.), because they think in units of wavelengths, will tend to plot instead ${\displaystyle \lambda F_{\lambda }}$ , where ${\displaystyle \lambda }$ (lambda) is the wavelength of the light. This is what we want to do here (because we have been thinking of the wavelengths of the Spitzer bands but not the frequencies). The catch here is that the units of ${\displaystyle F_{\lambda }}$ are NOT Janskys.

${\displaystyle \lambda \times \nu =c}$, the speed of light. In order to convert the Janskys into units of ${\displaystyle F_{\lambda }}$, you need to take into account the differentials (ah-HA, calculus being used here!), e.g., the fact that

   ${\displaystyle {\frac {dF}{d\lambda }}={\frac {dF}{d\nu }}{\frac {d\nu }{d\lambda }}}$ and ${\displaystyle d\nu ={\frac {c}{\lambda ^{2}}}d\lambda }$


So you need to multiply the ${\displaystyle F_{\nu }}$ by ${\displaystyle c/\lambda ^{2}}$ to convert it into ${\displaystyle F_{\lambda }}$. But we are not done yet! Recalling from above, the units of ${\displaystyle F_{\lambda }}$ are not an energy density. You need to get another factor of ${\displaystyle \lambda }$ in there to make the units work out to be energy density: calculate ${\displaystyle \lambda F_{\lambda }}$ to get units of ergs/s/cm^2.

SO, IN SUMMARY: Take your ${\displaystyle F_{\nu }}$ measurements that are in Jy. (Ensure they are in Jy! If they're in magnitudes, convert them to Jy first; see 'magnitude' discussion above.) Multiply by ${\displaystyle 10^{-23}}$ to get them into cgs units. Multiply these ${\displaystyle F_{\nu }}$ values by ${\displaystyle c/\lambda ^{2}}$ to get them into ${\displaystyle F_{\lambda }}$. Multiply them by ${\displaystyle \lambda }$ to get them into ${\displaystyle \lambda F_{\lambda }}$. WATCH YOUR UNITS. NB: c = 2.997924d10 cm/sec

In other words, in order to convert our data from photometry to SEDs, we need to do the following:

1. Read in the catalogs you have.
2. Convert any magnitudes (and errors) into flux densities (if necessary).
3. Make SED plots for individual objects, but converting numbers first into the right units:
1. Create an array of the wavelengths of each measurement, keeps a copy of the version in microns, and convert to cm.
2. For any real measurements, convert the flux densities (probably in microJanskys) into cgs units.
3. For any real measurements, convert ${\displaystyle F_{\nu }}$ into ${\displaystyle F_{\lambda }}$ by multiplying the ${\displaystyle F_{\nu }}$ values by the ${\displaystyle d\nu /d\lambda }$ corresponding to the wavelength of each bandpass.
4. For any real measurements, multiply ${\displaystyle F_{\lambda }}$ by the lambda corresponding to the wavelength of each bandpass to get ${\displaystyle \lambda F_{\lambda }}$.
4. For any real measurements, plot the log of the ${\displaystyle \lambda F_{\lambda }}$ data points (in cgs units) against the log of the lambda data points (in microns, only because that makes it easier to read). Label the axes (with units)! Plot the error bars on top of the data points (also converted from uJy).

# Notes on plotting

Why are we plotting ${\displaystyle \lambda F_{\lambda }}$ vs. ${\displaystyle \lambda }$ instead of ${\displaystyle \nu }$? Well, only because I think in wavelength, not frequency. I don't know off the top of my head the frequencies of the Spitzer bandpasses, but I do know their wavelengths.

Why are we plotting ${\displaystyle \lambda F_{\lambda }}$ instead of ${\displaystyle \nu F_{\nu }}$? Well, only for internal consistency. Since one axis is in wavelength units, it makes sense to have the other axis also in wavelength units.

If you have gotten this far using real data, you will find (if you have done the calculations correctly) that you have numbers that are very small, like 9.77237e-12, 1.99526e-11, etc. Any time you find yourself with these kinds of numbers, you should automatically plot in log space (or log/log space), NOT linear space. You want to actually plot log(${\displaystyle \lambda F_{\lambda }}$) vs. log(${\displaystyle \lambda }$).

# The next step

IF you are highly motivated and ready to go on to the next step... It can be useful, in the course of analysis of the Spitzer or Herschel or WISE data, to pretend that the contribution from the star is a blackbody. It's not really, but it's awful close, especially in the infrared. A blackbody's flux density is given by (where T is temperature, and other constants are given below)

  ${\displaystyle B_{\lambda }=\left({\frac {2hc^{2}/\lambda ^{5}}{\exp(hc/\lambda kT)-1)}}\right)}$       (eqn 3)


but of course we want to plot ${\displaystyle \lambda \times B_{\lambda }}$:

  ${\displaystyle \lambda B_{\lambda }=\left({\frac {2hc^{2}/\lambda ^{4}}{\exp(hc/\lambda kT)-1)}}\right)}$       (eqn 4)


Values of these constants all in cgs units:

• h = 6.6260755d-27 erg*sec
• c = 2.997924d10 cm/sec
• k = 1.380658d-16 erg/deg

So, in summary, for the list of things to do above, add this one:

1. For any real measurements, for any star with at least 2 fluxes, fit a model -- one very simple way to do that is to fit a blackbody to the energies derived from the three 2MASS and first 2 IRAC bands. There are two free parameters in this fit -- the temperature of the blackbody and an additive (in the log) offset related to the distance of the object. If we know the temperature of the star (via a spectral type) and the distance to the object, then we know the values for the temperature and the offset.