# Converting from $\mathrm{W \: m^{-2}}$ to $\mathrm{Jy \: km \: s^{-1}}$

Suppose I observe a spectral line with rest frequency $$\nu_0$$ and integrated flux density $$F = 10^{-26} \mathrm{W/m^2}$$. Is the following line of reasoning correct?

$$1 \: \mathrm{Jy} = 10^{-26} \: \mathrm{W/m^2/Hz}$$ therefore $$F = 1 \: \mathrm{Jy \: Hz}$$. This can be interpreted as a peak flux in $$\mathrm{Jy}$$ multiplied by a linewidth in $$\mathrm{Hz}$$. Thus to convert to $$\mathrm{Jy \: km \: s^{-1}}$$ I just need to multiply the linewidth by $$c/\nu_0$$ to convert it from $$\mathrm{Hz}$$ to $$\mathrm{km \: s^{-1}}$$. Hence, $$F = c/\nu_0 \: \mathrm{Jy \: km \: s^{-1}}$$ (where $$c$$ is in units of $$\mathrm{km \: s^{-1}}$$).

You are essentially correct, but here's a more in-depth explanation:

Both units are a measure of the total flux $$F$$ of a light source in some interval, e.g. a spectral line or a broader region given by a filter. The flux is obtained by integrating a spectrum’s flux density, i.e. the flux per "bin". A spectrum is a distribution of photons as a function of the photons' energy or, equivalently, wavelength or frequency. The bins are then whatever you choose on your $$x$$ axis.

Typically you will use frequency $$\nu$$ for low-energy photons (radio to microwaves, with units of e.g. MHz or GHz), wavelength $$\lambda$$ for mid-range photons (with units of µm for infrared, and Å for optical to UV), and energy $$E$$ for high-energy photons (keV for X-ray and MeV gamma-rays). The flux density can then be denoted $$F_\nu$$, $$F_\lambda$$, and $$F_E$$, respectively, although the latter notation is not common; these photons are often so infrequent that one simply uses "counts", i.e. the absolute number of individual photons.

Note though that astronomers often are quite slobby and uses the word "flux" when they actually mean flux density (as well as for intensity, radiance, etc.).

But there is another very common way to show a spectrum, namely as a function of $$\mathrm{km}\,\mathrm{s}^{-1}$$. If you consider a spectral line from some electronic transition, it is emitted at a particular frequency (almost a delta function). But because the atoms emitting the photon do not lie still, but have a random distribution of velocities $$v$$, they will broaden the line. To first order, an atom moving with a velocity $$dv$$ with respect to the systemic velocity Doppler-shifts its emitted photon by an amount $$d\nu = \frac{dv}{c}\nu_0,$$ where $$\nu_0$$ is the line center frequency, and $$c$$ is the speed of light. This means that we can also show a spectral line centered on $$0$$, with positive and negative velocities on the $$x$$ axis corresponding to higher and lower frequencies.

This is an informative way to show physical processes such as temperature, turbulence or bulk motion of the gas.

### Units

Although astronomers tend to like cgs units, we also dislike huge and tiny numbers with too many powers of tens, plus we're in general not very consistent. Historically, flux in optical and UV light is typically given in cgs ($$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$$), but wavelengths are measured in Ångström, not cm, so flux densities are measured in $$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$$.

On the other hand, radio astronomers tends to use SI units for flux ($$\mathrm{W}\,\mathrm{m}^{-2}$$) and flux density ($$\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}$$). However, since typical astronomical sources are extremely faint, radio astronomers often measure flux densities in jansky (Jy), where $$1\,\mathrm{Jy} = 10^{-26}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}.$$

### Converting from W m–2 to Jy km s–1

Because of the fear of small or large powers of ten, and because as described above spectra can be shown as a function of both frequency and velocity, you can actually integrate you flux density in jansky over velocity, and thus the total flux of radio sources is periodically measured in $$\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$$. With $$F_\nu d\nu = F_v dv$$, combining the equations above the relation is (beware of the similarity between $$\nu$$ and $$v$$) $$\begin{array}{rcl} \int F_\nu d\nu & = & \int F_\nu \frac{\nu_0}{c} dv\\ \int \frac{F_\nu}{\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{d\nu}{\mathrm{Hz}} & = & 10^{-26} \frac{\nu_0/\mathrm{Hz}}{c/\mathrm{km}\,\mathrm{s}^{-1}} \int \frac{F_\nu}{10^{-26}\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{dv}{\mathrm{km}\,\mathrm{s}^{-1}}\\ \frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} & = & 10^{-26} \frac{10^9\nu_0/\mathrm{GHz}}{3\times10^5} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}\\ \frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} & = & \frac{1}{3\times10^{22}}\frac{\nu_0}{\mathrm{GHz}} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}} \end{array}$$

For example, for the CO(1→0) line with a rest frequency of $$\nu_0 = 115.3\,\mathrm{GHz}$$, to convert from flux in $$\mathrm{W}\,\mathrm{m}^{-2}$$ to flux in $$\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$$ you multiply by $$2.6\times10^{20}$$.

You can read more about conversions in this document on the low-resolution spectrometer SPIRE on the Herschel telescope. Note though that this conversion differs from mine by a factor of $$\pi$$. I think this is because in frequency the instrumental resolution of SPIRE broadens a delta line to a sinc function which is normalized to $$\pi$$, not unity.