<p>You are essentially correct, but here's a more in-depth explanation:</p>
<p>Both units are a measure of the total flux <span class="math-container">$F$</span> 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 <em>density</em>, i.e. the flux per &quot;bin&quot;. 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 <span class="math-container">$x$</span> axis.</p>
<p>Typically you will use frequency <span class="math-container">$\nu$</span> for low-energy photons (radio to microwaves, with units of e.g. MHz or GHz), wavelength <span class="math-container">$\lambda$</span> for mid-range photons (with units of µm for infrared, and Å for optical to UV), and energy <span class="math-container">$E$</span> for high-energy photons (keV for X-ray and MeV gamma-rays). The flux density can then be denoted <span class="math-container">$F_\nu$</span>, <span class="math-container">$F_\lambda$</span>, and <span class="math-container">$F_E$</span>, respectively, although the latter notation is not common; these photons are often so infrequent that one simply uses &quot;counts&quot;, i.e. the absolute number of individual photons.</p>
<p>Note though that astronomers often are quite slobby and uses the word &quot;flux&quot; when they actually mean flux density (as well as for intensity, radiance, etc.).</p>
<h3>Doppler broadening</h3>
<p>But there is another very common way to show a spectrum, namely as a function of <span class="math-container">$\mathrm{km}\,\mathrm{s}^{-1}$</span>. 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 <span class="math-container">$v$</span>, they will broaden the line. To first order, an atom moving with a velocity <span class="math-container">$dv$</span> with respect to the systemic velocity Doppler-shifts its emitted photon by an amount
<span class="math-container">$$
d\nu = \frac{dv}{c}\nu_0,
$$</span>
where <span class="math-container">$\nu_0$</span> is the line center frequency, and <span class="math-container">$c$</span> is the speed of light. This means that we can also show a spectral line centered on <span class="math-container">$0$</span>, with positive and negative velocities on the <span class="math-container">$x$</span> axis corresponding to higher and lower frequencies.</p>
<p>This is an informative way to show physical processes such as temperature, turbulence or bulk motion of the gas.</p>
<h3>Units</h3>
<p>Although astronomers tend to like <a href="https://en.wikipedia.org/wiki/Centimetre%E2%80%93gram%E2%80%93second_system_of_units" rel="noreferrer">cgs</a> 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 (<span class="math-container">$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$</span>), but wavelengths are measured in Ångström, not cm, so flux <em>densities</em> are measured in <span class="math-container">$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$</span>.</p>
<p>On the other hand, radio astronomers tends to use SI units for flux (<span class="math-container">$\mathrm{W}\,\mathrm{m}^{-2}$</span>) and flux density (<span class="math-container">$\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}$</span>). However, since typical astronomical sources are extremely faint, radio astronomers often measure flux densities in <a href="https://en.wikipedia.org/wiki/Jansky" rel="noreferrer">jansky</a> (Jy), where
<span class="math-container">$$
1\,\mathrm{Jy} = 10^{-26}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}.
$$</span></p>
<h3>Converting from W m<sup>–2</sup> to Jy km s<sup>–1</sup></h3>
<p>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 <span class="math-container">$\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$</span>. With <span class="math-container">$F_\nu d\nu = F_v dv$</span>, combining the equations above the relation is (beware of the similarity between <span class="math-container">$\nu$</span> and <span class="math-container">$v$</span>)
<span class="math-container">$$
\begin{array}{rcl}
\int F_\nu d\nu &amp; = &amp; \int F_\nu \frac{\nu_0}{c} dv\\
\int \frac{F_\nu}{\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{d\nu}{\mathrm{Hz}} &amp; = &amp; 10^{-26} \frac{\nu_0/\mathrm{Hz}}{c/\mathrm{km}\,\mathrm{s}^{-1}} \int \frac{F_\nu}{10^{-26}\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{dv}{\mathrm{km}\,\mathrm{s}^{-1}}\\
\frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} &amp; = &amp; 10^{-26} \frac{10^9\nu_0/\mathrm{GHz}}{3\times10^5} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}\\
\frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} &amp; = &amp; \frac{1}{3\times10^{22}}\frac{\nu_0}{\mathrm{GHz}} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}
\end{array}
$$</span></p>
<p>For example, for the CO(1→0) line with a rest frequency of <span class="math-container">$\nu_0 = 115.3\,\mathrm{GHz}$</span>, to convert from flux in  <span class="math-container">$\mathrm{W}\,\mathrm{m}^{-2}$</span> to flux in <span class="math-container">$\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$</span> you multiply by <span class="math-container">$2.6\times10^{20}$</span>.</p>
<hr />
<p>You can read more about conversions in <a href="https://www.cosmos.esa.int/documents/12133/1035800/fts-line-flux-conv.pdf/" rel="noreferrer">this document</a> on the low-resolution spectrometer SPIRE on the <a href="https://en.wikipedia.org/wiki/Herschel_Space_Observatory" rel="noreferrer">Herschel telescope</a>. Note though that this conversion differs from mine by a factor of <span class="math-container">$\pi$</span>. I think this is because in frequency the instrumental resolution of SPIRE broadens a delta line to a <a href="https://en.wikipedia.org/wiki/Sinc_function" rel="noreferrer">sinc</a> function which is normalized to <span class="math-container">$\pi$</span>, not unity.</p>

CC BY-SA 4.0
2021-03-04T13:41:48.687Z
2021-03-05T10:35:52.397Z
2021-03-05T10:35:52.397Z
pela
<p>Suppose I observe a spectral line with rest frequency <span class="math-container">$\nu_0$</span> and integrated flux density <span class="math-container">$F = 10^{-26} \mathrm{W/m^2}$</span>. Is the following line of reasoning correct?</p>
<p><span class="math-container">$1 \: \mathrm{Jy} = 10^{-26} \: \mathrm{W/m^2/Hz}$</span> therefore <span class="math-container">$F = 1 \: \mathrm{Jy \: Hz}$</span>. This can be interpreted as a peak flux in <span class="math-container">$\mathrm{Jy}$</span> multiplied by a linewidth in <span class="math-container">$\mathrm{Hz}$</span>. Thus to convert to <span class="math-container">$\mathrm{Jy \: km \: s^{-1}}$</span> I just need to multiply the linewidth by <span class="math-container">$c/\nu_0$</span> to convert it from <span class="math-container">$\mathrm{Hz}$</span> to <span class="math-container">$\mathrm{km \: s^{-1}}$</span>. Hence, <span class="math-container">$F = c/\nu_0 \: \mathrm{Jy \: km \: s^{-1}}$</span> (where <span class="math-container">$c$</span> is in units of <span class="math-container">$\mathrm{km \: s^{-1}}$</span>).</p>

<spectra><units>
Converting from $\mathrm{W \: m^{-2}}$ to $\mathrm{Jy \: km \: s^{-1}}$
Framazu
# Converting from $\mathrm{W \: m^{-2}}$ to $\mathrm{Jy \: km \: s^{-1}}$

(Asked by: username_0 on 2020-09-20)

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 $\mathrm{km \: s^{-1}}$).

**Tags:** spectra, units


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## 1 answers

### Answer by username_1 on 2021-03-04. Score: 10 | Accepted answer

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.).

### Doppler broadening

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](https://en.wikipedia.org/wiki/Centimetre%E2%80%93gram%E2%80%93second_system_of_units) 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](https://en.wikipedia.org/wiki/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 $2.6\times10^{20}$.

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You can read more about conversions in [this document](https://www.cosmos.esa.int/documents/12133/1035800/fts-line-flux-conv.pdf/) on the low-resolution spectrometer SPIRE on the [Herschel telescope](https://en.wikipedia.org/wiki/Herschel_Space_Observatory). 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](https://en.wikipedia.org/wiki/Sinc_function) function which is normalized to $\pi$, not unity.


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