<p><span class="math-container">$\sigma = \delta\nu\sqrt{N}\sigma_{\mathrm{rms}}$</span></p>
<p>Channel width <span class="math-container">$\delta\nu$</span> is in km/s (spectral resolution) and rms is in Jy/beam (noise in channel): I imagine you already know these.</p>
<p><span class="math-container">$N$</span>, the number of independent measurements, is the number of pixels inside the mask divided by beam area in pixels. Physically, it means that you are counting the number of beams within your mask. These are your &quot;independent measurements&quot;. In fact, the paper you mention discusses this very briefly right after giving the formula for the upper limit.</p>
<p>Edits:</p>
<p><em>Why square root?</em>: Central limit theorem. The noise goes down as the square root of the number of samples. If you want a &quot;hard&quot; upper limit, yes, you wouldn't use the square root. But it is extremely unlikely that all the random fluctuations are just in one direction such that they all push the mean to a level well below the detection threshold. The probability of the event that the upper limit calculated acoording to the given formula is incorrect depends on the chosen <span class="math-container">$\sigma$</span>-level. <a href="https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule" rel="nofollow noreferrer"><span class="math-container">$3\sigma$</span> implies that that probability is <span class="math-container">$1-0.997=0.3\%$</span></a>.</p>
<p><em>On the units of upper limit</em>: The line intensities are in Jy/beam.km/s. This is because you are adding the flux intensities <strong>over each channel</strong>. Imagine the real spectrum of an emission (flux intensity versus velocity/frequency); you're finding out the area under the curve, ie, integrating <span class="math-container">$S_{\mathrm{int}}=\int{S\mathrm{d\nu}}$</span>. So line intensity is in Jy/beam.km/s, and naturally its upper limit also will be in the same units.</p>
<p>You can also build intuition of the units of line intensity in the following way. Imagine there is an emission with constant intensity of 1 Jy/beam from velocities -10 km/s to +10 km/s and 0 Jy/beam elsewhere. Should your line intensity be dependent on your spectral resolution? No. Whether you use channels of width 1 km/s or 10 km/s, the line intensity must be the same. (Finally, to get the line flux density, you would integrate the intensity Jy/beam.km/s over the area to get Jy.km/s)</p>

CC BY-SA 4.0
2022-07-25T19:59:45.193Z
Thanks very much for this. I am still having trouble understanding why this equation is correct in the first place... intuitively it seems like I should just measure the rms in Jy/beam, then multiply by number of beams in the mask to get flux density upper limit in Jy? Not sure why the channel width or square root are involved?
lucas
CC BY-SA 4.0
2022-07-25T22:59:28.513Z
@lucas I edited the answer to try to explain about the units, channel width, and the square root :)
terribleGuitarPlayer
CC BY-SA 4.0
2022-07-25T19:20:39.887Z
2022-07-25T22:58:23.957Z
2022-07-25T22:58:23.957Z
terribleGuitarPlayer
<p>I have ALMA data which are non-detections of some spectral lines in a protoplanetary disk. The data is in the form of spectral cubes. I am hoping to estimate an upper limit on the flux of each of the spectral lines, given the fact they are not detected.</p>
<p>What is the standard procedure for doing this?</p>
<p>So far, I have tried following the procedure outlined in <a href="https://arxiv.org/pdf/1901.02689.pdf" rel="nofollow noreferrer">Carney et al. (2019)</a>...</p>
<ol>
<li>Create a 'Keplerian mask', so that only pixels associated with the position and velocity of the disk are included</li>
<li>Apply the mask to the spectral cube, and generate an integrated intensity map</li>
<li>Measure the rms in the integrated intensity map</li>
<li>Translate the measured rms into an upper limit</li>
</ol>
<p>I am having trouble with step 4. In Carney et al. (2019), they use the formula:</p>
<p><span class="math-container">$\sigma = \delta \nu \sqrt{N} \sigma_{\text{rms}}$</span></p>
<p>where <span class="math-container">$\sigma$</span> is the flux upper limit, <span class="math-container">$\delta \nu$</span> is the channel width, <span class="math-container">$N$</span> is the number of independant measurements, and <span class="math-container">$\sigma_{\text{rms}}$</span> is the measured rms per channel.</p>
<p>I'm not sure exactly how to apply this to my data? Or is there another way to estimate an upper limit on the flux?</p>

<radio-astronomy><spectroscopy><interferometry><proto-planetary-disk>
Estimate upper limits on flux values in the case of a non-detection?
lucas
# Estimate upper limits on flux values in the case of a non-detection?

(Asked by: username_0 on 2021-10-21)

I have ALMA data which are non-detections of some spectral lines in a protoplanetary disk. The data is in the form of spectral cubes. I am hoping to estimate an upper limit on the flux of each of the spectral lines, given the fact they are not detected.

What is the standard procedure for doing this?

So far, I have tried following the procedure outlined in [Carney et al. (2019)](https://arxiv.org/pdf/1901.02689.pdf)...

1. Create a 'Keplerian mask', so that only pixels associated with the position and velocity of the disk are included
2. Apply the mask to the spectral cube, and generate an integrated intensity map
3. Measure the rms in the integrated intensity map
4. Translate the measured rms into an upper limit

I am having trouble with step 4. In Carney et al. (2019), they use the formula:

$\sigma = \delta \nu \sqrt{N} \sigma\_{\text{rms}}$

where $\sigma$ is the flux upper limit, $\delta \nu$ is the channel width, $N$ is the number of independant measurements, and $\sigma\_{\text{rms}}$ is the measured rms per channel.

I'm not sure exactly how to apply this to my data? Or is there another way to estimate an upper limit on the flux?

**Tags:** radio-astronomy, spectroscopy, interferometry, proto-planetary-disk

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## 1 answers

### Answer by username_1 on 2022-07-25. Score: 3 | Accepted answer

$\sigma = \delta\nu\sqrt{N}\sigma\_{\mathrm{rms}}$

Channel width $\delta\nu$ is in km/s (spectral resolution) and rms is in Jy/beam (noise in channel): I imagine you already know these.

$N$, the number of independent measurements, is the number of pixels inside the mask divided by beam area in pixels. Physically, it means that you are counting the number of beams within your mask. These are your "independent measurements". In fact, the paper you mention discusses this very briefly right after giving the formula for the upper limit.

Edits:

*Why square root?*: Central limit theorem. The noise goes down as the square root of the number of samples. If you want a "hard" upper limit, yes, you wouldn't use the square root. But it is extremely unlikely that all the random fluctuations are just in one direction such that they all push the mean to a level well below the detection threshold. The probability of the event that the upper limit calculated acoording to the given formula is incorrect depends on the chosen $\sigma$-level. [$3\sigma$ implies that that probability is $1-0.997=0.3\%$](https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule).

*On the units of upper limit*: The line intensities are in Jy/beam.km/s. This is because you are adding the flux intensities **over each channel**. Imagine the real spectrum of an emission (flux intensity versus velocity/frequency); you're finding out the area under the curve, ie, integrating $S\_{\mathrm{int}}=\int{S\mathrm{d\nu}}$. So line intensity is in Jy/beam.km/s, and naturally its upper limit also will be in the same units.

You can also build intuition of the units of line intensity in the following way. Imagine there is an emission with constant intensity of 1 Jy/beam from velocities -10 km/s to +10 km/s and 0 Jy/beam elsewhere. Should your line intensity be dependent on your spectral resolution? No. Whether you use channels of width 1 km/s or 10 km/s, the line intensity must be the same. (Finally, to get the line flux density, you would integrate the intensity Jy/beam.km/s over the area to get Jy.km/s)

> **Comments:**
> **username_0**: Thanks very much for this. I am still having trouble understanding why this equation is correct in the first place... intuitively it seems like I should just measure the rms in Jy/beam, then multiply by number of beams in the mask to get flux density upper limit in Jy? Not sure why the channel width or square root are involved?  
> **username_1**: @lucas I edited the answer to try to explain about the units, channel width, and the square root :)  

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