<p>I have a definitive solution for the first question. First of all, I realized that the conversion factor of <span class="math-container">$3.305×10^{-7}$</span> is not okay (sorry for the typo: it is <span class="math-container">$10^{-7}$</span>, not <span class="math-container">$10^7$</span>), but I would like to explain my procedure in a clearer way than before.</p>
<p>Let's start from the equation to convert the flux density in terms of frequency, <span class="math-container">$f_\nu$</span>, to flux density in terms of wavelength, <span class="math-container">$f_\lambda$</span>:</p>
<p><span class="math-container">$f_\nu = \frac{\lambda^2}{c} \space f_\lambda$</span> or equivalently <span class="math-container">$\frac{f_\nu}{Jy} = 3.33564×10^4 \space (\frac{\lambda}{Å})^2 \space \frac{f_\lambda}{erg \space s^{-1} \space cm^{-2} \space Å^{-1}}$</span></p>
<p>Now, we can obtain the information of the wavelength <span class="math-container">$\lambda$</span> and the flux density <span class="math-container">$f_\lambda$</span> directly from the header of the HST FITS file:</p>
<ul>
<li><p><span class="math-container">$\lambda = \rm{PHOTPLAM}$</span>, where PHOTPLAM represents the so called &quot;pivot wavelength&quot; in units of <span class="math-container">$Å$</span> (see <a href="https://en.wikipedia.org/wiki/AB_magnitude" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/AB_magnitude</a>).</p>
</li>
<li><p><span class="math-container">$\frac{f_\lambda}{erg \space s^{-1} \space cm^{-2} \space Å^{-1}} = \frac{\rm{PHOTFLAM}}{erg \space cm^{-2} \space Å^{-1} \space electrons^{-1}} \space\space \frac{X_{HST, pix}}{electrons \space s^{-1}}$</span>, where PHOTFLAM is the &quot;inverse sensitivity&quot; and <span class="math-container">$X_{HST, pix}$</span> is my way to denote the value of the pixel of the HST FITS image.</p>
</li>
</ul>
<p>So, we can transform the value of each pixel from units of <span class="math-container">$electrons \space s^{-1}$</span> to Jy, just substituting the header information into the following final and practical formula:</p>
<p><span class="math-container">$\frac{f_\nu}{Jy} = 3.33564×10^4 \space (\frac{\rm{PHOT\bf{P}}\rm{LAM}}{Å})^2 \space \frac{\rm{PHOT\bf{F}}\rm{LAM}}{erg \space cm^{-2} \space Å^{-1} \space electrons^{-1}} \space\space \frac{X_{HST, pix}}{electrons \space s^{-1}}$</span></p>
<p>Please notice that the values in the header can change due to the camera and the band used. In the case of my FITS file, I have</p>
<p>HST CAMERA AND FILTER | PHOT<strong>F</strong>LAM             | PHOT<strong>P</strong>LAM</p>
<p>ACS WFC1 F775W        | 9.91377802592592E-20 | 7693.411714814814</p>
<p>ACS WFC1 F850LP | 1.50169706333333E-19 |9036.597123333335</p>
<p>WFC3 IR F105W | 3.03865733593518E-20 | 10551.04690640578</p>
<p>WFC3 IR F110W | 1.52741283843628E-20 | 11534.45855553776</p>
<p>WFC3 IR F125W | 2.24834173714407E-20 | 12486.05978577568</p>
<p>WFC3 IR F160W | 1.92756031304868E-20 | 15369.17570896557</p>

CC BY-SA 4.0
2023-01-24T20:33:07.467Z
2023-01-24T21:07:59.527Z
2023-01-24T21:07:59.527Z
lol
<p>I am trying to figure out how to convert the units of some FITS files to Jy. These files refer to observations in different bands using different instruments. I list them below with my way of converting the units. Notice that the second procedure is similar to another question (here: <a href="https://astronomy.stackexchange.com/questions/36955/astronomical-data-convertion-from-jy-pixel-to-mjy-sr">Astronomical data convertion from Jy/pixel to MJy/sr?</a>).</p>
<ol>
<li><p><strong>HST observations with the camera WFC3 using the filter F160W and others.</strong> In this case, I found &quot;BUNIT   = 'ELECTRONS/S' / brightness units&quot; in the file header. So, to pass from e/s to Jy, I did the following: first, thanks to HST website info (<a href="https://www.stsci.edu/hst/instrumentation/acs/data-analysis/zeropoints" rel="nofollow noreferrer">https://www.stsci.edu/hst/instrumentation/acs/data-analysis/zeropoints</a>), I multiplied each pixel value for the PHOTFLAM=<span class="math-container">$9.9138 \times 10^{-20}$</span> (inverse sensitivity in units of <span class="math-container">$erg \space cm^{-2} Å^{-1} e^{-1}$</span>) to obtain the flux density in terms of wavelength (<span class="math-container">$erg \space s^{-1} cm^{-2} Å^{-1}$</span>), then I multiplied the result for the factor <span class="math-container">$\lambda^{2}/c$</span> to get the flux density in terms of the frequency (<span class="math-container">$erg \space s^{-1} cm^{-2} Hz^{-1}$</span>). Finally, I multiplied the result for <span class="math-container">$10^{23}$</span> to convert these units into Jy. All the procedure can be summarized by saying that, considering a wavelength in microns, to pass from HST units to Jy, I should multiply for the constant <span class="math-container">$3.305 \times 10^{7}$</span> and for the square of lambda, which I set as the central wavelength of the filter (e.g., 1.5396 microns fo F160W).</p>
</li>
<li><p><strong>Spitzer IRAC, channels from 1 to 4.</strong> I can read on the website (7th question, here: <a href="https://irsa.ipac.caltech.edu/docs/knowledgebase/spitzer_irac.html" rel="nofollow noreferrer">https://irsa.ipac.caltech.edu/docs/knowledgebase/spitzer_irac.html</a>):&quot;If you want to convert 'MJy/sr into flux density/pixel units, you can convert steradians into arcseconds squared, and then multiply by the area of the pixel. Remember that in BCDs the pixel area is approximately 1.22 arcseconds squared, whereas in the pipeline mosaic the pixelsize by default is 0.6 arcseconds squared exactly. So, for example, for the pipeline mosaic a pixel value needs to be multiplied by (1E12 micro-Jy)/(4.254517E10 arcsec**2) x 0.6 arcsec x 0.6 arcsec = 8.461595 to obtain micro-Jy/pixel flux densities.&quot;. So, I confirm the units of the files' header &quot;BUNIT  = 'MJy/sr' / Units of image data (MJy/sr)&quot; for all channels, and, given the above explanation, I understand that I do not need the FLUXCONV values. Just in case, they are channel 1: &quot;FLUXCONV= 0.1069 / Flux Conv. factor (MJy/sr per DN/sec)&quot;; channel 2: &quot;FLUXCONV= 0.1382 / Flux Conv. factor (MJy/sr per DN/sec)&quot;; channel 3: &quot;FLUXCONV= 0.5858 / Flux Conv. factor (MJy/sr per DN/sec)&quot;; channel 4: &quot;FLUXCONV= 0.2026 / Flux Conv. factor (MJy/sr per DN/sec)&quot;. I only need to measure the pixel size on the image, which is 0.6 arcseconds for all four images. So I should multiply the pixel value (in MJy/sr)  for the constant <span class="math-container">$8.461595 \times 10^2$</span> to obtain the value in Jy.</p>
</li>
</ol>
<p>What do you think about my solutions? What about using the HST filter's central wavelength as the lambda value? I would appreciate any idea/confirmation of the procedure. Thank you very much for your help.</p>

CC BY-SA 4.0
2023-01-06T08:51:41.853Z
Regarding your choice of the central lambda as the value, I don’t really know what the standard choice is there but one thing you could do is try the edge lambda values of the filter, see how much it changes your result, and then ask yourself if you’re comfortable with that amount of discrepancy for what you’re using this for; if yes, then great, otherwise it might be worthwhile trying to find some literature that makes this conversion and see if you can replicate it using the data they use
Justin T
CC BY-SA 4.0
2023-01-10T21:51:58.803Z
Thank you very much for your comments. @JustinT, I tried, but there is a difference that I need to know if I can ignore; probably not. On the other hand, I suppose it is hard to find a paper describing that conversion: my idea is that the conversion is assumed as "general knowledge," which is not worth mentioning. However, thanks for the idea, I will try papers describing catalogs, which should give more technical details.
lol
CC BY-SA 4.0
2023-01-12T00:07:25.663Z
@lol What I mean by replicate is that while they probably won’t elaborate on the exact way they use to do it, if it is something like a central wavelength you might find a paper where they have data before and after a conversion, and through fine tuning maybe you can sense a pattern as to what they do. Since spitzer and Hubble data is available to you in an unprocessed form, you should be able to compare and contrast with their processed data and try different things till your numbers line up (assuming they don’t do anything too sophisticated, but at least it will tell you how close you are)
Justin T
<observational-astronomy><hubble-telescope><units>
Conversion to Jy of Hubble WFC3 units (electrons/s) and of Spitzer IRAC (MJy/sr)
lol
# Conversion to Jy of Hubble WFC3 units (electrons/s) and of Spitzer IRAC (MJy/sr)

(Asked by: username_0 on 2022-12-27)

I am trying to figure out how to convert the units of some FITS files to Jy. These files refer to observations in different bands using different instruments. I list them below with my way of converting the units. Notice that the second procedure is similar to another question (here: [Astronomical data convertion from Jy/pixel to MJy/sr?](https://astronomy.stackexchange.com/questions/36955/astronomical-data-convertion-from-jy-pixel-to-mjy-sr)).

1. **HST observations with the camera WFC3 using the filter F160W and others.** In this case, I found "BUNIT = 'ELECTRONS/S' / brightness units" in the file header. So, to pass from e/s to Jy, I did the following: first, thanks to HST website info (<https://www.stsci.edu/hst/instrumentation/acs/data-analysis/zeropoints>), I multiplied each pixel value for the PHOTFLAM=$9.9138 \times 10^{-20}$ (inverse sensitivity in units of $erg \space cm^{-2} Å^{-1} e^{-1}$) to obtain the flux density in terms of wavelength ($erg \space s^{-1} cm^{-2} Å^{-1}$), then I multiplied the result for the factor $\lambda^{2}/c$ to get the flux density in terms of the frequency ($erg \space s^{-1} cm^{-2} Hz^{-1}$). Finally, I multiplied the result for $10^{23}$ to convert these units into Jy. All the procedure can be summarized by saying that, considering a wavelength in microns, to pass from HST units to Jy, I should multiply for the constant $3.305 \times 10^{7}$ and for the square of lambda, which I set as the central wavelength of the filter (e.g., 1.5396 microns fo F160W).
2. **Spitzer IRAC, channels from 1 to 4.** I can read on the website (7th question, here: <https://irsa.ipac.caltech.edu/docs/knowledgebase/spitzer_irac.html>):"If you want to convert 'MJy/sr into flux density/pixel units, you can convert steradians into arcseconds squared, and then multiply by the area of the pixel. Remember that in BCDs the pixel area is approximately 1.22 arcseconds squared, whereas in the pipeline mosaic the pixelsize by default is 0.6 arcseconds squared exactly. So, for example, for the pipeline mosaic a pixel value needs to be multiplied by (1E12 micro-Jy)/(4.254517E10 arcsec\*\*2) x 0.6 arcsec x 0.6 arcsec = 8.461595 to obtain micro-Jy/pixel flux densities.". So, I confirm the units of the files' header "BUNIT = 'MJy/sr' / Units of image data (MJy/sr)" for all channels, and, given the above explanation, I understand that I do not need the FLUXCONV values. Just in case, they are channel 1: "FLUXCONV= 0.1069 / Flux Conv. factor (MJy/sr per DN/sec)"; channel 2: "FLUXCONV= 0.1382 / Flux Conv. factor (MJy/sr per DN/sec)"; channel 3: "FLUXCONV= 0.5858 / Flux Conv. factor (MJy/sr per DN/sec)"; channel 4: "FLUXCONV= 0.2026 / Flux Conv. factor (MJy/sr per DN/sec)". I only need to measure the pixel size on the image, which is 0.6 arcseconds for all four images. So I should multiply the pixel value (in MJy/sr) for the constant $8.461595 \times 10^2$ to obtain the value in Jy.

What do you think about my solutions? What about using the HST filter's central wavelength as the lambda value? I would appreciate any idea/confirmation of the procedure. Thank you very much for your help.

**Tags:** observational-astronomy, hubble-telescope, units
> **Comments:**
> **username_1**: Regarding your choice of the central lambda as the value, I don’t really know what the standard choice is there but one thing you could do is try the edge lambda values of the filter, see how much it changes your result, and then ask yourself if you’re comfortable with that amount of discrepancy for what you’re using this for; if yes, then great, otherwise it might be worthwhile trying to find some literature that makes this conversion and see if you can replicate it using the data they use  
> **username_0**: Thank you very much for your comments. @JustinT, I tried, but there is a difference that I need to know if I can ignore; probably not. On the other hand, I suppose it is hard to find a paper describing that conversion: my idea is that the conversion is assumed as "general knowledge," which is not worth mentioning. However, thanks for the idea, I will try papers describing catalogs, which should give more technical details.  
> **username_1**: @lol What I mean by replicate is that while they probably won’t elaborate on the exact way they use to do it, if it is something like a central wavelength you might find a paper where they have data before and after a conversion, and through fine tuning maybe you can sense a pattern as to what they do. Since spitzer and Hubble data is available to you in an unprocessed form, you should be able to compare and contrast with their processed data and try different things till your numbers line up (assuming they don’t do anything too sophisticated, but at least it will tell you how close you are)  

---

## 1 answers

### Answer by username_0 on 2023-01-24. Score: 1

I have a definitive solution for the first question. First of all, I realized that the conversion factor of $3.305×10^{-7}$ is not okay (sorry for the typo: it is $10^{-7}$, not $10^7$), but I would like to explain my procedure in a clearer way than before.

Let's start from the equation to convert the flux density in terms of frequency, $f\_\nu$, to flux density in terms of wavelength, $f\_\lambda$:

$f\_\nu = \frac{\lambda^2}{c} \space f\_\lambda$ or equivalently $\frac{f\_\nu}{Jy} = 3.33564×10^4 \space (\frac{\lambda}{Å})^2 \space \frac{f\_\lambda}{erg \space s^{-1} \space cm^{-2} \space Å^{-1}}$

Now, we can obtain the information of the wavelength $\lambda$ and the flux density $f\_\lambda$ directly from the header of the HST FITS file:

* $\lambda = \rm{PHOTPLAM}$, where PHOTPLAM represents the so called "pivot wavelength" in units of $Å$ (see <https://en.wikipedia.org/wiki/AB_magnitude>).
* $\frac{f\_\lambda}{erg \space s^{-1} \space cm^{-2} \space Å^{-1}} = \frac{\rm{PHOTFLAM}}{erg \space cm^{-2} \space Å^{-1} \space electrons^{-1}} \space\space \frac{X\_{HST, pix}}{electrons \space s^{-1}}$, where PHOTFLAM is the "inverse sensitivity" and $X\_{HST, pix}$ is my way to denote the value of the pixel of the HST FITS image.

So, we can transform the value of each pixel from units of $electrons \space s^{-1}$ to Jy, just substituting the header information into the following final and practical formula:

$\frac{f\_\nu}{Jy} = 3.33564×10^4 \space (\frac{\rm{PHOT\bf{P}}\rm{LAM}}{Å})^2 \space \frac{\rm{PHOT\bf{F}}\rm{LAM}}{erg \space cm^{-2} \space Å^{-1} \space electrons^{-1}} \space\space \frac{X\_{HST, pix}}{electrons \space s^{-1}}$

Please notice that the values in the header can change due to the camera and the band used. In the case of my FITS file, I have

HST CAMERA AND FILTER | PHOT**F**LAM | PHOT**P**LAM

ACS WFC1 F775W | 9.91377802592592E-20 | 7693.411714814814

ACS WFC1 F850LP | 1.50169706333333E-19 |9036.597123333335

WFC3 IR F105W | 3.03865733593518E-20 | 10551.04690640578

WFC3 IR F110W | 1.52741283843628E-20 | 11534.45855553776

WFC3 IR F125W | 2.24834173714407E-20 | 12486.05978577568

WFC3 IR F160W | 1.92756031304868E-20 | 15369.17570896557


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