<p>In addition to @User123 answer, and assuming you are subtracting luminosities and not magnitudes, there is usually one more obstacle: the <a href="https://en.wikipedia.org/wiki/Catastrophic_cancellation" rel="nofollow noreferrer">catastrophic cancelation</a>.</p>
<p>In short, if the difference between the total and the source B luminosity is near the accuracy of either of them, you will get no meaningful estimate for the source A.</p>
<p>In practical terms, I wonder as well, how you got the separate magnitude for the source B while you are not able to resolve A and B in the first place. Of course, one can use their specific spectral and/or variational features, but this is outside of the scope of the question.</p>

CC BY-SA 4.0
2021-09-06T22:10:03.913Z
Thanks for your comment. In the scenario I'm curious about, the two sources were not originally together. You could separately resolve their individual magnitudes. They then moved nearer to one another such that they could not be individually resolved.
theta
CC BY-SA 4.0
2021-08-28T11:10:39.863Z
2021-08-28T11:10:39.863Z
fraxinus
<p>The magnitudes don't sum and subtract that way: consider two 0.00 magnitude stars closely together. Is their combined magnitude 0.00? Actually, it is -0.75.</p>
<hr />
<p><a href="https://www.google.com/url?sa=t&amp;source=web&amp;rct=j&amp;url=https://www.astro.keele.ac.uk/jkt/pubs/JKTeq-fluxsum.pdf&amp;ved=2ahUKEwjClOTijtLyAhVegv0HHdCDDOwQFnoECDYQAQ&amp;usg=AOvVaw1JfFAyT0-DYrcxifVJFM4X" rel="noreferrer">This</a> paper might help you with the derivation of the formula for addition of magnitudes, but you are interested in subtraction. You just need to rearrange the formula in the paper to
<span class="math-container">$$m_a=-2.5\log(10^{-0.4m_{TOT}}-10^{-0.4m_b})$$</span></p>
<p>If you want to discover other calculations regarding magnitude, search for Pogson's law.</p>

CC BY-SA 4.0
2021-08-27T21:17:50.883Z
2021-08-27T21:28:30.060Z
2021-08-27T21:28:30.060Z
User123
<p>Let's say I observe a source that is actually two sources (source A and source B) where each cannot be individually resolved. Source A is a variable source and source B is a constant source. I want to measure the magnitude of source A by &quot;subtracting&quot; the known constant magnitude of source B from the total measured magnitude. Can this be done through simple subtraction? I know with flux the subtraction is straightforward, but magnitude is logarithmic so I am unsure how to do this.</p>

<observational-astronomy><photometry><magnitude>
Can you subtract magnitude of constant source to get magnitude of variable source?
theta
# Can you subtract magnitude of constant source to get magnitude of variable source?

(Asked by: username_0 on 2021-08-27)

Let's say I observe a source that is actually two sources (source A and source B) where each cannot be individually resolved. Source A is a variable source and source B is a constant source. I want to measure the magnitude of source A by "subtracting" the known constant magnitude of source B from the total measured magnitude. Can this be done through simple subtraction? I know with flux the subtraction is straightforward, but magnitude is logarithmic so I am unsure how to do this.

**Tags:** observational-astronomy, photometry, magnitude


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## 2 answers

### Answer by username_2 on 2021-08-27. Score: 6 | Accepted answer

The magnitudes don't sum and subtract that way: consider two 0.00 magnitude stars closely together. Is their combined magnitude 0.00? Actually, it is -0.75.

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[This](https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.astro.keele.ac.uk/jkt/pubs/JKTeq-fluxsum.pdf&ved=2ahUKEwjClOTijtLyAhVegv0HHdCDDOwQFnoECDYQAQ&usg=AOvVaw1JfFAyT0-DYrcxifVJFM4X) paper might help you with the derivation of the formula for addition of magnitudes, but you are interested in subtraction. You just need to rearrange the formula in the paper to
$$m\_a=-2.5\log(10^{-0.4m\_{TOT}}-10^{-0.4m\_b})$$

If you want to discover other calculations regarding magnitude, search for Pogson's law.


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### Answer by username_1 on 2021-08-28. Score: 3

In addition to @username_2 answer, and assuming you are subtracting luminosities and not magnitudes, there is usually one more obstacle: the [catastrophic cancelation](https://en.wikipedia.org/wiki/Catastrophic_cancellation).

In short, if the difference between the total and the source B luminosity is near the accuracy of either of them, you will get no meaningful estimate for the source A.

In practical terms, I wonder as well, how you got the separate magnitude for the source B while you are not able to resolve A and B in the first place. Of course, one can use their specific spectral and/or variational features, but this is outside of the scope of the question.

> **Comments:**
> **username_0**: Thanks for your comment. In the scenario I'm curious about, the two sources were not originally together. You could separately resolve their individual magnitudes. They then moved nearer to one another such that they could not be individually resolved.  

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