I have two methods which allow for a detection of a signal that is at most of magnitude x. If method 2 can detect signal that are half a magnitude larger, i.e. fainter, let's say x.5 mag, is it correct to say that method 2 can detect signals that are about 60% fainter than what method 1 can detect?

Is the 60% correct? I got this number by converting the magnitudes $m_i$ to flux $f_i$, i.e. $$100^{(m_1-m_2)/2.5}=\frac{f_2}{f_1}\approx 0.40$$ $$\Rightarrow f_2=0.40 f_1\sim 40\% f_1$$

Thus, $f_2$ is about $40\% f_1$ which means about $60\%$ fainter. I have never worked with magnitudes before, so is this a valid statement?

  • $\begingroup$ Completely correct, except you should divide by 2.5, not by 5. But your result is correct, so I guess it's just a typo. $\endgroup$ – pela Jul 9 '19 at 9:50
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    $\begingroup$ Of course, you are right! I actually calculated it the wrong way, but I have updated the numbers so now it should be 60% fainter. Thanks a lot for your reply! $\endgroup$ – Philipp Jul 9 '19 at 11:00
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    $\begingroup$ @pela are you sure? $100^{1/5} = 10^{1/2.5}$ $\endgroup$ – Mike G Jul 9 '19 at 13:40
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    $\begingroup$ @MikeG Oh I'm sorry, you're absolutely right. I read "10" instead of "100", since that's usually how you write it. +1 for your answer below! $\endgroup$ – pela Jul 9 '19 at 20:43

Textbooks typically express the flux vs. magnitude relation something like this:

$$m_2 - m_1 = -2.5 \log_{10} \frac{f_2}{f_1}$$

which we can transform into this:

$$\frac{f_2}{f_1} = 10^{(m_1 - m_2) / 2.5}$$

so the original version of your formula was correct.

If m2 - m1 = 0.5, then f2 / f1 = 0.63. Some readers find comparisons like "x% fainter" confusing, so I would say method 2 is more sensitive than method 1 by a factor of 1.58.


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