<p>Textbooks typically express the flux vs. magnitude relation something like this:</p>

<p><span class="math-container">$$m_2 - m_1 = -2.5 \log_{10} \frac{f_2}{f_1}$$</span></p>

<p>which we can transform into this:</p>

<p><span class="math-container">$$\frac{f_2}{f_1} = 10^{(m_1 - m_2) / 2.5}$$</span></p>

<p>so the original version of your formula was correct.</p>

<p>If m<sub>2</sub> - m<sub>1</sub> = 0.5, then f<sub>2</sub> / f<sub>1</sub> = 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.</p>

CC BY-SA 4.0
2019-07-09T18:54:39.063Z
2019-07-09T20:31:19.437Z
2019-07-09T20:31:19.437Z
Mike G
<p>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?</p>

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

<p>Thus, <span class="math-container">$f_2$</span> is about <span class="math-container">$40\% f_1$</span> which means about <span class="math-container">$60\%$</span> fainter. I have never worked with magnitudes before, so is this a valid statement?</p>

CC BY-SA 4.0
2019-07-09T09:50:28.853Z
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.
pela
CC BY-SA 4.0
2019-07-09T11:00:19.333Z
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!
Philipp
CC BY-SA 4.0
2019-07-09T13:40:55.213Z
@pela are you sure? $100^{1/5} = 10^{1/2.5}$
Mike G
CC BY-SA 4.0
2019-07-09T20:43:11.570Z
@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!
pela
<apparent-magnitude>
Comparing different magnitudes. Is this statement correct?
Philipp
# Comparing different magnitudes. Is this statement correct?

(Asked by: username_0 on 2019-07-09)

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?

**Tags:** apparent-magnitude

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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!  
> **username_2**: @username_1 are you sure? $100^{1/5} = 10^{1/2.5}$  
> **username_1**: @username_2 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!  

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## 1 answers

### Answer by username_2 on 2019-07-09. Score: 5

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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