<p>Because you can't take the logarithm of something with dimensions only the log of a number.</p>
<p>Your magnitude would depend on what you measured f in. In other words, by choosing a set of units, you choose a zero point! Whether that unit is Jy, W/m<span class="math-container">$^2$</span> or the flux of some other object.</p>

CC BY-SA 4.0
2021-05-14T22:27:35.653Z
2021-05-14T22:27:35.653Z
ProfRob
<p>Let <span class="math-container">$f_*$</span> and <span class="math-container">$f_0$</span> be the observed flux of a star and a reference flux in a particular spectral band, and let <span class="math-container">$m_*$</span> and <span class="math-container">$m_0$</span> be their respective apparent magnitudes. Then the star's magnitude is given by <span class="math-container">$m_*-m_0=-2.5\log(f_*/f_0)$</span>.</p>
<p><a href="https://www.britannica.com/science/star-astronomy/Stellar-positions" rel="nofollow noreferrer"><em>Encyclopædia Britannica</em></a> maintains that</p>
<blockquote>
<p>All magnitude systems must have a reference, or zero, point.</p>
</blockquote>
<p>Why is this? It seems to me that one could just say <span class="math-container">$m_*=-2.5\log f_*$</span> and move on, no?</p>

CC BY-SA 4.0
2021-05-14T21:36:39.687Z
1. It's $m_{\star} - m_{0} = -2.5 \log(f_{\star} / f_{0})$.
2. How can you convert a magnitude back into a flux if you don't have a zero point?
3. Just saying $m_{\star} = -2.5 \log f_{\star}$ is equivalent to having a zero point with $f_{0} = 1$ in whatever the flux units are.
Peter Erwin
CC BY-SA 4.0
2021-05-14T21:37:30.470Z
@PeterErwin Fixed, just a brain glitch. thanks
gen-ℤ ready to perish
CC BY-SA 4.0
2021-05-14T21:45:48.853Z
4. For historical reasons, magnitudes have usually defined so that the star Vega has an apparent magnitude of zero, or else gives values not too broadly different. This means you have to set $f_{0}$ in the appropriate units so that, e.g., an observation of Vega will indeed give you a magnitude close to 0. (This is generically useful because if I tell you an object has an apparently magnitude of 10, you'll have a generally correct idea of how bright it is without having to look up how the magnitude was generated.)
Peter Erwin
<photometry><apparent-magnitude>
Why must all magnitude systems have a reference point?
gen-ℤ ready to perish
# Why must all magnitude systems have a reference point?

(Asked by: username_0 on 2021-05-14)

Let $f\_\*$ and $f\_0$ be the observed flux of a star and a reference flux in a particular spectral band, and let $m\_\*$ and $m\_0$ be their respective apparent magnitudes. Then the star's magnitude is given by $m\_\*-m\_0=-2.5\log(f\_\*/f\_0)$.

[*Encyclopædia Britannica*](https://www.britannica.com/science/star-astronomy/Stellar-positions) maintains that

> All magnitude systems must have a reference, or zero, point.

Why is this? It seems to me that one could just say $m\_\*=-2.5\log f\_\*$ and move on, no?

**Tags:** photometry, apparent-magnitude
> **Comments:**
> **username_1**: 1. It's $m_{\star} - m_{0} = -2.5 \log(f_{\star} / f_{0})$. 2. How can you convert a magnitude back into a flux if you don't have a zero point? 3. Just saying $m_{\star} = -2.5 \log f_{\star}$ is equivalent to having a zero point with $f_{0} = 1$ in whatever the flux units are.  
> **username_0**: @PeterErwin Fixed, just a brain glitch. thanks  
> **username_1**: 4. For historical reasons, magnitudes have usually defined so that the star Vega has an apparent magnitude of zero, or else gives values not too broadly different. This means you have to set $f_{0}$ in the appropriate units so that, e.g., an observation of Vega will indeed give you a magnitude close to 0. (This is generically useful because if I tell you an object has an apparently magnitude of 10, you'll have a generally correct idea of how bright it is without having to look up how the magnitude was generated.)  

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## 1 answers

### Answer by username_2 on 2021-05-14. Score: 3 | Accepted answer

Because you can't take the logarithm of something with dimensions only the log of a number.

Your magnitude would depend on what you measured f in. In other words, by choosing a set of units, you choose a zero point! Whether that unit is Jy, W/m$^2$ or the flux of some other object.


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