<p>Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name "distance modulus" when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for <A HREF="https://en.wikipedia.org/wiki/Radiative_transfer">spectral radiance</A> (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the inverse square law, though, so there's no risk that we're confusing them here.</p>

CC BY-SA 3.0
2016-10-19T22:39:04.237Z
Ok thank you. So since one distance has to be set to 10pc, does this mean that I'm not able to use this to calculate the apparent brightness of a planet within our solar system?
theoreticalmasterrace
CC BY-SA 3.0
2016-10-19T22:41:03.877Z
Not at all. The formula works whether one distance is set to 10 parsecs or not, it just has a special name when one of them is set there (I don't recall which one, because I never use magnitudes, myself).
Sean Lake
CC BY-SA 3.0
2016-10-19T22:49:47.363Z
Absolute magnitude $=$ the magnitude at 10 parsecs.
ProfRob
CC BY-SA 3.0
2016-10-19T22:08:59.653Z
2016-10-19T22:08:59.653Z
Sean Lake
<p>I know that the equation for the difference in apparent magnitudes can be represented as: 
$\Delta m=-2.5\log\left(\dfrac{I_{1}}{I_{2}}\right)$<br>
and the apparent brightness of a celestial object is shown as:<br>
$\ B=\dfrac{L}{4\pi\ D^2}$  </p>

<p>Some people in my class told me that$\ B$ and$\ I$ were the same thing but used in different cases. This explanation didn't help me though. While playing around with the equations, I found this nice relation:<br>
$\ \dfrac{B_{1}}{B_{2}}=\left[\dfrac {\left(\dfrac{L}{4\pi\ D_{2}^2}\right)}{\left(\dfrac{L}{4\pi\ D_{1}^2}\right)} \right]$<br>
$\ \space\ \space\ \space\ ={\dfrac{4\pi\ D_{1}L}{4\pi\ D_{2}^2L}}$<br>
$\ \space\ \space\ \space\ ={\dfrac{D_{1}^2}{D_{2}^2}}$  </p>

<p>So if$\ {B_{1}}={I_{1}}$ and$\ B_{2}=I_{2}$holds true, is this following equation also true?  </p>

<p>$\ {\dfrac{D_{1}^2}{D_{2}^2}} \space\ = 10^\left(-{\dfrac{\Delta {m}}{2.5}}\right)$<br>
$\ \space\ \space\ \space\ \space\ =\dfrac{1}{10^\left({\dfrac{\Delta {m}}{2.5}}\right)}$   </p>

<hr>

<p>However, if$\ B$ and$\ I$ are not the same, what is the difference between the two measurements? And can either one be used to accurately measure how bright a non-luminous celestial body (ie. a planet) appears from Earth?  </p>

<p>Also... does $\ m$ have an SI unit? I've never used any units when I solve for it</p>

<apparent-magnitude><magnitude>
What's the difference between apparent brightness and apparent magnitude? Are they the same?
theoreticalmasterrace
# What's the difference between apparent brightness and apparent magnitude? Are they the same?

(Asked by: username_0 on 2016-10-19)

I know that the equation for the difference in apparent magnitudes can be represented as:
$\Delta m=-2.5\log\left(\dfrac{I\_{1}}{I\_{2}}\right)$  
and the apparent brightness of a celestial object is shown as:  
$\ B=\dfrac{L}{4\pi\ D^2}$

Some people in my class told me that$\ B$ and$\ I$ were the same thing but used in different cases. This explanation didn't help me though. While playing around with the equations, I found this nice relation:  
$\ \dfrac{B\_{1}}{B\_{2}}=\left[\dfrac {\left(\dfrac{L}{4\pi\ D\_{2}^2}\right)}{\left(\dfrac{L}{4\pi\ D\_{1}^2}\right)} \right]$  
$\ \space\ \space\ \space\ ={\dfrac{4\pi\ D\_{1}L}{4\pi\ D\_{2}^2L}}$  
$\ \space\ \space\ \space\ ={\dfrac{D\_{1}^2}{D\_{2}^2}}$

So if$\ {B\_{1}}={I\_{1}}$ and$\ B\_{2}=I\_{2}$holds true, is this following equation also true?

$\ {\dfrac{D\_{1}^2}{D\_{2}^2}} \space\ = 10^\left(-{\dfrac{\Delta {m}}{2.5}}\right)$  
$\ \space\ \space\ \space\ \space\ =\dfrac{1}{10^\left({\dfrac{\Delta {m}}{2.5}}\right)}$

---

However, if$\ B$ and$\ I$ are not the same, what is the difference between the two measurements? And can either one be used to accurately measure how bright a non-luminous celestial body (ie. a planet) appears from Earth?

Also... does $\ m$ have an SI unit? I've never used any units when I solve for it

**Tags:** apparent-magnitude, magnitude

---

## 1 answers

### Answer by username_1 on 2016-10-19. Score: 5 | Accepted answer

Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name "distance modulus" when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for [spectral radiance](https://en.wikipedia.org/wiki/Radiative_transfer) (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the inverse square law, though, so there's no risk that we're confusing them here.

> **Comments:**
> **username_0**: Ok thank you. So since one distance has to be set to 10pc, does this mean that I'm not able to use this to calculate the apparent brightness of a planet within our solar system?  
> **username_1**: Not at all. The formula works whether one distance is set to 10 parsecs or not, it just has a special name when one of them is set there (I don't recall which one, because I never use magnitudes, myself).  
> **username_2**: Absolute magnitude $=$ the magnitude at 10 parsecs.  

---

