<p>Here is my attempt to reconcile your calculations.</p>
<p>If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by
<span class="math-container">$$f_{\nu} = 10^{(-48.6-22.5)/2.5} =  3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$</span></p>
<p>If the distance is 1991 Mpc, then the absolute g magnitude is
<span class="math-container">$$ M_g = m_g - 5\log d + 5 = -19.0$$</span></p>
<p>The absolute AB magnitude of the Sun <em>in the g-band</em> (from your source) is 5.12.</p>
<p>The latter tells us that the ratio of the &quot;<em>g-band</em> luminosites&quot; of the galaxy and the Sun are given by
<span class="math-container">$$\frac{L_g}{L_{\odot,g}} = 10^{0.4(M_{{\odot},g} - M_g)} = 4.45\times 10^{9}$$</span></p>
<p>One cannot say more than this, in particular one cannot calculate the luminosity of the galaxy, without knowing more about its spectrum.</p>
<p><strong>Also note that the equation above cannot be used to find the ratio of flux in one band to bolometric flux, as I think you are trying to do</strong>. To see this, consider that the absolute V-band magnitude and bolometric magnitude of the Sun are almost the same. This does <em>not</em> mean that all the flux from the Sun emerges in the V band!</p>
<p>Your second method requires a figure for <span class="math-container">$\Delta \nu$</span>, but you haven't said what you have used. The g-filter has a width of around <span class="math-container">$\Delta \lambda =100$</span> nm. Using <span class="math-container">$\lambda \nu = c$</span> we can say
<span class="math-container">$$ \Delta \nu  = |c\Delta \lambda/\lambda^2| = 1.28\times 10^{14}\ {\rm Hz}$$</span>
<span class="math-container">$$ L_g = 4\pi d^2 f_{\nu} \Delta \nu = 2.2 \times 10^{42} {\rm erg/s}$$</span></p>
<p>The two figures would agree precisely if the &quot;g-band luminosity&quot; of the Sun were <span class="math-container">$4.9 \times 10^{32}$</span> erg/s, or in other words if 13% of the solar luminosity emerged in the g-band as defined above. This does not sound unreasonable, but requires a detailed integration of the solar spectrum over the actual g-band filter profile.</p>
<p><strong>Additional:</strong> Just to check the numbers above, we can estimate the flux density of the Sun (in wavelength units, above the atmosphere) to be about 1.7 W m<span class="math-container">$^{-2}$</span> nm<span class="math-container">$^{-1}$</span> at 467 nm (just search on 'solar spectrum' for many examples) at a distance of <span class="math-container">$1.5 \times 10^{11}$</span> m. Putting these numbers together, I calculate a &quot;g-band luminosity&quot; for the Sun of
<span class="math-container">$$ L_{\odot,g}=4\pi d^2 f_{\lambda}\Delta \lambda = 4.8\times 10^{32}\ {\rm erg/s} \ ,$$</span>
which is indeed 13% of the solar bolometric luminosity, and thus
<span class="math-container">$$\frac{L_g}{L_{\odot}} = 0.13\times 4.45\times 10^{9} = 5.79\times 10^8\ .$$</span></p>

CC BY-SA 3.0
2018-02-25T04:13:22.437Z
13% is too much. This is supposed to be exact since it's almost doing the same calculation with two different methods. I suspect that $L_g$ via flux is correct given we use luminosity distance (distance adjusted for redshift), but further adjustments are needed to use magnitude and we might not be fine using $d_L$ for both methods. I guess what I'm saying is the problem might be due to redshift, but I have to think about it more.
Miladiouss
CC BY-SA 3.0
2018-02-25T04:16:35.517Z
Also, people don't do $L_g/L_{\odot,g}$, they just do $L_g/L_{\odot}$ which is just an assumed constant of $3.848 \times 10^26$ W. The point is unit having simple units not comparison necessarily.
Miladiouss
CC BY-SA 3.0
2018-02-25T06:21:24.517Z
@Miladiouss on what basis do say that 13% is too much? The "flux" method is not exact, since you apparently don't know the galaxy spectrum and assuming a top hat filter function is incorrect.
ProfRob
CC BY-SA 4.0
2018-02-22T11:25:24.393Z
2023-04-17T17:40:50.723Z
2023-04-17T17:40:50.723Z
ProfRob
<p>How can I calculate the (non-bolometric) luminosity $L$ of a galaxy (or a star for that matter) over a given band from its AB apparent magnitude $m_{AB}$ over that band and its luminosity distance $d_L$?</p>

<p>For instance, consider the g-band which typically has a $\lambda_{eff} = 467 \text{ nm}$ and a $\Delta \lambda = 100 \text{ nm}$. Given this galaxy has an apparent AB magnitude of $m_g = 22.5$ and luminosity distance of $1991 \text{ Mpc}$ (i.e. $z = 0.355$ if you are curious), what is its luminosity? I know I shold use the following equation, but I don't know what value to pick for $M_\odot$.</p>

<p>$$L/L_\odot = 10^{0.4(M_\odot - M)}$$</p>

<p>I tried 5.12 from <a href="http://www.astronomy.ohio-state.edu/~martini/usefuldata.html" rel="noreferrer">a website</a> which is the value of $M_\odot$ in g-band but it gives me a different answer compared to when I calculate luminosity using flux:</p>

<p>$$L = 4 \pi d_{L}^2 f_\nu \Delta \nu$$</p>

<p>where $f_\nu$ is flux density and $\Delta \nu$ is the frequency width of the band (this is an approximation to the integration over frequency, assuming bands are step functions). So, how can I find luminosity using absolute magnitude?</p>

<p>Please provide your references.</p>

CC BY-SA 3.0
2018-02-16T13:13:24.267Z
How different were your two answers?  In addition, 100nm is a pretty wide band to assume uniform flux density.
Carl Witthoft
CC BY-SA 3.0
2018-02-21T07:06:55.657Z
You are confused. $L$ is usually reserved for a luminosity integrated over all wavelengths. This *cannot* be calculated simply from the g-band magnitude and distance without knowing more about the galaxy.
ProfRob
CC BY-SA 3.0
2018-02-21T18:45:49.837Z
@CarlWitthoft, I get magnitude wrong by 1.
Miladiouss
CC BY-SA 3.0
2018-02-21T18:45:53.893Z
@RobJeffries, In the above question L is the luminosity over the g-band. And it doesn't matter if the flux is not constant because a telescope measures the average flux over a band. So, in this question, L is the luminosity over the g-band and it is not equal to total luminosity and I have made my notation clear by stating "non-bolometric luminosity". Feel free to edit it to $L_{g-band}$.
Miladiouss
CC BY-SA 3.0
2018-02-21T18:52:26.450Z
Then what are $L_{\odot}$ and $M_{\odot}$ supposed to be? These have very clear defined meanings, and they are bolometric quantities. You also need to tell us what flux (density) and $\Delta \nu$ you used if we are meant to be looking for errors.
ProfRob
CC BY-SA 3.0
2018-02-21T19:01:42.227Z
@RobJeffries, $L_\odot$ is a constant equal to $3.848 \times 10^{26}$ W. $M_\odot$ on the other hand, should be band dependent. What's its value then? That's my question! I have stated $\Delta \nu$ and $f$ can be calculated from $m_g$.
Miladiouss
<luminosity><instruments><absolute-magnitude>
How to calculate luminosity in g-band from absolute AB magnitude and luminosity distance?
Miladiouss
# How to calculate luminosity in g-band from absolute AB magnitude and luminosity distance?

(Asked by: username_0 on 2018-02-16)

How can I calculate the (non-bolometric) luminosity $L$ of a galaxy (or a star for that matter) over a given band from its AB apparent magnitude $m\_{AB}$ over that band and its luminosity distance $d\_L$?

For instance, consider the g-band which typically has a $\lambda\_{eff} = 467 \text{ nm}$ and a $\Delta \lambda = 100 \text{ nm}$. Given this galaxy has an apparent AB magnitude of $m\_g = 22.5$ and luminosity distance of $1991 \text{ Mpc}$ (i.e. $z = 0.355$ if you are curious), what is its luminosity? I know I shold use the following equation, but I don't know what value to pick for $M\_\odot$.

$$L/L\_\odot = 10^{0.4(M\_\odot - M)}$$

I tried 5.12 from [a website](http://www.astronomy.ohio-state.edu/~martini/usefuldata.html) which is the value of $M\_\odot$ in g-band but it gives me a different answer compared to when I calculate luminosity using flux:

$$L = 4 \pi d\_{L}^2 f\_\nu \Delta \nu$$

where $f\_\nu$ is flux density and $\Delta \nu$ is the frequency width of the band (this is an approximation to the integration over frequency, assuming bands are step functions). So, how can I find luminosity using absolute magnitude?

Please provide your references.

**Tags:** luminosity, instruments, absolute-magnitude
> **Comments:**
> **username_1**: How different were your two answers?  In addition, 100nm is a pretty wide band to assume uniform flux density.  
> **username_2**: You are confused. $L$ is usually reserved for a luminosity integrated over all wavelengths. This *cannot* be calculated simply from the g-band magnitude and distance without knowing more about the galaxy.  
> **username_0**: @CarlWitthoft, I get magnitude wrong by 1.  
> **username_0**: @RobJeffries, In the above question L is the luminosity over the g-band. And it doesn't matter if the flux is not constant because a telescope measures the average flux over a band. So, in this question, L is the luminosity over the g-band and it is not equal to total luminosity and I have made my notation clear by stating "non-bolometric luminosity". Feel free to edit it to $L_{g-band}$.  
> **username_2**: Then what are $L_{\odot}$ and $M_{\odot}$ supposed to be? These have very clear defined meanings, and they are bolometric quantities. You also need to tell us what flux (density) and $\Delta \nu$ you used if we are meant to be looking for errors.  
> **username_0**: @RobJeffries, $L_\odot$ is a constant equal to $3.848 \times 10^{26}$ W. $M_\odot$ on the other hand, should be band dependent. What's its value then? That's my question! I have stated $\Delta \nu$ and $f$ can be calculated from $m_g$.  

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## 1 answers

### Answer by username_2 on 2018-02-22. Score: 4

Here is my attempt to reconcile your calculations.

If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by
$$f\_{\nu} = 10^{(-48.6-22.5)/2.5} = 3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$

If the distance is 1991 Mpc, then the absolute g magnitude is
$$ M\_g = m\_g - 5\log d + 5 = -19.0$$

The absolute AB magnitude of the Sun *in the g-band* (from your source) is 5.12.

The latter tells us that the ratio of the "*g-band* luminosites" of the galaxy and the Sun are given by
$$\frac{L\_g}{L\_{\odot,g}} = 10^{0.4(M\_{{\odot},g} - M\_g)} = 4.45\times 10^{9}$$

One cannot say more than this, in particular one cannot calculate the luminosity of the galaxy, without knowing more about its spectrum.

**Also note that the equation above cannot be used to find the ratio of flux in one band to bolometric flux, as I think you are trying to do**. To see this, consider that the absolute V-band magnitude and bolometric magnitude of the Sun are almost the same. This does *not* mean that all the flux from the Sun emerges in the V band!

Your second method requires a figure for $\Delta \nu$, but you haven't said what you have used. The g-filter has a width of around $\Delta \lambda =100$ nm. Using $\lambda \nu = c$ we can say
$$ \Delta \nu = |c\Delta \lambda/\lambda^2| = 1.28\times 10^{14}\ {\rm Hz}$$
$$ L\_g = 4\pi d^2 f\_{\nu} \Delta \nu = 2.2 \times 10^{42} {\rm erg/s}$$

The two figures would agree precisely if the "g-band luminosity" of the Sun were $4.9 \times 10^{32}$ erg/s, or in other words if 13% of the solar luminosity emerged in the g-band as defined above. This does not sound unreasonable, but requires a detailed integration of the solar spectrum over the actual g-band filter profile.

**Additional:** Just to check the numbers above, we can estimate the flux density of the Sun (in wavelength units, above the atmosphere) to be about 1.7 W m$^{-2}$ nm$^{-1}$ at 467 nm (just search on 'solar spectrum' for many examples) at a distance of $1.5 \times 10^{11}$ m. Putting these numbers together, I calculate a "g-band luminosity" for the Sun of
$$ L\_{\odot,g}=4\pi d^2 f\_{\lambda}\Delta \lambda = 4.8\times 10^{32}\ {\rm erg/s} \ ,$$
which is indeed 13% of the solar bolometric luminosity, and thus
$$\frac{L\_g}{L\_{\odot}} = 0.13\times 4.45\times 10^{9} = 5.79\times 10^8\ .$$

> **Comments:**
> **username_0**: 13% is too much. This is supposed to be exact since it's almost doing the same calculation with two different methods. I suspect that $L_g$ via flux is correct given we use luminosity distance (distance adjusted for redshift), but further adjustments are needed to use magnitude and we might not be fine using $d_L$ for both methods. I guess what I'm saying is the problem might be due to redshift, but I have to think about it more.  
> **username_0**: Also, people don't do $L_g/L_{\odot,g}$, they just do $L_g/L_{\odot}$ which is just an assumed constant of $3.848 \times 10^26$ W. The point is unit having simple units not comparison necessarily.  
> **username_2**: @Miladiouss on what basis do say that 13% is too much? The "flux" method is not exact, since you apparently don't know the galaxy spectrum and assuming a top hat filter function is incorrect.  

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