<p>If the normalised filter response function is $R_{\nu}$ then the measured  flux is
$$ F = \int f_{\nu} R_{\nu}\ d\nu  $$
The integration is done over the frequency range of the filter.</p>

<p>If you measure a flux through a filter then the process cannot be inverted exactly. However the <em>average</em> flux density can be found by dividing the total flux by the effective frequency range
$$\langle f_{\nu} \rangle = F/ \int R_{\nu} d\nu $$ </p>

CC BY-SA 3.0
2016-10-13T08:14:46.857Z
I'm still struggling with this. This measured flux depends on the filter curve -- how do we measure a true flux, or a true luminosity, at all, then? That is, something that is actually properly normalized to the true luminosity of the object? Will this not always depend on the filter?
user1991
CC BY-SA 3.0
2016-10-13T11:01:06.030Z
@John I don't understand your supplementary question. You cannot measure the luminosity of an object by observing through a single filter. All you can obtain is the average flux from the object in a particular wavelength range and that is done with eqn 2 of my answer. $F$ is what you measure and it depends on the true flux distribution of the object and the instrument response function $R_{\nu}$.
ProfRob
CC BY-SA 3.0
2016-10-13T14:01:39.527Z
I guess my question boils down to this: From this average $f_{\nu}$ that we have for a filter, how would one go about calculating the corresponding luminosity *in this band*, $L_{\nu}$? These are often used in literature.
user1991
CC BY-SA 3.0
2016-10-13T14:53:38.647Z
@John ?? $L_\nu = 4\pi d^2 f_{\nu}$, where $d$ is the distance and this assumes no extinction.
ProfRob
CC BY-SA 3.0
2016-10-17T07:04:31.397Z
I'm sorry for being so unclear - it's mainly because my issue was not quite clear to me either.
It boils down to this: Suppose authors in a paper are talking about integrating a spectrum in a particular filter to obtain the luminosity of a galaxy, *in units of $L_{\odot}$. What does that mean? Is that an approximation of the bolometric luminosity based on the average flux in this filter? Not clear to me.
user1991
CC BY-SA 3.0
2016-10-17T08:40:45.133Z
@John Give an example and ask a different question.
ProfRob
CC BY-SA 3.0
2016-10-10T17:44:04.620Z
2016-10-10T18:08:13.407Z
2016-10-10T18:08:13.407Z
ProfRob
<p>The definition of flux and flux density is fairly straightforward. However, I'm musing over the practical usage of it. </p>

<p>For one thing, if you observe a source through a filter, how would one go from the total observed energy/number of photons to the flux <em>per unit wavelength</em>? It's not a matter of simply dividing by the width of the filter -- a filter can have any exotic form of course.</p>

<p>So, how is this derived?
Related, <em>given</em> a $f_{\nu}$ in some filter, what would be the way to go to total flux within that filter?</p>

<observational-astronomy><luminosity>
Flux density to flux
user1991
# Flux density to flux

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

The definition of flux and flux density is fairly straightforward. However, I'm musing over the practical usage of it.

For one thing, if you observe a source through a filter, how would one go from the total observed energy/number of photons to the flux *per unit wavelength*? It's not a matter of simply dividing by the width of the filter -- a filter can have any exotic form of course.

So, how is this derived?
Related, *given* a $f\_{\nu}$ in some filter, what would be the way to go to total flux within that filter?

**Tags:** observational-astronomy, luminosity

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## 1 answers

### Answer by username_1 on 2016-10-10. Score: 3 | Accepted answer

If the normalised filter response function is $R\_{\nu}$ then the measured flux is
$$ F = \int f\_{\nu} R\_{\nu}\ d\nu $$
The integration is done over the frequency range of the filter.

If you measure a flux through a filter then the process cannot be inverted exactly. However the *average* flux density can be found by dividing the total flux by the effective frequency range
$$\langle f\_{\nu} \rangle = F/ \int R\_{\nu} d\nu $$

> **Comments:**
> **username_0**: I'm still struggling with this. This measured flux depends on the filter curve -- how do we measure a true flux, or a true luminosity, at all, then? That is, something that is actually properly normalized to the true luminosity of the object? Will this not always depend on the filter?  
> **username_1**: @John I don't understand your supplementary question. You cannot measure the luminosity of an object by observing through a single filter. All you can obtain is the average flux from the object in a particular wavelength range and that is done with eqn 2 of my answer. $F$ is what you measure and it depends on the true flux distribution of the object and the instrument response function $R_{\nu}$.  
> **username_0**: I guess my question boils down to this: From this average $f_{\nu}$ that we have for a filter, how would one go about calculating the corresponding luminosity *in this band*, $L_{\nu}$? These are often used in literature.  
> **username_1**: @John ?? $L_\nu = 4\pi d^2 f_{\nu}$, where $d$ is the distance and this assumes no extinction.  
> **username_0**: I'm sorry for being so unclear - it's mainly because my issue was not quite clear to me either. It boils down to this: Suppose authors in a paper are talking about integrating a spectrum in a particular filter to obtain the luminosity of a galaxy, *in units of $L_{\odot}$. What does that mean? Is that an approximation of the bolometric luminosity based on the average flux in this filter? Not clear to me.  
> **username_1**: @John Give an example and ask a different question.  

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