<p>When doing actual calculations, astrophysicists tend to normalise and approximate everything.</p>
<p>Instead of talking about brightness, we think of radiation output in terms of some spectral flux (Watt, erg/s, Jansky, number of photons per unit time...) depending on what you're studying.</p>
<p>At any rate, I believe the answer you're looking for is simple luminosity (Watts): radiant power emitted by an object over time.</p>
<p>It is proportional to the temperature and surface area of the object:</p>
<p><span class="math-container">$L=4\pi\sigma R^2 T^4$</span></p>
<p>As you already intuited, the luminosities of two different objects are related to their absolute magnitudes:</p>
<p><span class="math-container">$M_1-M_2=-2.5\log\frac{L_1}{L_2}$</span></p>

CC BY-SA 4.0
2022-12-02T16:43:52.997Z
2022-12-02T16:43:52.997Z
Marco Leonardi
<p>I know the surface brightness is calculated by taking the apparent magnitude of an object and then divide by the size of the thing. But this is more of &quot;observed&quot; surface brightness, how do you calculate a sort of &quot;absolute&quot; surface brightness? I was thinking about calculating the absolute magnitude (taking into account the cosmology, K correction, etc) and then calculating the physical size of the object from the apparent size in the sky (i.e. arcsec-&gt;kpc, using the angular diameter distance).</p>
<p>The surface brightness calculated this way would have units of absolute magnitudes/kpc, which is more reasonable to me (especially for comparing sources at different redshifts), but I have never seen it, only using apparent magnitudes and physical size, or simply correcting by <span class="math-container">$(1+z)^4$</span> for cosmological dimming. Why is this?</p>

CC BY-SA 4.0
2022-12-01T18:46:55.667Z
Surface brightness is determined using exclusively our observations of brightness of the object: it is "antropocentric", If you want to include suggested corrections, you enter into realm of astrophysics, where the objects are studied absolutely with no reference to our frame of reference. Here, our direction is not special in any way. The problem with your absolute surface brightness is that it is not absolute at all, since it would be calculated from the surface brightness, which is still linked to our direction of observation. As such, it would most probably be of no interest.
User123
<galaxy><photometry><apparent-magnitude><size><absolute-magnitude>
Surface Brightness Definition
JOAQUÍN HERNÁNDEZ
# Surface Brightness Definition

(Asked by: username_0 on 2022-12-01)

I know the surface brightness is calculated by taking the apparent magnitude of an object and then divide by the size of the thing. But this is more of "observed" surface brightness, how do you calculate a sort of "absolute" surface brightness? I was thinking about calculating the absolute magnitude (taking into account the cosmology, K correction, etc) and then calculating the physical size of the object from the apparent size in the sky (i.e. arcsec->kpc, using the angular diameter distance).

The surface brightness calculated this way would have units of absolute magnitudes/kpc, which is more reasonable to me (especially for comparing sources at different redshifts), but I have never seen it, only using apparent magnitudes and physical size, or simply correcting by $(1+z)^4$ for cosmological dimming. Why is this?

**Tags:** galaxy, photometry, apparent-magnitude, size, absolute-magnitude
> **Comments:**
> **username_1**: Surface brightness is determined using exclusively our observations of brightness of the object: it is "antropocentric", If you want to include suggested corrections, you enter into realm of astrophysics, where the objects are studied absolutely with no reference to our frame of reference. Here, our direction is not special in any way. The problem with your absolute surface brightness is that it is not absolute at all, since it would be calculated from the surface brightness, which is still linked to our direction of observation. As such, it would most probably be of no interest.  

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## 1 answers

### Answer by username_2 on 2022-12-02. Score: 2

When doing actual calculations, astrophysicists tend to normalise and approximate everything.

Instead of talking about brightness, we think of radiation output in terms of some spectral flux (Watt, erg/s, Jansky, number of photons per unit time...) depending on what you're studying.

At any rate, I believe the answer you're looking for is simple luminosity (Watts): radiant power emitted by an object over time.

It is proportional to the temperature and surface area of the object:

$L=4\pi\sigma R^2 T^4$

As you already intuited, the luminosities of two different objects are related to their absolute magnitudes:

$M\_1-M\_2=-2.5\log\frac{L\_1}{L\_2}$


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