<p>(P.S. this is naked eyed)</p>
<p>Magnitudes are easily defined for point-like objects i.e. stars. However, for extended objects, the &quot;magnitudes&quot; quoted are actually integrated magnitudes.</p>
<p>For instance, Wikipedia says that M7 (Ptolemy's cluster) is of visual magnitude 3.3. Does this mean that we perceive M7 as a 3.3 magnitude star? Definitely not.</p>
<p>More quantitatively, when given integrated magnitudes, are we assuming that the individual elements are not clearly resolvable? (e.g. stars in open/globular cluster; what if they are?)</p>
<p>How do we know if it is visible? Is there any figure equivalent to the 6th apparent magnitude of the dimmest star observable? It might possibly be calculated from the integrated magnitude.</p>
<p>Thanks in advance!</p>

CC BY-SA 4.0
2022-07-28T05:32:53.970Z
The problem with "how do we know if it's visible?" is that most extended objects vary in surface brightness considerably.  So you might be able to see parts of it, but not all.
Greg Miller
CC BY-SA 4.0
2022-07-28T15:26:46.943Z
In theory, you could look at the visual magnitude for a grid at the limit of human visual acuity, but I'm not sure that would work. It also differs per person, and this would imply a person with lower visual acuity could see fainter objects, which seems counterintuitive.
Barry Carter
CC BY-SA 4.0
2022-07-28T22:14:56.680Z
You would need to know the surface brightness profile, and your eye's angular resolution. If at any point (likely in the center) the SB of your object, measured in magnitudes per area, times the area of your resolution, exceeds your eye's limiting threshold, you can see the object.
pela
CC BY-SA 4.0
2024-02-06T02:44:19.483Z
In the specific case you mention, for M 7 being of magnitude 3.3, it means that when you combine the magnitudes of its individual components, you get a total magnitude of 3.3. Please note that adding magnitudes is not as simple as 1+1=2, as the values are logarithmic.
Pierre Paquette
<observational-astronomy><amateur-observing><apparent-magnitude><naked-eye>
How do stellar magnitude and integrated magnitudes compare, perceptually?
Cheng
# How do stellar magnitude and integrated magnitudes compare, perceptually?

(Asked by: username_0 on 2022-07-28)

(P.S. this is naked eyed)

Magnitudes are easily defined for point-like objects i.e. stars. However, for extended objects, the "magnitudes" quoted are actually integrated magnitudes.

For instance, Wikipedia says that M7 (Ptolemy's cluster) is of visual magnitude 3.3. Does this mean that we perceive M7 as a 3.3 magnitude star? Definitely not.

More quantitatively, when given integrated magnitudes, are we assuming that the individual elements are not clearly resolvable? (e.g. stars in open/globular cluster; what if they are?)

How do we know if it is visible? Is there any figure equivalent to the 6th apparent magnitude of the dimmest star observable? It might possibly be calculated from the integrated magnitude.

Thanks in advance!

**Tags:** observational-astronomy, amateur-observing, apparent-magnitude, naked-eye
> **Comments:**
> **username_1**: The problem with "how do we know if it's visible?" is that most extended objects vary in surface brightness considerably.  So you might be able to see parts of it, but not all.  
> **username_2**: In theory, you could look at the visual magnitude for a grid at the limit of human visual acuity, but I'm not sure that would work. It also differs per person, and this would imply a person with lower visual acuity could see fainter objects, which seems counterintuitive.  
> **username_3**: You would need to know the surface brightness profile, and your eye's angular resolution. If at any point (likely in the center) the SB of your object, measured in magnitudes per area, times the area of your resolution, exceeds your eye's limiting threshold, you can see the object.  
> **username_4**: In the specific case you mention, for M 7 being of magnitude 3.3, it means that when you combine the magnitudes of its individual components, you get a total magnitude of 3.3. Please note that adding magnitudes is not as simple as 1+1=2, as the values are logarithmic.  

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## 0 answers

