Skip to content
Jot
Home
Trending
Top
Tags
Groups
Theme
Privacy·Terms
Sign in
Home
Search
Trending
Sign in
Limiting magnitude from first principles — Jot

Limiting magnitude from first principles

T
Tomas Vrba
·November 30, 2025·1,621 words·9 min read
astrophotographylight-pollutionmathsmeasurement
☆stars: 13⑂forks: 1
Part 1 of 3·Backyard photons→

Magnitude is a ratio wearing the costume of a quantity. Nearly every error I have made in twelve years of measuring faint things came from forgetting that sentence.

What the number is

A magnitude is defined against a reference flux, never against zero:

mx=−2.5log⁡10 ⁣(FxFx,0)m_x = -2.5\log_{10}\!\left(\frac{F_x}{F_{x,0}}\right)mx​=−2.5log10​(Fx,0​Fx​​)

FxF_xFx​ is the flux received in some passband xxx, and Fx,0F_{x,0}Fx,0​ is the zero point flux for that same band. Change the band, change the number. The minus sign is inherited from the Greeks, who called the brightest stars first class and the faintest sixth, and we have been apologising for it since.

The form you actually use is the difference, because the zero point cancels and you no longer need to know it:

m1−m2=−2.5log⁡10 ⁣(F1F2)m_1 - m_2 = -2.5\log_{10}\!\left(\frac{F_1}{F_2}\right)m1​−m2​=−2.5log10​(F2​F1​​)

Where the fifth root of 100 comes from

Pogson, in 1856, wanted the six inherited classes kept and the scale made arithmetic. He fixed a first magnitude star as one hundred times as bright as a sixth magnitude star. Five steps, factor of one hundred.

Then require that the scale be geometric, which is what "equal steps look equal" means for a logarithmic sense organ. One step is a fixed ratio rrr, and five steps multiply:

r5=100⟹r=1001/5=102/5=2.511886…r^{5} = 100 \quad\Longrightarrow\quad r = 100^{1/5} = 10^{2/5} = 2.511886\ldotsr5=100⟹r=1001/5=102/5=2.511886…

That is the entire origin of the 2.512 you see quoted everywhere. It is also where the 2.5 in the definition comes from, since log⁡10r=2/5\log_{10} r = 2/5log10​r=2/5 exactly, so −2.5log⁡10-2.5\log_{10}−2.5log10​ of a flux ratio counts steps of rrr and nothing else.

Two consequences worth writing on the inside of your eyelids.

Magnitudes add when fluxes multiply, so you can never average magnitudes. To combine two sources you convert both to flux, add the fluxes, convert back. People do this wrong with stacked frames and with sky readings constantly.

And 0.1 magnitudes is a flux ratio of 100.1/2.5=100.04=1.09610^{0.1/2.5} = 10^{0.04} = 1.096100.1/2.5=100.04=1.096. Nine and a half percent. So 0.1 mag/arcsec^2 of extra sky brightness is not a rounding error, it is a tenth of your exposure time.

Surface brightness, which is a different animal

mag/arcsec^2 is a magnitude per unit solid angle. It is not a flux and it is not a total. An object's integrated magnitude and its mean surface brightness are related through its area:

μ=m+2.5log⁡10A\mu = m + 2.5\log_{10} Aμ=m+2.5log10​A

with AAA in square arcseconds. This is why a magnitude 10 galaxy can be far harder than a magnitude 12 star. The star puts all of its photons in one place and the galaxy spreads them across ninety thousand square arcseconds.

The naked eye

The best modelling I know of for visual thresholds is Crumey (2014)1, who replaced the Hecht formula that light pollution studies had used since 1947. For a sky background of surface brightness μsky\mu_{\text{sky}}μsky​ in mag/arcsec^2 above 21:

m=0.4260 μsky−2.3650−2.5log⁡Fm = 0.4260\,\mu_{\text{sky}} - 2.3650 - 2.5\log Fm=0.4260μsky​−2.3650−2.5logF

and for the two lower ranges,

m=0.383 μsky−1.44−2.5log⁡F(20<μsky<22)m = 0.383\,\mu_{\text{sky}} - 1.44 - 2.5\log F \qquad (20 < \mu_{\text{sky}} < 22)m=0.383μsky​−1.44−2.5logF(20<μsky​<22)

m=0.27 μsky+0.8−2.5log⁡F(18<μsky<20)m = 0.27\,\mu_{\text{sky}} + 0.8 - 2.5\log F \qquad (18 < \mu_{\text{sky}} < 20)m=0.27μsky​+0.8−2.5logF(18<μsky​<20)

FFF is a field factor covering the observer and the viewing situation. Crumey speculated it lies between about 1.4 and 2.4 for most people, with 2 typical. Larger FFF gives a shallower limit.

First thing I did was check the three expressions against each other in their overlaps, because a set of piecewise fits that disagree at the seams is a set of fits I am not going to trust. At μsky=21.3\mu_{\text{sky}} = 21.3μsky​=21.3 the first two give 5.956 and 5.965. At 20.0 the last two give 5.447 and 5.467. They agree to about two hundredths. Good.

Second thing I did was put them next to the naked eye limits in the Bortle table, and this is where it gets awkward. Bortle published his scale in the February 2001 issue of Sky and Telescope. The sky brightness column that circulates alongside it now comes from a nomogram, not from meters pointed at the same skies at the same time as the star counts. Here is the comparison at F=2F = 2F=2, using each formula in its own valid range and taking the midpoint of each Bortle class:

Bortle classSQM (mag/arcsec^2)Bortle's naked eye limitCrumey, F = 2Gap
1 excellent21.76 to 22.07.6 to 8.06.201.4 to 1.8
2 truly dark21.6 to 21.757.1 to 7.56.121.0 to 1.4
3 rural21.3 to 21.66.6 to 7.06.020.6 to 1.0
4 brighter rural20.8 to 21.36.3 to 6.55.870.4 to 0.6
4.5 transition20.3 to 20.86.1 to 6.35.680.4 to 0.6
5 suburban19.25 to 20.35.6 to 6.05.390.2 to 0.6

Even taking the most generous F=1.4F = 1.4F=1.4, which adds 0.39 to every model figure, class 1 still comes out more than a magnitude short of the claimed 7.6 to 8.0. I am not going to tell you Bortle was wrong. I will tell you that the two columns are measuring different acts. One is a physical threshold for a target you are looking for. The other is a summary of what experienced observers, dark adapted for an hour, on fields they know by heart, using averted vision, report as their faintest star. Those are not the same number and the difference between them is somewhere between half a magnitude and nearly two.

Aperture

The 5 log D everybody quotes falls straight out of Pogson. Flux collected is proportional to entrance pupil area, area is proportional to D2D^{2}D2, so substitute into the difference form:

Δm=2.5log⁡10 ⁣(D22D12)=5log⁡10 ⁣(D2D1)\Delta m = 2.5\log_{10}\!\left(\frac{D_2^{2}}{D_1^{2}}\right) = 5\log_{10}\!\left(\frac{D_2}{D_1}\right)Δm=2.5log10​(D12​D22​​)=5log10​(D1​D2​​)

which is why the standard shape is

m=5log⁡D+Nm = 5\log D + Nm=5logD+N

with DDD the entrance pupil in centimetres. NNN is empirical and it is where all the honesty lives. Values from 6.8 to 8.7 appear in the literature. Crumey, choosing parameters he considered typical of normal dark site observing (eye pupil 0.7 cm, F=2F = 2F=2), derived N=7.69N = 7.69N=7.69.

The table, and how much of it to believe

Aperture5 log Dm at N = 6.8m at N = 7.69m at N = 8.7
50 mm3.4910.2911.1912.20
80 mm4.5211.3212.2113.22
100 mm5.0011.8012.6913.70
150 mm5.8812.6813.5714.58
200 mm6.5113.3114.2015.21
250 mm6.9913.7914.6815.69
300 mm7.3914.1915.0816.09
400 mm8.0114.8115.7016.71

Read the row for your telescope and then read it across. The literature band is 1.9 magnitudes wide, a flux factor of 5.8. Anyone who tells you their 200 mm reaches magnitude 14.2 has picked a value of NNN and not mentioned it.

The middle column is optimistic and here is the list of what it assumes, all of which I have failed at least once. A dark site, because NNN silently contains the sky brightness and 7.69 was derived for dark site conditions, so under my class 4 sky it does not apply at all. A fully dark adapted observer, meaning an hour with no phone. A known field, because you find a marginal star far more easily when you know where it is. Transmittance, which the black sky form of the equation makes explicit as m=5log⁡D+8−2.5log⁡(p2F/T)m = 5\log D + 8 - 2.5\log(p^{2}F/T)m=5logD+8−2.5log(p2F/T) with TTT around 0.75 typical, and a refractor with a star diagonal and six months of pollen on the objective is not at 0.75. Seeing good enough that the star stays a point when you push the magnification, which on my median 2.7 arcsecond nights it does not.

This is a point source table. It predicts nothing whatsoever about extended objects, and treating it as though it does is the single most common error I see. Raising magnification spreads the sky background over more retina, so its surface brightness in the eyepiece drops, while a star stays essentially a point delivering the same total flux. Contrast improves, which is why high power finds faint stars. An extended object dims at exactly the same rate as the sky does, so the contrast ratio between galaxy and background is invariant under magnification. What changes is the target's angular size, and the eye's contrast threshold falls as targets get larger, up to a limit. So there is an optimum power for a galaxy, not a monotonic gain, and the aperture table has nothing to say about it. Crumey's own paper makes the point about Hecht's older formula: it "is applicable only to point sources and is shown to be of limited accuracy".

Against my own garden

On 3 October 2025 my meter read 20.90 at the zenith and I counted stars in a field I have my own photometry for, from my own frames. Faintest I could hold with averted vision: 6.1.

Put that into the applicable expression and solve for the field factor rather than for mmm:

2.5log⁡F=0.383(20.90)−1.44−6.1=0.465⟹F=1.532.5\log F = 0.383(20.90) - 1.44 - 6.1 = 0.465 \quad\Longrightarrow\quad F = 1.532.5logF=0.383(20.90)−1.44−6.1=0.465⟹F=1.53

Which sits inside Crumey's stated range. So the model needed no defending and neither did I. My FFF is 1.53 because my garden gives an unobstructed zenith and because I have memorised where the faint stars are, and memorising them is cheating in precisely the direction the model already allows for.

None of the above tells you what you will see tonight. It gives you a ceiling. The only number actually worth collecting is the gap between that ceiling and your real result, because the gap is made of specific, nameable, fixable things. Mine is about 0.4 magnitudes and I can list all four of them. I spent three winters finding that out on a single galaxy: @dust/two-hundred-and-eighteen-hours-on-one-galaxy.

Footnotes

  1. Crumey, A. "Human contrast threshold and astronomical visibility", Monthly Notices of the Royal Astronomical Society 442(3), 2600 to 2619, 2014. doi:10.1093/mnras/stu992. The abstract is worth reading even if the fits are not what you came for. ↩

Next →218 hours on one galaxy
All 3 parts in “Backyard photons”

⑂Forked once

  • ⑂
    Mapping my own light pollution
    by @nullisland · 01 Aug 2026

On this page

  • What the number is
  • Where the fifth root of 100 comes from
  • Surface brightness, which is a different animal
  • The naked eye
  • Aperture
  • The table, and how much of it to believe
  • Against my own garden