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The sizes of the stars in the picture below have more to do with their brigtness than the actual size of the star. I'm looking for a formula that relates the viewing size of a star with its brightness.

The Central Stars of Orion

This is what I've worked on so far:

From Wikipedia on Apparent Magnitude, History

The viewing brightness can be related to the apparent magnitude by

$$B=0.4^A$$

Before I sit down and count pixels and try to do a curve fit to find the the relationship, I'm guessing this work has been done before. I wonder if anyone has any leads that would get me to a good equation that I can use to generate the view sizes of stars.


Edit:

My main goal is to make a reasonable 'cartoon' of the various constellations for the astronomy classes I'm teaching. My rough work so far looks like this. (I'm fairly certain the vertical scale is off.)

This is a cartoon of some of the stars of Orion.

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    $\begingroup$ This is a function of the medium on which the image is recorded. There will be no simple relationship between the size of the stellar image and brightness. $\endgroup$ – adrianmcmenamin Sep 2 '17 at 21:45
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    $\begingroup$ This is a good question, since stars have effectively zero angular diameter. I've now asked a version of this at photo.stackexchange.com/questions/92387/… -- @adrianmcmenamin is correct in saying the actual size will depend on the medium, but the relation between sizes may be constant. $\endgroup$ – user21 Sep 3 '17 at 13:16
  • $\begingroup$ @barrycarter, Thanks! There's a good lead there, the Airy disk might answer this, I'll look at the details today. $\endgroup$ – David Elm Sep 3 '17 at 16:27
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    $\begingroup$ I added more links to the photo.stackexchange.com/questions/92387/… that might be helpful to your specific question. $\endgroup$ – user21 Sep 4 '17 at 2:49
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    $\begingroup$ Stellarium (which shows the sky as viewed optically, not photographically) appears to display stars with a pixel count (not diameter) proportional to their brightness (not magnitude). For example, a magnitude 0 star would have 100 times as many pixels as magnitude 5 star, and thus a diameter 10 times as large. $\endgroup$ – user21 Sep 5 '17 at 1:03
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Based on the math in the Wikipedia article about the Airy Disk:

The intensity of the starlight as a function of angular distance $\theta$ from the center is reasonably approximated by a Gaussian $$ I(\theta) = I_0 \exp(-\theta^2/(2\sigma^2)) $$ where $\sigma$ is a measure for the width of the intensity profile, which for an ideal optical system and diffraction-limited images is $$ \sigma \approx 0.42 \lambda N $$ (where $\lambda$ is the wavelength and $N$ is the f-number) and which is independent of the peak intensity $I_0$. For a non-ideal or not-diffraction-limited system, the Gaussian should still provide a reasonable approximation, but then $\sigma$ is (much) larger than for the ideal system.

The apparent edge of a star in an image occurs where the intensity $I(\theta)$ drops below a certain minimum $I_\text{limit}$, which (from the first equation) corresponds to an angular distance $$ \theta_\text{edge} = \sigma\sqrt{2\log(I_0/I_\text{limit})}$$

We can rewrite this in terms of visual magnitude $V$. For fixed $\sigma$, $I_0$ is a measure for the brightness of the star. Brightness is related to magnitude $V$ through $$ V = -2.5 \log_{10}(I/I_\text{ref}) $$ (with $I_\text{ref}$ the reference intensity for the magnitude scale) so $$\log(I_0/I_\text{limit}) = 0.4 \log(10) (V_\text{limit} - V_0)$$ and \begin{eqnarray} \theta_\text{edge} & = & \sigma \sqrt{0.8 \log(10) (V_\text{limit} - V_0)} \approx 1.36 \sigma \sqrt{V_\text{limit} - V_0} \\ V_\text{limit} - V_0 & = & \frac{\theta_\text{edge}^2}{0.8 \log(10) \sigma^2} \approx 0.543 \left( \frac{\theta_\text{edge}}{\sigma} \right)^2 \end{eqnarray}

So, the difference between the limiting magnitude and the star's magnitude is proportional to the apparent angular area of the star in the image.

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  • $\begingroup$ This prescription is reasonable, but has nothing to do with diffraction really. The angular size of stellar images is actually the same because of atmospheric "seeing", which would normally be much bigger than the Airy disk $\sigma$ you define. With CCDs there are other effects like charge transfer efficiency and saturation to consider; and there are serious non-linearities in the response of photographic plates. $\endgroup$ – Rob Jeffries Sep 5 '17 at 14:09
  • $\begingroup$ My description does not apply only to diffraction-limited cases. Both diffraction-limited and seeing-limited cases are reasonably described using the Gaussian. Saturation and non-linearities in the response are only important if they noticeably change the size of the stellar image defined by the $I_\text{limit}$ contour, which implies significant transfer by the recording device or medium of signal across where the star image's edge would have otherwise been. It would be interesting to see a test of the final relationship based on the picture. $\endgroup$ – Louis Strous Sep 5 '17 at 15:00
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This is the problem faced by all star map makers and has been "solved" by each map-maker. The solution has some aesthetic considerations if you use purely area Sirius becomes huge ... So instead of using a formula measure up the symbol sizes used in your favorite star atlas for each magnitude range. Check copy-right for your use.

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  • $\begingroup$ Are you saying Stellarium does not use area? I think it does, even for Sirius. $\endgroup$ – user21 Sep 7 '17 at 17:47
  • $\begingroup$ I am not referring to any particular star chart or set of software. The point is that it's not a straight-forward problem but has been solved in slightly different ways by different star map makers. So why not use what other people have done rather than re-invent the wheel. $\endgroup$ – TazAstroSpacial Sep 10 '17 at 23:02
  • $\begingroup$ OK, I am disagreeing with your statement "if you use purely area Sirius becomes huge". That's not true. It's quite reasonable to use area, even for Sirius. $\endgroup$ – user21 Sep 11 '17 at 1:27

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