# Why do stars appear as circles, not points?

Excluding the Sun, stars are so far away that their angular diameter is effectively zero. However, when you take pictures of them, brighter stars appear as circles, not points. Why?

In theory, any star, regardless of brightness, should hit at most one small point of whatever medium is being used to take the photograph. Why do nearby points of the medium also respond? Does excessive light "bleed" into nearby points, and, if so, is the "bleeding" the same for digital and non-digital cameras?

Does it have something to do with the lens? Does the lens expand a single point of light into a small circle, depending on brightness?

I ran into this while trying to answer https://astronomy.stackexchange.com/questions/22474/how-to-find-the-viewing-size-of-a-star which effectively asks: what's the function (if any) that relates star brightness to the size of a star's disk on photographic film (or digital media)?

Note: I do realize that a star's visual and photographic magnitudes can be different, and am assuming the answer will be based on photographic magnitude.

• User1118321 mentions (reason 1.) another likely mechanical reason for the effect irrespective of the optical issues. I would add that practical reason to the theory-based ones of mine and others.
– Stan
Sep 3, 2017 at 19:10
– user2844
Sep 4, 2017 at 2:49
• "In theory, any star, regardless of brightness, should hit at most one small point of whatever medium is being used to take the photograph." I am unaware of any such theory, and since it disagrees with observation, any such theory must be wrong. What is this theory, and how did you come to believe it? I am interested to learn how people come to believe false things. Sep 5, 2017 at 17:16
• @EricLippert That's a little harsh... I'm saying the angular diameter of a star is effectively zero, so if light from the star hit photographic media directly, and the photographic media were "pixellized", the star's direct light would light up at most one pixel. Does that help?
– user2844
Sep 5, 2017 at 18:18
• Not intending to be harsh; text-only media can make casual inquiries sound like interrogations, unfortunately. That does help; now we can consider consequences of your theory. First: if the angular diameter is "effectively zero" then how can it be lighting any number of pixels? A thing of size zero is infinitely smaller than any pixel. So already something seems fishy about this theory. Second: if the angular diameter is extremely small, then the ratio of the camera aperture diameter to the object's perceived diameter is enormous; that seems like it ought to be a factor. Is it? Sep 5, 2017 at 18:58

Whenever light passes a boundary, it diffracts, or bends, due to the wavelike property of light interacting with that boundary. An aperture in an optical system, typically circular or circle-like, is one such boundary.

How light interacts with the aperture is described by the point spread function (PSF), or how much and to what degree a point source of light spreads as a result of passing through the optical system. The PSF is determined by the geometry of the system (including the shape and size of the aperture; the shape(s) of the lenses; etc.) and the wavelength of light passing through the optical system. The PSF is essentially the impulse response of the optical system to an impulse function, a point of light of some unit amount of energy that is infinitesimally narrow or tightly bounded in 2D space.

The convolution of light from the subject with the point spread function results in a produced image that appears more spread out than the original object. By Wikipedia user Default007, from Wikimedia Commons. Public Domain.

For a perfectly round aperture in a theoretical optically-perfect imaging system, the PSF function is described by an Airy disk, which is a bullseye-target-like pattern of concentric rings of alternating regions of constructive interference (where the light's waves interact constructively to "add up") and destructive interference (where the light's waves interact so as to cancel themselves out).

It's important to note that the Airy disk pattern is not a result of imperfect lens qualities, or errors in tolerances in manufacturing, etc. It is strictly a function of the shape and size of the aperture and the wavelength of light passing through it. Thus, the Airy disk is a sort of upper-bound on the quality of a single image that can be produced by the optical system1.

A point source of light passing through a round aperture will spread to produce an Airy disk pattern. By Sakurambo, from Wikimedia Commons. Public Domain.

When the aperture is sufficiently large, such that most of the light passing through the lens does not interact with the aperture edge, we say the image is no longer diffraction limited. Any non-perfect images produced at that point are not due to the diffraction of the light by the aperture edge. In real (non-ideal) imaging systems, these imperfections include (but limited to): noise (thermal, pattern, read, shot, etc.); quantization errors (which can be considered another form of noise); optical aberrations of the lens; calibration and alignment errors.

Notes:

1. There are techniques to improve the images produced, such that the apparent optical quality of the imaging system is better than the Airy disk –limit. Image stacking techniques, such as lucky imaging, increase the apparent quality by stacking multiple (often hundreds) different images of the same subject together. While the Airy disk looks like a fuzzy set of concentric circles, it really represents a probability of where a point source of light entering the camera system will land on the imager. The resulting increase in quality produced by image stacking is due to increasing the statistical knowledge of the locations of the photons. That is, image stacking reduces the probabilistic uncertainty produced by diffraction of the light through the aperture as described by the PSF, by throwing a surplus of redundant information at the problem.

2. Regarding the relation in apparent size to brightness of the star or point source: a brighter source of light increases the intensity ("height") of the PSF, but does not increase its diameter. But increased light intensity coming into an imaging system means that more photons illuminate the boundary pixels of the region illuminated by the PSF. This is a form of "light blooming", or apparently "spilling" of light into neighboring pixels. This increases the apparent size of the star.

• Slight defocusing (a/k/a real world physical hardware instead of theoretical lens designs) also spreads light over an even larger area than a theoretically perfect lens would. The more intense the point source of light is, the further the spread will be before the intensity on the edge falls below the limits of sensitivity for the recording medium. It's called the 'noise floor' for digital, but in chemically based film there is also a minimum amount of photon energy needed to strike each grain of the photographic emulsion to cause the required chemical reaction in each grain's molecules. Sep 4, 2017 at 7:57
• @MichaelClark Very good point. Yeah, I kinda glossed over the diffusion, reflection, and other light spreading caused by all the real-world effects, such as what you describe.
– scottbb
Sep 4, 2017 at 23:38
• As an extension to Note 2, it's probably worth noting that many astrophotographic sensors also lack any anti-blooming protection to prevent "full" pixels from overflowing into adjacent ones. This is a deliberate tradeoff that requires the user to be more aware of when the sensor saturates, but allows significantly faster light collection. Most of the time its impact can be kept minimal by choosing appropriate exposure times for each frame in the image stack. The occasional exception involves a very bright star next to a very faint object, ex nightsky.at/Photo/Neb/B33_Newton.jpg Sep 5, 2017 at 13:40
• This is an authoritative discussion of lenses, but I'm not sure this really zeros in on a conclusive explanation for stars in photos are extended blobs. Are the spots Airy patterns? If so, where are the oscillations? They may be washed out since each wavelength has a different period. If not, is it "blooming"? If so, is it a sensor problem (it seems to happen in photographic emulsion as well) or is the blooming caused by imperfections in the glass or coating?
– uhoh
Aug 29, 2019 at 0:40
• @uhoh if the image is undersampled (the Airy disc is several times smaller than a single pixel), there is nowhere near enough resolution to see the Airy disc as anything other than roughly a square (and maybe some neighboring pixels getting a little bit of signal if the star was overexposed. Only if the image is very oversampled will an Airy disc appear as the graphic from Wikipedia. There's just not enough resolution in a camera to make a star appear as 50+ (just picking a significant number) pixels across in order to resolve the faint tings from the idealized Airy disk.
– scottbb
Aug 29, 2019 at 1:34

The size of the "point" is affected by the wavelength dependent "Point-Spread Function" (PSF) of the lens system you're using.

Diffraction of light, which determines the system’s resolution limit, blurs out any point-like object to a certain minimal size and shape called the Point Spread Function. The PSF, then, is the three-dimensional image of a point-like object at the image plane. The PSF is usually taller than it is wide (like an American football standing on its tip), because optical systems have worse resolution in the depth direction than in the lateral direction.

The PSF varies depending on the wavelength of the light you are viewing: shorter wavelengths of light (such as blue light, 450nm) result in a smaller PSF, while longer wavelengths (such as red light, 650nm) result in a larger PSF and therefore worse resolution. Also, the Numerical Aperture (NA) of the objective lens that you use affects the size and shape of the PSF: a high-NA objective gives you a nice small PSF and therefore better resolution.

Surprisingly the PSF is independent of the intensity of the point. This is true for both astrophotography and microscopy.

• Wait. If "PSF is independent of the intensity of the point", shouldn't that mean all red stars are the same size, regardless of brightness? That's not what actually happens, though.
– user2844
Sep 3, 2017 at 15:35
• @barrycarter: The (optical) PSF is independent of the intensity of the point. However, the PSF of a properly focused camera tends to be very sharply peaked (by design -- if it weren't, the entire image would look blurry), and for faint stars only the central peak of the PSF is actually detectable. The brighter the star, the more clearly the faint outlying parts of the PSF can be seen, while the central peak quickly becomes bright enough to saturate the sensor (or film). Sep 3, 2017 at 20:50
• The ideal PSF is independent of intensity. The quantized PSF, which is what any digital camera measures, is not.
– E.P.
Sep 3, 2017 at 23:30

There are a few reasons I can think of:

1. The most common is the lens. Getting a lens to focus at infinity can be tricky on some lenses that let you focus "past" infinity. But even if you can get it exact, the lens itself may still spread it out some.
2. Another reason is that it is possible for the light to actually hit more than one sensor site, either because the sensor site (or film grains) are not perfectly aligned with every star, or because the projection of the star onto the sensor or film is actually larger than a single sensor site or film grain.
3. The atmosphere also spreads out the light coming from the stars which leads to a bigger circle for each one.
• Thanks! Quick comment on 3: astrophotography taken from airless space shows the same effect, so I don't think it's that.
– user2844
Sep 3, 2017 at 15:34
• It could be a minimal effect. I mention it because I know that it is an issue for scientific astrophotography. I know that in some cases they even go so far as to shoot a laser into the sky to see how the atmosphere is distorted and they adjust their lenses or mirrors to compensate. But maybe for artistic shots it's not as big a concern? It may also be a bigger effect when using a longer lens (esp. like a telescope) due to the smaller field of view? I don't really know, but I've heard it mentioned so included it. Sep 3, 2017 at 19:58
• Astrophotography taken from space is often taken at narrow enough angles of view that stars are no longer dimensionless points. Sep 5, 2017 at 3:15

I took a small area from your photo and enlarged it (resampled by a factor of 10).

I marked two interesting regions. Region A indicates a star, blurred by the optics approximately into an 3x3 pixel area with a peak of diameter 2-3 pixel, I would say. This is the blurring effect as described in scottbb's answer.

However, the bright star at position B is much wider and also shows saturation in the center. My guess is that this additional broadening is caused by pixel bleed through or just by saturation.

is the "bleeding" the same for digital and non-digital cameras?

Probably not. Non-digital cameras have a much higher contrast range, so saturation may be less of an issue and pixel bleeding which is an electronic effect might not occur at all.

However, with a HDR recording scheme within a digital camera one should be able correct for the additional broadening and make spot B look like spot A only much brighter.

To change the size of the blurring effect you could play around with the aperture of your camera and image stars (or printed dots on paper, if stars are not available or a small hole in dark cardboard with a light source behind far away).

Well studied by George Airy, Astronomer Royal, published in 1830. Now called the Airy disk or Airy pattern, a point source star images with an alternating light and dark rings surrounding a central disk. The diameter of the first dark ring is 2.44 wave lengths for a well corrected lens with circular aperture. This is a key fact when regarding the resolving power of a lens. It is difficult, but not impossible to image these concentric rings. Most images amalgamate these rings.

John Strutt, 3rd Baron Rayleigh (Astronomer Royal) further published what is now called the Rayleigh Criterion covering the theoretical maximum resolving power of a lens. “Resolving Power in lines millimeter is 1392 ÷ f-number. Thus f/1 = 1392 lines per millimeter maximum. For f/2 = 696 lines per millimeter. For f/8 = 174 lines per millimeter. Please note: Resolving power for apertures larger than f/8 is higher than film intended to be pictorially useful, can exploit. Also, resolving power is measured by imaging parallel lines with white spaces between. When finally ruled lines are seen to merge, their spacing is the limit of resolution for that imaging system. Few if any lenses have bested the Rayleigh Criterion.

• While interesting, this answer would profit from a few more explanations in layman's terms. Especially the quote in the second paragraph contains information that is probably not very useful as it is. Sep 5, 2017 at 8:12

There are two factors here.

The first is optical, in any imagine system various affects come into play, including diffraction, imperfect focus and for terrestrial telescopes the affects of the atmosphere. These affects come together to spread a point of light over an area on the film or sensor and they can be represented by a mathematical function known as a "point spread function". These functions do not have a hard edge, as you get further from the center the light will get weaker (possibly with some oscillation) but it doesn't go away completely.

The second is in how the film or sensor (or even your retina) reacts to light, there is a relatively narrow range of light intensities that will be usefully recorded, light that is too strong will saturate the film or sensor, light that is too weak will be lost in the noise. The brightest objects in the sky are hundreds of times brighter than the dimmest that are visible to the naked eye.

This means that the visible size of of an object depends on it's brightness, for a dim object you only see the center of the point spread function, the rest is lost in the noise.

For a bright object the center of the point spread function is overexposed and appears as a white blob while the color of the object is only revealed in the border.